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Convolution help.by seang
Tags: convolution 
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#1
Mar2806, 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(t1). So for example, if h(t) = exp(at)u(t) an x(t) = u(t1). Is this even the right set up? I don't think it is.
[tex]\int_o^t e^{a(tT)}u(t1)dT[/tex] the other idea I had was [tex]\int_1^t e^{a(tT)}u(t)dT[/tex] but I doubt this too. Help? 


#2
Mar2806, 03:33 PM

P: 113

lemme just be clear about your notation. Is this right?
[tex]x(t) = u(t1)[/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] 


#3
Mar2806, 06:00 PM

P: 185

[tex] h(t  T) = e^{a(tT)} [/tex]
[tex]x(T) = u(T1) [/tex] Maybe? 


#4
Mar2806, 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(tT)[/itex]?
Thats the correct expression for x(T). What's [itex]u(t)[/itex]? Is it the unit step function? 


#5
Mar2806, 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?



#6
Mar2806, 10:55 PM

P: 113

yes. recall from algebra that the graph of [itex]u(t1)[/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(ta) = g(ta)h(ta)[/itex]. Use this to come up with an expression for [itex]h(tT)[/itex]. 


#7
Mar2806, 11:04 PM

P: 185

I feel like a n00bie.
So [tex] h(t  T) = e^{a(t1)} [/tex]? 


#8
Mar2806, 11:13 PM

P: 113

[tex]h(t) = e^{at}u(t)[/tex]
to get [itex]h(tT)[/itex] replace all instances of [itex]t[/itex] with [itex]tT[/itex] 


#9
Mar2806, 11:19 PM

P: 185

so h(tT):
[tex]h(tT) = e^{a(tT)}u(tT)}[/tex] forgive me its late and I'm having a really hard time understanding this. 


#10
Mar2806, 11:32 PM

P: 113

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


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