# "I'm trying to graph a simple response function: 1/(1-0.5s^-1) Now, I know that the function can also be written as: s/(s-0.5) So 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?"

demitereur 2022-09-22 Answered
I'm trying to graph a simple response function: 1/(1-0.5s^-1)
Now, I know that the function can also be written as: s/(s-0.5)
So 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?
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## Answers (1)

Zackary Galloway
Answered 2022-09-23 Author has 17 answers
The problem is, you have a highly unstable transfer function. So you can't expect a decaying impulse/step-response. In other words:
$H\left(s\right)=\frac{s}{s-0.5}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}h\left(t\right)={\mathcal{L}}^{-1}\left\{H\right\}=\delta \left(t\right)+\frac{1}{2}{e}^{t/2}$
For impulse-response:
$y\left(t\right)=h\left(t\right)\ast \delta \left(t\right)=h\left(t\right)$
and for step-response:
$y\left(t\right)={\int }_{0}^{t}h\left(\tau \right)d\tau =\text{u}\left(t\right)+{e}^{t/2}$
where ∗ stands for convolution. As you see, both responses have exponential growth in time.
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