【 MATLAB 】freqz 函数介绍（数字滤波器的频率响应）

Frequency response of digital filter

[h,w] = freqz(b,a,n)

[h,w] = freqz(d,n)

[h,w] = freqz(___,n,'whole')

freqz(___)

[h,f] = freqz(___,n,fs)

[h,f] = freqz(___,n,'whole',fs)

h = freqz(___,w)

h = freqz(___,f,fs)

[h,w] = freqz(sos,n)

[h,w] = freqz(b,a,n)​ returns the n​-point frequency response vector, h​, and the corresponding angular frequency vector, w​, for the digital filter with numerator and denominator polynomial coefficients stored in b​ and a, respectively.

[h，w] = freqz（b，a，n）返回数字滤波器的n点频率响应矢量h和相应的角频率矢量w，其中分子和分母多项式系数分别存储在b和a中。

[h,w] = freqz(___,n,'whole')​ returns the frequency response at n sample points around the entire unit circle.

[h,w] = freqz(___,n,'whole') 返回整个单位圆周围的n个采样点的频率响应。

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

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)')

[h,w] = freqz(d,n)​ returns the n​-point complex frequency response for the digital filter, d.

Design an FIR bandpass filter with passband between  and  rad/sample and 3 dB of ripple. The first stopband goes from  0 to  rad/sample and has an attenuation of 40 dB. The second stopband goes from  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)')

freqz(___) with no output arguments plots the frequency response of the filter.

Design an FIR lowpass filter of order 80 using a Kaiser window with . Specify a normalized cutoff frequency of  rad/sample. Display the magnitude and phase responses of the filter.

b = fir1(80,0.5,kaiser(81,8));

freqz(b,1)

Design the same filter using designfilt. Display its magnitude and phase responses using fvtool.

d = designfilt('lowpassfir','FilterOrder',80, ...

               'CutoffFrequency',0.5,'Window',{'kaiser',8});

freqz(d)

Note:   If the input to freqz​ is single precision, the frequency response is calculated using single-precision arithmetic. The output, h, is single precision.

[h,w] = freqz(sos,n)​ returns the n​-point complex frequency response corresponding to the second-order sections matrix, sos.

[h,f] = freqz(___,n,fs)​ returns the frequency response vector, ​h​, and the corresponding physical frequency vector, f​, for the digital filter with numerator and denominator polynomial coefficients stored in b​ and a​, respectively, given the sample rate, fs.[h,f] = freqz(___,n,'whole',fs)​ returns the frequency at ​n​ points ranging between 0 and fs.h = freqz(___,w)​ returns the frequency response vector, ​h​, at the normalized frequencies supplied in w.h = freqz(___,f,fs)​ returns the frequency response vector, ​h​, at the physical frequencies supplied in f.