# Convolution & Systems

1. Oct 13, 2015

### Zondrina

Hi everyone, I'm having some trouble understanding how convolutions are applied to systems.

Suppose I'm given the impulse response of a system as $g(t) = u(t)$ and I'm also given the system input $x(t) = u(t) - u(t-2)$.

The system output should then be given by:

$$y(t) = g(t) * x(t) = \int_0^t g(t - x) x(x) \space dx = \int_0^t u(t-x)u(x) \space dx - \int_0^t u(t-x)u(x-2) \space dx$$

The integrals should reduce to this I think:

$$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx$$

How do I finish this integration? I thought about setting $v = t - x$, but I am unsure.

2. Oct 13, 2015

### Svein

Your convolution formula is not according to the standard (https://en.wikipedia.org/wiki/Convolution). It should be $y(t)=g(t)*x(t)=\int_{-\infty}^{\infty}g(z)x(t-z)dz=\int_{-\infty}^{\infty}u(z)(u(t-z)-u(t-z-2))dz$.

3. Oct 13, 2015

### Zondrina

Why would this matter? The convolution is a commutative operation: $f*g = g*f$.

Also, in the context of systems it is usually taken for granted that $0 \leq t < \infty$ and the convolution limits are taken between $0$ and $t$.

I believe I have figured out the next logical step in the integration:

$$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx = \int_0^2 u(t-x) \space dx$$

Okay wait then the area under this function for $t \geq 2$ would simply be equal to $2$. It would be equal to $t$ for $0 \leq t \leq 2$.

Therefore:

$$\int_0^2 u(t-x) \space dx = t[u(t) - u(t - 2)] + 2u(t - 2) = tu(t) + [2 - t]u(t - 2)$$

I guess the answer was to think about the area under the function.

Last edited: Oct 13, 2015
4. Oct 14, 2015

### Svein

I do not question that. What I did was to use the definition of convolution on your problem.
Maybe - but you need to show that this is applicable to the problem.