Frequency Response - MATLAB & Simulink - MathWorks United Kingdom (2024)

Frequency Response from Transfer Function

Compute and display the magnitude response of the third-order IIR lowpass filter described by the following transfer function:

$$H(z)=\frac{0.05634(1+z^{-1})(1-1.0166z^{-1}+z^{-2})}{(1-0.683z^{-1})(1-1.4461z^{-1}+0.7957z^{-2})}.$$

Express the numerator and denominator as polynomial convolutions. Find the frequency response at 2001 points spanning the complete unit circle.

b0 = 0.05634;b1 = [1 1];b2 = [1 -1.0166 1];a1 = [1 -0.683];a2 = [1 -1.4461 0.7957];b = b0*conv(b1,b2);a = conv(a1,a2);[h,w] = freqz(b,a,'whole',2001);

Plot the magnitude response expressed in decibels.

plot(w/pi,20*log10(abs(h)))ax = gca;ax.YLim = [-100 20];ax.XTick = 0:.5:2;xlabel('Normalized Frequency (\times\pi rad/sample)')ylabel('Magnitude (dB)')

Frequency Response of an FIR Bandpass Filter

Design an FIR bandpass filter with passband between $$0.35\pi$$ and $$0.8\pi$$ rad/sample and 3 dB of ripple. The first stopband goes from $$0$$ to $$0.1\pi$$ rad/sample and has an attenuation of 40 dB. The second stopband goes from $$0.9\pi$$ rad/sample to the Nyquist frequency and has an attenuation of 30 dB. Compute the frequency response. Plot its magnitude in both linear units and decibels. Highlight the passband.

sf1 = 0.1;pf1 = 0.35;pf2 = 0.8;sf2 = 0.9;pb = linspace(pf1,pf2,1e3)*pi;bp = designfilt('bandpassfir', ... 'StopbandAttenuation1',40, 'StopbandFrequency1',sf1,... 'PassbandFrequency1',pf1,'PassbandRipple',3,'PassbandFrequency2',pf2, ... 'StopbandFrequency2',sf2,'StopbandAttenuation2',30);[h,w] = freqz(bp,1024);hpb = freqz(bp,pb);subplot(2,1,1)plot(w/pi,abs(h),pb/pi,abs(hpb),'.-')axis([0 1 -1 2])legend('Response','Passband','Location','South')ylabel('Magnitude')subplot(2,1,2)plot(w/pi,db(h),pb/pi,db(hpb),'.-')axis([0 1 -60 10])xlabel('Normalized Frequency (\times\pi rad/sample)')ylabel('Magnitude (dB)')

Magnitude Response of a Highpass Filter

Design a 3rd-order highpass Butterworth filter having a normalized 3-dB frequency of $$0.5\pi$$ rad/sample. Compute its frequency response. Express the magnitude response in decibels and plot it.

[b,a] = butter(3,0.5,'high');[h,w] = freqz(b,a);dB = mag2db(abs(h));plot(w/pi,dB)xlabel('\omega / \pi')ylabel('Magnitude (dB)')ylim([-82 5])

Comparison of Analog IIR Lowpass Filters

Design a 5th-order analog Butterworth lowpass filter with a cutoff frequency of 2 GHz. Multiply by $$2\pi$$ to convert the frequency to radians per second. Compute the frequency response of the filter at 4096 points.

n = 5;fc = 2e9;[zb,pb,kb] = butter(n,2*pi*fc,"s");[bb,ab] = zp2tf(zb,pb,kb);[hb,wb] = freqs(bb,ab,4096);

Design a 5th-order Chebyshev Type I filter with the same edge frequency and 3 dB of passband ripple. Compute its frequency response.

[z1,p1,k1] = cheby1(n,3,2*pi*fc,"s");[b1,a1] = zp2tf(z1,p1,k1);[h1,w1] = freqs(b1,a1,4096);

Design a 5th-order Chebyshev Type II filter with the same edge frequency and 30 dB of stopband attenuation. Compute its frequency response.

[z2,p2,k2] = cheby2(n,30,2*pi*fc,"s");[b2,a2] = zp2tf(z2,p2,k2);[h2,w2] = freqs(b2,a2,4096);

Design a 5th-order elliptic filter with the same edge frequency, 3 dB of passband ripple, and 30 dB of stopband attenuation. Compute its frequency response.

[ze,pe,ke] = ellip(n,3,30,2*pi*fc,"s");[be,ae] = zp2tf(ze,pe,ke);[he,we] = freqs(be,ae,4096);

Design a 5th-order Bessel filter with the same edge frequency. Compute its frequency response.

[zf,pf,kf] = besself(n,2*pi*fc);[bf,af] = zp2tf(zf,pf,kf);[hf,wf] = freqs(bf,af,4096);

Plot the attenuation in decibels. Express the frequency in gigahertz. Compare the filters.

plot([wb w1 w2 we wf]/(2e9*pi), ... mag2db(abs([hb h1 h2 he hf])))axis([0 5 -45 5])gridxlabel("Frequency (GHz)")ylabel("Attenuation (dB)")legend(["butter" "cheby1" "cheby2" "ellip" "besself"])

The Butterworth and Chebyshev Type II filters have flat passbands and wide transition bands. The Chebyshev Type I and elliptic filters roll off faster but have passband ripple. The frequency input to the Chebyshev Type II design function sets the beginning of the stopband rather than the end of the passband. The Bessel filter has approximately constant group delay along the passband.