e, Phasor, and
Sinusoidal Steady-State Analysis
Chien-Jung Li
Department of Electronic Engineering
National Taipei University of Technology

Department of Electronic Engineering, NTUT
Compound Interest
• 複利公式: 本金P, 年利率r, 一年複利n次,
t年後本金加利息之總和為
 
  
 
1
nt
r
S P
n
• Let P=1, r=1, and t=1
 
  
 
1
1
n
S
n
When n goes to infinite, S converges to 2.718… (= e)
 Let P=10萬, r/n=10%/12, t=1 S=11,0471
 Let P=10萬, r/n=10%, and n=36, t=1 S=3,091,268
2/33

Department of Electronic Engineering, NTUT
Development of Logarithm
• Michael Stifel (1487-1567)
• John Napier (1550-1617)
• 利用對數而將乘法變成加法的特性，刻卜勒成功
計算了火星繞日的軌道。
 
  
2 52 5 7
m m m m
 
 
7 7 4 3
4
m m m
m
  
  
2 2 3 1
3
1m m m
mm
  
3 2 1 0 1 2 3
, , , , 1, , , ,m m m m m m m
3/33

Department of Electronic Engineering, NTUT
Definition of dB (分貝)
• , where
• Power gain
• Voltage gain
• Power (dBW)
• Power (dBm)
• Voltage (dBV)
• Voltage (dBuV)
  10 logdB G   aG
b
    
 
2
1
10 log
P
P
    
 
2
1
20 log
V
V
  10 log
1-W
P
  10 log
1-mW
P
  20 log
1-Volt
V
 
 20 log
1- V
V
相對量 (比例, 比值, 無單位, dB)
絕對量 (因相對於一絕對單位,
因此可表示一絕對量. 有單位,
單位即為dBW, dBm, dBV…)
4/33

Department of Electronic Engineering, NTUT
In some textbooks, phasor may be
represented as
Euler’s Formula
• Euler’s Formula cos sinjx
e x j x 
     
      
  
      cos Re Re
j t j j t
p p pv t V t V e V e e

   
def
j
p pV V e V
• Phasor (相量)
 Don’t be confused with Vector (向量) which is commonly
denoted as A
(How it comes?)
取實部 (即cosine部分) phasor
Consider a real signal v(t) that can be represented as:
V
V
5/33

Department of Electronic Engineering, NTUT
Definition of e
lim 1
n
x
n
x
e
n
 
  
 
2 3
lim 1 1
1! 2! 3!
n
x
n
x x x x
e
n
 
       
 
x jx
   
2 3
1
1! 2! 3!
jx jx jxjx
e     
• Euler played a trick let , where 1j  
1
lim 1
n
n
e
n
 
  
 
6/33

Department of Electronic Engineering, NTUT
• Since , , ,
How It Comes…
1j   2
1j   3
1j    4
1j 
   
            
   
2 4 3 5
1
2! 4! 3! 5!
x x x x
j x
2 4
cos 1
2! 4!
x x
x     
3 5
sin
3! 5!
x x
x x    
cos sinjx
e x j x 
cos sinjx
e x j x
 
cos
2
jx jx
e e
x





sin
2
jx jx
e e
x
j
       
2 3
1
1! 2! 3!
jx jx jxjx
e
• Use and
we have
(姊妹式)
7/33

Department of Electronic Engineering, NTUT
Coordinate Systems
x-axis
y-axis
x-axis
y-axis
P(r,θ)
θ
r
P(x,y)
2 2
r x y 
1
tan
y
x
 

cosx r 
siny r 
 Cartesian Coordinate System
(笛卡兒座標系, 直角座標系)
 Polar Coordinate System
(極坐標系)
(x,0)
(0,y)
 cos ,0r 
 0, sinr 
Projection
on x-axis
Projection
on y-axis
8/33

Department of Electronic Engineering, NTUT
Sine Waveform
x-axis
y-axis
P(x,y)
x
y
r
θ θθ
y
θ
0 π/2 π 3π/2 2π
 Go along the circle, the projection on y-axis results in a sine wave.
9/33

Department of Electronic Engineering, NTUT
x
θ
0
π/2
π
3π/2
Cosine Waveform
x-axis
y-axis
θ
 Go along the circle, the projection
on x-axis results in a cosine wave.
 Sinusoidal waves relate to a Circle
very closely.
 Complete going along the circle to
finish a cycle, and the angle θ
rotates with 2π rads and you are
back to the original starting-point
and. Complete another cycle
again, sinusoidal waveform in one
period repeats again. Keep going
along the circle, the waveform will
periodically appear.
10/33

Department of Electronic Engineering, NTUT
Complex Plan (I)
It seems to be the same thing with x-y plan, right?
• Carl Friedrich Gauss (1777-1855) defined the complex plan.
He defined the unit length on Im-axis is equal to “j”.
A complex Z=x+jy can be denoted as (x, yj) on the complex plan.
(sometimes, ‘j’ may be written as ‘i’ which represent imaginary)
Re-axis
Im-axis
Re-axis
Im-axis
P(r,θ)
θ
r
P(x,yj)
2 2
r x y 
1
tan
y
x
 

cosx r 
siny r 
(x,0j)
(0,yj)
 cos ,0r 
 0, sinr 
 1j  
11/33

Department of Electronic Engineering, NTUT
Complex Plan (II)
Re-axis
Im-axis
1
Every time you multiply something by j, that thing will rotate
90 degrees.
1j   2
1j   3
1j    4
1j 
1*j=j
j
j*j=-1
-1
-j
-1*j=-j -j*j=1
(0.5,0.2j)
(-0.2, 0.5j)
(-0.5, -0.2j)
(0.2, -0.5j)
• Multiplying j by j and so on:
12/33

Department of Electronic Engineering, NTUT
Sine Waveform
Re-axis
Im-axis
P(x,y)
x
y
r
θ θθ
y=rsinθ
θ
0 π/2 π 3π/2 2π
To see the cosine waveform, the same operation can be applied
to trace out the projection on Re-axis.
13/33

Department of Electronic Engineering, NTUT
Phasor Representation (I) – Sine Basis
          
    sin Im Imj j t j j
sv t A t Ae e Ae e
Re-axis
Im-axis
P(A,ф)
y=Asin ф
θ
0 π/2 π 3π/2 2π
ф
t 
Given the phasor denoted as a point on the complex-plan, you
should know it represents a sinusoidal signal. Keep this in
mind, it is very very important!
time-domain waveform
14/33

Department of Electronic Engineering, NTUT
Phasor Representation (II) – Cosine Basis
          
    cos Re Rej j t j j
sv t A t Ae e Ae e
Re-axis
Im-axis
P(A,ф)
y=Acos ф
θ
0 π/2 π 3π/2 2π
ф
t 
time-domain waveform
15/33

Department of Electronic Engineering, NTUT
Phasor Representation (III)
      
    1
1 1 1 1sin Im j j t
v t A t A e e
Re-axis
Im-axis
P(A1,ф1)
ф1
P(A2,ф2)
P(A3,ф3)
θ
0 π/2 π 3π/2 2π
t 
A1sin ф1
      
    2
2 2 2 2sin Im j j t
v t A t A e e
      
    3
3 3 3 3sin Im j j t
v t A t A e e
A2sin ф2
A3sin ф3
16/33

Department of Electronic Engineering, NTUT
Mathematical Operation
j t
j tde
j e
dt


 
1j t j t
e dt e
j
 

 
   0
1 t
v t i t dt
C
 
  

  0
1 1t
j t j t j t
Ve Ie dt I e
C j C

  
1
CV I Z I
j C
 
 di t
v t L
dt

 
 
  
j t
j t j t
d Ie
Ve L j LI e
dt
   LV j L I Z I

 
1 1
CZ
j C sC
 LZ j L sL
• L and C: from time-domain to phasor-domain analysis
(s is the Laplace operator)
     , here let 0s j
17/33

Department of Electronic Engineering, NTUT
Phasor is what you always face with
• 電路學、電子學: Phasor 常見為一個固定值 (亦可為變量)
• 電磁學、微波工程: Phasor 常見為變動量, 隨傳播方向變化
• 通訊系統: Phasor 常見為變動量, 隨時間變化
 此變動的phasor也經常被稱作複數波包(complex envelope)、波包
(envelope)，或帶通訊號的等效低通訊號(equivalent lowpass signal of
the bandpass signal)。Phasor如果被拆成正交兩成分，常稱作I/Q訊
號，而在數位通訊裡表示I/Q訊號的複數平面(座標系)也被稱為星座
圖(constellation)。
• Don’t be afraid of phasor, you will see it many times in your
E.E. life. It just appears with different names, and it is just a
representation or an analysis technique.
• Keep in mind that a phasor represents a signal, it’s like a
head on your body.
18/33

Department of Electronic Engineering, NTUT
Simple Relation Between Sine and Cosine
• Sine Cosine
π/2 π 3π/2 2π
sinθ
θ
0
cosθ
• Negative sine or cosine
   cos sin 90
   sin cos 90
    cos cos 180
    sin sin 180
Try to transform into sine-form:cos
               cos sin 90 sin 270 sin 90
19/33

Department of Electronic Engineering, NTUT
Cosine as a Basis
   
 cos Re j t
pv t V t Ve
 0pV V
   
 
 
    
 
sin cos Re
2
j t
p pv t V t V t Ve
   90pV V
     
      cos cos Re j t
p pv t V t V t Ve
 180pV V
   
 
 
     
 
sin cos Re
2
j t
p pv t V t V t Ve
 90pV V
 cosine
 sine
 negative cosine
 negative sine
Phasor
Phasor
Phasor
Phasor
20/33

Department of Electronic Engineering, NTUT
Sine as a Basis
   
 sin Im j t
pv t V t Ve
 0pV V
   
 
 
    
 
cos sin Im
2
j t
p pv t V t V t Ve
 90pV V
     
      sin sin Im j t
p pv t V t V t Ve
 180pV V
   
 
 
     
 
cos sin Im
2
j t
p pv t V t V t Ve
   90pV V
Phasor
Phasor
Phasor
Phasor
 cosine
 sine
 negative cosine
 negative sine
21/33

Department of Electronic Engineering, NTUT
Addition of Sinusoidal
 A basic property of sinusoidal functions is that the sum of an arbitrary
number of sinusoids of the same frequency is equivalent to a single
sinusoid of the given frequency. It must be emphasized that all sinusoids
must be of the same frequency.
     sinpv t V t
 1 1 1pV V
 2 2 2pV V
 n pn nV V
   1 2 nV V V V
                  1 1 2 2sin sin sinp p pn nv t V t V t V t
 1v t  2v t  nv t
22/33

Department of Electronic Engineering, NTUT
Example
      0 1 2v t v t v t
    1 20cos 100 120v t t      2 15sin 100 60v t t
    1 20 30 17.3205 10V j
     2 15 120 7.5 12.9904V j
       0 17.3205 10 7.5 12.9904V j j
    0 25sin 100 66.87v t t
    9.8205 22.9904 25 66.87j
     1 20 120 10 17.321V j
    2 15 150 12.9904 7.5V j
        0 10 17.321 12.9904 7.5V j j
    22.9904 9.8205 25 23.13j
    0 25cos 100 23.13v t t
  25sin 100 66.87t
Choose the basis you like, and the results are identical.
and For
calculate
 use sine function as a basis  use cosine function as a basis
23/33

Department of Electronic Engineering, NTUT
Steady-state Impedance
  
V
Z R jX
I
• Steady-state impedance
resistance
reactance
  
I
Y G jB
Z
• Steady-state admittance
conductance
susceptance
 30 40Z j
 30R
 40X
  

1
0.012 0.016
30 40
Y j
j
 0.012G S
 0.016X S
24/33

Department of Electronic Engineering, NTUT
Conversion to Phasor-domain
 i t
 v t V
I
RR
 i t
 v t
 i t
 v t
C
L

1
j C
V
I
j LV
I
 V R I

 
1
V I
j C
 V j L I
V
I
V
I
V
I
V and I are in-phase
V lags I by 90o
V leads I by 90o
R
C
L
25/33

Department of Electronic Engineering, NTUT
Frequency Response
Frequency-independent
All pass
Frequency-dependent
High-pass
Frequency-dependent
Low-pass
V
I
R

1
j C
V
I
j LV
I
  Z R jX R

  
1
Z R jX
C
  2 f
  2 f
  2 f
  Z R jX L
26/33

Department of Electronic Engineering, NTUT
Calculate the Impedance (I)

1
j C
V
• Calculate the impedance of a 0.01-uF capacitor at (a) f=50Hz
(b) 1kHz (c) 1MHz
  
      
  6
1
0 318.309 k
2 50 0.01 10
Z R jX j
j
  318.309 kX  318.309 kZ
I
(a) f = 50 Hz
  
      
  3 6
1
0 15.92 k
2 1 10 0.01 10
Z R jX j
j
  15.92 kX  15.92 kZ
(b) f = 1 kHz
  
      
  6 6
1
0 15.92
2 1 10 0.01 10
Z R jX j
j
  15.92X  15.92Z
(c) f = 1 MHz
 0.01 μFC
27/33

Department of Electronic Engineering, NTUT
Calculate the Impedance (II)
• Calculate the impedance of a 100-mH inductor at (a) f=50Hz
(b) 1kHz (c) 1MHz
  
       3
0 2 50 100 10 31.42Z R jX j j
 31.42X  31.42Z
(a) f = 50 Hz
  
        3 3
0 2 1 10 100 10 628.32Z R jX j j
 628.32X  628.32Z
(b) f = 1 kHz
  
        6 3
0 2 1 10 100 10 628.32 kZ R jX j j
 628.32 kX  628.32 kZ
(c) f = 1 MHz
j LV
I
 100 mHL
28/33

Department of Electronic Engineering, NTUT
Calculate the Impedance (III)
• Calculate the impedance of following circuit at (a) f=50Hz
(b) 1kHz (c) 1MHz
 
 
 
      
  6
1
200 0.2 318.309 k
2 50 0.01 10
Z R jX j
j
 318.309 kZ
(a) f = 50 Hz
 
 
 
      
  3 6
1
200 0.2 15.92 k
2 1 10 0.01 10
Z R jX j
j
 15.92 kZ
(b) f = 1 kHz
 
 
 
      
  6 6
1
200 200 15.92
2 1 10 0.01 10
Z R jX j
j
 200.63Z
(c) f = 1 MHz

1
j C
 0.01 μFC
R
 200R
   318.309k 89.96Z
   15.92k 89.26Z
  200.63 -4.55Z
29/33

Department of Electronic Engineering, NTUT
Calculate the Impedance (IV)
• Calculate the impedance of following circuit at (a) f=50Hz
(b) 1kHz (c) 1MHz
    
        3
200 2 50 100 10 200 31.42Z R jX j j
 202.45Z
(a) f = 50 Hz
    
         3 3
200 2 1 10 100 10 200 628.32Z R jX j j
 659.38Z
(b) f = 1 kHz
    
         6 3
200 2 1 10 100 10 0.2 628.32 kZ R jX j j
 628.32 kZ
(c) f = 1 MHz
j L
 100 mHL
R
 200R
  202.45 8.93Z
  659.38 72.34Z
  628.32 k 89.98Z
30/33