# Convolution help.

1. Mar 28, 2006

### seang

I'm having trouble understanding convolution. In particular, in convolving h(t) and x(t), I have no idea what to do when x(t) = u(t-1). So for example, if h(t) = exp(-at)u(t) an x(t) = u(t-1). Is this even the right set up? I don't think it is.

$$\int_o^t e^{-a(t-T)}u(t-1)dT$$

the other idea I had was

$$\int_1^t e^{-a(t-T)}u(t)dT$$

but I doubt this too.
Help?

Last edited: Mar 28, 2006
2. Mar 28, 2006

### nocturnal

$$x(t) = u(t-1)$$

where $u(t)$ is the unit step function, ie:

$$u(x) = \left \lbracket \begin{array}{cc} 0 & \mbox{if } x<0 \\ 1 & \mbox{if } x \geq 0 \\ \end{array} \right.$$

and $$h(t) = e^{-at}u(t)$$.

the convolution of x and h, denoted $x \ast h$, is given by,
$$(x \ast h)(t) = \int_{-\infty}^{\infty}x(T)h(t - T)dT$$

Also, convolution is commutative so $x \ast h = h \ast x$. Therefore choose the one that makes your calculation easier. From your post it looks like you chose $x \ast h$

your first goal is to come up with the correct expressions for

$$h(t - T) = \ ?? [/itex] [tex]x(T) = \ ??$$

Last edited: Mar 28, 2006
3. Mar 28, 2006

### seang

$$h(t - T) = e^{-a(t-T)}$$
$$x(T) = u(T-1)$$

Maybe?

4. Mar 28, 2006

### nocturnal

In your first post you said $h(t) = e^{-at}u(t)$. What happened to $u$ in $h(t-T)$?

Thats the correct expression for x(T).

What's $u(t)$? Is it the unit step function?

Last edited: Mar 28, 2006
5. Mar 28, 2006

### seang

yeah it is, and that's what's catching me a little i think; is the u(t) in the h(t) related to the u(t) which x(t) is equal to?

6. Mar 28, 2006

### nocturnal

yes. recall from algebra that the graph of $u(t-1)$ is the graph of $u(t)$ shifted to the right 1 unit.

If $f,g,$ and $h$ are functions such that $f(t) = g(t)h(t)$, and $a$ is a real number, then $f(t-a) = g(t-a)h(t-a)$. Use this to come up with an expression for $h(t-T)$.

Last edited: Mar 28, 2006
7. Mar 28, 2006

### seang

I feel like a n00bie.

So
$$h(t - T) = e^{-a(t-1)}$$?

8. Mar 28, 2006

### nocturnal

$$h(t) = e^{-at}u(t)$$

to get $h(t-T)$ replace all instances of $t$ with $t-T$

9. Mar 28, 2006

### seang

so h(t-T):

$$h(t-T) = e^{-a(t-T)}u(t-T)}$$

forgive me its late and I'm having a really hard time understanding this.

10. Mar 28, 2006

### nocturnal

yes

now you can substitute the expresions for $x(T)$ and $h(t-T)$ in the convolution integral.

Last edited: Mar 28, 2006