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Convolution help.

by seang
Tags: convolution
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seang
#1
Mar28-06, 01:18 PM
P: 185
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.

[tex]\int_o^t e^{-a(t-T)}u(t-1)dT[/tex]

the other idea I had was

[tex]\int_1^t e^{-a(t-T)}u(t)dT[/tex]

but I doubt this too.
Help?
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nocturnal
#2
Mar28-06, 03:33 PM
P: 113
lemme just be clear about your notation. Is this right?

[tex]x(t) = u(t-1)[/tex]

where [itex]u(t)[/itex] is the unit step function, ie:

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

and [tex]h(t) = e^{-at}u(t)[/tex].

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

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

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

[tex] h(t - T) = \ ?? [/itex]
[tex]x(T) = \ ?? [/tex]
seang
#3
Mar28-06, 06:00 PM
P: 185
[tex] h(t - T) = e^{-a(t-T)} [/tex]
[tex]x(T) = u(T-1) [/tex]

Maybe?

nocturnal
#4
Mar28-06, 09:19 PM
P: 113
Convolution help.

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

Thats the correct expression for x(T).

What's [itex]u(t)[/itex]? Is it the unit step function?
seang
#5
Mar28-06, 10:09 PM
P: 185
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
#6
Mar28-06, 10:55 PM
P: 113
yes. recall from algebra that the graph of [itex]u(t-1)[/itex] is the graph of [itex]u(t)[/itex] shifted to the right 1 unit.

If [itex]f,g,[/itex] and [itex]h[/itex] are functions such that [itex]f(t) = g(t)h(t)[/itex], and [itex]a[/itex] is a real number, then [itex]f(t-a) = g(t-a)h(t-a)[/itex]. Use this to come up with an expression for [itex]h(t-T)[/itex].
seang
#7
Mar28-06, 11:04 PM
P: 185
I feel like a n00bie.

So
[tex] h(t - T) = e^{-a(t-1)} [/tex]?
nocturnal
#8
Mar28-06, 11:13 PM
P: 113
[tex]h(t) = e^{-at}u(t)[/tex]

to get [itex]h(t-T)[/itex] replace all instances of [itex]t[/itex] with [itex]t-T[/itex]
seang
#9
Mar28-06, 11:19 PM
P: 185
so h(t-T):

[tex]h(t-T) = e^{-a(t-T)}u(t-T)}[/tex]

forgive me its late and I'm having a really hard time understanding this.
nocturnal
#10
Mar28-06, 11:32 PM
P: 113
yes

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


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