ECE307 Frequency Response of a Circuit 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 ECE 307-4 2 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 ECE 307-4 3 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 ECE 307-4 5 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 ECE 307-4 7 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 ECE 307-4 8 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 ECE 307-4 9 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 ECE 307-4 11 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 ECE 307-4 14 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 22R 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)') ECE 307-4 16 8 Frequency Response of a Circuit R1 Example V V 10k V1 1Vac C1 0Vdc 1.99n 0 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 ECE 307-4 18 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 ECE 307-4 19 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 ECE 307-4 20 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 ECE 307-4 21 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 22R 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 2 0 1.