Plotting impulse and step responses on Matlab

Question

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?

It just doesn't seem right that both graphs have the same shape.

Answer 1

The problem is, you have a highly unstable transfer function. So you can't expect a decaying impulse/step-response. In other words: $$H(s)=\frac s{s-0.5}\implies h(t)=\mathcal L^{-1}\{H\}=\delta(t)+\frac 12 e^{t/2}$$ For impulse-response: $$y(t)=h(t)*\delta(t)=h(t)$$ 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.