Frequency Response of a Circuit

Home , Cutoff frequency, Frequency response, RC circuit, RL circuit

ECE307

Z. Aliyazicioglu

Electrical and Computer Engineering Department Cal Poly Pomona

FrequencyThe LaplaceResponse Transform of a Circuit

Some Preliminaries

Analysis of a circuit with varying frequency of a sinusoidal sources is called the frequency response of a circuit

, Frequency selection in the circuits are called filters because of their ability to filter out certain input signals on the basis of frequency

Output signal Input signal FilterFilter

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1 Frequency Response of a Circuit

Some Preliminaries

We remember that the transfer function is the output voltage to the input voltage of a circuit in s-domain is. Vs() Hs()= 0 Vsi ()

Using sinusoidal source, the transfer function will be the magnitude and phase of output voltage to the magnitude and phase of input voltage of a circuit .

In this case we will use (jω) instead of s . Vj()ω Hj()ω = 0 Vji ()ω

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Frequency Response of a Circuit

Frequency Response Using transfer function of circuit, we plot a frequency response of the circuit for both amplitude and phase with changing source frequency

One graph of |H(jω)| versus frequency jω. It is called the Magnitude plot. One graph of θ(jω) versus frequency ω. It is called the Phase Angle plot

|H(jω)| θ(jω) ω

ωc

Passband Stopband

ω θ(jωc)

ωc ω : Cutoff frequency c ECE 307-4 4

2 Frequency Response of a Circuit

Filter

A Low-Pass filter passes A High-Pass filter passes signals at frequencies lower signals at frequencies higher than the cutoff frequency than the cutoff frequency from from the input to the output the input to the output

|H(jω)| ωc Cutoff frequency |H(jω)|

Passband Stopband Passband ω Stopband 0 ω θ(jω) ωc 0 θ(jω) ωc

θ(jωc) θ(jωc)

Ideal Low-Pass Filter Ideal High-Pass Filter

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Frequency Response of a Circuit

Filter A Band-Pass filter passes A Band-reject filter passes signals within the band defined signals outside the band defined by two cutoff frequencies from by two cutoff frequencies from the input to the output the input to the output ω Cutoff frequency |H(jω)| c |H(jω)| Stop Passband Stopband band Stop ω Pass Pass θ(jω) band band band ωc1 ωc2 ω θ(jω ) θ(jω) c1 ω ω θ(jωc1) c1 c2

0 θ(jωc1)

θ(jωc1) Ideal Band-Pass Filter Ideal Band-Reject Filter ECE 307-4 6

3 Frequency Response of a Circuit

Cutoff Frequency The transfer function magnitude is decreased by the factor 1/√2 from its maximum value is called cutoff frequency

1 Hj()ω = H c 2 max

|Hmax | is the maximum magnitude of the transfer function

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Frequency Response of a Circuit

Low-Pass Filter A Serial RL Circuit

sL 1 2 Vs() R R + 0 = Vs() sLR Vi(s) R i + Hs()= L Vo(s) R s + - L

To find frequency response, substitute s=jω in equation

R Magnitude Response Hj()ω = L Phase Response R R jω + L L Hj()ω = −1 ωL 2 θω()j =− tan 2 R R ω +  L

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4 Frequency Response of a Circuit

A Serial RL Circuit

When ω=0 When ω=∞ R R Hj(0)==L 1 L 2 Hj()∞ == 0 2 R 2 0 +  2 R L ∞+ L

−1 0L D θ(0)j =− tan = 0 −1 ∞L D  θ()j∞=− tan =− 90 R R To find Cutoff Frequency Result R 1 R Hj()ω ==L ω = c 2 2 c L 2 R ωc +  L

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Frequency Response of a Circuit

Example R=1KΩ F=1KHz. L=? Plot H(jω). 1000 R 1000 LH== =0.159 Hj()ω = 0.159 ωπc 2 * * 1000 1000 jω + 0.159 >> R=1000; >> L=0.159; >> f=0:25:10000; >> w=2*pi*f; >> subplot (2,1,1) >> h=abs((R/L)./(j*w+(R/L))); >> semilogx(w,h) >> grid on >> title('|H(j\omega)|') >> xlabel ('\omega') >> ylabel ('|H(j\omega)|') >> theta=angle((R/L)./(j*w+(R/L))); >> subplot (2,1,2) >> degree=theta*180/pi; >> semilogx(w,degree) >> grid on >> title('\theta(j\omega)') >> xlabel('\omega') >> ylabel('\theta(j\omega)')ECE 307-4 10

5 Frequency Response of a Circuit

Example R=1KΩ F=1KHz. L=? Plot H(jω).

RL/ Hs()= sRL+ / Matlab

>> syms s >> n=[0 1000/0.159]; >> d=[1 1000/0.159]; >> g=tf(n,d)

Transfer function: 6289 ------s + 6289

>> bode (g) >> grid on

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Frequency Response of a Circuit

L1 Example 1 2 0.159H V V V1 1Vac R1 0Vdc 1k

0 1. 0V

fc=1KHz

0. 5V

0V 100Hz 300Hz 1. 0KHz 3. 0KHz 10KHz 3 0 KHz 100KHz V( L1 : 2 ) V( V1 : + ) Fr equency ECE 307-4 12

6 Frequency Response of a Circuit

Low-Pass Filter A Serial RC Circuit

R 1 1 Vs() RC + 0 = sC Hs()= 1 1 Vi(s) 1/sC Vo(s) Vsi () s + R + RC - sC

To find frequency response, substitute s=jω in equation

1 Magnitude Response Hj()ω = RC Phase Response 1 jω + 1 RC θω()j =− tan−1 ( ωRC) Hj()ω = RC 2 2 1 ω +  RC ECE 307-4 13

Frequency Response of a Circuit

A Serial RC Circuit

When ω=0 When ω=∞ 1 1 Hj(0)==RC 1 RC 2 Hj()∞ == 0 2 1 2 0 +  2 1 RC ∞+ RC

−1 D θ(0)jRC=− tan() 0 = 0 θ()jRC∞=− tan−1 ( ∞) =− 90D

To find Cutoff Frequency Result 1 1 1 Hj()ω ==RC ω = c 2 2 c RC 2 1 ωc +  RC

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7 Frequency Response of a Circuit

Example A series RC low-pass filter cutoff frequency is 8KHz. R=10KΩ Find the capacitor value

1 1 1 C = CnF==1.99 ωc = RC ωcR 2 *π * 8000x 10000

Example A series RL low-pass filter cutoff frequency is 2KHz. R=5KΩ Find the inductor value. Find |H(jω)| at 50 KHz?

R R 5000 ωc = L = LH==2.5 L ωc 2000 R 5000 Hj(ω )==L 2.5 = 0.0635 50KHz 22 22R 5000  ωπ++(2 * * 50000)   L 2.5  ECE 307-4 15

Frequency Response of a Circuit

Example A series RC low-pass filter cutoff frequency is 8KHz. R=10KΩ, C=1.99 nF

r=10000; c=19.9*10^-9;

f=0:25:100000; w=2*pi*f; h=abs((1/(r*c))./(j*w+1/(r*c))); subplot (2,1,1) semilogx(w,h) grid on title('|H(j\omega)') xlabel ('\omega') ylabel ('|H(j\omega)') theta=angle((1/(r*c))./(j*w+1/(r*c)) ); subplot (2,1,2) degree=theta*180/pi; semilogx (w, degree) grid on title('\theta(j\omega)') xlabel('\omega') ylabel('\theta(j\omega)')

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8 Frequency Response of a Circuit

R1 Example V V 10k V1 1Vac C1 0Vdc 1.99n

1. 0V

1 f =8KHz Hf()= c c 2

0. 5V

0V 100Hz 1. 0KHz 1 0 KHz 100KHz 1. 0MHz V( R1 : 1 ) V( V1 : + ) Fr equency ECE 307-4 17

Frequency Response of a Circuit

High-Pass Filter A Serial RC Circuit

1/sC Vs() R s + 0 = Hs()= Vs() 1 1 Vi(s) R Vo(s) i R + s + - sC RC

To find frequency response, substitute s=jω in equation

jω Hj()ω = Magnitude Response 1 Phase Response jω + RC ω Hj()ω = D −1 2 θω()90tanj =− ( ωRC) 2 1 ω +  RC

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9 Frequency Response of a Circuit

A Serial RC Circuit

When ω=0 When ω=∞ 0 Hj(0)== 0 ∞ 2 Hj()∞ == 1 2 2 1 0 +  2 1 RC ∞+ RC

DD−1 0 θ(j 0)=− 90 tan = 90 DD−1 ∞  θ()90tanj∞ =− = 0 RC RC To find Cutoff Frequency Result ω 1 Hj()ω ==c 1 c 2 2 ωc = 2 1 RC ωc +  RC

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Frequency Response of a Circuit

High-Pass Filter A Serial RL Circuit R

1 + Vs() sL s 0 = Hs()= Vi(s) sL R Vsi () sLR+ s + Vo(s) L 2 -

To find frequency response, substitute s=jω in equation

jω Hj()ω = Magnitude Response Phase Response R jω + L ω Hj()ω = D −1 ωL 2 θω()90tanj =−  2 R R ω +  L

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10 Frequency Response of a Circuit

A Serial RL Circuit

When ω=0 When ω=∞ 0 Hj(0)== 0 ∞ 2 Hj()∞ == 1 2 2 R 0 +  2 R L ∞+ L

DD−1 0L θ(j 0)=− 90 tan = 90 DD−1 ∞L  θ()90tanj∞ =− = 0 R R To find Cutoff Frequency Result ω 1 Hj()ω ==c R c 2 2 ωc = 2 R L ωc +  L

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Frequency Response of a Circuit

Example Define R and L values for a high pass filter with a cutoff frequency of 10KHz. Find |H(jω)|at 5 KHz

R We can’t calculate R and L values ωc = L independently. We can select R or L values then define the other

Let RK=Ω1 Result R 1000 L = LmH==15.9 ωc 2 *π * 10000

ω 2 *π * 5000 Hj(ω )== = 0.4469 5KHz 22 22R 1000  ωπ++(2 * * 5000)   L 0.0159  ECE 307-4 22

11 Frequency Response of a Circuit

Example A RL high pass filter with a cutoff frequency of 10KHz. 1000 RK= 1 Ω LmH==15.9 2 *π * 10000

R=1000; L=15.9*10^-3;

f=0:25:70000; w=2*pi*f; h=abs((j*w)./(j*w+R/L)); subplot (2,1,1) semilogx(w,h) grid on title('|H(j\omega)') xlabel ('\omega') ylabel ('|H(j\omega)') theta=angle((j*w)./(j*w+R/L)); subplot (2,1,2) plot (w, theta) degree=theta*180/pi; semilogx (f, degree) grid on title('\theta(j\omega)') xlabel('\omega') ylabel('\theta(j\omega)') ECE 307-4 23

Frequency Response of a Circuit

Example R1

V 1 V 1K V1 L2 1Vac 0Vdc 15.9mH

1. 0V

1 f =8KHz Hf()= c c 2

0. 5V

0V 100Hz 1 . 0 KHz 10KHz 1 0 0 KHz 1. 0MHz V( R1 : 2 ) V( V1 : + ) ECE 307-4 24 Fr equency

12 Frequency Response of a Circuit Plotting phase: R1 Take the probe Phase of Voltage, which is under Pspice, Markers, and Advanced. 1 VP 1K VP Marked the node you want to see phase V1 L2 1Vac 0Vdc 15.9mH

100d 0

(Hf()= 45D 50d fc=8KHz c

0d 100Hz 1 . 0 KHz 10KHz 1 0 0 KHz 1. 0MHz VP ( V1 : + ) VP ( R1 : 2 ) Fr equency ECE 307-4 25

Frequency Response of a Circuit

Example Let’s place load resister in parallel to inductor in RL high-pass filter shown in the figure a. Find the transfer function

b. Rs=RL=1KΩ, find L value for cutoff frequency at 10KHz. R

1 RsLL + Vs0() RL + sL Vi(s) sL RL Vo(s) = Vs() RsL i R + L 2 - RsLL +

R L s RR+ Ks where RL 1 Hs()==L K = K ==0.5 RR R RR+ 11+ ssK++L L RRL+ L L Result R ω = K R 1 c L LK==0.5 = 7.95 mH ωπc 2* *10 ECE 307-4 26

13 Frequency Response of a Circuit

Example Rs=RL=1KΩ, L=7.95 mH High-pass filter cutoff frequency at 10KHz. jKω Hj()ω = R=1000; R RL=1000; jKω + L=7.95*10^-3; L K=RL/(R+RL)

f=0:25:70000; w=2*pi*f; h=abs((j*w*K)./(j*w+K*R/L)); subplot (2,1,1) semilogx(w,h) grid on title('|H(j\omega)') xlabel ('\omega') ylabel ('|H(j\omega)') theta=angle((j*w*K)./(j*w+K*R/L)); subplot (2,1,2) plot (w, theta) degree=theta*180/pi; semilogx (f, degree) grid on title('\theta(j\omega)') xlabel('\omega') ylabel('\theta(j\omega)') grid on ECE 307-4 27