# Help proving if a system is LTI?

1. Feb 2, 2015

### asdf12312

1. The problem statement, all variables and given/known data
A DT system is defined by the input-output relationship y[n] = x[n] ∗ h[n],
where x[n] is input, y[n] is output, '*' is convolution, and h[n] = u[n+1].
Is this an LTI system? Explain.

2. Relevant equations
x[n] - input, h[n] - impulse response
y[n]=h[n]*x[n] = Σ (k=-inf to inf) x[k]h[n-k] = Σ (k=-inf to inf) h[k]x[n-k] (convolution sum for DT)
Rule for y[n]=h[n]*x[n]: if x[n]=u[n], y[n]=Σ (k=-inf to n) h[k]

3. The attempt at a solution
I'm trying to prove if it is LTI system or not. I can prove that convolution sum is linear, y[k]=Σ(k=-inf to inf) A(x[k]h[n-k])= Ay[k] but not sure its time invariant. if I use the rule in the book, I guess x[n] and h[n] are interchangeable in convolution sum so h[n] can be used as input. Then using the rule, y[n]=Σ (k=-inf to n) x[k+1] which I suppose means its time invariant? y[n-n0]=Σ (k=-inf to n) x[k+1-n0], pretty sure this is true but is this enough to prove its LTI system?

Last edited: Feb 2, 2015
2. Feb 7, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Feb 7, 2015

### jssamp

If it is time invariant the output you get from a time shifted input is the same as if you time shifted the output the same amount.

You have system H and its output y such that the system is described by y[n] = H{ x[n] },
if you shift input x1[n] so you get a new input x2[n] = x1[n-1] and you put this new input into H you get y2[n] = H{ x1[n-1] }
and you shift the output so y1[n-1] = H{ x1[n] },
y2[n] = y1[n-1] for time invariant system.

if it is linear then if the inputs x[n] = x1[n] produce output y[n] = y1[n] and x[n] = x2[n] produces y[n] = y2[n] then
x[n] = a*x1[n] + b*x2[n] will have output y[n] = ay1[n] + by2[n]