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