bat365 Homepage - Welcome

[b,a] = butter(n,Wn) returns the transfer function coefficients of an nth-order lowpass digital Butterworth filter with normalized cutoff frequency Wn.

[b,a] = butter(n,Wn,ftype) designs a lowpass, highpass, bandpass, or bandstop Butterworth filter, depending on the value of ftype and the number of elements of Wn. The resulting bandpass and bandstop designs are of order 2n.

Note: See Limitations for information about numerical issues that affect forming the transfer function.

[z,p,k] = butter(___) designs a lowpass, highpass, bandpass, or bandstop digital Butterworth filter and returns its zeros, poles, and gain. This syntax can include any of the input arguments in previous syntaxes.

[A,B,C,D] = butter(___) designs a lowpass, highpass, bandpass, or bandstop digital Butterworth filter and returns the matrices that specify its state-space representation.

[___] = butter(___,'s') designs a lowpass, highpass, bandpass, or bandstop analog Butterworth filter with cutoff angular frequency Wn.

Examples

Lowpass Butterworth Transfer Function

Design a 6th-order lowpass Butterworth filter with a cutoff frequency of 300 Hz, which, for data sampled at 1000 Hz, corresponds to $0.6 π$ rad/sample. Plot its magnitude and phase responses. Use it to filter a 1000-sample random signal.

fc = 300;
fs = 1000;

[b,a] = butter(6,fc/(fs/2));

freqz(b,a,[],fs)

subplot(2,1,1)
ylim([-100 20])

Figure contains 2 axes objects. Axes object 1 with title Phase, xlabel Frequency (Hz), ylabel Phase (degrees) contains an object of type line. Axes object 2 with title Magnitude, xlabel Frequency (Hz), ylabel Magnitude (dB) contains an object of type line.

dataIn = randn(1000,1);
dataOut = filter(b,a,dataIn);

Bandstop Butterworth Filter

Design a 6th-order Butterworth bandstop filter with normalized edge frequencies of $0.2 π$ and $0.6 π$ rad/sample. Plot its magnitude and phase responses. Use it to filter random data.

[b,a] = butter(3,[0.2 0.6],'stop');
freqz(b,a)

$Figure contains 2 axes objects. Axes object 1 with title Phase, xlabel Normalized Frequency (\times\pi rad/sample), ylabel Phase (degrees) contains an object of type line. Axes object 2 with title Magnitude, xlabel Normalized Frequency (\times\pi rad/sample), ylabel Magnitude (dB) contains an object of type line.$

dataIn = randn(1000,1);
dataOut = filter(b,a,dataIn);

Highpass Butterworth Filter

Design a 9th-order highpass Butterworth filter. Specify a cutoff frequency of 300 Hz, which, for data sampled at 1000 Hz, corresponds to $0.6 π$ rad/sample. Plot the magnitude and phase responses. Convert the zeros, poles, and gain to second-order sections. Display the frequency response of the filter.

[z,p,k] = butter(9,300/500,"high");
sos = zp2sos(z,p,k);
freqz(sos)

Bandpass Butterworth Filter

Design a 20th-order Butterworth bandpass filter with a lower cutoff frequency of 500 Hz and a higher cutoff frequency of 560 Hz. Specify a sample rate of 1500 Hz. Use the state-space representation. Design an identical filter using designfilt.

fs = 1500;
[A,B,C,D] = butter(10,[500 560]/(fs/2));

d = designfilt("bandpassiir",FilterOrder=20, ...
    HalfPowerFrequency1=500,HalfPowerFrequency2=560, ...
    SampleRate=fs);

Convert the state-space representation to second-order sections. Visualize the frequency responses.

sos = ss2sos(A,B,C,D);

fvt = fvtool(sos,d,Fs=fs);
legend(fvt,["butter" "designfilt"])

Figure Figure 1: Magnitude Response (dB) contains an axes object. The axes object with title Magnitude Response (dB), xlabel Frequency (Hz), ylabel Magnitude (dB) contains 2 objects of type line. These objects represent butter, designfilt.

Comparison of Analog IIR Lowpass Filters

Design a 5th-order analog Butterworth lowpass filter with a cutoff frequency of 2 GHz. Multiply by $2 π$ 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])
grid
xlabel("Frequency (GHz)")
ylabel("Attenuation (dB)")
legend(["butter" "cheby1" "cheby2" "ellip" "besself"])

Figure contains an axes object. The axes object with xlabel Frequency (GHz), ylabel Attenuation (dB) contains 5 objects of type line. These objects represent 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.

Input Arguments

`n` — Filter order
integer scalar

Filter order, specified as an integer scalar. For bandpass and bandstop designs, n represents one-half the filter order.

Data Types: double

`Wn` — Cutoff frequency
scalar | two-element vector

Cutoff frequency, specified as a scalar or a two-element vector. The cutoff frequency is the frequency at which the magnitude response of the filter is 1 / √2.

If Wn is scalar, then butter designs a lowpass or highpass filter with cutoff frequency Wn.
If Wn is the two-element vector [w1 w2], where w1 < w2, then butter designs a bandpass or bandstop filter with lower cutoff frequency w1 and higher cutoff frequency w2.
For digital filters, the cutoff frequencies must lie between 0 and 1, where 1 corresponds to the Nyquist rate—half the sample rate or π rad/sample.
For analog filters, the cutoff frequencies must be expressed in radians per second and can take on any positive value.

Data Types: double

`ftype` — Filter type
`'low'` | `'bandpass'` | `'high'` | `'stop'`

Filter type, specified as one of the following:

'low' specifies a lowpass filter with cutoff frequency Wn. 'low' is the default for scalar Wn.
'high' specifies a highpass filter with cutoff frequency Wn.
'bandpass' specifies a bandpass filter of order 2n if Wn is a two-element vector. 'bandpass' is the default when Wn has two elements.
'stop' specifies a bandstop filter of order 2n if Wn is a two-element vector.

Output Arguments

`b,a` — Transfer function coefficients
row vectors

Transfer function coefficients of the filter, returned as row vectors of length n + 1 for lowpass and highpass filters and 2n + 1 for bandpass and bandstop filters.

For digital filters, the transfer function is expressed in terms of b and a as

$H (z) = \frac{B (z)}{A (z)} = \frac{b(1) + b(2) z^{- 1} + \dots + b(n+1) z^{- n}}{a(1) + a(2) z^{- 1} + \dots + a(n+1) z^{- n}} .$
For analog filters, the transfer function is expressed in terms of b and a as

$H (s) = \frac{B (s)}{A (s)} = \frac{b(1) s^{n} + b(2) s^{n - 1} + \dots + b(n+1)}{a(1) s^{n} + a(2) s^{n - 1} + \dots + a(n+1)} .$

Data Types: double

`z,p,k` — Zeros, poles, and gain
column vectors, scalar

Zeros, poles, and gain of the filter, returned as two column vectors of length n (2n for bandpass and bandstop designs) and a scalar.

For digital filters, the transfer function is expressed in terms of z, p, and k as
$H (z) = k \frac{(1 - z(1) z^{- 1}) (1 - z(2) z^{- 1}) \dots (1 - z(n) z^{- 1})}{(1 - p(1) z^{- 1}) (1 - p(2) z^{- 1}) \dots (1 - p(n) z^{- 1})} .$
For analog filters, the transfer function is expressed in terms of z, p, and k as
$H (s) = k \frac{(s - z(1)) (s - z(2)) \dots (s - z(n))}{(s - p(1)) (s - p(2)) \dots (s - p(n))} .$

Data Types: double

`A,B,C,D` — State-space matrices
matrices

State-space representation of the filter, returned as matrices. If m = n for lowpass and highpass designs and m = 2n for bandpass and bandstop filters, then A is m × m, B is m × 1, C is 1 × m, and D is 1 × 1.

For digital filters, the state-space matrices relate the state vector x, the input u, and the output y through

$\begin{matrix} x (k + 1) = A x (k) + B u (k) \\ y (k) = C x (k) + D u (k) . \end{matrix}$
For analog filters, the state-space matrices relate the state vector x, the input u, and the output y through

$\begin{array}{l} \dot{x} = A x + B u \\ y = C x + D u . \end{array}$

Data Types: double

More About