202 Lec 16

8/7/2019 202 Lec 16

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FREQUENCY RESPONSE OF

SIMPLE CIRCUITS

Ray DeCarlo School of ECE

Purdue University

West Lafayette, IN [email protected]

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EE-202, Frequency Response p 2 © R. A. DeCarlo

I. MEANING OF FREQUENCY RESPONSE

1. Recall Sinusoidal Steady State Analysis

(a) M (! ) = K " H ( j ! )

(b) ! (" ) = # H ( j " )+ $

CONCLUSION: the magnitude or gain, H ( j ! ) , and angle or phase,

! H ( j " ) , specify the effect of the circuit/system has on input

sinusoids, K cos(! t +" ) .

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2. DEFINITION: the frequency response of a circuit/system

is the transfer function evaluated along the imaginary axis, i.e.,

H ( j ! ) = H (s)]s= j ! . For single-input single-output circuits/systems,

for each ω, H ( j ! ) is a complex number:

Output(s)

Input(s)

!

"#

s=

j $

= H ( j $ )% H ( j $ ) & H ( j $ )

(a) ! H ( j " ) is the phase response , and

(b) H ( j ! ) is the magnitude response /GAIN.

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EXAMPLE 1. A band pass (BP) transfer function is one that

passes frequencies in a band and eliminates those outside the band.

Two such BP transfer functions are:

H 1(s) =0.25s

s2+ 0.25s +1

=0.25s

(s + 0.125)2+ (0.992)

2

and

H 2(s) =0.0625s

2

s4+ 0.35355s

3+ 2.0625s

2+ 0.35355s +1

=0.0625s

(s + 0.0962)2+ (1.0884)

2!

0.0625s

(s + 0.08058)2+ (0.91164)

2

Remark: s =! = 0 and s =! = " are the two most importantfrequencies:

(a) H 1(0) = H 1(!) = 0

(b) H 2 (0) = H 2 (!) = 0

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(a) Plot H ( j ! ) using the following MATLAB code:

»n1 = 0.25*[1 0];

»d1 = [1 0.25 1];

»n2 = 0.0625*[1 0 0];

»d2 = [1 3.5355e-01 2.0625 3.5355e-01 1];

»w=0.2:0.005:2;

»h1 = freqs(n1,d1,w);


»plot(w,abs(h1),w,abs(h2))

»grid

»xlabel('Frequency r/s')

»ylabel('Magnitude response')

»gtext('2nd Order BP')»gtext('4th Order BP')

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency r/s

M a g n i t u d e r e s p o n s e

TextEnd

2nd Order BP

4th Order BP

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(b) Qualitative Analysis Using Pole-zero plot.

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(c) The Idea of Frequency Scaling:

»Kf = 1000;

»n1 = 0.25*[1/Kf 0];

»d1 = [1/Kf^2 0.25/Kf 1];

»n2 = 0.0625*[1/Kf^2 0 0];»d2 = [1/Kf^4 3.5355e-01/Kf^3 2.0625/Kf^2 3.5355e-01/Kf 1];

»w=0.2*Kf:1:2*Kf;



»plot(w,abs(h1),w,abs(h2))

»grid

»xlabel('Frequency r/s')

»ylabel('Magnitude response')

»gtext('2nd Order BP')»gtext('4th Order BP')

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EXAMPLE 2. Magnitude response H ( j ! ) of three normalized low

pass Butterworth filter transfer functions:

(a) First Order Normalized Butterworth

H (s) =1

s +1

(b) 2nd Order Normalized Butterworth

H (s) =

1

LC

s2+ Rs

Ls +

1

LC

=1

s2+ 2 s +1

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(c) 3rd Order Normalized Butterworth

H (s) =1

s3+ 2s

2+ 2s +1

(d) MATLAB Code:

»w = logspace(-2,2,800);

»n1 = 1; d1 = [1 1];

»n2 = 1; d2 = [1 sqrt(2) 1];

»n3 = 1; d3 = [1 2 2 1];




»semilogx(w,abs(h1),w,abs(h2),w,abs(h3))

»grid

»xlabel('Normalized rad frequency')

»ylabel('Magnitude Response')

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(e) Magnitude Response Plots

10-2

10-1

100

101

102

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized rad frequency

M a g n i t u d e R e s p o n s e

TextEnd

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(f) Pole-Zero Plot of 2nd and 3rd Order Filters

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EXAMPLE 3. Plot of H ( j ! ) of three low pass Butterworth

transfer functions of frequency scaled by K f = 1000 :

H new(s) = H old sK f

!

" #$

% & =

1

s

1000

! " #

$ % & +1

H new(s) = H old s

K f

!

" #

$

% & =

1

s

1000

! " #

$ % &

2

+ 2

s

1000

! " #

$ % & +1

H new(s) = H old s

K f

!

" #

$

% & =

1

s

1000

! " #

$ % & 3

+ 2s

1000

! " #

$ % & 2

+ 2s

1000

! " #

$ % & +1

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100

101

102

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized rad frequency

M a g n i t u d e R e s p o n s e

TextEnd

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EXAMPLE 4. Frequency response of the (High Pass) RC circuit; by

V-division H (s) =s

s +1

C

=s

s +100

STEP 1: IMPORTANT FREQUENCIES:

ω H(jω)

0 H(j0) = 0∠90o

∞ H(j∞) = 1∠0o

100 H(j100) = 0.707 ∠ 45o

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STEP 2. ASYMPTOTIC BEHAVIOR.

H ( j ! ) = R

R +1

j ! C

= j ! CR

j ! CR +1=

j !

100

j

!

100 +1

|H(jω)| --> 1 as ω--> ∞

|H(jω)| --> 0 as ω --> 0

∠H(jω) --> 0 as ω-->∞

∠H(jω) --> 90o

as ω --> 0

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STEP 3. PLOTS FROM MATLAB:

»z = 0;

»p = -100;

»zplane(z,p)»grid

-100 -80 -60 -40 -20 0

-40

-30

-20

-10

0

10

20

30

40

Real part

I m a g i n a r y p a r t

TextEnd

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»w = logspace(0,4,500);

»H = freqs([1 0],[1 100],w);»semilogx(w,abs(H))

»grid

»xlabel('Frequency in rad/sec')»ylabel('Magnitude') »

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»semilogx(w,180*angle(H)/pi)

»grid

»xlabel('Frequency rad/sec')

»ylabel('Angle in degrees')

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101

102

103

104

105

-200

-150

-100

-50

0

Frequency in rad/sec

A n g l e i n d e g r e e s

2nd Order LP Frequency Response

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EXAMPLE 5. Frequency response of the (Band Pass) RLC circuit

STEP 1: CALCULATION OF TRANSFER FUNCTION: By V-division,

H (s) =Vout (s)

Vin(s)=

R

R + Ls +1

Cs

=

R

Ls

s2+ R

Ls +

1

LC

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STEP 2: IMPORTANT FREQUENCIES

H ( j ! ) =

j !

L1

LC "!

2+ j

!

L

= j 10!

104"!

2+ j 10!

ω H(jω)

0 H(j0) = 0∠90o

∞ H(j∞) = 0∠0o

100 1

?? 0.707 ∠±45o

?? 0.707 ∠±45o

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»w = logspace(0,4,1000);

»R = 10; L = 0.1; C = 1e-3;

»n = [R/L 0];»d = [1 R/L 1/(L*C)];

»h = freqs(n,d,w);

»semilogx(w,abs(h))»grid

»xlabel('Frequency in rad/s')

»ylabel('Magnitude Response')

100

101

102

103

104

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in rad/s

M a g n i t u d e R e s p

o n s e

TextEnd

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»semilogx(w,angle(h)*180/pi)

»grid

»xlabel('Frequency in rad/s')

»ylabel('Phase Response')

»

100

101

102

103

104

-100

-80

-60

-40

-20

0

20

40

60

80

100

Frequency in rad/s

P h a s e

R e s p o n s e

TextEnd

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EXAMPLE 6. Frequency response of the (Band Reject) RLCcircuit

STEP 1: CALCULATION OF TRAMSFER FUNCTION: By V-division,

H (s) = R

R +1

Cs +1

Ls

=

s2+

1

LC

s2+

1

RC s +

1

LC

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101

102

103

0

0.2

0.4

0.6

0.8

1


M a g n i t u d e

Band Reject Response

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85 90 95 100 105 110 1150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


M a g n i t u d e

Close-up Band Reject Response

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»

101

102

103

-100

-50

0

50

100


P h a s e

i n

d e g r e e s

Band Reject Response

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