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
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
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
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
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
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
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).
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 )
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
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.
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
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
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