"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 ( s ) = s s 0.5 h ( t ) = L 1 { H } = δ ( t ) + 1 2 e t / 2
For impulse-response:
y ( t ) = h ( t ) δ ( t ) = h ( t )
and for step-response:
y ( t ) = 0 t h ( τ ) d τ = u ( t ) + e t / 2
where ∗ stands for convolution. As you see, both responses have exponential growth in time.
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