INC 112 Basic Circuit Analysis

Week 13

Frequency Response

Frequency Response

AC

+

-

Asin(ωt) L

R+

vL(t)-

What is the change of vL(t)when the frequency of thevoltage source change?

AC

+

-

Asin(ωt) L

R+

vL(t)-

ω

|vL| (magnitude)

A

Log ω

|vL| (magnitude)

Log scale of frequency

A

Linear Scale & Log Scale

1 2 3 4 5 6

0.1 1 10 100 1000 1000

Linear Scale

Log Scale

AC

+

-

Asin(ωt)

R+

vC(t)-

CWhat is the frequency responseof vc(t)

Log ω

|vc| (magnitude)

A

Input & Output

SystemInput Output

Input is usually what we can control.

Output is usually what we are interested in.

AC

+

-

Asin(ωt)

R+

vC(t)-

CWhat is the frequency responseof vc(t)

Input = voltage source

Output = vc(t)

Frequency Domain

Frequency domain is another point of view of things in the world.

Some analysis are easier done in frequency domain than time domain.

Metaphor of Frequency Domain

(1,1,1)

(2,-3,0)

Time Domain

Frequency Domain

4

22sin

2 s

tTime Domain Frequency Domain

R L C R sL 1/sC

Transfer Function

Transfer function = ratio of output and input in frequency domain

System

H(s)

System

H(s)

Input Output

U(s) Y(s)

)(

)()(

sU

sYsH Transfer function

Note: Transfer function describes characteristic of a system.

Example: Obtaining Transfer Function

input is v(t)output is vc(t)

Change all components to Phasor(frequency domain)

Kirchoff’s Voltage Law

CssIsV

sV

CsRLs

sI

Cs

sIsRIsLsIsV

C

1)()(

)(1

1)(

)()()()(

LCs

LR

s

LCsV

sVC1

1

)(

)(

2 Transfer

function

Frequency Response

SystemInput Output

Freq 10HzAmplitude = 1 Phase = 0

Freq 100HzAmplitude = 1 Phase = 0

Freq 800HzAmplitude = 1 Phase = 0

Freq 10HzAmplitude = 0.96 Phase = 12

Freq 100HzAmplitude = 0.82 Phase = 44

Freq 800HzAmplitude = 0.53 Phase = 56

Frequency Response Plot

Note: log frequency and log magnitude

Magnitude

Phase

The HP 35670A Dynamic Signal Analyzer obtains

frequency responsedata from a

physicalsystem.

Fourier Series

“Any periodic signal can be written in the sum of sine wave signalsat different frequency.”

Filter

AC

+

-2sin(t)+3sin(9t)

R +vC(t)

-C

Log ω

|vc| (magnitude)

We want to get rid of the signal 3sin(9t)

AC

+

-

2sin(t)+3sin(9t)

ω=5

Cut-off frequency

Example

AC

+

-2sin(t)+2sin(9t)

1Ω +vC(t)

-0.2f

Find vc(t)

Use superposition to consider the effect of two different frequencies

AC

+

-

2sin(t)

1Ω +vC(t)

-0.2f AC

+

-

2sin(9t)

1Ω +vC(t)

-0.2f

AC

+

-

2∟0

1 +vC(t)

--j5

For 2sin(t)

55

2.01

11j

jjCj

31.1196.1

0269.781.5

90502

15

5

j

jVc

)31.11sin(96.1)( ttvC

AC

+

-

2∟0

1 +vC(t)

--j0.556

For 3sin(9t)

556.0556.0

2.09

11j

jjCj

93.60972.0

0207.29144.1

90556.002

1556.0

556.0

j

jVc

)93.609sin(972.0)( ttvC

Finally, we add vc(t) from both frequencies

)93.609sin(972.0)31.11sin(96.1)( tttvC

AC

+

-2sin(t)+2sin(9t)

1Ω +vC(t)

-0.2f

Power Line Systems

3 phases systems have three wire called “line” which aresine wave with phase shift 120 from each other

AC

AC

AC

Power Plant

Line 1

Line 2

Line 3

02220

2402220

1202220

3 phases, 3 wires

AC

AC

AC

Power Plant

Line 1

Line 2

Line 3

02220

2402220

1202220

3 phases, 4 wires

Neutral

L1-N = 220Vrms, L2-N = 220Vrms L3-N = 220Vrms

L1-L2 = 381Vrms, L2-L3 = 381Vrms, L1-L3 = 381Vrms

Star Connection ( Y-connection)

AC

AC

AC

Z

Load

Each load get 220 Vrms

Delta Connection ( Δ-connection)

AC

AC

AC Z

Load

Each load get 381 Vrms