Understanding Convolution: h(t) & x(t) w/ x(t)=u(t-1)

[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?

 

[nocturnal]

lemme just be clear about your notation. Is this right?

x(t) = u(t-1)

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

<br /> u(x) = \left \lbracket<br /> \begin{array}{cc}<br /> 0 &amp; \mbox{if } x&lt;0 \\<br /> 1 &amp; \mbox{if } x \geq 0 \\<br /> \end{array}<br /> \right.<br />

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]<br /> x(T) = \ ??

 

[seang]

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

Maybe?

 

[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?

 

[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?

 

[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).

 

[seang]

I feel like a n00bie.

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

 

[nocturnal]

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

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

 

[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.

 

[nocturnal]

yes

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