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?

Lustyku8

Lustyku8

Answered question

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?

Answer & Explanation

espovilham7

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 ) = 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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