Circuit Network Analysis - [Chapter2] Sinusoidal Steady-state Analysis

Network Analysis

Chapter 2 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

ntr

S Pn

• Let P=1, r=1, and t=1

= +

11

n

Sn

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


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 7m m m m

( )−= =7 7 4 3

4m m m

m

( )− −= = =2 2 3 1

31m m m mm

− − − =⋯ ⋯3 2 1 0 1 2 3, , , , 1, , , , m m m m m m m

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Definition of dB (分貝)

• , where

• Power gain

• Voltage gain

• Power (dBW)

• Power (dBm)

• Voltage (dBV)

• Voltage (dBuV)

( )= ⋅10 logdB G ( )= aG b = ⋅

2

110 log P

P

= ⋅

2

120 log V

V

( )= ⋅10 log 1-WP

( )= ⋅10 log 1-mWP

( )= ⋅20 log 1-VoltV

( )µ= ⋅20 log 1- VV

相對量 ((((比例, , , , 比值, , , , 無單位, dB), dB), dB), 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dBWdBWdBWdBW, , , , dBmdBmdBmdBm, , , , dBVdBVdBVdBV…)…)…)…)

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In some textbooks, phasor may be

represented as

Euler’s Formula

• Euler’s Formula cos sinjxe x j x= +

( ) ( ) ( ) ω φ φ ωω φ += ⋅ + = ⋅ = ⋅cos Re Rej t j j tp p pv t V t V e V e e

φ φ= ⋅ = ∠def

jp pV V e V

• Phasor (相量)

Don’t be confused with VectorVectorVectorVector (向量) which is commonly

denoted as A

(How it comes?)

取實部 (即cosine部分) phasor

A real sinusoidal signal v(t) that can be represented as:

VV

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Definition of e

lim 1n

x

n

xe

n→∞

= +

2 3

lim 1 11! 2! 3!

nx

n

x x x xe

n→∞

= + = + + + +

…

x jx=

( ) ( )2 3

11! 2! 3!

jx jx jxjxe = + + + +…

• Euler played a trick let , where 1j = −

1lim 1

n

ne

n→∞

= +

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• Since , , ,

How It Comes…

1j = − 2 1j = − 3 1j = − − 4 1j =

= − + − + + − + − +

… …

2 4 3 5

12! 4! 3! 5!x x x x

j x

2 4

cos 12! 4!x x

x = − + − +…3 5

sin3! 5!x x

x x= − + − +…

cos sinjxe x j x= +

cos sinjxe x j x− = −

cos2

jx jxe ex

−+=

−−=sin2

jx jxe ex

j

( ) ( )= + + + +…

2 3

11! 2! 3!

jx jx jxjxe

• Use and

we have

(姊妹式)

7/33


Coordinate Systems

x-axis

y-axis

x-axis

y-axis

P(r,θ)

θ

r

P(x,y)

2 2r x y= +1tan

yx

θ −=

cosx r θ=siny r θ=

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 θ

( )0, sinr θ

Projection

on x-axis

Projection

on y-axis

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Sine Waveform

x-axis

y-axis

P(x,y)

x

yr

θ θθ

y

θ0 π/2 π 3π/2 2π

Go along the circle, the projection on y-axis results in a sine wave.

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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 CircleCircleCircleCircle

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.

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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 ImImImIm-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 2r x y= +1tan

yx

θ −=

cosx r θ=siny r θ=

(x,0j)

(0,yj)

( )cos ,0r θ

( )0, sinr θ

( )1j = −

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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=jj

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:

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Sine Waveform

Re-axis

Im-axis

P(x,y)

x

yr

θ θθ

y=rsinθ

θ0 π/2 π 3π/2 2π

To see the cosine waveform, the same operation can be applied

to trace out the projection on ReReReRe-axis.

13/33


Phasor Representation (I) – Sine Basis

( ) ( ) φ ω φ θω φ= + = =sin Im Imj j t j jsv t A t Ae e Ae e

Re-axis

Im-axis

P(A,ф)

y=Asin ф

θ0 π/2 π 3π/2 2π

ф

tθ ω=

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

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Phasor Representation (II) – Cosine Basis

( ) ( ) φ ω φ θω φ= + = =cos Re Rej j t j jsv t A t Ae e Ae e

Re-axis

Im-axis

P(A,ф)

y=Acos ф

θ0 π/2 π 3π/2 2π

ф

tθ ω=

time-domain waveform

15/33


Phasor Representation (III)

( ) ( ) φ ωω φ= + = 11 1 1 1sin Im j j tv t A t A e e

Re-axis

Im-axis

P(A1,ф1)

ф1

P(A2,ф2)

P(A3,ф3)

θ0 π/2 π 3π/2 2π

tθ ω=

A1sin ф1



A2sin ф2

A3sin ф3

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Mathematical Operation

j tj tde

j edt

ωωω= ⋅ 1j t j te dt e

jω ω

ω= ⋅∫

( ) ( )0

1 tv t i t dt

C= ∫

ω= = ⋅1

CV I Z Ij C

( ) ( )di tv t L

dt=

ω= ⋅ = ⋅LV j L I Z I

ω= =1 1

CZj C sC

ω= =LZ j L sL

• LLLL and CCCC: from time-domain to phasor-domain analysis

(s is the Laplace operator)( )σ ω σ= + =, here let 0s j

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Phasor Everywhere

• 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)。

• You will see “Phasor” 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.

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Simple Relation Between Sine and Cosine

• Sine CosineSine CosineSine CosineSine Cosine

π/2 π 3π/2 2π

sinθ

θ0

cosθ

• Negative sine or cosineNegative sine or cosineNegative sine or cosineNegative sine or cosine

( )θ θ= + cos sin 90

( )θ θ= − sin cos 90

( )θ θ− = + cos cos 180

( )θ θ− = + sin sin 180

Try to transform into sine-form:θ−cos

( ) ( ) ( )θ θ θ θ− = − + = + = − cos sin 90 sin 270 sin 90

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Cosine as a Basis

( ) ωω= =cos Re j tpv t V t Ve= ∠ 0pV V

( ) ωπω ω = = − =

sin cos Re2

j tp pv t V t V t Ve

= ∠ − 90pV V

( ) ( ) ωω ω π= − = + =cos cos Re j tp pv t V t V t Ve

= ∠ 180pV V

( ) ωπω ω = − = + =

sin cos Re2


= ∠ 90pV V

cosinecosinecosinecosine

sinesinesinesine

negative cosinenegative cosinenegative cosinenegative cosine

negative sinenegative sinenegative sinenegative sine

Phasor

Phasor

Phasor

Phasor

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Sine as a Basis

( ) ωω= =sin Im j tpv t V t Ve= ∠ 0pV V

( ) ωπω ω = = + =

cos sin Im2


= ∠ 90pV V

( ) ( ) ωω ω π= − = + =sin sin Im j tp pv t V t V t Ve

= ∠ 180pV V

( ) ωπω ω = − = − =

cos sin Im2


= ∠ − 90pV V

Phasor

Phasor

Phasor

Phasor

cosinecosinecosinecosine

sinesinesinesine

negative cosinenegative cosinenegative cosinenegative cosine

negative sinenegative sinenegative sinenegative sine

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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.

( ) ( )ω θ= +sinpv t V t

θ= ∠1 1 1pV V

θ= ∠2 2 2pV V

θ= ∠n pn nV V

= + + +⋯1 2 nV V V V

( ) ( ) ( ) ( )ω θ ω θ ω θ= + + + + + +⋯1 1 2 2sin sin sinp p pn nv t V t V t V t

( )1v t ( )2v t ( )nv t

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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 25 203.13j

( ) ( )= +

0 25cos 100 203.13v 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

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Steady-state Impedance

= = +VZ R jX

I

• Steady-state impedance

resistance

reactance

= = +IY G jB

Z

• Steady-state admittance

conductance

susceptance

= +30 40Z j

= Ω30R= Ω40X

= = −+1

0.012 0.01630 40

Y jj

= 0.012G S

= −0.016X S

24/33


Conversion to Phasor-domain

( )i t

( )v t V

I

RR

( )i t

( )v t

( )i t

( )v t

C

L

ω1

j CV

I

ωj LV

I

= ⋅V R I

ω= ⋅1

V Ij C

ω= ⋅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


Frequency Response

Frequency-independent

All pass

Frequency-dependent

High-pass

Frequency-dependent

Low-pass

V

I

R

ω1

j CV

I

ωj LV

I

= + =Z R jX R

ω= + = 1

Z R jXC

ω π= 2 f

ω π= 2 f

ω π= 2 f

ω= + =Z R jX L

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Calculate the Impedance (I)

ω1

j CV

• Calculate the impedance of a 0.01-uF capacitor at (a) f=50Hz

(b) 1kHz (c) 1MHz

( )π −= + = + = − Ω⋅ × 6

10 318.309 k

2 50 0.01 10Z R jX j

j

= − Ω318.309 kX = Ω318.309 kZI

(a) f = 50 Hz

( )π −= + = + = − Ω

× ⋅ ×3 6

10 15.92 k

2 1 10 0.01 10Z R jX j

j

= − Ω15.92 kX = Ω15.92 kZ

(b) f = 1 kHz

( )π −= + = + = − Ω

× ⋅ ×6 6

10 15.92

2 1 10 0.01 10Z R jX j

j

= − Ω15.92 X = Ω15.92 Z

(c) f = 1 MHz= 0.01 µFC

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Calculate the Impedance (II)

• Calculate the impedance of a 100-mH inductor at (a) f=50Hz

(b) 1kHz (c) 1MHz

( )π −= + = + ⋅ × = Ω30 2 50 100 10 31.42 Z R jX j j

= Ω31.42 X = Ω31.42 Z

(a) f = 50 Hz

( )π −= + = + × ⋅ × = Ω3 30 2 1 10 100 10 628.32 Z R jX j j

= Ω628.32 X = Ω628.32 Z

(b) f = 1 kHz

( )π −= + = + × ⋅ × = Ω6 30 2 1 10 100 10 628.32 kZ R jX j j

= Ω628.32 kX = Ω628.32 kZ

(c) f = 1 MHz

ωj LV

I

= 100 mHL

28/33


Calculate the Impedance (III)

• Calculate the impedance of following circuit at (a) f=50Hz

(b) 1kHz (c) 1MHz

( ) ( )π −= + = + = − Ω

⋅ × 6

1200 0.2 318.309 k

2 50 0.01 10Z R jX j

j

= Ω318.309 kZ

(a) f = 50 Hz

( ) ( )π −

= + = + = − Ω× ⋅ ×3 6

1200 0.2 15.92 k

2 1 10 0.01 10Z R jX j

j

= Ω15.92 kZ

(b) f = 1 kHz

( ) ( )π −

= + = + = − Ω× ⋅ ×6 6

1200 200 15.92

2 1 10 0.01 10Z R jX j

j

= Ω200.63 Z

(c) f = 1 MHz

ω1

j C

= 0.01 µFC

R

= Ω200 R

= ∠ − Ω318.309k 89.96 Z

= ∠ − Ω15.92k 89.26 Z

= ∠ Ω200.63 -4.55 Z

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Calculate the Impedance (IV)

• Calculate the impedance of following circuit at (a) f=50Hz

(b) 1kHz (c) 1MHz

( ) ( )π −= + = + ⋅ × = + Ω3200 2 50 100 10 200 31.42 Z R jX j j

= Ω202.45 Z

(a) f = 50 Hz

( ) ( )π −= + = + × ⋅ × = + Ω3 3200 2 1 10 100 10 200 628.32 Z R jX j j

= Ω659.38 Z

(b) f = 1 kHz

( ) ( )π −= + = + × ⋅ × = + Ω6 3200 2 1 10 100 10 0.2 628.32 kZ R jX j j

= Ω628.32 kZ

(c) f = 1 MHz

ωj L

= 100 mHL

R

= Ω200 R

= ∠ Ω202.45 8.93 Z

= ∠ Ω659.38 72.34 Z

= ∠ Ω628.32 k 89.98 Z

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Power in AC Circuits

( ) ( )ω φ= +sinpi t I t

( ) ( )ω φ θ= + +sinpv t V t

Instantaneous power absorbed by the circuit:

( ) ( ) ( ) ( ) ( )ω φ θ ω φ= = + + +sin sinp pp t v t i t V I t t

( ) ( ) ( )= =∫ ∫0 0

1 1T TP p t dt v t i t dt

T T

Average power:

( ) ( )= − − +1 1sin sin cos cos

2 2A B A B A B

Steady-state

AC circuit

( )i t

( )v t

( ) ( )ω φ θ ω φ= + + +∫01

sin sinT

p pV I t t dtT

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Power in AC Circuits

Average power:

( )θ ω φ θ = − + + ∫ ∫0 0

cos cos 2 22

T Tp pV Idt t dt

T

( ) ( ) θθ θ ∗= = = =

0

cos 1cos cos Re

2 2 2 2

T

p p p p p pV I V I V It T VI

T T

Steady-state

AC circuit

( )i t

( )v t( ) ( )ω φ θ ω φ= + + +∫0

1sin sin

T

p pP V I t t dtT

V

Iθ φ

( )φ θ+= jpV V e

φ∗ −= jpI I e

θ=* jp pVI V I e

θ=*Re cosp pVI V I

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Root-Mean-Square (RMS) Value

θθ θ= = =

coscos cos

22 2p p p p

rms rms

V I V IP V I

(RMS value is also called the effective value)

When the circuit contains L and C, the current and voltage may not be

in-phase (they can be in-phase if effects of L and C cancelled at the given frequency),

and hence the apparent power may not be totally absorbed by the circuit.

Define RMS voltage and current as

=2p

rms

VV =

2p

rms

II

power factor (PF)

is define as the power factor (功率因子/因素)θ≤ ≤0 cos 1

×Actual power = Apparent power Power factorθ= cosrms rmsV I

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Circuit Network Analysis - [Chapter2] Sinusoidal Steady-state Analysis

Engineering

Transcript of Circuit Network Analysis - [Chapter2] Sinusoidal Steady-state Analysis