Lustyku8

2022-09-26

I tried plotting the step and impulse responses in Matlab:

sys = tf([1 0],[1 -0.5])

figure(1);

step(sys);

figure(2);

impulse(sys);

However, both graphs look the same (can't post images of my graphs, I need more rep to do it).

Both graphs have exponential growth, but shouldn't the impulse response look like exponential decay?

sys = tf([1 0],[1 -0.5])

figure(1);

step(sys);

figure(2);

impulse(sys);

However, both graphs look the same (can't post images of my graphs, I need more rep to do it).

Both graphs have exponential growth, but shouldn't the impulse response look like exponential decay?

espovilham7

Beginner2022-09-27Added 10 answers

The problem is, you have a highly unstable transfer function. So you can't expect a decaying impulse/step-response. In other words:

$$H(s)=\frac{s}{s-0.5}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}h(t)={\mathcal{L}}^{-1}\{H\}=\delta (t)+\frac{1}{2}{e}^{t/2}$$

For impulse-response:

$$y(t)=h(t)\ast \delta (t)=h(t)$$

and for step-response:

$$y(t)={\int}_{0}^{t}h(\tau )d\tau =\text{u}(t)+{e}^{t/2}$$

where ∗ stands for convolution. As you see, both responses have exponential growth in time.

$$H(s)=\frac{s}{s-0.5}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}h(t)={\mathcal{L}}^{-1}\{H\}=\delta (t)+\frac{1}{2}{e}^{t/2}$$

For impulse-response:

$$y(t)=h(t)\ast \delta (t)=h(t)$$

and for step-response:

$$y(t)={\int}_{0}^{t}h(\tau )d\tau =\text{u}(t)+{e}^{t/2}$$

where ∗ stands for convolution. As you see, both responses have exponential growth in time.

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