RF Circuit Design - [Ch1-1] Sinusoidal Steady-state Analysis

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

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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 34

m m mm

2 2 3 1

31m m mmm

3 2 1 0 1 2 3, , , , 1, , , , m m m m m m m

3/33


Definition of dB (分貝)

• , where

• Power gain

• Voltage gain

• Power (dBW)

• Power (dBm)

• Voltage (dBV)

• Voltage (dBuV)

10 logdB G aGb

2

1

10 logP

P

2

1

20 logV

V

10 log1-W

P

10 log1-mW

P

20 log1-Volt

V

20 log

1- VV

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…)

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In some textbooks, phasor may be

represented as

Euler’s Formula

• Euler’s Formula cos sinjxe x j x

cos Re Rej 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

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Definition of e

lim 1

n

x

n

xe

n

2 3

lim 1 11! 2! 3!

n

x

n

x x x xe

n

x jx

2 3

11! 2! 3!

jxjx jxjx

e

• 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 xj x

2 4

cos 12! 4!

x xx

3 5

sin3! 5!

x xx x

cos sinjxe x j x

cos sinjxe x j x

cos2

jx jxe ex

sin

2

jx jxe ex

j

2 3

11! 2! 3!

jxjx jxjx

e

• Use and

we have

(姊妹式)

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Coordinate Systems

x-axis

y-axis

x-axis

y-axis

P(r,θ)

θ

r

P(x,y)

2 2r x y

1tany

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

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Sine Waveform

x-axis

y-axis

P(x,y)

x

y r

θ θ θ

y

θ 0 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 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.

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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 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 2r x y

1tany

x

cosx r

siny r

(x,0j)

(0,yj)

cos ,0r

0, sinr

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=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:

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Sine Waveform

Re-axis

Im-axis

P(x,y)

x

y r

θ θ θ

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.

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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


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

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Phasor Representation (III)

1

1 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)

θ 0 π/2 π 3π/2 2π

t

A1sin ф1

2


3


A2sin ф2

A3sin ф3

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Mathematical Operation

j tj tde

j edt

1j t j te dt ej

0

1 t

v t i t dtC

0

1 1tj t j t j tVe Ie dt I e

C j C

1CV I Z I

j C

di t

v t Ldt

j t

j t j td Ie

Ve L j LI edt

LV j L I Z I

1 1CZ

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

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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.

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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

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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

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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

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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.

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

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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 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

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Steady-state Impedance

V

Z R jXI

• Steady-state impedance

resistance

reactance

I

Y G jBZ

• Steady-state admittance

conductance

susceptance

30 40Z j

30R

40X

10.012 0.016

30 40Y j

j

0.012G S

0.016X S

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Conversion to Phasor-domain

i t

v t V

I

RR

i t

v t

i t

v t

C

L

1

j CV

I

j LV

I

V R I

1V 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


Frequency Response

Frequency-independent

All pass

Frequency-dependent

High-pass

Frequency-dependent

Low-pass

V

I

R

1

j CV

I

j LV

I

Z R jX R

1Z R jX

C

2 f

2 f

2 f

Z R jX L

26/33


Calculate the Impedance (I)

1

j CV

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

(b) 1kHz (c) 1MHz

6

10 318.309 k

2 50 0.01 10Z R jX j

j

318.309 kX 318.309 kZI

(a) f = 50 Hz

3 6

10 15.92 k

2 1 10 0.01 10Z R jX j

j

15.92 kX 15.92 kZ

(b) f = 1 kHz

6 6

10 15.92

2 1 10 0.01 10Z R jX j

j

15.92 X 15.92 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

30 2 50 100 10 31.42 Z R jX j j

31.42 X 31.42 Z

(a) f = 50 Hz

3 30 2 1 10 100 10 628.32 Z R jX j j

628.32 X 628.32 Z

(b) f = 1 kHz

6 30 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


Calculate the Impedance (III)

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

(b) 1kHz (c) 1MHz

6

1200 0.2 318.309 k

2 50 0.01 10Z R jX j

j

318.309 kZ

(a) f = 50 Hz

3 6

1200 0.2 15.92 k

2 1 10 0.01 10Z R jX j

j

15.92 kZ

(b) f = 1 kHz

6 6

1200 200 15.92

2 1 10 0.01 10Z R jX j

j

200.63 Z

(c) f = 1 MHz

1

j C

0.01 μFC

R

200 R

318.309k 89.96 Z

15.92k 89.26 Z

200.63 -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

3200 2 50 100 10 200 31.42 Z R jX j j

202.45 Z

(a) f = 50 Hz

3 3200 2 1 10 100 10 200 628.32 Z R jX j j

659.38 Z

(b) f = 1 kHz

6 3200 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

200 R

202.45 8.93 Z

659.38 72.34 Z

628.32 k 89.98 Z

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