# Convolution Interval

Updated: Jan 14, 2020

Calculate the output of this system given that the input signal and the system’s impulse response are:

One thing that should be addressed before we start the problem is what is u(t).

u(t) is the step function, or Heaviside step function. It stands for:

When multiplied with another function it just means that the function will start at 0, even when it would normally have values below 0. This is why the $$h(t) = e^(-t)u(t)$$ starts at $t = 0$. This also means that if you have an integral from negative infinity to infinity, then it would change to and integral from zero to infinity (since the function equals zero when t = 0)

Next, we should address what convolution even is. Convolution integrals are denoted as y(t) = x(t)*h(t). They can be used to compute the output to a linear, time invariant system with no initial conditions (I’ll go over what those are at the end). Its formula is:

So how does convolution actually work?

Basically, you are breaking up the input signal into many individual impulses (lines) at points lambda. You then calculate the output for each one of those lines over time by multiplying the broken up input with the shifted output (shifted by those same values/points, lambda, so the line at lambda 1 is multiplied by the response h(t) shifted to lambda 1). To get the most accurate outcome, we want to have as many of these shifted responses as possible (an infinite amount) added together, so we take the integral of those multiplied components with respect to the shifts (lambda). This represents x(t)*h(t)

Note that you could have done the opposite. You could have broken up the impulse response into lines and shifted the input, so h(t)*x(t):

We will be using the second way for this problem.

Note: subtracting a constant, like in x(t-k), shifts the graph right, and adding, x(t+k) to the shifts left.