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THE LAPLACE TRANSFORM
  IN CIRCUIT ANALYSIS
A Resistor in the s Domain


+
        R i
v
                   v=Ri (Ohm’s Law).
    +
         R I
    V

                   V(s)=RI(s
An Inductor in the s Domain
                                      di
Initial current of I 0            v=L
                                      dt
    +
                                      V(s)=L[sI(s)-i(0-)]=sLI(s)-LI0
    v   L i

                                          V ( s) I0
                                 I ( s) =       +
                                           sL     s          I
                       +
                               sL I                 +
                   V                                    sL       I0
                                                    V
                                                                 s
                           +     LI0
A Capacitor in the s Domain
                                            dv
    Initially charged to V0 volts.      i=C
                                            dt
                              −
    I ( s ) = C [ sV ( s ) − v (0 )] = sCV ( s ) − CV0
          1            V0
V ( s) =     ÷I ( s ) +
          sC           I
                           s
                                              +
                                          I           1/sC
    +       i             +
                   1/sC           CV0
v       C                 V                   V
                                                  +
                                                      V0/s
The Natural Response of an RC Circuit

            t=0     +       1/sC             I                 +
 C    +               R
                  i v                                  R
       V0                   V0/s   +                           V



  V0   1
     =   I ( s ) + RI ( s )
   s sC
                                       V0
               CV0                          R
     I ( s) =        =
              RCs + 1 s + (                 1
                                              RC   )
     V0 −                    −
  i = e u( t ) ⇒ v = Ri = V0e u( t )
                   t                                       t
                       RC                                      RC


     R
The Step Response of a Parallel Circuit
?
The Step Response of a Parallel Circuit

     V V    I dc
sCV + +   =
     R sL    s
                    I dc
                           C
V=
       s +(
       2       1
                   RC ) s + ( 1 LC )
                         I dc
                                LC
IL =
       s[ s + (
           2         1
                         RC ) s + ( 1 LC ) ]
384 × 105
IL =
     s( s 2 + 64000 s + 16 × 108 )
                        384 × 105
IL =
     s( s + 32000 − j 24000)( s + 32000 + j 24000)
                                        *
     K1           K2                  K2
IL =    +                    +
     s    s + 32000 − j 24000 s + 32000 + j 24000
     384 × 105
K1 =           = 24 × 10−3
     16 × 108
              384 × 105
K2 =                               = 20 × 10−3 126.870
     ( −32000 + j 24000)( j 48000)
i L ( t ) = [24 + 40e −32000 t cos(24000t + 126.870 )]u( t )mA
Transient Response of a
                Parallel RLC Circuit
Replacing the dc current source with a sinusoidal current source
                                                          sI m
       i g = I m cos ω t A ⇒ I g ( s ) =              2            2
                                                     s +ω

      V ( s) =
                          ( 1C ) s                  I g ( s)
                  2
                 s + ( 1 RC ) s + ( 1 LC )

      V ( s) =
                                ( )
                                  Im
                                       C   s2
                 ( s 2 + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )]

       I L ( s) =
                  V ( s)
                         = 2
                                            LC s(    Im
                                                               )
                   sL     ( s + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )]
I m = 24mA,       ω = 40000rad / s
                         384 × 105 s
I L (s) = 2
         ( s + 16 × 108 )( s 2 + 64000 s + 16 × 108 )
                             *                                  *
              K1          K1              K2                  K2
I L (s) =            +           +                   +
          s − j 40000 s + j 40000 s + 32000 − j 24000 s + 32000 + j 24000
                  384 × 105 ( j 40000)
K1 =                                             = 7.5 × 10−3 − 900
     ( j 80000)(32000 + j16000)(32000 + j 64000)
              384 × 105 ( −32000 + j 24000)
K2 =                                                 = 12.5 × 10−3 900
     ( −32000 − j16000)( −32000 + j 64000)( j 48000)
i L ( t ) = (15sin 40000t − 25e −32000 t sin 24000t )u( t )mA
i Lss = 15sin 40000t mA
Mesh Analysis
?
Mesh Analysis (cont.)


336
     = (42 + 8.4 s ) I1 − 42 I 2
 s
   0 = −42 I1 + (90 + 10 s ) I 2
          40( s + 9)      15   14      1
 I1 =                   =    −     −
      s( s + 2)( s + 12) s s + 2 s + 12
             168          7 8.4      1.4
 I2 =                   = −       +
      s( s + 2)( s + 12) s s + 2 s + 12
−2 t        −12 t
i1 = (15 − 14e          −e           )u( t ) A,
                 −2 t            −12 t
i2 = (7 − 8.4e          + 1.4e           )u( t ) A.
           336(90)
i1 (∞ ) =          = 15 A,
           42(48)
           15(42)
i2 ( ∞ ) =        =7A
             90
Thevenin’s Theorem
Use the Thevenin’s theorem to find v c (t)
?
Thevenin’s Theorem (cont.)

      (480 / s )(0.002 s )     480
VTh =                      =
        20 + 0.002 s         s + 10 4

            0.002 s(20) 80( s + 7500)
ZTh = 60 +                 =
            20 + 0.002 s         s + 10 4
4
                   480 /( s + 10 )
IC =                         4             5
     [80( s + 7500) /( s + 10 )] + [(2 × 10 ) / s]
                6s                      6s
IC = 2                       6
                               =              2
     s + 10000 s + 25 × 10        ( s + 5000)
        −30000           6
IC =               +
     ( s + 5000) 2
                     s + 5000
                         −5000 t            −5000 t
ic ( t ) = ( −30000te                + 6e             )u( t ) A
                           5                                      5
      1      2 × 10     6s           12 × 10
Vc =    IC =                  2
                                =
     sC         s ( s + 5000)     ( s + 5000)2
                 5      −5000 t
vc ( t ) = 12 × 10 te             u( t )V
MUTUAL INDUCTANCE EXAMPLE



                             i2(t)=?


       −     60             −
   i1 (0 ) =    = 5 A, i2 (0 ) = 0
             12
?
MUTUAL INDUCTANCE EXAMPLE




     Using the T-equivalent of the
     inductors, and s-domain
     equivalent gives the following circuit
(3 + 2 s ) I1 + 2 sI 2 = 10
                                2 sI1 + (12 + 8 s ) I 2 = 10



            2.5        1.25 1.25
I2 =                 =     −
     ( s + 1)( s + 3) s + 1 s + 3
                    −t        −3 t
i2 ( t ) = 1.25(e        −e          )u( t ) A
THE TRANSFER FUNCTION


  The transfer function is defined as the ratio of the Laplace
  transform of the output to the Laplace transform of the input
  when all the initial conditions are zero.



         Y ( s)
H ( s) =                   Y(s) is the Laplace transform of the output,
         X ( s)            X(s) is the Laplace transform of the input.
THE TRANSFER FUNCTION
        (cont.)

                      I ( s)        1
          H1 ( s ) =         =
                     Vg ( s ) R + sL + 1/ sC
                             sC
                   = 2
                     s LC + RCs + 1
                     V ( s)         1
          H 2 ( s) =         = 2
                     Vg ( s ) s LC + RCs + 1
EXAMPLE

Find the transfer function V 0 /V g and
determine the poles and zeros of
H(s).
?
EXAMPLE

V0 − Vg         V0         V0 s
         +              + 6 =0
 1000       250 + 0.05 s 10
         1000( s + 5000)
V0 = 2                       V
                           6 g
      s + 6000 s + 25 × 10
         V0       1000( s + 5000)
H ( s) =     = 2
         Vg s + 6000 s + 25 × 106
p1 = −3000 + j 4000,
p2 = −3000 − j 4000
z1 = −5000
Assume that v g (t)=50tu(t). Find v 0 (t). Identify the
 transient and steady-state components of v 0 (t).

                                 1000( s + 5000)      50
V0 ( s ) = H ( s )V g ( s ) = 2                    6    2
                             ( s + 6000 s + 25 × 10 ) s
                             *
         K1                K1         K2 K3
=                  +                 + 2 +
  s + 3000 − j 4000 s + 3000 + j 4000 s    s
                  −4          0                                −4
K1 = 5 5 × 10          79.7 , K 2 = 10, K 3 = −4 × 10
                    −4 −3000 t                             0
v0 = [10 5 × 10 e                 cos(4000t + 79.7 )
                         −4
       + 10t − 4 × 10 ]u( t )V
The transient component is generated by the poles of
  the transfer function and it is:


              −4 −3000 t                                 0
10 5 × 10 e                 cos(4000t + 79.7 )
  The steady-state components are generated by the poles
  of the driving function (input):


                                −4
    (10t − 4 × 10 )u( t )
Time Invariant Systems

 If the input delayed by a seconds, then

L{ x ( t − a )u( t − a )} = e       − as
                                           X (s)
                             − as
Y ( s ) = H ( s ) X ( s )e
 y( t ) = L−1
                { Y ( s )} = y( t − a )u( t − a )
Therefore, delaying the input by a seconds simply delays the
response function by a seconds. A circuit that exhibits this
characteristic is said to be time invariant.
Impulse Response

If a unit impulse source drives the circuit, the response of the
circuit equals the inverse transform of the transfer function.



        x( t ) = δ (t ) ⇒ X ( s ) = 1
        Y ( s) = H ( s)
        y( t ) = L   −1
                          { H ( s )} = h(t )
      Note that this is also the natural response of the circuit
      because the application of an impulsive source is equivalent
      to instantaneously storing energy in the circuit.
CONVOLUTION INTEGRAL

            x(t)                                                 y(t)
                            x(t)             y(t)
                    t                N
   a            b                                                               t
                                                      a                 b

Circuit N is linear with no initial stored energy. If we know the
form of x(t), then how is y(t) described? To answer this
question, we need to know something about N. Suppose we know
the impulse response of the system.
                                                      y( t ) = h( t )
   x(t ) = δ (t )

            (1)               x(t)             y(t)
                        t                N                                  t
Instead of applying the unit impulse at t=0, let us suppose that it is
applied at t=λ. The only change in the output is a time delay.

              δ (t − λ )        N       h( t − λ )

  Next, suppose that the unit impulse has some strength other than
  unity. Let the strength be equal to the value of x(t) when t= λ. Since
  the circuit is linear, the response should be multiplied by the same
  constant x(λ)

            x (λ )δ ( t − λ )       N      x (λ )h( t − λ )
Now let us sum this latest input over all possible values of λ and
  use the result as a forcing function for N. From the linearity, the
  response is the sum of the responses resulting from the use of all
  possible values of λ
     +∞                                +∞
    ∫−∞   x( λ )δ ( t − λ )d λ   N    ∫−∞ x(λ )h( t − λ )d λ
From the sifting property of the unit impulse, we see that the input is
simply x(t)

                                       +∞
                        X(t)     N    ∫−∞ x(λ )h( t − λ )d λ
Our question is now answered. When x(t) is known, and h(t), the
unit impulse response of N is known, the response is expressed by
                         +∞
       y( t ) = ∫−∞ x ( λ )h( t − λ )d λ
 This important relation is known as the convolution integral . It
 is often abbreviated by means of

   y( t ) = x ( t ) * h( t )
 Where the asterisk is read “convolved with”. If we let z=t-λ,
 then dλ=-dz, and the expression for y(t) becomes
             −∞                                  +∞
 y( t ) = ∫∞ − x ( t − z )h( z )dz = ∫−∞ x ( t − z )h( z )dz
                             ∞                          ∞
  y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
∞                           ∞
y( t ) = x( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
Convolution and Realizable Systems


For a physically realizable system, the response of the system
cannot begin before the forcing function is applied. Since
h(t) is the response of the system when the unit impulse is applied at
t=0, h(t) cannot exist for t<0. It follows that, in the second integral, the
integrand is zero when z<0; in the first integral, the integrand is zero
when (t-z) is negative, or when z>t. Therefore, for realizable
systems the convolution integral becomes


                           t                           ∞
y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫0 x ( t − z )h( z )dz
EXAMPLE

 x(t)
                                               −t
        1   t
                h(t)    y(t)
                                   h( t ) = 2e u( t )
x ( t ) = u( t ) − u( t − 1)
                               ∞
y( t ) = x ( t ) * h( t ) = ∫0 x ( t − z )h( z )dz
        ∞                                 −z
= ∫0 [u( t − z ) − u( t − z − 1)][2e u( z )]dz
Graphical Method of Convolution
Since h(z) does not exist prior to t=0 and vi(t-z) does not exist
for z>t, product of these functions has nonzero values only in
the interval of 0<z<t for the case shown where t<1.
                t
     y( t ) = ∫0 2e − z dz = 2(1 − e − t )        0≤ t ≤1
When t>1, the nonzero values for the product are obtained in the
interval (t-1)<z<t.
                 t       −z                       −t
    y( t ) = ∫t −1 2e dz = 2(e − 1)e                       t >1
EXAMPLE


Apply a unit-step function, x(t)=u(t), as the input to a system whose
impulse response is h(t) and determine the corresponding output
y(t)=x(t)*h(t).
?
h(t)=u(t)-2u(t-1)+u(t-2),
When t<0, there is no overlap and y(t)=0 for t<0
For 0<t<1, the curves overlap from z=0 to z=t and product is 1.
Thus,
                          t
            y( t ) = ∫0 1dz = t           0< t <1
 When 1<t<2, h(t-z) has slid far enough to the right to bring under
 the step function that part of the negative square extending from 0
 to z=t-1. Thus,
         t −1         t            t −1    t
y( t ) = ∫0 −1dz + ∫t −11dz = − z 0 + z t −1 = 2 − t ,     1< t < 2
Finally, when t>2, h(t-z) has slid far enough to the right so that it lies
entirely to the right of z=0
                 t −1               t
     y( t ) = ∫t − 2 −1dz + ∫t −11dz = 0                  t>2
Convolution and the Laplace Transform
Let F 1 (s) and F 2 (s) be the Laplace transforms of f1(t) and f2(t),
respectively. Now, consider the laplace transform of f1(t)*f2(t),


           L{ f1 ( t ) * f 2 ( t )} = L   {∫   ∞
                                              f (λ ) f 2 (t
                                            −∞ 1
                                                              − λ )d λ   }
  Since we are dealing with the time functions that do not exist
  prior to t=0-, the lower limit can be changed to 0-
                                   ∞      ∞ − st
       L{ f1 ( t ) * f 2 ( t )} = ∫0 [ ∫
                                    −
                                          0−
                                             e f1 ( λ ) f 2 ( t   − λ )dt ]d λ
f1(λ) does not depend on t, and it can be moved outside the inner
integral
                                   ∞            ∞ − st
        L{ f1 ( t ) * f 2 ( t )} = ∫0 f1 ( λ )[ ∫
                                    −
                                                0−
                                                   e f 2 (t    − λ )dt ]d λ
                                   ∞            ∞ − s( x + λ )
                               = ∫0 f1 (λ )[ ∫
                                    −
                                                0−
                                                   e           f 2 ( x )dx ]d λ
                                   ∞            − sλ    ∞ − sx
                               = ∫0 f1 (λ )e
                                    −                  [∫
                                                        0−
                                                           e f 2 ( x )dx ]d λ
                                   ∞
                               = ∫0 f1 (λ )e − sλ [ F2 ( s )]d λ
                                    −


                                           ∞
                               = F2 ( s )∫0 f1 (λ )e − sλ d λ
                                            −



                               = F1 ( s ) ×F2 ( s )
STEADY-STATE SINUSOIDAL RESPONSE

If the input of a circuit is a sinusoidal function   x ( t ) = A cos(ω t + θ )

      x ( t ) = A cos ω t cosθ − A sin ω t sin θ
                ( A cosθ ) s ( A sin θ )ω
      X ( s) = 2          2
                             − 2       2
                  s +ω          s +ω
               A(s cosθ − ω sinθ )
            =         2     2
                     s +ω
                       A(s cosθ − ω sinθ )
      Y ( s) = H ( s)         2    2
                             s +ω
The partial fraction expansion of Y(s) is
                    *
           K1     K1
Y ( s) =       +       + ∑ terms generated by the poles of H(s)
         s − jω s + jω

    If the poles of H(s) lie in the left half of the s plane, the corresponding
    time-domain terms approach zero as t increases and they do not
    contribute to the steady-state response. Thus only the first two terms
    determine the steady-state response.
H ( s ) A( s cosθ − ω sinθ )
K1 =
                 s + jω           s = jω

     H ( jω ) A( jω cosθ − ω sinθ )
   =
                   2 jω
     H ( jω ) A(cosθ + j sinθ ) 1
   =                            = H ( jω ) Ae jθ
                  2               2

H ( jω ) = H ( jω ) e jφ (ω )
         A            j[θ +φ (ω )]
K1 = H ( jω ) e
         2
yss ( t ) = A H ( jω ) cos[ω t + θ + φ (ω )]
EXAMPLE
If the input is 120 cos(5000t+300)V, find the steady-state expression for v0


                1000( s + 5000)
    H ( s) = 2
             s + 6000 s + 25 × 106
                          1000(5000 + j 5000)
    H ( j 5000) =          6                        6
                  −25 × 10 + j 5000(6000) + 25 × 10
         1 + j1    2
       =        =    − 450
           j6     6
              120 2
    v0 ss   =       cos(5000t + 300 − 450 )
                6
        =20 2 cos(5000t − 150 )V
THE IMPULSE FUNCTION
 IN CIRCUIT ANALYSIS
       The capacitor is charged to an initial voltage
       V0 at the time the switch is closed. Find the
       expression for i(t) as R     0
                      V0                     V0
                           s
        I=                             =           R
             R + ( 1 sC ) + ( 1 sC )
                       1          2
                                           s+(    1
                                                    RC e )

             C1C 2             V0 − t / RC            
       Ce =          i (t ) =  e                  e
                                                       ÷u( t )
            C1 + C 2           R                      
        As R decreases, the initial current (V0/R)
        increases and the time constant (RCe)
        decreases. Apparently i is approaching an
        impulse function as R approaches zero.
The total area under the i versus t curve represents the total charge
transferred to C2 after the switch is closed.

                     ∞ V0
     Area=q = ∫0     −       e − t / RC dt = V0C e
                                     e

                         R
Thus, as R approaches zero, the current approaches an impulse
strength V0Ce.   i → V0C eδ ( t )
Series Inductor Circuit
        Find v0. Note that opening the switch forces
        an instantaneous change in the current L2.

             i1 (0− ) = 10 A,   i2 (0− ) = 0

            V0       V0 − [(100 s ) + 30]
                  +                       =0
         2 s + 15          3 s + 10
               40( s + 7.5) 12( s + 7.5)
         V0 =                +
                 s( s + 5)          s+5
               60           10
         V0 =     + 12 +
                s          s+5
                                               −5 t
          v0 ( t ) = 12δ ( t ) + (60 + 10e            )u( t )V
Does this solution make sense? To answer this question, first let
        us determine the expression for the current.

                   (100 s ) + 30 4  2
                I=              = +
                     5 s + 25    s s+5
                                    −5 t
                i ( t ) = (4 + 2e          )u( t ) A
Before the switch is opened, current through L1 is 10A and in L2 is 0 A,
after the switch is opened both currents are 6A. Then the current in L 1
changes instantaneously from 10 A to 6 A, while the current in L 2 changes
instantaneously from 0 to 6 A. How can we verify that these instantaneous
jumps in the inductor current make sense in terms of the physical behavior
of the circuit?
Switching operation places two inductors in series. Any impulsive voltage
appearing across the 3H inductor must be balanced by an impulsive
voltage across the 2H inductor. Faraday’s law states that the induced
voltage is proportional to the change in flux linkage ( v = d λ )
before switching                                                dt

        λ = L1i1 + L2i2 = 3(10) + 2(0) = 30 Wb-turns
After switching

        λ = ( L1 + L2 )i (0+ ) = 5i (0+ )
            +    30
        i (0 ) =    = 6A
                 5
  Thus the solution agrees with the principle of the conservation
  of flux linkage.
Impulsive Sources
                       When the voltage source is applied, the initial
                       energy in the inductor is zero; therefore the initial
                       current is zero. There is no voltage drop across R,
                       so the impulsive source appears directly across L

                           1 t                   +    V0
                        i = ∫0 V0δ ( x )dx ⇒ i (0 ) =
                                 −                       A
                           L                          L

Thus, in an infinitesimal             Current in the circuit decays to
moment, the impulsive voltage         zero in accordance with the
source has stored                     natural response of the circuit
                 2                         V0 − (
    1  V0           1 V02              i= e
                                                      R
                                                              )t
                                                                   u( t )
w=    L ÷ =
                                                          L
                 J
     2  L  2 L                           L
EXAMPLE
   Find i(t) and v0(t) for t>0

       50 + ( 100 s ) + 30
  I=
        25 + 5 s
     12   4
  =     +
    s+5 s
  i ( t ) = (12e −5 t + 4)u( t ) A
                             60   60
  V0 = (15 + 2 s ) I = 32 +     +
                            s+5 s
                                 −5 t
  v0 ( t ) = 32δ ( t ) + (60e           + 60)u( t )V
Transfer Functions
1. The Laplace transform
2. Solution of linear differential equations
3. Transient response example
The Laplace Transform
• Definition
                               ∞
       F ( s ) = L[ f (t )] = ∫ f (t )e − st dt
                               0

   – Time (t) is replaced by a new independent variable (s)
   – We call s the Laplace transform variable
• The Laplace domain
   – Often more convenient to work in Laplace domain than time
     domain
   – Time domain  ordinary differential equations in t
   – Laplace domain  algebraic equations in s
• General solution approach
   –    Formulate model in time domain
   –    Convert model to Laplace domain
   –    Solve problem in Laplace domain
   –    Invert solution back to time domain
Laplace Transform of Selected
                     Functions
  • Constant function: f(t) = a
                                                                 ∞
                                ∞         a                                a a
        F ( s ) = L(a ) = ∫ ae − st dt = − e − st                    = 0 −−  =
                           0              s                      0         s s
  • Exponential function: f(t) = e-bt

F ( s ) = L (e   −bt
                          ∞
                       ) = ∫ e e dt = ∫ e
                          0
                              −bt   − st
                                            0
                                             ∞
                                                 −( b + s ) t
                                                                dt = −
                                                                        1
                                                                       b+s
                                                                               [
                                                                           e −( b + s ) t   ]   ∞

                                                                                                0
                                                                                                    =
                                                                                                          1
                                                                                                        s +b
  • Derivatives and integrals

         df  ∞ df − st              ∞                            ∞
       L  = ∫         e dt = ∫ f (t )e − st sdt + f (t )e − st = sF ( s ) − f (0)
         dt  0 dt                  0                             0

      dn f 
     L n  = s n F ( s ) − s n −1 f (0) − s n −2 f (1) (0) −  − sf ( n −2 ) (0) − f ( n −1) (0)
       dt 
           
                t f (t*)dt * = ∞  t f (t*)dt *e − st dt = 1 F ( s )
              L ∫
               0
                                  ∫0 ∫0
                                                          
                                                                       s
Properties of Laplace Transforms
• Superposition
    L[af (t ) + bg (t )] = aF ( s ) + bG ( s )
• Final value theorem
     If lim y (t ) exists ⇒ lim y (t ) = lim[ sY ( s )]
         t →∞                          t →∞      s →0
• Initial value theorem

      lim y (t ) = lim[ sY ( s )]
       t →0              s →∞
• Time delay


  f d (t ) = f (t − t0 ) S (t − t0 )    L
                                        →              Fd ( s ) = e − st0 F ( s )
                     − st 0               L−1
      Fd ( s ) = e            F ( s)   →       f d (t ) = f (t − t0 ) S (t − t0 )
Linear ODE Example
• ODE
        dy
    5      + 4y = 2      y ( 0) = 1
        dt
• Laplace transform
                                      2
    5[ sY ( s ) − y (0)] + 4Y ( s ) =
                                      s
• Substitute y(0) & rearrange

               5s + 2     s + 0.4
   Y (s) =             =
             s (5s + 4) s ( s + 0.8)
• Inverse Laplace transform

             −1              s + 0.4 
                            −1
   y (t ) = L [Y ( s )] = L 
                             s ( s + 0.8) 
                                           
Linear ODE Example cont.
• Table 3.1
      
     −1       s + b3          b3 − b1 −b1t b3 − b2 −b2t
    L                       =        e +          e
       ( s + b1 )( s + b2 )  b2 − b1      b1 − b2

• Our problem

             −1 s + 0.4 
    y (t ) = L                ⇒ b1 = 0 b2 = 0.8 b3 = 0.4
                s ( s + 0.8) 

• Substitute and simplify

     
     −1      s + 0.4       0.4 − 0 −0t 0.4 − 0.8 −0.8t
   L                       =       e +          e      = 0.5 + 0.5e −0.8t
      ( s + 0)( s + 0.8)  0.8 − 0
                                        0 − 0.8
General ODE Solution Procedure
              • Procedure
                 – Transform to Laplace
                   domain
                 – Solve resulting
                   algebraic equations
                 – Transform solution
                   back to time domain
              • Partial fraction
                expansion
                 – Necessary when
                   inverse Laplace
                   transform not
                   tabularized
                 – Break complex
                   functions into simpler
                   tabularized functions
Partial Fraction Example
• Partial fraction expansion
                 s +5           s +5        α     α
   Y (s) =               =                = 1 + 2
             s 2 + 5s + 4 ( s + 1)( s + 4) s + 1 s + 4


• Determination of coefficients
        s +5         4      s +5           1
   α1 =            =   α2 =             =−
        s + 4 s =−1 3       s + 1 s =−4    3

• Inverse Laplace transform
             −1           −1 4 / 3 1 / 3  4 −t 1 − 4 t
   y (t ) = L [Y ( s )] = L       −       = 3e −3e
                             s +1 s + 4 
Repeated Factor Example

                         s +1        α1       α2       α3
             Y (s) =              =      +           +
                     s ( s + 2) 2
                                    s + 2 ( s + 2) 2
                                                        s

           s +1       1         s +1                 1        1
      α2 =          =   α3 =                       =   α1 = −
             s s =−2 2       ( s + 2) 2     s =0
                                                     4        4

             s +1        α1       α2       α3 −1 / 4     1/ 2      1/ 4
Y ( s) =              =      +           +    =      +           +
         s ( s + 2) 2
                        s + 2 ( s + 2) 2
                                            s   s + 2 ( s + 2) 2
                                                                    s

                    −1              1 −2t 1 −2t 1
            y (t ) = L [Y ( s )] = − e + te + S (t )
                                    4     2     4
Quadratic Factor Example
                                        s +1        α α2 α s + α4
                     Y (s) =                       = 1 + 2 + 23
                               s 2 ( s 2 + 4 s + 5) s s     s + 4s + 5

                s + 1 = α1s ( s 2 + 4 s + 5) + α 2 ( s 2 + 4 s + 5) + (α 3 s + α 4 ) s 2

       (α1 + α 3 ) s 3 + (4α1 + α 2 + α 4 ) s 2 + (5α1 + 4α 2 − 1) s + (5α 2 − 1) = 0

                  s +1      α α2 α s + α4        0.04 0.2 − 0.04s − 0.36
Y ( s) =                   = 1 + 2 + 23        =     + 2 + 2
           s ( s + 4 s + 5) s s
            2   2
                                    s + 4s + 5     s  s     s + 4s + 5

                       − 0.04 s − 0.36 − 0.04( s + 2)    − 0.28
                                      =               +
                         s + 4s + 5
                          2
                                        ( s + 2) + 1 ( s + 2) 2 + 1
                                                2




           y (t ) = L−1[Y ( s )] = 0.04S (t ) + 0.2t − 0.04e −2t cos t − 0.28e −2t sin t
Transient Response Example




• Component balance
    dc1                    dc1
  V     = q (ci − c1 ) ⇒ 4     + 2c1 = 2ci   c1 (0) = 0
    dt                     dt
• Step input
              0 t ≤ 0              5
    ci (t ) =         ⇒ Ci ( s ) =
              5 t > 0              s
Transient Response Example cont.
• Laplace transform
                                                              2
   4[ sC1 ( s ) − 0] + 2C1 ( s ) = 2Ci ( s ) ⇒ C1 ( s ) =          Ci ( s )
                                                            4s + 2
• Substitute input

                 2 5         5
   C1 ( s ) =          =
              4 s + 2 s s (2 s + 1)
• Inverse Laplace transform


                            
              −1
   c1 (t ) = L 
                      5
                 s (2 s + 1) 
                                    (
                               = 5 1 − e −t / 2   )
                            
Simulink Solution
                                                                                inlet

                                                                               Inlet 1


                                                           2
                                                                               outlet
                                                         4s+2
                                      Step               Tank               To Workspace


                          6

                                                                      >> plot(tout,inlet)
                                                        Input
                                                        Output

                                                                      >> hold
                          5



                                                                      >> plot(tout,outlet,'r')
Concentration (kmol/m )
                     3




                          4



                          3
                                                                      >> axis([0 15 0 6])
                                                                      >> ylabel('Concentration
                          2                                             (kmol/m^3)')
                          1
                                                                      >> xlabel('Time (min)')
                                                                      >> legend(‘Input','Output')
                          0
                              0   5                10            15
                                      Time (min)
Engineering Circuit Analysis



Ch3 Basic RL and RC Circuits

    3.1 First-Order RC Circuits
    3.2 First-Order RL Circuits
    3.3 Examples




 References: Hayt-Ch5, 6; Gao-Ch5;
 References
Ch3 Basic RL and RC Circuits

3.1 First-Order RC Circuits




Key Words:
    Words
  Transient Response of RC Circuits, Time constant
Ch3 Basic RL and RC Circuits

3.1 First-Order RC Circuits




Key Words:
    Words
  Transient Response of RC Circuits, Time constant
Ch3 Basic RL and RC
                              Circuits
    3.1 First-Order RC Circuits


•    Used for filtering signal by blocking certain frequencies and
     passing others. e.g. low-pass filter
•    Any circuit with a single energy storage element, an arbitrary
     number of sources and an arbitrary number of resistors is a
     circuit of order 1.
•    Any voltage or current in such a circuit is the solution to a 1st
     order differential equation.

    Ideal Linear Capacitor
                                           dq    dv
                                i (t ) =      =c       vc (t +) =v C (t )
                                           dt    dt
                                                                  1 2
                             Energy stored w = ∫ pdt = ∫ cvdv =     cv
                                                                  2
        A capacitor is an energy storage device → memory
        device.
Ch3 Basic RL and RC
                          Circuits
3.1 First-Order RC Circuits

                            vr(t)
                        +           -

                             R          +
                    +
            vs(t)                   C       vc(t)
                    -
                                        -




 •   One capacitor and one resistor
 •   The source and resistor may be equivalent to a circuit with
     many resistors and sources.
Ch3 Basic RL and RC
                          Circuits
3.1 First-Order RC Circuits

Transient Response of RC Circuits
                                                                      E − vc
                                                               ic =
     Switch is thrown to 1
                                                                        R
                                                   KVL around the loop: ic R + vC = E
                      R
         C
                                                                   dvc
                                                               C       R + vc = E
 •            ο
                  2                                                dt
                                                                               t
                                                                          −
              ο
                       K                                        vC = Ae       RC
                                                                                   +E
                  1
          E
                      Initial condition vC (0+) =v C (0−) = 0                           A = −E
                                t                      t
                           −                       −
        vC = E (1 − e          RC
                                    ) = E (1 − e   τ
                                                           )
                                                                         τ = RC
                 dvc E                −
                                        t
                                                                   Called time constant
          ic = C    = e                 τ

                 dt R
Ch3 Basic RL and RC Circuits
      3.1 First-Order RC Circuits

       Time                          τ = RC
       Constant
5Ω                                                              −
                                                                    t
                                                                                   dvc E −t / τ
                                 R             vC = E (1 − e    τ       )             = e
                   C                                                               dt  τ
                             2                 dvc           E       E
         •               ο
                                                     t =0   = →τ =
             10V         ο
                                 K             dt            τ     dvc
                             1                                         t =0
∧
                   E                                               dt


              5V                                                                     R=2k
                                                                                     C=0.1µF


         SEL > >
            0V
                0s     RC               1 ms     2 ms                       3 ms           4 ms
                       V( 2 )
Ch3 Basic RL and RC
                          Circuits
3.1 First-Order RC Circuits

Transient Response of RC Circuits
                                      vc + ic R = 0
     Switch to 2                            dvc
                                       ic = C
                     R
                                            dt
        C                                   dv
                                     vc + RC c = 0
                                             dt
                 2                                   t
 •           ο                                  −
                     K                vc = Ae       RC
             ο
                 1
         E               Initial condition vC (0+) =v C (0−) = E


                                 vc = Ee −t / RC = Ee −t / τ
                                              E −t / τ
                                     ic = −     e
                                              R
Ch3 Basic RL and RC Circuits
          3.1 First-Order RC Circuits

              Time                            τ = RC                       −
                                                                                t
                                                                                           −
                                                                                               t
                                                                                               τ
              Constant                                     vC (t ) = Ee        RC   = Ee
    5Ω
                                          R                 dvC                E
                              C
                                                                         =−
Ω
                                                             dt   t =0         τ
Ω                                     2
                     •            ο
                                                                         E
                                  ο
                                           K                 τ =−
           10V                        1                              dvC
IS ∧                          E
                                                                      dt     t =0




            5V                                                                                 R=2k
                                                                                               C=0.1µF


         SEL > >
            0V
                0s                        1 . 0 ms     2 . 0 ms                     3 . 0 ms             4 . 0 ms
                     V( 2 )
                                                                                                         Ti me
Ch3 Basic RL and RC Circuits
 3.1 First-Order RC Circuits
Ch3 Basic RL and RC
                          Circuits
3.2 First-Order RL Circuits




 Key Words:
     Words
   Transient Response of RL Circuits, Time constant
Ch3 Basic RL and RC
                          Circuits
3.2 First-Order RL Circuits


 Ideal Linear Inductor
                                                                           t
               i(t)                         dψ      di (t )           1
      The                 +         v(t ) =     =L            i (t ) = ∫ v( x)dx
      rest                                   dt      dt               L −∞
        of            L   v(t)                   di
       the                           P = iv = Li                i L (t +) =i L (t −)
                          -                      dt
     circuit
                                                                               1
                                 Energy stored:(t ) = ∫ pdt = ∫ Lidi = Li 2 (t )
                                            wL
                                                                               2

 •     One inductor and one resistor
 •     The source and resistor may be equivalent to a circuit with
       many resistors and sources.
Ch3 Basic RL
3.2 First-OrderRC Circuits
    and RL Circuits
Transient Response of RL Circuits
                                               di
                                              vL = L
     Switch to 1                               dt
                       R
                                 KVL around the loop: iR + vL = E
        L
                                                       di
                   2
                                               E=L        + iR
 •             ο                                       dt
               ο
                       K    Initial condition t = 0, i (0 + ) = i (0 − ) = 0
                   1
         E        R
          E      − t     E
     → i = (1 − e ) = (1 − e −t /τ )
                  L
          R              R                                 τ = L/ R
       → vR = iR = E (1 − e −t /τ )                     Called time
                                                        constant
                         d E    − t 
                                   R
                   di                        E R −Rt
             vL = L = L ⋅  1 − e L  = L ⋅ ⋅ ⋅ e L = Ee −t /τ
                   dt    dt  R 
                             
                                      
                                          R L
Ch3 Basic RL and RC Circuits
 3.2 First-Order RL Circuits


 Time constant
                    . i (t)



                                              t
                   0               τ
  • Indicate how fast i (t) will drop to zero.
  • It is the amount of time for i (t) to drop to zero if it is
                   dit
    dropping at the initial rate          .
                       dt   t =0
Ch3 Basic RL and RC Circuits
 3.2 First-Order RL Circuits


 Time constant
                    . i (t)



                                              t
                   0               τ
  • Indicate how fast i (t) will drop to zero.
  • It is the amount of time for i (t) to drop to zero if it is
                   dit
    dropping at the initial rate          .
                       dt   t =0
Ch3 Basic RL
               and RC
              Circuits
3.2 First-Order RL Circuits

Transient Response of RL Circuits
                                 di                 t′ : 0 → t
                              L + iR = 0
      Switch to 2                dt                 i′ : I 0 → i ( t )
                     R        di    R                     1
                                                      i( t )          t R
                                 = − dt             ∫I0 i′di′ = ∫0 − L dt ′
         L

                              i     L
                 2
  •          ο                           R                            R
                             → i = Ae
                                        − t
                                         L          ln i ′ iI(0t ) = − × ′ t0
                                                                        t
             ο
                     K                                                L
                 1                                                    R
         E                      i (t )   R                           − t
                             ln        =− t          i (t ) = I 0   e L
                                 I0      L

                                                               E
                         Initial condition    t = 0, I 0 =
                                                               R

                                        E − R t E −t / τ
                                      i= e L = e
                                        R       R
Ch3 Basic RL and RC Circuits
 3.2 First-Order RL Circuits
                            SEL > >
Transient Response of RL Circuits                     Input energy to L
                           4 . 0 mA




                   R
       L
                           2 . 0 mA


               2
 •         ο
                   K
           ο                    0A
               1                      0s              1 ms     2 ms       3 ms   4 ms
       E                                   I ( L1)

                                L export its energy , dissipated by R
                          4 . 0 mA




                          2 . 0 mA




                          SEL > >
                             0A
                                 0s                  1 ms     2 ms    3 ms       4 ms
Ch3 Basic RL and RC Circuits
                        Summary


                                         Steady        time
                        Initial Value   Value (t →   constant
                         (t=0)              ∞)           τ

     RL      Source        i0 = 0         iL =
                                                 E
                                                 R
                                                      L/R
 Circuits   (0 state)
            Source-
              free                E
                           i0 =           i=0         L/R
            (0 input)             R

             Source
    RC      (0 state)      v0 = 0         v=E          RC
 Circuits
             Source-
              free         v0 = E         v=0          RC
            (0 input)
Ch3 Basic RL and RC Circuits
                           Summary


     The Time
     Constant
 •   For an RC circuit, τ = RC
 •   For an RL circuit, τ = L/R
 •   -1/τ is the initial slope of an exponential with an initial value
     of 1
 •   Also, τ is the amount of time necessary for an exponential to
     decay to 36.7% of its initial value
Ch3 Basic RL and RC Circuits
                                Summary


•   How to determine initial conditions for a transient circuit. When a sudden
    change occurs, only two types of quantities will remain the same as
    before the change.
     – IL(t), inductor current
     – Vc(t), capacitor voltage
•   Find these two types of the values before the change and use them as
    the initial conditions of the circuit after change.
Ch3 Basic RLanand RC Circuits
 3.3 Examples (Analyzing RC circuit or RL circuit)

 Method 1
  1) Thévenin Equivalent.(Draw out C or L)


     Simplify the circuit

               Veq , Req

  2) Find Leq(Ceq), and τ = Leq/Req (τ = CeqReq)

  3) Substituting Leq(Ceq) and τ to the previous solution of differential
  equation for RC (RL) circuit .
Ch3 Basic RL
3.3 Examples RC Circuits or RL circuit)
     and (Analyzing an RC circuit
Method 2
 1) KVL around the loop → the differential equation

 2) Find the homogeneous solution.
 3) Find the particular solution.

 4) The total solution is the sum of    the particular and homogeneou
 solutions.
Ch3 Basic RL and RC Circuits
 3.3 Examples

         About Calculation for The Initial Value
 →i
                                  ↓iC               ↓iL                vC ( 0+ ) = vC ( 0− )
                        t=0
                                                                       ( R1 / / R3 ) = 2Ω
                                                                                     2Ω
                                                                   vC ( 0 ) = 8V ×         = 4V
                                                                                   2Ω + 2Ω

 i(0+)
                                      iC(0+)             iL(0+)        iL ( 0+ ) = iL ( 0− )
                                                                                   8V
                                                                      i ( 0) =           = 2A
                                  +                       +
                                                                                 2Ω + 2Ω
                                        vC(0+)=4V         vL(0+)                       4Ω
                                  _
                                                          -        iL ( 0 ) = 2A ×           = 1A
                                                     ∨
                                                         1A                          4Ω + 4Ω
Ch3 Basic RL
3.3 Examples RC Circuits or RL circuit)
     and (Analyzing an RC circuit
Method 3 (step-by-
                                    t
step)                             −
   In general, f (t ) = f (∞) + Ae τ           Given f(0+) , thus A = f(0+) – f(∞)
                                                         t
                                                     −
              f (t ) = f (∞) + [ f (0+ ) − f (∞)]e       τ

                               Initial   Stead
                                         y
  1) Draw the circuit for t = 0- and find v(0-) or i(0-)
  2) Use the continuity of the capacitor voltage, or inductor current,
        draw the circuit for t = 0+ to find v(0+) or i(0+)
  3) Find v(∞), or i(∞) at steady state
  4) Find the time constant τ
      –    For an RC circuit, τ = RC
      –    For an RL circuit, τ = L/R
  5) The solution is:     f (t ) = f (∞) + [ f (0+ ) − f (∞)]e −t /τ
Ch3 Basic RL and RC Circuits
               3.3 Examples
                   − −                   5Ω        5Ω
US                  −
      _
                   P3.1 vC (0)= 0, Find vC (t) for t ≥ 0.                      ο
                                                                                         i1   6k

                                           5Ω                                      t=0        R1    i2     i3
I1   I1       I1   I1                                                                                ∨      ∨
     Method 3:                                5Ω
                                                                         +                         R2 3k
                         +                                       t   E        9V                                C=1000PF
                                                               −
          vc ( t ) =       ( ) +  vc ( 0 ) I− ∧ c ( ∞ )  e
                         µU                                              _
                         vc 1 ∞
                         -                    v         
                                                                 τ                                              pf
                                               S
          E
                                                     3K
          vc ( 0 ) = 0, vc ( ∞ ) = 9V ×                    = 3V
                                  βI1              6K + 3K

      Apply Thevenin theorem :
                                                                                              −1
                                                            1    1 
                                                    RTh   =    +   ÷ =2KΩ
                                                            6KΩ 3KΩ
                                                    τ =RThC =2KΩ 1000pF =2 × −6
                                                                ×           10                                     s
                                                                                 t
                                                                             −
                                                    vc ( t ) =3 −3e           2× −
                                                                                10 6
                                                                                         V
Ch3 Basic RL and RC Circuits

              −
                  3.3 Examples 5Ω                5Ω
                    −
                  −
 _
                                                                                                   C=1000PF
              P3.2 vC (0)= 0, Find vC (t) for t ≥ 0.
                                5Ω                                                                  + -
I1       I1   I1                                                                                     vC
                                            5Ω                                          ο
                                                                                             t=0    R1=10k
         vc ( 0 ) = 0
                  +
                                                                             +
                      µ U1
                      -     10KΩ ∧                                       v              6V
         vc ( ∞ ) = 6V ×
                               IS
     E                              = 4.615V                                 _                                R2 3k
                         10KΩ + 3KΩ                                                                R1=20k

                                 β I1
          Apply Thevenin’s theorem :
          ↑
                             ↓
                                                                    −1
                                               1       1   30
                                        RTh =        +   ÷ = KΩ
                                               10KΩ 3KΩ    13
                                                   30
                                        τ = RThC = KΩ × 1000pF = 2.31× 10−6 s
                                                   13
                                                                             t
                                                                    −
                                        vc ( t ) = 4.615 − 4.615e       2.31×10−6
                                                                                    V
Differential Equation
         Solutions of Transient
                Circuits
                Dr. Holbert
               March 3, 2008




Lect12               99           EEE 202
1st Order Circuits
• Any circuit with a single energy storage
  element, an arbitrary number of sources,
  and an arbitrary number of resistors is a
  circuit of order 1
• Any voltage or current in such a circuit is
  the solution to a 1st order differential
  equation


Lect12               EEE 202                    100
RLC Characteristics
Element        V/I Relation             DC Steady-State
Resistor        vR (t ) = R iR (t )         V=IR
Capacitor                  d vC (t )      I = 0; open
               iC (t ) = C
                             dt

Inductor                   d iL (t )      V = 0; short
               vL (t ) = L
                             dt
              ELI and the ICE man




Lect12                        EEE 202                     101
A First-Order RC Circuit
                               vr(t)
                         +             –

                                R          +

             vs(t)   +                 C       vc(t)
                     –
                                           –



• One capacitor and one resistor in series
• The source and resistor may be
  equivalent to a circuit with many resistors
  and sources
Lect12                       EEE 202                   102
The Differential Equation
                                 vr(t)
                          +              –

                                  R          +

              vs(t)   +                  C
                      –                          vc(t)
                                             –




KVL around the loop:
             vr(t) + vc(t) = vs(t)


Lect12                        EEE 202                    103
RC Differential Equation(s)
                                     t
                            1
  From KVL:       R i (t ) + ∫ i ( x)dx = vs (t )
                            C −∞
Multiply by C;        di (t )              dvs (t )
take derivative    RC         + i (t ) = C
                       dt                    dt
Multiply by R;       dvr (t )                dvs (t )
note vr=R·i       RC          + vr (t ) = RC
                       dt                      dt

 Lect12                    EEE 202                      104
A First-Order RL Circuit
                                   +


             is(t)   R         L       v(t)


                                   –



• One inductor and one resistor in parallel
• The current source and resistor may be
  equivalent to a circuit with many resistors
  and sources
Lect12               EEE 202                    105
The Differential Equations
                                         +


              is(t)        R         L       v(t)


                                         –



KCL at the top node:
                       t
             v(t ) 1
                  + ∫ v( x)dx = is (t )
              R    L −∞

Lect12                     EEE 202                  106
RL Differential Equation(s)
                                 t
                  v(t ) 1
 From KCL:             + ∫ v( x)dx = is (t )
                   R    L −∞

Multiply by L;    L dv(t )             dis (t )
take derivative
                           + v(t ) = L
                  R dt                   dt




Lect12                 EEE 202                    107
1st Order Differential Equation
Voltages and currents in a 1st order circuit
satisfy a differential equation of the form
             dx(t )
                    + a x(t ) = f (t )
              dt

where f(t) is the forcing function (i.e., the
independent sources driving the circuit)


Lect12                 EEE 202                  108
The Time Constant (τ)
• The complementary solution for any first
  order circuit is
                               −t /τ
                vc (t ) = Ke

• For an RC circuit, τ = RC
• For an RL circuit, τ = L/R
• Where R is the Thevenin equivalent
  resistance
Lect12               EEE 202                 109
What Does vc(t) Look Like?


                        τ = 10-4




Lect12              EEE 202           110
Interpretation of τ
• The time constant, τ, is the amount of time
  necessary for an exponential to decay to
  36.7% of its initial value
• -1/τ is the initial slope of an exponential
  with an initial value of 1




Lect12              EEE 202                111
Applications Modeled by
          a 1st Order RC Circuit
• The windings in an electric motor or
  generator
• Computer RAM
     – A dynamic RAM stores ones as charge on a
       capacitor
     – The charge leaks out through transistors
       modeled by large resistances
     – The charge must be periodically refreshed


Lect12                  EEE 202                    112
Important Concepts
• The differential equation for the circuit
• Forced (particular) and natural
  (complementary) solutions
• Transient and steady-state responses
• 1st order circuits: the time constant (τ)
• 2nd order circuits: natural frequency (ω0)
  and the damping ratio (ζ)

Lect12               EEE 202                   113
The Differential Equation
• Every voltage and current is the solution to
  a differential equation
• In a circuit of order n, these differential
  equations have order n
• The number and configuration of the
  energy storage elements determines the
  order of the circuit
• n ≤ number of energy storage elements

Lect12               EEE 202                114
The Differential Equation
• Equations are linear, constant coefficient:
            d n x(t )         d n −1 x(t )
         an      n
                      + an −1       n −1
                                           + ... + a0 x(t ) = f (t )
              dt                dt
• The variable x(t) could be voltage or current
• The coefficients an through a0 depend on the
  component values of circuit elements
• The function f(t) depends on the circuit elements
  and on the sources in the circuit


Lect12                            EEE 202                              115
The Differential Equation
• Equations are linear, constant coefficient:
            d n x(t )         d n −1 x(t )
         an      n
                      + an −1       n −1
                                           + ... + a0 x(t ) = f (t )
              dt                dt
• The variable x(t) could be voltage or current
• The coefficients an through a0 depend on the
  component values of circuit elements
• The function f(t) depends on the circuit elements
  and on the sources in the circuit


Lect12                            EEE 202                              116
Building Intuition
• Even though there are an infinite number
  of differential equations, they all share
  common characteristics that allow intuition
  to be developed:
   – Particular and complementary solutions
   – Effects of initial conditions



Lect12               EEE 202                117
Differential Equation Solution
• The total solution to any differential
  equation consists of two parts:
                 x(t) = xp(t) + xc(t)
• Particular (forced) solution is xp(t)
   – Response particular to a given source
• Complementary (natural) solution is xc(t)
   – Response common to all sources, that
     is, due to the “passive” circuit elements

Lect12                EEE 202                    118
Forced (or Particular) Solution
• The forced (particular) solution is the solution to
  the non-homogeneous equation:
            d n x(t )         d n −1 x(t )
         an      n
                      + an −1       n −1
                                           + ... + a0 x(t ) = f (t )
              dt                dt
• The particular solution usually has the form of a
  sum of f(t) and its derivatives
     – That is, the particular solution looks like the forcing
       function
     – If f(t) is constant, then x(t) is constant
     – If f(t) is sinusoidal, then x(t) is sinusoidal


Lect12                             EEE 202                             119
Natural/Complementary Solution

• The natural (or complementary) solution is
  the solution to the homogeneous
  equation:
          n           n −1
            d x(t )         d x(t )
         an     n
                    + an −1     n −1
                                     + ... + a0 x(t ) = 0
             dt              dt


• Different “look” for 1st and 2nd order ODEs


Lect12                       EEE 202                        120
First-Order Natural Solution
• The first-order ODE has a form of
              dxc (t ) 1
                      + xc (t ) = 0
                dt     τ
• The natural solution is
                                         −t /τ
                         xc (t ) = Ke
• Tau (τ) is the time constant
          • For an RC circuit, τ = RC
          • For an RL circuit, τ = L/R

Lect12                         EEE 202           121
Second-Order Natural
               Solution
• The second-order ODE has a form of
            d 2 x(t )         dx(t )
                 2
                      + 2ζω 0        + ω 0 x(t ) = 0
                                         2

              dt               dt
• To find the natural solution, we solve the
  characteristic equation:
                 s + 2ζω 0 s + ω = 0
                   2                   2
                                       0
    which has two roots: s1 and s2
• The complementary solution is (if we’re lucky)
             xc (t ) = K1e + K 2 e
                          s1t      s2t


Lect12                      EEE 202                    122
Initial Conditions
• The particular and complementary solutions
  have constants that cannot be determined
  without knowledge of the initial conditions
• The initial conditions are the initial value of the
  solution and the initial value of one or more of its
  derivatives
• Initial conditions are determined by initial
  capacitor voltages, initial inductor currents, and
  initial source values

Lect12                   EEE 202                    123
2nd Order Circuits
• Any circuit with a single capacitor, a single
  inductor, an arbitrary number of sources,
  and an arbitrary number of resistors is a
  circuit of order 2
• Any voltage or current in such a circuit is
  the solution to a 2nd order differential
  equation


Lect12               EEE 202                 124
A 2nd Order RLC Circuit
                         i (t)


                                    R

             vs(t)   +                     C
                     –
                                    L




The source and resistor may be equivalent
to a circuit with many resistors and sources

Lect12                           EEE 202       125
The Differential Equation
                       i(t)       + vr(t)    –


                                      R          +
                          +                  C       vc(t)
                          –
               vs(t)
                                     vl(t)       –
                              –              +


                                      L


  KVL around the loop:
           vr(t) + vc(t) + vl(t) = vs(t)

Lect12                            EEE 202                    126
RLC Differential Equation(s)
From KVL:              t
                    1               di (t )
          R i (t ) + ∫ i ( x)dx + L         = vs (t )
                    C −∞             dt
Divide by L, and take the derivative
                                       2
         R di (t ) 1            d i (t ) 1 dvs (t )
                  +    i (t ) +     2
                                        =
         L dt       LC           dt       L dt


Lect12                       EEE 202                    127
The Differential Equation
Most circuits with one capacitor and inductor
are not as easy to analyze as the previous
circuit. However, every voltage and current
in such a circuit is the solution to a
differential equation of the following form:
          2
         d x(t )         dx(t )
             2
                 + 2ζω 0        + ω0 x(t ) = f (t )
                                   2

          dt              dt

Lect12                      EEE 202                   128
Class Examples
• Drill Problems P6-1, P6-2




• Suggestion: print out the two-page “First
  and Second Order Differential Equations”
  handout from the class webpage


Lect12              EEE 202                   129

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Unit i

  • 1. THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS
  • 2. A Resistor in the s Domain + R i v v=Ri (Ohm’s Law). + R I V V(s)=RI(s
  • 3. An Inductor in the s Domain di Initial current of I 0 v=L dt + V(s)=L[sI(s)-i(0-)]=sLI(s)-LI0 v L i V ( s) I0 I ( s) = + sL s I + sL I + V sL I0 V s + LI0
  • 4. A Capacitor in the s Domain dv Initially charged to V0 volts. i=C dt − I ( s ) = C [ sV ( s ) − v (0 )] = sCV ( s ) − CV0  1  V0 V ( s) =  ÷I ( s ) +  sC  I s + I 1/sC + i + 1/sC CV0 v C V V + V0/s
  • 5. The Natural Response of an RC Circuit t=0 + 1/sC I + C + R i v R V0 V0/s + V V0 1 = I ( s ) + RI ( s ) s sC V0 CV0 R I ( s) = = RCs + 1 s + ( 1 RC ) V0 − − i = e u( t ) ⇒ v = Ri = V0e u( t ) t t RC RC R
  • 6. The Step Response of a Parallel Circuit
  • 7. ?
  • 8. The Step Response of a Parallel Circuit V V I dc sCV + + = R sL s I dc C V= s +( 2 1 RC ) s + ( 1 LC ) I dc LC IL = s[ s + ( 2 1 RC ) s + ( 1 LC ) ]
  • 9. 384 × 105 IL = s( s 2 + 64000 s + 16 × 108 ) 384 × 105 IL = s( s + 32000 − j 24000)( s + 32000 + j 24000) * K1 K2 K2 IL = + + s s + 32000 − j 24000 s + 32000 + j 24000 384 × 105 K1 = = 24 × 10−3 16 × 108 384 × 105 K2 = = 20 × 10−3 126.870 ( −32000 + j 24000)( j 48000) i L ( t ) = [24 + 40e −32000 t cos(24000t + 126.870 )]u( t )mA
  • 10. Transient Response of a Parallel RLC Circuit Replacing the dc current source with a sinusoidal current source sI m i g = I m cos ω t A ⇒ I g ( s ) = 2 2 s +ω V ( s) = ( 1C ) s I g ( s) 2 s + ( 1 RC ) s + ( 1 LC ) V ( s) = ( ) Im C s2 ( s 2 + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )] I L ( s) = V ( s) = 2 LC s( Im ) sL ( s + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )]
  • 11. I m = 24mA, ω = 40000rad / s 384 × 105 s I L (s) = 2 ( s + 16 × 108 )( s 2 + 64000 s + 16 × 108 ) * * K1 K1 K2 K2 I L (s) = + + + s − j 40000 s + j 40000 s + 32000 − j 24000 s + 32000 + j 24000 384 × 105 ( j 40000) K1 = = 7.5 × 10−3 − 900 ( j 80000)(32000 + j16000)(32000 + j 64000) 384 × 105 ( −32000 + j 24000) K2 = = 12.5 × 10−3 900 ( −32000 − j16000)( −32000 + j 64000)( j 48000) i L ( t ) = (15sin 40000t − 25e −32000 t sin 24000t )u( t )mA i Lss = 15sin 40000t mA
  • 13. ?
  • 14. Mesh Analysis (cont.) 336 = (42 + 8.4 s ) I1 − 42 I 2 s 0 = −42 I1 + (90 + 10 s ) I 2 40( s + 9) 15 14 1 I1 = = − − s( s + 2)( s + 12) s s + 2 s + 12 168 7 8.4 1.4 I2 = = − + s( s + 2)( s + 12) s s + 2 s + 12
  • 15. −2 t −12 t i1 = (15 − 14e −e )u( t ) A, −2 t −12 t i2 = (7 − 8.4e + 1.4e )u( t ) A. 336(90) i1 (∞ ) = = 15 A, 42(48) 15(42) i2 ( ∞ ) = =7A 90
  • 16. Thevenin’s Theorem Use the Thevenin’s theorem to find v c (t)
  • 17. ?
  • 18. Thevenin’s Theorem (cont.) (480 / s )(0.002 s ) 480 VTh = = 20 + 0.002 s s + 10 4 0.002 s(20) 80( s + 7500) ZTh = 60 + = 20 + 0.002 s s + 10 4
  • 19. 4 480 /( s + 10 ) IC = 4 5 [80( s + 7500) /( s + 10 )] + [(2 × 10 ) / s] 6s 6s IC = 2 6 = 2 s + 10000 s + 25 × 10 ( s + 5000) −30000 6 IC = + ( s + 5000) 2 s + 5000 −5000 t −5000 t ic ( t ) = ( −30000te + 6e )u( t ) A 5 5 1 2 × 10 6s 12 × 10 Vc = IC = 2 = sC s ( s + 5000) ( s + 5000)2 5 −5000 t vc ( t ) = 12 × 10 te u( t )V
  • 20. MUTUAL INDUCTANCE EXAMPLE i2(t)=? − 60 − i1 (0 ) = = 5 A, i2 (0 ) = 0 12
  • 21. ?
  • 22. MUTUAL INDUCTANCE EXAMPLE Using the T-equivalent of the inductors, and s-domain equivalent gives the following circuit
  • 23. (3 + 2 s ) I1 + 2 sI 2 = 10 2 sI1 + (12 + 8 s ) I 2 = 10 2.5 1.25 1.25 I2 = = − ( s + 1)( s + 3) s + 1 s + 3 −t −3 t i2 ( t ) = 1.25(e −e )u( t ) A
  • 24. THE TRANSFER FUNCTION The transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input when all the initial conditions are zero. Y ( s) H ( s) = Y(s) is the Laplace transform of the output, X ( s) X(s) is the Laplace transform of the input.
  • 25. THE TRANSFER FUNCTION (cont.) I ( s) 1 H1 ( s ) = = Vg ( s ) R + sL + 1/ sC sC = 2 s LC + RCs + 1 V ( s) 1 H 2 ( s) = = 2 Vg ( s ) s LC + RCs + 1
  • 26. EXAMPLE Find the transfer function V 0 /V g and determine the poles and zeros of H(s).
  • 27. ?
  • 28. EXAMPLE V0 − Vg V0 V0 s + + 6 =0 1000 250 + 0.05 s 10 1000( s + 5000) V0 = 2 V 6 g s + 6000 s + 25 × 10 V0 1000( s + 5000) H ( s) = = 2 Vg s + 6000 s + 25 × 106 p1 = −3000 + j 4000, p2 = −3000 − j 4000 z1 = −5000
  • 29. Assume that v g (t)=50tu(t). Find v 0 (t). Identify the transient and steady-state components of v 0 (t). 1000( s + 5000) 50 V0 ( s ) = H ( s )V g ( s ) = 2 6 2 ( s + 6000 s + 25 × 10 ) s * K1 K1 K2 K3 = + + 2 + s + 3000 − j 4000 s + 3000 + j 4000 s s −4 0 −4 K1 = 5 5 × 10 79.7 , K 2 = 10, K 3 = −4 × 10 −4 −3000 t 0 v0 = [10 5 × 10 e cos(4000t + 79.7 ) −4 + 10t − 4 × 10 ]u( t )V
  • 30. The transient component is generated by the poles of the transfer function and it is: −4 −3000 t 0 10 5 × 10 e cos(4000t + 79.7 ) The steady-state components are generated by the poles of the driving function (input): −4 (10t − 4 × 10 )u( t )
  • 31. Time Invariant Systems If the input delayed by a seconds, then L{ x ( t − a )u( t − a )} = e − as X (s) − as Y ( s ) = H ( s ) X ( s )e y( t ) = L−1 { Y ( s )} = y( t − a )u( t − a ) Therefore, delaying the input by a seconds simply delays the response function by a seconds. A circuit that exhibits this characteristic is said to be time invariant.
  • 32. Impulse Response If a unit impulse source drives the circuit, the response of the circuit equals the inverse transform of the transfer function. x( t ) = δ (t ) ⇒ X ( s ) = 1 Y ( s) = H ( s) y( t ) = L −1 { H ( s )} = h(t ) Note that this is also the natural response of the circuit because the application of an impulsive source is equivalent to instantaneously storing energy in the circuit.
  • 33. CONVOLUTION INTEGRAL x(t) y(t) x(t) y(t) t N a b t a b Circuit N is linear with no initial stored energy. If we know the form of x(t), then how is y(t) described? To answer this question, we need to know something about N. Suppose we know the impulse response of the system. y( t ) = h( t ) x(t ) = δ (t ) (1) x(t) y(t) t N t
  • 34. Instead of applying the unit impulse at t=0, let us suppose that it is applied at t=λ. The only change in the output is a time delay. δ (t − λ ) N h( t − λ ) Next, suppose that the unit impulse has some strength other than unity. Let the strength be equal to the value of x(t) when t= λ. Since the circuit is linear, the response should be multiplied by the same constant x(λ) x (λ )δ ( t − λ ) N x (λ )h( t − λ )
  • 35. Now let us sum this latest input over all possible values of λ and use the result as a forcing function for N. From the linearity, the response is the sum of the responses resulting from the use of all possible values of λ +∞ +∞ ∫−∞ x( λ )δ ( t − λ )d λ N ∫−∞ x(λ )h( t − λ )d λ From the sifting property of the unit impulse, we see that the input is simply x(t) +∞ X(t) N ∫−∞ x(λ )h( t − λ )d λ
  • 36. Our question is now answered. When x(t) is known, and h(t), the unit impulse response of N is known, the response is expressed by +∞ y( t ) = ∫−∞ x ( λ )h( t − λ )d λ This important relation is known as the convolution integral . It is often abbreviated by means of y( t ) = x ( t ) * h( t ) Where the asterisk is read “convolved with”. If we let z=t-λ, then dλ=-dz, and the expression for y(t) becomes −∞ +∞ y( t ) = ∫∞ − x ( t − z )h( z )dz = ∫−∞ x ( t − z )h( z )dz ∞ ∞ y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
  • 37. ∞ y( t ) = x( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
  • 38. Convolution and Realizable Systems For a physically realizable system, the response of the system cannot begin before the forcing function is applied. Since h(t) is the response of the system when the unit impulse is applied at t=0, h(t) cannot exist for t<0. It follows that, in the second integral, the integrand is zero when z<0; in the first integral, the integrand is zero when (t-z) is negative, or when z>t. Therefore, for realizable systems the convolution integral becomes t ∞ y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫0 x ( t − z )h( z )dz
  • 39. EXAMPLE x(t) −t 1 t h(t) y(t) h( t ) = 2e u( t ) x ( t ) = u( t ) − u( t − 1) ∞ y( t ) = x ( t ) * h( t ) = ∫0 x ( t − z )h( z )dz ∞ −z = ∫0 [u( t − z ) − u( t − z − 1)][2e u( z )]dz
  • 40. Graphical Method of Convolution
  • 41. Since h(z) does not exist prior to t=0 and vi(t-z) does not exist for z>t, product of these functions has nonzero values only in the interval of 0<z<t for the case shown where t<1. t y( t ) = ∫0 2e − z dz = 2(1 − e − t ) 0≤ t ≤1 When t>1, the nonzero values for the product are obtained in the interval (t-1)<z<t. t −z −t y( t ) = ∫t −1 2e dz = 2(e − 1)e t >1
  • 42. EXAMPLE Apply a unit-step function, x(t)=u(t), as the input to a system whose impulse response is h(t) and determine the corresponding output y(t)=x(t)*h(t).
  • 43. ?
  • 45. When t<0, there is no overlap and y(t)=0 for t<0 For 0<t<1, the curves overlap from z=0 to z=t and product is 1. Thus, t y( t ) = ∫0 1dz = t 0< t <1 When 1<t<2, h(t-z) has slid far enough to the right to bring under the step function that part of the negative square extending from 0 to z=t-1. Thus, t −1 t t −1 t y( t ) = ∫0 −1dz + ∫t −11dz = − z 0 + z t −1 = 2 − t , 1< t < 2
  • 46. Finally, when t>2, h(t-z) has slid far enough to the right so that it lies entirely to the right of z=0 t −1 t y( t ) = ∫t − 2 −1dz + ∫t −11dz = 0 t>2
  • 47. Convolution and the Laplace Transform Let F 1 (s) and F 2 (s) be the Laplace transforms of f1(t) and f2(t), respectively. Now, consider the laplace transform of f1(t)*f2(t), L{ f1 ( t ) * f 2 ( t )} = L {∫ ∞ f (λ ) f 2 (t −∞ 1 − λ )d λ } Since we are dealing with the time functions that do not exist prior to t=0-, the lower limit can be changed to 0- ∞ ∞ − st L{ f1 ( t ) * f 2 ( t )} = ∫0 [ ∫ − 0− e f1 ( λ ) f 2 ( t − λ )dt ]d λ
  • 48. f1(λ) does not depend on t, and it can be moved outside the inner integral ∞ ∞ − st L{ f1 ( t ) * f 2 ( t )} = ∫0 f1 ( λ )[ ∫ − 0− e f 2 (t − λ )dt ]d λ ∞ ∞ − s( x + λ ) = ∫0 f1 (λ )[ ∫ − 0− e f 2 ( x )dx ]d λ ∞ − sλ ∞ − sx = ∫0 f1 (λ )e − [∫ 0− e f 2 ( x )dx ]d λ ∞ = ∫0 f1 (λ )e − sλ [ F2 ( s )]d λ − ∞ = F2 ( s )∫0 f1 (λ )e − sλ d λ − = F1 ( s ) ×F2 ( s )
  • 49. STEADY-STATE SINUSOIDAL RESPONSE If the input of a circuit is a sinusoidal function x ( t ) = A cos(ω t + θ ) x ( t ) = A cos ω t cosθ − A sin ω t sin θ ( A cosθ ) s ( A sin θ )ω X ( s) = 2 2 − 2 2 s +ω s +ω A(s cosθ − ω sinθ ) = 2 2 s +ω A(s cosθ − ω sinθ ) Y ( s) = H ( s) 2 2 s +ω
  • 50. The partial fraction expansion of Y(s) is * K1 K1 Y ( s) = + + ∑ terms generated by the poles of H(s) s − jω s + jω If the poles of H(s) lie in the left half of the s plane, the corresponding time-domain terms approach zero as t increases and they do not contribute to the steady-state response. Thus only the first two terms determine the steady-state response.
  • 51. H ( s ) A( s cosθ − ω sinθ ) K1 = s + jω s = jω H ( jω ) A( jω cosθ − ω sinθ ) = 2 jω H ( jω ) A(cosθ + j sinθ ) 1 = = H ( jω ) Ae jθ 2 2 H ( jω ) = H ( jω ) e jφ (ω ) A j[θ +φ (ω )] K1 = H ( jω ) e 2 yss ( t ) = A H ( jω ) cos[ω t + θ + φ (ω )]
  • 52. EXAMPLE If the input is 120 cos(5000t+300)V, find the steady-state expression for v0 1000( s + 5000) H ( s) = 2 s + 6000 s + 25 × 106 1000(5000 + j 5000) H ( j 5000) = 6 6 −25 × 10 + j 5000(6000) + 25 × 10 1 + j1 2 = = − 450 j6 6 120 2 v0 ss = cos(5000t + 300 − 450 ) 6 =20 2 cos(5000t − 150 )V
  • 53. THE IMPULSE FUNCTION IN CIRCUIT ANALYSIS The capacitor is charged to an initial voltage V0 at the time the switch is closed. Find the expression for i(t) as R 0 V0 V0 s I= = R R + ( 1 sC ) + ( 1 sC ) 1 2 s+( 1 RC e ) C1C 2  V0 − t / RC  Ce = i (t ) =  e e ÷u( t ) C1 + C 2  R  As R decreases, the initial current (V0/R) increases and the time constant (RCe) decreases. Apparently i is approaching an impulse function as R approaches zero.
  • 54. The total area under the i versus t curve represents the total charge transferred to C2 after the switch is closed. ∞ V0 Area=q = ∫0 − e − t / RC dt = V0C e e R Thus, as R approaches zero, the current approaches an impulse strength V0Ce. i → V0C eδ ( t )
  • 55. Series Inductor Circuit Find v0. Note that opening the switch forces an instantaneous change in the current L2. i1 (0− ) = 10 A, i2 (0− ) = 0 V0 V0 − [(100 s ) + 30] + =0 2 s + 15 3 s + 10 40( s + 7.5) 12( s + 7.5) V0 = + s( s + 5) s+5 60 10 V0 = + 12 + s s+5 −5 t v0 ( t ) = 12δ ( t ) + (60 + 10e )u( t )V
  • 56. Does this solution make sense? To answer this question, first let us determine the expression for the current. (100 s ) + 30 4 2 I= = + 5 s + 25 s s+5 −5 t i ( t ) = (4 + 2e )u( t ) A Before the switch is opened, current through L1 is 10A and in L2 is 0 A, after the switch is opened both currents are 6A. Then the current in L 1 changes instantaneously from 10 A to 6 A, while the current in L 2 changes instantaneously from 0 to 6 A. How can we verify that these instantaneous jumps in the inductor current make sense in terms of the physical behavior of the circuit?
  • 57. Switching operation places two inductors in series. Any impulsive voltage appearing across the 3H inductor must be balanced by an impulsive voltage across the 2H inductor. Faraday’s law states that the induced voltage is proportional to the change in flux linkage ( v = d λ ) before switching dt λ = L1i1 + L2i2 = 3(10) + 2(0) = 30 Wb-turns After switching λ = ( L1 + L2 )i (0+ ) = 5i (0+ ) + 30 i (0 ) = = 6A 5 Thus the solution agrees with the principle of the conservation of flux linkage.
  • 58. Impulsive Sources When the voltage source is applied, the initial energy in the inductor is zero; therefore the initial current is zero. There is no voltage drop across R, so the impulsive source appears directly across L 1 t + V0 i = ∫0 V0δ ( x )dx ⇒ i (0 ) = − A L L Thus, in an infinitesimal Current in the circuit decays to moment, the impulsive voltage zero in accordance with the source has stored natural response of the circuit 2 V0 − ( 1  V0  1 V02 i= e R )t u( t ) w= L ÷ = L J 2  L 2 L L
  • 59. EXAMPLE Find i(t) and v0(t) for t>0 50 + ( 100 s ) + 30 I= 25 + 5 s 12 4 = + s+5 s i ( t ) = (12e −5 t + 4)u( t ) A 60 60 V0 = (15 + 2 s ) I = 32 + + s+5 s −5 t v0 ( t ) = 32δ ( t ) + (60e + 60)u( t )V
  • 60. Transfer Functions 1. The Laplace transform 2. Solution of linear differential equations 3. Transient response example
  • 61. The Laplace Transform • Definition ∞ F ( s ) = L[ f (t )] = ∫ f (t )e − st dt 0 – Time (t) is replaced by a new independent variable (s) – We call s the Laplace transform variable • The Laplace domain – Often more convenient to work in Laplace domain than time domain – Time domain  ordinary differential equations in t – Laplace domain  algebraic equations in s • General solution approach – Formulate model in time domain – Convert model to Laplace domain – Solve problem in Laplace domain – Invert solution back to time domain
  • 62. Laplace Transform of Selected Functions • Constant function: f(t) = a ∞ ∞ a  a a F ( s ) = L(a ) = ∫ ae − st dt = − e − st = 0 −−  = 0 s 0  s s • Exponential function: f(t) = e-bt F ( s ) = L (e −bt ∞ ) = ∫ e e dt = ∫ e 0 −bt − st 0 ∞ −( b + s ) t dt = − 1 b+s [ e −( b + s ) t ] ∞ 0 = 1 s +b • Derivatives and integrals  df  ∞ df − st ∞ ∞ L  = ∫ e dt = ∫ f (t )e − st sdt + f (t )e − st = sF ( s ) − f (0)  dt  0 dt 0 0 dn f  L n  = s n F ( s ) − s n −1 f (0) − s n −2 f (1) (0) −  − sf ( n −2 ) (0) − f ( n −1) (0)  dt     t f (t*)dt * = ∞  t f (t*)dt *e − st dt = 1 F ( s ) L ∫ 0   ∫0 ∫0     s
  • 63. Properties of Laplace Transforms • Superposition L[af (t ) + bg (t )] = aF ( s ) + bG ( s ) • Final value theorem If lim y (t ) exists ⇒ lim y (t ) = lim[ sY ( s )] t →∞ t →∞ s →0 • Initial value theorem lim y (t ) = lim[ sY ( s )] t →0 s →∞ • Time delay f d (t ) = f (t − t0 ) S (t − t0 ) L → Fd ( s ) = e − st0 F ( s ) − st 0 L−1 Fd ( s ) = e F ( s) → f d (t ) = f (t − t0 ) S (t − t0 )
  • 64. Linear ODE Example • ODE dy 5 + 4y = 2 y ( 0) = 1 dt • Laplace transform 2 5[ sY ( s ) − y (0)] + 4Y ( s ) = s • Substitute y(0) & rearrange 5s + 2 s + 0.4 Y (s) = = s (5s + 4) s ( s + 0.8) • Inverse Laplace transform −1  s + 0.4  −1 y (t ) = L [Y ( s )] = L   s ( s + 0.8)  
  • 65. Linear ODE Example cont. • Table 3.1  −1 s + b3  b3 − b1 −b1t b3 − b2 −b2t L  = e + e  ( s + b1 )( s + b2 )  b2 − b1 b1 − b2 • Our problem −1 s + 0.4  y (t ) = L   ⇒ b1 = 0 b2 = 0.8 b3 = 0.4  s ( s + 0.8)  • Substitute and simplify  −1 s + 0.4  0.4 − 0 −0t 0.4 − 0.8 −0.8t L  = e + e = 0.5 + 0.5e −0.8t  ( s + 0)( s + 0.8)  0.8 − 0  0 − 0.8
  • 66. General ODE Solution Procedure • Procedure – Transform to Laplace domain – Solve resulting algebraic equations – Transform solution back to time domain • Partial fraction expansion – Necessary when inverse Laplace transform not tabularized – Break complex functions into simpler tabularized functions
  • 67. Partial Fraction Example • Partial fraction expansion s +5 s +5 α α Y (s) = = = 1 + 2 s 2 + 5s + 4 ( s + 1)( s + 4) s + 1 s + 4 • Determination of coefficients s +5 4 s +5 1 α1 = = α2 = =− s + 4 s =−1 3 s + 1 s =−4 3 • Inverse Laplace transform −1 −1 4 / 3 1 / 3  4 −t 1 − 4 t y (t ) = L [Y ( s )] = L  −  = 3e −3e  s +1 s + 4 
  • 68. Repeated Factor Example s +1 α1 α2 α3 Y (s) = = + + s ( s + 2) 2 s + 2 ( s + 2) 2 s s +1 1 s +1 1 1 α2 = = α3 = = α1 = − s s =−2 2 ( s + 2) 2 s =0 4 4 s +1 α1 α2 α3 −1 / 4 1/ 2 1/ 4 Y ( s) = = + + = + + s ( s + 2) 2 s + 2 ( s + 2) 2 s s + 2 ( s + 2) 2 s −1 1 −2t 1 −2t 1 y (t ) = L [Y ( s )] = − e + te + S (t ) 4 2 4
  • 69. Quadratic Factor Example s +1 α α2 α s + α4 Y (s) = = 1 + 2 + 23 s 2 ( s 2 + 4 s + 5) s s s + 4s + 5 s + 1 = α1s ( s 2 + 4 s + 5) + α 2 ( s 2 + 4 s + 5) + (α 3 s + α 4 ) s 2 (α1 + α 3 ) s 3 + (4α1 + α 2 + α 4 ) s 2 + (5α1 + 4α 2 − 1) s + (5α 2 − 1) = 0 s +1 α α2 α s + α4 0.04 0.2 − 0.04s − 0.36 Y ( s) = = 1 + 2 + 23 = + 2 + 2 s ( s + 4 s + 5) s s 2 2 s + 4s + 5 s s s + 4s + 5 − 0.04 s − 0.36 − 0.04( s + 2) − 0.28 = + s + 4s + 5 2 ( s + 2) + 1 ( s + 2) 2 + 1 2 y (t ) = L−1[Y ( s )] = 0.04S (t ) + 0.2t − 0.04e −2t cos t − 0.28e −2t sin t
  • 70. Transient Response Example • Component balance dc1 dc1 V = q (ci − c1 ) ⇒ 4 + 2c1 = 2ci c1 (0) = 0 dt dt • Step input 0 t ≤ 0 5 ci (t ) =  ⇒ Ci ( s ) = 5 t > 0 s
  • 71. Transient Response Example cont. • Laplace transform 2 4[ sC1 ( s ) − 0] + 2C1 ( s ) = 2Ci ( s ) ⇒ C1 ( s ) = Ci ( s ) 4s + 2 • Substitute input 2 5 5 C1 ( s ) = = 4 s + 2 s s (2 s + 1) • Inverse Laplace transform   −1 c1 (t ) = L  5 s (2 s + 1)  ( = 5 1 − e −t / 2 )  
  • 72. Simulink Solution inlet Inlet 1 2 outlet 4s+2 Step Tank To Workspace 6 >> plot(tout,inlet) Input Output >> hold 5 >> plot(tout,outlet,'r') Concentration (kmol/m ) 3 4 3 >> axis([0 15 0 6]) >> ylabel('Concentration 2 (kmol/m^3)') 1 >> xlabel('Time (min)') >> legend(‘Input','Output') 0 0 5 10 15 Time (min)
  • 73. Engineering Circuit Analysis Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits 3.2 First-Order RL Circuits 3.3 Examples References: Hayt-Ch5, 6; Gao-Ch5; References
  • 74. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Key Words: Words Transient Response of RC Circuits, Time constant
  • 75. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Key Words: Words Transient Response of RC Circuits, Time constant
  • 76. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits • Used for filtering signal by blocking certain frequencies and passing others. e.g. low-pass filter • Any circuit with a single energy storage element, an arbitrary number of sources and an arbitrary number of resistors is a circuit of order 1. • Any voltage or current in such a circuit is the solution to a 1st order differential equation. Ideal Linear Capacitor dq dv i (t ) = =c vc (t +) =v C (t ) dt dt 1 2 Energy stored w = ∫ pdt = ∫ cvdv = cv 2 A capacitor is an energy storage device → memory device.
  • 77. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits vr(t) + - R + + vs(t) C vc(t) - - • One capacitor and one resistor • The source and resistor may be equivalent to a circuit with many resistors and sources.
  • 78. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Transient Response of RC Circuits E − vc ic = Switch is thrown to 1 R KVL around the loop: ic R + vC = E R C dvc C R + vc = E • ο 2 dt t − ο K vC = Ae RC +E 1 E Initial condition vC (0+) =v C (0−) = 0 A = −E t t − − vC = E (1 − e RC ) = E (1 − e τ ) τ = RC dvc E − t Called time constant ic = C = e τ dt R
  • 79. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Time τ = RC Constant 5Ω − t dvc E −t / τ R vC = E (1 − e τ ) = e C dt τ 2 dvc E E • ο t =0 = →τ = 10V ο K dt τ dvc 1 t =0 ∧ E dt 5V R=2k C=0.1µF SEL > > 0V 0s RC 1 ms 2 ms 3 ms 4 ms V( 2 )
  • 80. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Transient Response of RC Circuits vc + ic R = 0 Switch to 2 dvc ic = C R dt C dv vc + RC c = 0 dt 2 t • ο − K vc = Ae RC ο 1 E Initial condition vC (0+) =v C (0−) = E vc = Ee −t / RC = Ee −t / τ E −t / τ ic = − e R
  • 81. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Time τ = RC − t − t τ Constant vC (t ) = Ee RC = Ee 5Ω R dvC E C =− Ω dt t =0 τ Ω 2 • ο E ο K τ =− 10V 1 dvC IS ∧ E dt t =0 5V R=2k C=0.1µF SEL > > 0V 0s 1 . 0 ms 2 . 0 ms 3 . 0 ms 4 . 0 ms V( 2 ) Ti me
  • 82. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits
  • 83. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Key Words: Words Transient Response of RL Circuits, Time constant
  • 84. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Ideal Linear Inductor t i(t) dψ di (t ) 1 The + v(t ) = =L i (t ) = ∫ v( x)dx rest dt dt L −∞ of L v(t) di the P = iv = Li i L (t +) =i L (t −) - dt circuit 1 Energy stored:(t ) = ∫ pdt = ∫ Lidi = Li 2 (t ) wL 2 • One inductor and one resistor • The source and resistor may be equivalent to a circuit with many resistors and sources.
  • 85. Ch3 Basic RL 3.2 First-OrderRC Circuits and RL Circuits Transient Response of RL Circuits di vL = L Switch to 1 dt R KVL around the loop: iR + vL = E L di 2 E=L + iR • ο dt ο K Initial condition t = 0, i (0 + ) = i (0 − ) = 0 1 E R E − t E → i = (1 − e ) = (1 − e −t /τ ) L R R τ = L/ R → vR = iR = E (1 − e −t /τ ) Called time constant d E  − t  R di E R −Rt vL = L = L ⋅  1 − e L  = L ⋅ ⋅ ⋅ e L = Ee −t /τ dt dt  R      R L
  • 86. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Time constant . i (t) t 0 τ • Indicate how fast i (t) will drop to zero. • It is the amount of time for i (t) to drop to zero if it is dit dropping at the initial rate . dt t =0
  • 87. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Time constant . i (t) t 0 τ • Indicate how fast i (t) will drop to zero. • It is the amount of time for i (t) to drop to zero if it is dit dropping at the initial rate . dt t =0
  • 88. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Transient Response of RL Circuits di t′ : 0 → t L + iR = 0 Switch to 2 dt i′ : I 0 → i ( t ) R di R 1 i( t ) t R = − dt ∫I0 i′di′ = ∫0 − L dt ′ L i L 2 • ο R R → i = Ae − t L ln i ′ iI(0t ) = − × ′ t0 t ο K L 1 R E i (t ) R − t ln =− t i (t ) = I 0 e L I0 L E Initial condition t = 0, I 0 = R E − R t E −t / τ i= e L = e R R
  • 89. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits SEL > > Transient Response of RL Circuits Input energy to L 4 . 0 mA R L 2 . 0 mA 2 • ο K ο 0A 1 0s 1 ms 2 ms 3 ms 4 ms E I ( L1) L export its energy , dissipated by R 4 . 0 mA 2 . 0 mA SEL > > 0A 0s 1 ms 2 ms 3 ms 4 ms
  • 90. Ch3 Basic RL and RC Circuits Summary Steady time Initial Value Value (t → constant (t=0) ∞) τ RL Source i0 = 0 iL = E R L/R Circuits (0 state) Source- free E i0 = i=0 L/R (0 input) R Source RC (0 state) v0 = 0 v=E RC Circuits Source- free v0 = E v=0 RC (0 input)
  • 91. Ch3 Basic RL and RC Circuits Summary The Time Constant • For an RC circuit, τ = RC • For an RL circuit, τ = L/R • -1/τ is the initial slope of an exponential with an initial value of 1 • Also, τ is the amount of time necessary for an exponential to decay to 36.7% of its initial value
  • 92. Ch3 Basic RL and RC Circuits Summary • How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. – IL(t), inductor current – Vc(t), capacitor voltage • Find these two types of the values before the change and use them as the initial conditions of the circuit after change.
  • 93. Ch3 Basic RLanand RC Circuits 3.3 Examples (Analyzing RC circuit or RL circuit) Method 1 1) Thévenin Equivalent.(Draw out C or L) Simplify the circuit Veq , Req 2) Find Leq(Ceq), and τ = Leq/Req (τ = CeqReq) 3) Substituting Leq(Ceq) and τ to the previous solution of differential equation for RC (RL) circuit .
  • 94. Ch3 Basic RL 3.3 Examples RC Circuits or RL circuit) and (Analyzing an RC circuit Method 2 1) KVL around the loop → the differential equation 2) Find the homogeneous solution. 3) Find the particular solution. 4) The total solution is the sum of the particular and homogeneou solutions.
  • 95. Ch3 Basic RL and RC Circuits 3.3 Examples About Calculation for The Initial Value →i ↓iC ↓iL vC ( 0+ ) = vC ( 0− ) t=0 ( R1 / / R3 ) = 2Ω 2Ω vC ( 0 ) = 8V × = 4V 2Ω + 2Ω i(0+) iC(0+) iL(0+) iL ( 0+ ) = iL ( 0− ) 8V i ( 0) = = 2A + + 2Ω + 2Ω vC(0+)=4V vL(0+) 4Ω _ - iL ( 0 ) = 2A × = 1A ∨ 1A 4Ω + 4Ω
  • 96. Ch3 Basic RL 3.3 Examples RC Circuits or RL circuit) and (Analyzing an RC circuit Method 3 (step-by- t step) − In general, f (t ) = f (∞) + Ae τ Given f(0+) , thus A = f(0+) – f(∞) t − f (t ) = f (∞) + [ f (0+ ) − f (∞)]e τ Initial Stead y 1) Draw the circuit for t = 0- and find v(0-) or i(0-) 2) Use the continuity of the capacitor voltage, or inductor current, draw the circuit for t = 0+ to find v(0+) or i(0+) 3) Find v(∞), or i(∞) at steady state 4) Find the time constant τ – For an RC circuit, τ = RC – For an RL circuit, τ = L/R 5) The solution is: f (t ) = f (∞) + [ f (0+ ) − f (∞)]e −t /τ
  • 97. Ch3 Basic RL and RC Circuits 3.3 Examples − − 5Ω 5Ω US − _ P3.1 vC (0)= 0, Find vC (t) for t ≥ 0. ο i1 6k 5Ω t=0 R1 i2 i3 I1 I1 I1 I1 ∨ ∨ Method 3: 5Ω + R2 3k + t E 9V C=1000PF − vc ( t ) = ( ) +  vc ( 0 ) I− ∧ c ( ∞ )  e µU _ vc 1 ∞ -  v  τ pf S E 3K vc ( 0 ) = 0, vc ( ∞ ) = 9V × = 3V βI1 6K + 3K Apply Thevenin theorem : −1  1 1  RTh = + ÷ =2KΩ  6KΩ 3KΩ τ =RThC =2KΩ 1000pF =2 × −6 × 10 s t − vc ( t ) =3 −3e 2× − 10 6 V
  • 98. Ch3 Basic RL and RC Circuits − 3.3 Examples 5Ω 5Ω − − _ C=1000PF P3.2 vC (0)= 0, Find vC (t) for t ≥ 0. 5Ω + - I1 I1 I1 vC 5Ω ο t=0 R1=10k vc ( 0 ) = 0 + + µ U1 - 10KΩ ∧ v 6V vc ( ∞ ) = 6V × IS E = 4.615V _ R2 3k 10KΩ + 3KΩ R1=20k β I1 Apply Thevenin’s theorem : ↑ ↓ −1  1 1  30 RTh =  + ÷ = KΩ  10KΩ 3KΩ  13 30 τ = RThC = KΩ × 1000pF = 2.31× 10−6 s 13 t − vc ( t ) = 4.615 − 4.615e 2.31×10−6 V
  • 99. Differential Equation Solutions of Transient Circuits Dr. Holbert March 3, 2008 Lect12 99 EEE 202
  • 100. 1st Order Circuits • Any circuit with a single energy storage element, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 1 • Any voltage or current in such a circuit is the solution to a 1st order differential equation Lect12 EEE 202 100
  • 101. RLC Characteristics Element V/I Relation DC Steady-State Resistor vR (t ) = R iR (t ) V=IR Capacitor d vC (t ) I = 0; open iC (t ) = C dt Inductor d iL (t ) V = 0; short vL (t ) = L dt ELI and the ICE man Lect12 EEE 202 101
  • 102. A First-Order RC Circuit vr(t) + – R + vs(t) + C vc(t) – – • One capacitor and one resistor in series • The source and resistor may be equivalent to a circuit with many resistors and sources Lect12 EEE 202 102
  • 103. The Differential Equation vr(t) + – R + vs(t) + C – vc(t) – KVL around the loop: vr(t) + vc(t) = vs(t) Lect12 EEE 202 103
  • 104. RC Differential Equation(s) t 1 From KVL: R i (t ) + ∫ i ( x)dx = vs (t ) C −∞ Multiply by C; di (t ) dvs (t ) take derivative RC + i (t ) = C dt dt Multiply by R; dvr (t ) dvs (t ) note vr=R·i RC + vr (t ) = RC dt dt Lect12 EEE 202 104
  • 105. A First-Order RL Circuit + is(t) R L v(t) – • One inductor and one resistor in parallel • The current source and resistor may be equivalent to a circuit with many resistors and sources Lect12 EEE 202 105
  • 106. The Differential Equations + is(t) R L v(t) – KCL at the top node: t v(t ) 1 + ∫ v( x)dx = is (t ) R L −∞ Lect12 EEE 202 106
  • 107. RL Differential Equation(s) t v(t ) 1 From KCL: + ∫ v( x)dx = is (t ) R L −∞ Multiply by L; L dv(t ) dis (t ) take derivative + v(t ) = L R dt dt Lect12 EEE 202 107
  • 108. 1st Order Differential Equation Voltages and currents in a 1st order circuit satisfy a differential equation of the form dx(t ) + a x(t ) = f (t ) dt where f(t) is the forcing function (i.e., the independent sources driving the circuit) Lect12 EEE 202 108
  • 109. The Time Constant (τ) • The complementary solution for any first order circuit is −t /τ vc (t ) = Ke • For an RC circuit, τ = RC • For an RL circuit, τ = L/R • Where R is the Thevenin equivalent resistance Lect12 EEE 202 109
  • 110. What Does vc(t) Look Like? τ = 10-4 Lect12 EEE 202 110
  • 111. Interpretation of τ • The time constant, τ, is the amount of time necessary for an exponential to decay to 36.7% of its initial value • -1/τ is the initial slope of an exponential with an initial value of 1 Lect12 EEE 202 111
  • 112. Applications Modeled by a 1st Order RC Circuit • The windings in an electric motor or generator • Computer RAM – A dynamic RAM stores ones as charge on a capacitor – The charge leaks out through transistors modeled by large resistances – The charge must be periodically refreshed Lect12 EEE 202 112
  • 113. Important Concepts • The differential equation for the circuit • Forced (particular) and natural (complementary) solutions • Transient and steady-state responses • 1st order circuits: the time constant (τ) • 2nd order circuits: natural frequency (ω0) and the damping ratio (ζ) Lect12 EEE 202 113
  • 114. The Differential Equation • Every voltage and current is the solution to a differential equation • In a circuit of order n, these differential equations have order n • The number and configuration of the energy storage elements determines the order of the circuit • n ≤ number of energy storage elements Lect12 EEE 202 114
  • 115. The Differential Equation • Equations are linear, constant coefficient: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt • The variable x(t) could be voltage or current • The coefficients an through a0 depend on the component values of circuit elements • The function f(t) depends on the circuit elements and on the sources in the circuit Lect12 EEE 202 115
  • 116. The Differential Equation • Equations are linear, constant coefficient: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt • The variable x(t) could be voltage or current • The coefficients an through a0 depend on the component values of circuit elements • The function f(t) depends on the circuit elements and on the sources in the circuit Lect12 EEE 202 116
  • 117. Building Intuition • Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition to be developed: – Particular and complementary solutions – Effects of initial conditions Lect12 EEE 202 117
  • 118. Differential Equation Solution • The total solution to any differential equation consists of two parts: x(t) = xp(t) + xc(t) • Particular (forced) solution is xp(t) – Response particular to a given source • Complementary (natural) solution is xc(t) – Response common to all sources, that is, due to the “passive” circuit elements Lect12 EEE 202 118
  • 119. Forced (or Particular) Solution • The forced (particular) solution is the solution to the non-homogeneous equation: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt • The particular solution usually has the form of a sum of f(t) and its derivatives – That is, the particular solution looks like the forcing function – If f(t) is constant, then x(t) is constant – If f(t) is sinusoidal, then x(t) is sinusoidal Lect12 EEE 202 119
  • 120. Natural/Complementary Solution • The natural (or complementary) solution is the solution to the homogeneous equation: n n −1 d x(t ) d x(t ) an n + an −1 n −1 + ... + a0 x(t ) = 0 dt dt • Different “look” for 1st and 2nd order ODEs Lect12 EEE 202 120
  • 121. First-Order Natural Solution • The first-order ODE has a form of dxc (t ) 1 + xc (t ) = 0 dt τ • The natural solution is −t /τ xc (t ) = Ke • Tau (τ) is the time constant • For an RC circuit, τ = RC • For an RL circuit, τ = L/R Lect12 EEE 202 121
  • 122. Second-Order Natural Solution • The second-order ODE has a form of d 2 x(t ) dx(t ) 2 + 2ζω 0 + ω 0 x(t ) = 0 2 dt dt • To find the natural solution, we solve the characteristic equation: s + 2ζω 0 s + ω = 0 2 2 0 which has two roots: s1 and s2 • The complementary solution is (if we’re lucky) xc (t ) = K1e + K 2 e s1t s2t Lect12 EEE 202 122
  • 123. Initial Conditions • The particular and complementary solutions have constants that cannot be determined without knowledge of the initial conditions • The initial conditions are the initial value of the solution and the initial value of one or more of its derivatives • Initial conditions are determined by initial capacitor voltages, initial inductor currents, and initial source values Lect12 EEE 202 123
  • 124. 2nd Order Circuits • Any circuit with a single capacitor, a single inductor, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 2 • Any voltage or current in such a circuit is the solution to a 2nd order differential equation Lect12 EEE 202 124
  • 125. A 2nd Order RLC Circuit i (t) R vs(t) + C – L The source and resistor may be equivalent to a circuit with many resistors and sources Lect12 EEE 202 125
  • 126. The Differential Equation i(t) + vr(t) – R + + C vc(t) – vs(t) vl(t) – – + L KVL around the loop: vr(t) + vc(t) + vl(t) = vs(t) Lect12 EEE 202 126
  • 127. RLC Differential Equation(s) From KVL: t 1 di (t ) R i (t ) + ∫ i ( x)dx + L = vs (t ) C −∞ dt Divide by L, and take the derivative 2 R di (t ) 1 d i (t ) 1 dvs (t ) + i (t ) + 2 = L dt LC dt L dt Lect12 EEE 202 127
  • 128. The Differential Equation Most circuits with one capacitor and inductor are not as easy to analyze as the previous circuit. However, every voltage and current in such a circuit is the solution to a differential equation of the following form: 2 d x(t ) dx(t ) 2 + 2ζω 0 + ω0 x(t ) = f (t ) 2 dt dt Lect12 EEE 202 128
  • 129. Class Examples • Drill Problems P6-1, P6-2 • Suggestion: print out the two-page “First and Second Order Differential Equations” handout from the class webpage Lect12 EEE 202 129

Notas do Editor

  1. Dr. Holbert Lecture 12 EEE 202