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IMPULSE AND STEP RESPONSE

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Ananthu S
Ananthu S 2017 年 1 月 17 日
回答済み: Girma 2024 年 1 月 11 日
I know that 'stepz' and 'impz' can be used to find the step and impulse response of a digital filter when numerator and denominator coefficients are given as arguments. How I can I find the above responses for a continuos time system using the numerator and denominator coefficients? Please clarify

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Richard Zappulla
Richard Zappulla 2017 年 1 月 17 日
編集済み: Richard Zappulla 2017 年 1 月 17 日
Hi,
Quite simply, you can use the step() and impulse() commands given a continuous-time system. The inputs to these commands is a transfer function generated using the tf(numerator_coeffs, denomenator_coeffs). If you do not specify an output to these functions, they will generate a plot with impulse or step response respectively. If you specify an output, then it is left you to plot the results.
I would encourage you to review the help files for each command, but below is a code snippet to get you going.
transFcn = tf([1], [1,1]) % Transfer function for the system 1/(s+1)
impulse(transFcn); % Generates impulse response
[y, t] = step(transFcn); % Generates response and time
figure(1), clf, plot(t, y), xlabel('Time'), ylabel('Response'), title('Step Response');
Hope this helps!
EDIT: swapped arguments in the plot() fcn call to match axis labels
  1 件のコメント
Ananthu S
Ananthu S 2017 年 1 月 17 日
Thanks for the information. I think step(b,a) also works. I got same answers when I tried both the methods.

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Girma
Girma 2024 年 1 月 11 日
transFcn = tf([1], [1,1]) % Transfer function for the system 1/(s+1)
impulse(transFcn); % Generates impulse response
[y, t] = step(transFcn); % Generates response and time
figure(1), clf, plot(t, y), xlabel('Time'), ylabel('Response'), title('Step Response');

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