How to calculate step and impulse response at multiple sampling rates I'm struggling with a basic concept.  Suppose I have a system $y(t)=e^{-at}u(t)\ast x(t)$ and I want to plot impulse response and step response using Matlab conv() function at various sampling rates.  I'm having trouble understanding what gain I should apply and why.  I tried:

a = 0.05;
cutoffTime = 250;
T = 1/samplingRates(i);  %assume this is defined somewhere [1, 10, 20]
timeRange = 0:T:cutoffTime;
step = ones(1, length(timeRange));
impulse = zeros(1, length(timeRange));
impulse(1) = 1;
exponential = exp(-a*timeRange);
stepResponse = conv(exponential, step, 'full');  
impulseResponse = conv(exponential, step, 'full');


I understand that mathematically sampling is modeled as multiplication by a pulse train, which introduces gain of $1/T$ to the discrete system.  So the reconstruction filter needs gain of T.  In the above code, for different sampling rates (say 1, 10, 20 samples/second) if I don't apply the gain of T to step response, I get different sized step response.  However if I apply the gain T to impulse response, I don't get the impulse response I expect for all 3 sampling rates.  Can anyone clarify where I'm going wrong in my implementation?

Thank you.
This is not the full answer you want, but the problem is that you can't permute sampling and convolution. Let's say that the $D(\cdot)$ operator takes a continuous signal and returns its sampled version. What you're doing is

$$y'(n) = D\left(e^{-at}u(t)\right) * D\left(x(t)\right)$$

What you want to get is

$$D\left(y(t)\right) = D\left(e^{-at}u(t) * x(t)\right) \neq y'(n).$$

It is in general impossible to do what you're trying to do, but I'm sure that good approximations exist. I'm not an expert on what they are or how to use them.