Documentation

stepz

Step response of digital filter

Syntax

[h,t] = stepz(b,a)
[h,t] = stepz(sos)
[h,t] = stepz(d)
[h,t] = stepz(...,n)
[h,t] = stepz(...,n,fs)
stepz(...)

Description

[h,t] = stepz(b,a) returns the step response of the filter with numerator coefficients, b, and denominator coefficients, a. stepz chooses the number of samples and returns the response in the column vector h and sample times in the column vector t (where t = [0:n-1]', and n = length(t) is computed automatically).

[h,t] = stepz(sos) returns the step response of the second order sections matrix, sos. 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, stepz considers the input to be the numerator vector, b. Each row of sos corresponds to the coefficients of a second-order (biquad) filter. The ith row of the sos matrix corresponds to [bi(1) bi(2) bi(3) ai(1) ai(2) ai(3)].

[h,t] = stepz(d) returns the step response of the digital filter, d. Use designfilt to generate d based on frequency-response specifications.

[h,t] = stepz(...,n) computes the first n samples of the step response when n is an integer (t = [0:n-1]'). If n is a vector of integers, the step response is computed only at those integer values with 0 denoting the time origin.

[h,t] = stepz(...,n,fs) computes n samples and produces a vector t of length n so that the samples are spaced 1/fs units apart. fs is assumed to be in Hz.

stepz(...) with no output arguments plots the step response of the filter. If you input the filter coefficients or second order sections matrix, the current figure window is used. If you input a digitalFilter, the step response is displayed in FVTool.

Note

If the input to stepz is single precision, the step response is calculated using single-precision arithmetic. The output, h, is single precision.

Examples

collapse all

Create a third-order Butterworth filter with normalized half-power frequency $0.4\pi$ rad/sample. Display its step response.

[b,a] = butter(3,0.4);
stepz(b,a)
grid Create an identical filter using designfilt and display its step response using fvtool.

d = designfilt('lowpassiir','FilterOrder',3,'HalfPowerFrequency',0.4);
stepz(d) Design a fourth-order lowpass elliptic filter with normalized passband frequency $0.4\pi$ rad/sample. Specify a passband ripple of 0.5 dB and a stopband attenuation of 20 dB. Plot the first 50 samples of the filter's step response.

[b,a] = ellip(4,0.5,20,0.4);
stepz(b,a,50)
grid Create the same filter using designfilt and display its step response using fvtool.

d = designfilt('lowpassiir','FilterOrder',4,'PassbandFrequency',0.4, ...
'PassbandRipple',0.5,'StopbandAttenuation',20, ...
'DesignMethod','ellip');
stepz(d,50) Algorithms

stepz filters a length n step sequence using

filter(b,a,ones(1,n))

and plots the results using stem.

To compute n in the auto-length case, stepz either uses n = length(b) for the FIR case or first finds the poles using p = roots(a), if length(a) is greater than 1.

If the filter is unstable, n is chosen to be the point at which the term from the largest pole reaches 106 times its original value.

If the filter is stable, n is chosen to be the point at which the term due to the largest amplitude pole is 5 × 10–5 of its original amplitude.

If the filter is oscillatory (poles on the unit circle only), stepz computes five periods of the slowest oscillation.

If the filter has both oscillatory and damped terms, n is chosen to equal five periods of the slowest oscillation or the point at which the term due to the pole of largest nonunit amplitude is 5 × 10–5 times its original amplitude, whichever is greater.

stepz also allows for delays in the numerator polynomial. The number of delays is incorporated into the computation for the number of samples.