Frequency Response of a Circuit - Cal Poly Pomonazaliyazici/ece307/Frequencyresponse-2.pdf1...

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Frequency Response of a Circuit

Z. Aliyazicioglu

Electrical and Computer Engineering DepartmentCal Poly Pomona

ECE 307 Network Analysis III

Band-Pass Filter

ECE 307-5 2


0 1 2c cω ω ω=

Three important parameters

Band-Pass Filter

Center frequency (or resonance frequency), ω0 is defined as the frequency for which a the transfer function of a circuit is purely real

Bandwidth, β is the width of the passband

Qualty factor is the ration of the center frequency ω0 to the bandwidth β.

0

2 1c c

fQf f

=−

ωω ω

=−

0

2 1c c

Q

β ω ω= −2 1c c

ωβ

= 0Q

β = −2 1c cf f

2

ECE 307-5 3


A Serial RLC CircuitBand-Pass Filter

2( ) 1

R sLH s Rs sL LC

=+ +

0( )1( )i

V s RV s sL R

sC

=+ +

2( ) 1

R jLH j R jL LC

ωω

ω ω=− + +

To find frequency response, substitute s=jω in equation

2 22

( )1

RLH j

RLC L

ωω

ω ω

= − +

1

2( ) 90 tan 1

RLj

LC

ωθ ω

ω

−

= − −

Phase ResponseMagnitude Response

-

sL1 2

1/sC

RVi(s)Vo(s)

+

ECE 307-5 4


At resonance frequency, the transfer function will be real. Or system total impedance will be real

00

1 0j Lj C

ωω

+ =

Hmax will be at |H(jω0)| substitute ω0=√1/LC

cRL

ω =

A Serial RLC Circuit

Result

01LC

ω =0

12

fLCπ

=

0

0 2 220 0

( )1

RLH j

RLC L

ωω

ω ω

= − +

0 22

1

( ) 11 1 1

RL LCH j

RLC LC L LC

ω = = − +

3

ECE 307-5 5


Set (1/√2)Hmax to find cutoff frequencies


Result

2 22

12 1

c

c c

RL

RLC L

ω

ω ω

= − +

2

11

2 2cR RL L LC

ω = − + +

2 2 221 1

2 c c cR R

LC L Lω ω ω

− + =

2 2 221 2c c c

R RLC L L

ω ω ω − = −

21c c

RLC L

ω ω − =

∓

2 1 0c cRL LC

ω ω+ − =

2

21

2 2cR RL L LC

ω = + +

0 1 21

c c LCω ω ω= =

Confirm

2 1 0c cRL LC

ω ω− − =

ECE 307-5 6


The Bandwidth β is


2 1

2 21 12 2 2 2

c c

R R R RL L LC L L LC

β ω ω= −

= + + − − + +

RL

β =

The Quality factor Q is

0

1LCQ RL

ωβ

= =2

LQCR

= 0

2 1c c

fQf f

=−

4

ECE 307-5 7


The cutoff frequencies in terms of β and ω0


( )2

21 02 2c

β βω ω = − + +

( )2

22 02 2c

β βω ω = + +

The cutoff frequencies in terms of Q and ω0

2

1 01 11

2 2c Q Qω ω

= − + +

2

1 01 11

2 2c Q Qω ω

= + +

ECE 307-5 8


Example Using serial RLC circuit, design band pass filter that select 1-10KHz frequency band for a graphic equalizer in your amplifier

We can use different approaches. Let’s find the center frequency first.

0 1 21

c c LCω ω ω= =

0 1 2 1000 *10000 3162.28c cf f f Hz= = =

01LC

ω =

Choose capacitor value as 1µF

2 2 60

1 1 2.533mH(2 3162.28) (10 )

LCω π −= = =

The Quality factor is 0

2 1

3162.28 0.351410000 1000c c

fQf f

= = =− −

2LQCR

= 2 6 20.00253 143.24

10 (0.3514)LRCQ −= = = ΩThe Resistor is

5

ECE 307-5 9


>> R=143.25;>> L=0.002533;>> C=0.000001;>> f=0:50:100000;>> w=2*pi*f;>> h=abs((R/L).*(j*w)./(-w.^2+(R/L)*(j*w)+(1/(L*C))));>> 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((R/L).*(j*w)./(-w.^2+(R/L)*(j*w)+(1/(L*C))));>> degree=theta*180/pi;>> semilogx(w,degree)>> grid on>> title('\theta(j\omega)')>> xlabel('\omega')>> ylabel('\theta(j\omega)')

R=143.25 Ω , L=2.533 mH, C=1 µF, Plot F=50 – 100 KHz.

2( ) 1

R jLH j R jL LC

ωω

ω ω=− + +

ωc1=6283 rad/s

ωc2=62,831 rad/s

ECE 307-5 10


Edit Simulation Profile

R1

143

V

V11Vac1Vdc

V

C1

1u

0

L1

0.002531 2

OrCad CaptureExample

6

ECE 307-5 11


Fr e q u e n c y

1 0 Hz 1 0 0 Hz 1 . 0 KHz 1 0 KHz 1 0 0 KHzV( L1 : 1 ) V( R1 : 2 )

0 V

0 . 4 V

0 . 8 V

1 . 2 V

( 1 0 . 0 0 0 K, 7 0 6 . 9 8 5 m)( 1 . 0 0 0 0 K, 7 0 6 . 4 6 8 m)

PSpice Result

Frequency Response of a Circuit - Cal Poly Pomonazaliyazici/ece307/Frequencyresponse-2.pdf1...

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