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Transcript of Frequency Response of a Circuit - Cal Poly Pomonazaliyazici/ece307/Frequencyresponse-4.pdf ·...
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Frequency Response of a Circuit
Z. Aliyazicioglu
Electrical and Computer Engineering DepartmentCal Poly Pomona
Parallel Resonance circuit
ECE 307-7
ECE 307-7 2
Frequency Response of a Circuit
Parallel Resonance circuit
( )( ) 1( ) ( ) 1 ( )
R
V jI j R
RI j V j Y j CRL
= =+
1 1( )Y j CR L
= +
To find frequency response, substitute s=j in equation
1( ) tan Rj CRL
=
Phase ResponseMagnitude Response
L
1
2
0
CI1
0Adc1Aac R
2
1( )1 ( )
H jRCRL
=+
-
2
ECE 307-7 3
=Frequency Response of a Circuit
At resonance frequency, the transfer function will be real. Or system total impedance will be real
00
1 0j Cj L
+ =
Hmax will be at |H(j0)| substitute
Result
01
LC =
01
2f
LC=
Parallel Resonance circuit
02
00
1( )1 ( )
H jRCR
L
=+
02
1( ) 111 ( )
H jLCRCRLLC
= =
+
01LC
=
ECE 307-7 4
Frequency Response of a Circuit
Set (1/2)Hmax to find cutoff frequencies
Result
2
11 1 1
2 2c RC RC LC = + +
2 1 1 0c cRC LC + =
2
21 1 1
2 2c RC RC LC = + +
0 1 2
1c c LC
= =
Confirm
2 1 1 0c cRC LC =
2
1 1( )2 1 ( )
c
cc
H jRCR
L
= =+
Parallel Resonance circuit
-
3
ECE 307-7 5
Frequency Response of a Circuit
The Bandwidth is
2 1
2 21 1 1 1 1 12 2 2 2
c c
RC RC LC RC RC LC
=
= + + + +
1RC
=
The Quality factor Q is
00
1Q RC RCLC
= = = CQ RL
= 02 1c c
fQf f
=
Parallel Resonance circuit
ECE 307-7 6
Frequency Response of a Circuit
The cutoff frequencies in terms of is 0
( )2
21 02 2c
= + +
( )2
22 02 2c
= + +
The cutoff frequencies in terms of Q is 02
1 01 11
2 2c Q Q
= + +
2
1 01 11
2 2c Q Q
= + +
Parallel Resonance circuit
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4
ECE 307-7 7
Frequency Response of a Circuit
Example Using parallel RLC circuit, design band pass filter that bandwidth 750 Hz and a center frequency of 1Khz Hz. Use a 10 mH inductor, Find R,C,c1, c2,and Q
Lets find L first.
0 1 21
c c LC = =
61 1
2 750(2.53)1083.75
RC
= =
=
Calculate R The Quality factor is
The cutoff frequencies
2 2 30
1 1 2.53(2 1000) 10 10
C FL x
= = =
0 1.33Q
= =
( )2
21 0 692.9 rad/s2 2c
= + + =
( )2
22 0 1443 rad/s2 2c
= + + =
ECE 307-7 8
Frequency Response of a Circuit
>> R=83.75;>> L=0.01;>> C=2.5330e-006;>> f=100:10:6000;>> w=2*pi*f;>> h=abs(1./(1+j*((w*C*R)+R./(w*L))));>> subplot (2,1,1)>> semilogx(w,h)>> grid on>> title('|H(j\omega)|')>> xlabel ('\omega')>> ylabel ('|H(j\omega)|')>> subplot (2,1,2)>> theta=angle(1./(1+j*((w*C*R)+R./(w*L))));>> degree=theta*180/pi;>> semilogx(w,degree)>> grid on>> title('\theta(j\omega)')>> xlabel('\omega')>> ylabel('\theta(j\omega)')
R=83.75 , L=10 mH, C=2.53 F, Plot F=100 6 KHz.
1( )1 ( )
H j Rj CRL
=+
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5
ECE 307-7 9
Frequency Response of a Circuit
OrCad CaptureExample
0V
R1
83.75
R2
0.1C1
2.53uF
I1
0Adc1Aac
II
0V
0
L1
10MH
1
2
Fr equency
100Hz 300Hz 1. 0KHz 3. 0KHz 10KHz- I ( R1) I ( I 1)
0A
0. 5A
1. 0A
( 695. 685, 709. 203m) ( 1. 4439K, 707. 063m)
ECE 307-7 10
R=2 K , L=2 H, C=2 F, Plot F=1 Hz 1 KHz.
Example
L1
2H
1 2
0
R1
2K
V0V
V21Vac0Vdc
V
C1
2u
Fr e q u e n c y
1 . 0 Hz 3 . 0 Hz 1 0 Hz 3 0 Hz 1 0 0 Hz 3 0 0 Hz 1 . 0 KHzV( V2 : +) V( R1 : 2 )
0 V
1 . 0 V
2 . 0 V
3 . 0 V
Frequency Response of a Circuit
