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FREQUENCY RESPONSE OF
SIMPLE CIRCUITS
Ray DeCarlo School of ECE
Purdue University
West Lafayette, IN 47907-1285decarlo@ecn.purdue.edu
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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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EE-202, Frequency Response p 3 © R. A. DeCarlo
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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EE-202, Frequency Response p 4 © R. A. DeCarlo
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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EE-202, Frequency Response p 5 © R. A. DeCarlo
(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);
»h2 = freqs(n2,d2,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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EE-202, Frequency Response p 6 © R. A. DeCarlo
(b) Qualitative Analysis Using Pole-zero plot.
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EE-202, Frequency Response p 7 © R. A. DeCarlo
(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;
»h1 = freqs(n1,d1,w);
»h2 = freqs(n2,d2,w);
»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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EE-202, Frequency Response p 8 © R. A. DeCarlo
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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EE-202, Frequency Response p 9 © R. A. DeCarlo
(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];
»h1 = freqs(n1,d1,w);
»h2 = freqs(n2,d2,w);
»h3 = freqs(n3,d3,w);
»semilogx(w,abs(h1),w,abs(h2),w,abs(h3))
»grid
»xlabel('Normalized rad frequency')
»ylabel('Magnitude Response')
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EE-202, Frequency Response p 10 © R. A. DeCarlo
(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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EE-202, Frequency Response p 11 © R. A. DeCarlo
(f) Pole-Zero Plot of 2nd and 3rd Order Filters
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EE-202, Frequency Response p 12 © R. A. DeCarlo
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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EE-202, Frequency Response p 13 © R. A. DeCarlo
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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EE-202, Frequency Response p 14 © R. A. DeCarlo
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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EE-202, Frequency Response p 15 © R. A. DeCarlo
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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EE-202, Frequency Response p 16 © R. A. DeCarlo
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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EE-202, Frequency Response p 17 © R. A. DeCarlo
»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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EE-202, Frequency Response p 18 © R. A. DeCarlo
»semilogx(w,180*angle(H)/pi)
»grid
»xlabel('Frequency rad/sec')
»ylabel('Angle in degrees')
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EE-202, Frequency Response p 19 © R. A. DeCarlo
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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EE-202, Frequency Response p 20 © R. A. DeCarlo
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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EE-202, Frequency Response p 21 © R. A. DeCarlo
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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EE-202, Frequency Response p 22 © R. A. DeCarlo
STEP 4. PLOTS FROM MATLAB:
»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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EE-202, Frequency Response p 23 © R. A. DeCarlo
»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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EE-202, Frequency Response p 24 © R. A. DeCarlo
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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EE-202, Frequency Response p 25 © R. A. DeCarlo
STEP 4. PLOTS FROM MATLAB:
101
102
103
0
0.2
0.4
0.6
0.8
1
Frequency in rad/sec
M a g n i t u d e
Band Reject Response
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EE-202, Frequency Response p 26 © R. A. DeCarlo
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
Frequency in rad/sec
M a g n i t u d e
Close-up Band Reject Response
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EE-202, Frequency Response p 27 © R. A. DeCarlo
»
101
102
103
-100
-50
0
50
100
Frequency in rad/sec
P h a s e
i n
d e g r e e s
Band Reject Response