Fundamentals of Electric CircuitsChapter 14

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Overview• The idea of the transfer function: a

means of describing the relationship between the input and output of a circuit.

• Bode plots and their utility in describing the frequency response of a circuit.

• The concept of resonance as applied to LRC circuits .

• Frequency filters.2

Frequency Response

• Frequency response is the variation in a circuit’s behavior with change in signal frequency.

• Filters play critical roles in blocking or passing specific frequencies or ranges of frequencies.

• Without them, it would be impossible to have multiple channels of data in radio communications.

3

Transfer Function

• One useful way to analyze the frequency response of a circuit is the concept of the transfer function H(ω).

• It is the frequency dependent ratio of a forced function Y(ω) to the forcing function X(ω).

4

YH

X

Transfer Function

• There are four possible input/output combinations:

5

Voltage gain

Current gain

Transfer impedance

Transfer admittance

o

i

o

i

o

i

o

i

VH

V

IH

I

VH

I

IH

V

ExampleFor the RC circuit, obtain the transfer function and its frequency response. Let

6

o sV V coss mv V t

1

2

1 1( )

1 1

1, tan

1 oo

jwCw

R jwC jwRC

wH

ww w

o

s

VH

V

1ow RC

7

ExampleFor the RL circuit, obtain the transfer function and its frequency response. Let

8

o sV V coss mv V t

1

2 2 2

( )

, 90 tano

jwLw

R jwL

wL wLH

RR w L

o

s

VH

V

o

Rw

L

Zeros and Poles• To obtain H(ω), we first convert to frequency

domain equivalent components in the circuit.• H(ω) can be expressed as the ratio of

numerator N(ω) and denominator D(ω) polynomials.

• Zeros are where the transfer function goes to zero.

• Poles are where it goes to infinity.• They can be related to the roots of N(ω) and

D(ω)9

NH

D

ExampleFor the RL circuit, calculate the gain andits poles and zeros

10

( ) ( )w wo iI I

2

4 2( ) ( )

4 2 1 0.5

( ) 4 2 ( 2),

( ) 4 2 2 2 1

j ww w

j w j w

w S S SS jw

w S S S S

o i

o

i

I I

I

I

Decibel Scale• The transfer function can be seen as an

expression of gain.

• Gain expressed in log form is typically expressed in bels, or more commonly decibels (1/10 of a bel)

11

2 210 10

1 1

10log 20logdB

P VG

P V

1 2 1 2

1 2 1 2

Properties of logarithms

1. log log log , 3. log log

2. log log log , 4. log 1=0

nPP P P P n P

P P P P

Bode Plots• One problem with the transfer function is

that it needs to cover a large range in frequency. Plotting the frequency response on a semilog plot makes the task easier.

• These plots are referred to as Bode plots.• Bode plots either show magnitude (in

decibels) or phase (in degrees) as a function of frequency.

1210

ln ln ln = ln

The real part of ln is a function of the magnitude

while the imaginary part is the phase. In a Bode plot

20log (the magnitude)

j j

dB

H He H e H j

H H

H H

H

13

Standard Form• The transfer function may be written in terms of

factors with real and imaginary parts. For example:

• This standard form may include the following seven factors in various combinations:– A gain K– A pole (jω)-1 or a zero (jω)

– A simple pole 1/(1+jω/p1) or a simple zero (1+jω/z1)

– A quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] or zero 1/[1+j21ω/ ωn+ (jω/ ωk)2]

14

1 2

1 1

2

1 2

1 / 1 2 / /

1 / 1 2 / /

k k

n n

K j j z j jH

j p j j

Bode Plots• In a bode plot, each of these factors is

plotted separately and then added graphically.

• Gain, K: the magnitude is 20log10K and the phase is 0°. Both are constant with frequency.

• Pole/zero at the origin: For the zero (jω), the slope in magnitude is 20 dB/decade and the phase is 90°. For the pole (jω)-1 the slope in magnitude is -20 dB/decade and the phase is -90°

15

16

Bode plot for gain K Bode plot for a zero( ) at the originjw

Bode Plots• Simple pole/zero: For the simple zero, the

magnitude is 20log10|1+jω/z1| and the phase is tan-1 ω/z1.

• Where:

• This can be approximated as a flat line and sloped line that intersect at ω=z1.

• This is called the corner or break frequency

17

10 101 1

20log 1 20logdB

as

jH

z z

Bode Plots

• The phase can be plotted as a series straight lines

• From ω=0 to ω≤z1/10, we let =0

• At ω=z1 we let =45°

• For ω≥10z1, we let = 90°

• The pole is similar, except the corner frequency is at ω=p1, the magnitude has a negative slope

18

19

1Bode plot of zero(1+ )jw z

Bode Plots

• Quadratic pole/zero: The magnitude of the quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] is -20log10 [1+j22ω/ ωn+ (jω/ ωn)2]

• This can be approximated as:

• Thus the magnitude plot will be two lines, one with slope zero for ω<ωn and the other with slope -40dB/decade, with ωn as the corner frequency

20

10as 0

as

0 40logdBn

H

Bode Plots

• The phase can be expressed as:

• This will be a straight line with slope of -90°/decade starting at ωn/10 and ending at 10 ωn.

• For the quadratic zero, the plots are inverted.

21

1 22 2

0 02 /

tan 901 /

180

nn

n

22

2 2 -1Bode plot of quadratic (1+ 2 )n nj w w w w

Bode Plots

23

Bode Plots

24

Resonance

• The most prominent feature of the frequency response of a circuit may be the sharp peak in the amplitude characteristics.

• Resonance occurs in any system that has a complex conjugate pair of poles.

• It enables energy storage in the firm of oscillations

• It requires at least one capacitor and inductor.

25

ExampleConstruct the Bode plots for the transfer function

26

200( )

( 2)( 10)

jww

jw jw

H

o 1 1

10 10 10 10

o 1 1

200 10( )

( 2)( 10) (1 2)(1 10)

10 90 tan 2 tan 10

1 2 1 10

20log 10 20log 20log 1 20log 12 10

90 tan tan2 10

dB

jw jww

jw jw jw jw

jww w

jw jw

jw jwH jw

w w

H

27

ExampleDraw the Bode plots for the transfer function

28

5( 2)( )

( 10)

jwH w

jw jw

ExampleDraw the Bode plots for the transfer function

29

2

( 10)( )

( 5)

jwH w

jw jw

ExampleDraw the Bode plots for the transfer function

30

2

2000000( 5)( )

( 10)( 100)

SH s

S S

Series Resonance• A series resonant circuit

consists of an inductor and capacitor in series.

• Consider the circuit shown.• Resonance occurs when the

imaginary part of Z is zero.• The value of ω is called the

resonant frequency

31

0

1rad/s

1

2

LC

HzLC

1 1( ) ( )

1 1 1Im( ) 0

s

o oo

w R jwL R j wLjwC wC

wL w L wwC w C LC

VZ H

I

Z

Series Resonance• At resonance:

– The impedance is purely resistive

– The voltage Vs and the current I are in phase

– The magnitude of the transfer function is minimum.– The inductor and capacitor voltages can be much more than

the source.

• There are two frequencies above and below resonance where the dissipated power is half the max:

32

2 2

1 2

1 1

2 2 2 2

R R R R

L L LC L L LC

22 1

1

m m

m mL o m C

o

V VI

RR wL wC

V VV w L QV V

R R w C

I

1 2ow w w

Quality Factor

• The “sharpness” of the resonance is measured quantitatively by the quality factor, Q.

• It is a measure of the peak energy stored divided by the energy dissipated in one period at resonance.

• It is also a measure of the ratio of the resonant frequency to its bandwidth, B

33

0

0

1 Peak energy stored in the circuit2

Energy dissipated by the ckt in one period at resonance

LQ

R CR

02 1

RB w w

L Q

34

02 1

RB w w

L Q

1 2

It is said to be a circuithigh-Q while Q 10.

, 2 2o o

B Bw w w w

ExampleIn the circuit, (1)Find . (2)Calculate Q and B

(3) Determine the amplitude of the current at

35

2 , 1 , 0.4R L mH C F

1 2, and ow w w

1 2, and ow w w

3 9

3 -3

1

2

1 2

1 150 / ,

10 0.4 10

50 10 10= =25 10,

250

2 / , 50 1 49 / ,25 2

50 1 51 /2

20At , 10A

210

At , 7.071A2 2

o

o

oo

o

mo

m

w krad sLC

w LQ

Rw B

B krad s w w krad sQ

Bw w krad s

Vw w I

RV

w w w IR

Parallel Resonance

• The parallel RLC circuit shown here is the dual of the series circuit shown previously.

• Resonance here occurs when the imaginary part of the admittance is zero.

• This results in the same resonant frequency as in the series circuit.

36

Parallel Resonance

• The relevant equations for the parallel resonant circuit are:

37

2 2

1 2

1 1 1 1 1 1

2 2 2 2RC RC LC RC RC LC

2 1

1B w w

RC 0

0

ow RQ RC

B L

2 2

1 2

1 2

1 1 2 , 1 1 22 2

For high-Q circuit (Q 10), , 2 2

o o

o o

B Bw w Q w w Q

Q Q

B Bw w w w

38

ExampleDetermine the resonant frequency of the circuit.

39

2

2

1 1 2 2Y=j0.1w + 0.1 j0.1w+

10 2 2 4 4

2At resonance Im( ) 0 0.1 - 0 2 /

4 4 o

j w

j w w

ww w w rad s

w

Y

Passive Filters

• A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.

• A filter is passive if it consists only of passive elements, R, L, and C.

• They are very important circuits in that many technological advances would not have been possible without the development of filters.

40

Passive Filters

• There are four types of filters:– Lowpass passes only low

frequencies and blocks high frequencies.

– Highpass does the opposite of lowpass

– Bandpass only allows a range of frequencies to pass through.

– Bandstop does the opposite of bandpass

41

Lowpass Filter• A typical lowpass filter is formed

when the output of a RC circuit is taken off the capacitor.

• The half power frequency is:

• This is also referred to as the cutoff frequency.

• The filter is designed to pass from DC up to ωc

42

1c RC

2 2 2

1( )

1

1 1 1( )

21

i

o

c c

c

wjwRC

w wRCw R C

VH

V

H

Highpass Filter• A highpass filter is also

made of a RC circuit, with the output taken off the resistor.

• The cutoff frequency will be the same as the lowpass filter.

• The difference being that the frequencies passed go from ωc to infinity.

43

2 2 2

( )1

1 1( )

21

i

o

cc c

c

jwRCw

jwRC

w RCw w

RCw R C

VH

V

H

Bandpass Filter• The RLC series resonant

circuit provides a bandpass filter when the output is taken off the resistor.

• The center frequency is:

• The filter will pass frequencies from ω1 to ω2.

• It can also be made by feeding the output from a lowpass to a highpass filter.

44

0

1

LC

( )( 1 )

i

o Rw

R j wL wC

V

HV

Bandstop Filter• A bandstop filter can be

created from a RLC circuit by taking the output from the LC series combination.

• The range of blocked frequencies will be the same as the range of passed frequencies for the bandpass filter.

45

2

1 2

1 2

1 2

( ) ,highpass filter

25 krad/s( )

i

o

c

c

R jww

R R jw w

R Rw

R R L

VH

V

ExampleDetermine what type of filter is shown in the figure. Calculate the corner or cutoff freqency.

46

R 2k , L 2H, C 2 F

2

222 2

22 2 2

2 2 22 6 3 2 2

4 2 2

//(1 )( )

//(1 )

(1 )

(1 )

12 1

2

2 1 4 10 10 2 1 4 , Let be krad/s

16 7 1 0 0.5509

o

i

cc

c c

c c c c c

c c c c

R sCs

sL R sC

R sRC R

sL R sRC s RLC sL R

w LRH w LC

RR w RLC w L

w w w w w

w w w w

VH

V

0.742 / 742 /krad s rad s

The corner frequency is the same as

the half-power frequency, i.e., 1 2H

ExampleDetermine what type of filter is shown in the figure. Calculate the corner or cutoff freqency.

47

1 2R =100 , L 2mHR

2

1 2

1 2

1 2

( ) ,highpass filter

25 krad/s( )

i

o

c

c

R jww

R R jw w

R Rw

R R L

VH

V

ExampleIf the bandpass filter in the figure is to reject a 200 Hz sinusoid while passing other freq. ,calculate the values of L, C and .

48

R 150

22

2 2 100 200 rad/s

1500.2387

2002 2 200 400 rad/s

1 1 12.653

400 0.2387

o o

oo

B f

R RB L H

L Bw f

w C Fw LLC

The bandwidth is 100 Hz

Active Filters

• Passive filters have a few drawbacks.– They cannot create gain greater than 1.– They do not work well for frequencies below the

audio range.– They require inductors, which tend to be bulky

and more expensive than other components.

• It is possible, using op-amps, to create all the common filters.

• Their ability to isolate input and output also makes them very desirable.

49

First Order Lowpass• The corner frequency will be:

50

1c

f fR C

( )

1 and //

1

1( )

1

1The dc gain is and the corner freq.

fo

i i

fi i f f

f f f

f f

i i f f

fc

i f f

w

RR R

jwC jwR C

Rw

R jwR C

Rw

R R C

ZVH

V Z

Z Z

ZH

Z

First Order Highpass

• The corner frequency will be:

51

1c

i iRC

( )

11 and

( )1 1

1The gain is as and the corner freq.

fo

i i

i ii i f f

i i

f f f i

i i i i i

fc

i i i

w

jwRCR R

jwC jwC

R jwR Cw

R jwC jwRC

Rw w

R RC

ZVH

V Z

Z Z

ZH

Z

Bandpass• To avoid the use of an inductor, it is possible

to use a cascaded series of lowpass active filter into a highpass active filter.

• To prevent unwanted signals passing, their gains are set to unity, with a final stage for amplification.

52

Figure 14.45

The analysis of the bandpass filter

54

2 2

1 2 1 2

21

12

1 1( ) ( ) ( )( )

1 1 1 1

1The lowpass section sets the upper corner freq. as

1The highpass section sets the lower corner freq. as

The center fr

f fo

i i i

R RjwRC jwRCw

jwRC jwRC R R jwRC jwRC

wRC

wRC

VH

V

1 2 2 1

1 2

2 1 1 2

2 21 2

1 2 1 2

2

1 2

eq., bandwidth and quality factor

, ,

( )(1 )(1 ) ( )( )

( ) , ( )( )

The passband gain

oo

f f

i i

f fo o

i i

f

i

ww w w B w w Q

BR Rjw w jww

wR jw w jw w R w jw w jw

R Rjww ww w w w

R w jw w jw R w w

R wK

R w w

H

H

Bandreject

• Creating a bandstop filter requires using a lowpass and highpass filter in parallel.

• Both output are fed into a summing amplifier.• It will function by amplifying the desired

signals compared to the signal to be rejected.

55

Figure 14.47

The analysis of the bandreject filter

57

21 1 12

1 2 2 1

21

12

(1 2 )1( )

1 1 (1 )(1 )

1The highpass section sets the upper corner freq. as

1The lowpass section sets the lower corner freq. as

T

f fo

i i i

R R j w w jw w wjwRCw

R jwRC jwRC R jw w jw w

wRC

wRC

VH

V

1 2 2 1

2

1 1 1 11 2

2 1 1 2

he center freq., bandwidth and quality factor

, ,

(1 2 ) 2( ) ,

(1 )(1 )

oo

f fo oo o

i o o i

ww w w B w w Q

B

R Rj w w jw w w ww w w w

R jw w jw w R w w

H

ExampleDesign a lowpass active filter with a dc gain of 4 and a corner frequency of 500 Hz.

58

6

12 2 500

The dc gain is (0) 4

If we select 0.2 , then

11.59 1.6

2 500 0.2 10

397.5 4004

c cf f

f

i

f

f

fi

w fR C

RH

R

C F

R k k

RR

ExampleDesign a highpass active filter with a high frequency gain of 5 and a corner frequency of 2kHz. Use a 0.1 capacitor in your design

59

F

6

12 2 2000

1795.7 800

2 2000 0.1 10

The gain of ( ) 5 5 4

c ci i

i

ff i

i

w fRC

R

RH R R k

R