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# freqz

Frequency response of digital filter

## Syntax

[h,w] = freqz(b,a,n)
[h,w] = freqz(sos,n)
[h,w] = freqz(d,n)
[h,w] = freqz(___,n,'whole')
[h,f] = freqz(___,n,fs)
[h,f] = freqz(___,n,'whole',fs)
h = freqz(___,w)
h = freqz(___,f,fs)
freqz(___)

## Description

example

[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.

example

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

example

[h,w] = freqz(d,n) returns the n-point complex frequency response for the digital filter, d.
[h,w] = freqz(___,n,'whole') returns the frequency response at n sample points around the entire unit circle.
[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 sampling frequency, 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.

example

freqz(___) with no output arguments plots the frequency response of the filter.Note:   If the input to freqz is single precision, the frequency response is calculated using single-precision arithmetic. The output, h, is single precision.

## Examples

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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)')

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

Express the transfer function in terms of second-order sections. 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]; sos1 = [b0*[b1 0] [a1 0]]; sos2 = [b2 a2]; [h,w] = freqz([sos1;sos2],'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)')

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)

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

## Input Arguments

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Transfer function coefficients, specified as vectors. Express the transfer function in terms of b and a as

$H\left({e}^{j\omega }\right)=\frac{B\left({e}^{j\omega }\right)}{A\left({e}^{j\omega }\right)}=\frac{\text{b(1)}+\text{b(2)}\text{\hspace{0.17em}}{e}^{-j\omega }+\text{b(3)}\text{\hspace{0.17em}}{e}^{-j2\omega }+\cdots +\text{b(M)}\text{\hspace{0.17em}}{e}^{-j\left(M-1\right)\omega }}{\text{a(1)}+\text{a(2)}\text{\hspace{0.17em}}{e}^{-j\omega }+\text{a(3)}\text{\hspace{0.17em}}{e}^{-j2\omega }+\cdots +\text{a(N)}\text{\hspace{0.17em}}{e}^{-j\left(N-1\right)\omega }}.$

Example: b = [1 3 3 1]/6 and a = [3 0 1 0]/3 specify a third-order Butterworth filter with normalized 3-dB frequency 0.5π rad/sample.

Data Types: double | single
Complex Number Support: Yes

Number of evaluation points, specified as a positive integer scalar no less than 2. When n is absent, it defaults to 512. For best results, set n to a value greater than the filter order.

Data Types: double

Second-order section coefficients, specified as a matrix. sos is a K-by-6 matrix, where the number of sections, K, must be greater than or equal to 2. If the number of sections is less than 2, freqz considers the input to be a numerator vector. Each row of sos corresponds to the coefficients of a second-order (biquad) filter. The ith row of sos corresponds to [bi(1) bi(2) bi(3) ai(1) ai(2) ai(3)].

Example: s = [2 4 2 6 0 2;3 3 0 6 0 0] specifies a third-order Butterworth filter with normalized 3-dB frequency 0.5π rad/sample.

Data Types: double | single
Complex Number Support: Yes

Digital filter, specified as a digitalFilter object. Use designfilt to generate a digital filter based on frequency-response specifications.

Example: d = designfilt('lowpassiir','FilterOrder',3,'HalfPowerFrequency',0.5) specifies a third-order Butterworth filter with normalized 3-dB frequency 0.5π rad/sample.

Sampling rate, specified as a positive scalar. When the unit of time is seconds, fs is expressed in hertz.

Data Types: double

Angular frequencies, specified as a vector and expressed in radians/sample. w must have at least two elements. w = π corresponds to the Nyquist frequency.

Frequencies, specified as a vector. f must have at least two elements. When the unit of time is seconds, f is expressed in hertz.

Data Types: double

## Output Arguments

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Frequency response, returned as a vector. If you specify n, h has length n. If you do not specify n, or specify n as the empty vector, h has length 512.

Angular frequencies, returned as a vector. w has values ranging from 0 to π. If you specify 'whole' in your input, the values in w range from 0 to 2π. If you specify n, w has length n. If you do not specify n, or specify n as the empty vector, w has length 512.

Frequencies, returned as a vector expressed in hertz. f has values ranging from 0 to fs/2 Hz. If you specify 'whole' in your input, the values in f range from 0 to fs Hz. If you specify n, f has length n. If you do not specify n, or specify n as the empty vector, f has length 512.

## Algorithms

The frequency response of a digital filter can be interpreted as the transfer function evaluated at z = e [1].

freqz determines the transfer function from the (real or complex) numerator and denominator polynomials you specify and returns the complex frequency response, H(e), of a digital filter. The frequency response is evaluated at sample points determined by the syntax that you use.

freqz generally uses an FFT algorithm to compute the frequency response whenever you don’t supply a vector of frequencies as an input argument. It computes the frequency response as the ratio of the transformed numerator and denominator coefficients, padded with zeros to the desired length.

When you do supply a vector of frequencies as an input argument, freqz evaluates the polynomials at each frequency point using Horner’s method of nested polynomial evaluation, dividing the numerator response by the denominator response.

## References

[1] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. 2nd Ed. Upper Saddle River, NJ: Prentice Hall, 1999.