# Convolution help.

by seang
Tags: convolution
 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. $$\int_o^t e^{-a(t-T)}u(t-1)dT$$ the other idea I had was $$\int_1^t e^{-a(t-T)}u(t)dT$$ but I doubt this too. Help?
 P: 113 lemme just be clear about your notation. Is this right? $$x(t) = u(t-1)$$ 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 $$h(t - T) = \ ?? [/itex] [tex]x(T) = \ ??$$
 P: 185 $$h(t - T) = e^{-a(t-T)}$$ $$x(T) = u(T-1)$$ Maybe?
 P: 113 Convolution help. 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?
 P: 113 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 $h(t-T)$.
 P: 185 I feel like a n00bie. So $$h(t - T) = e^{-a(t-1)}$$?
 P: 113 $$h(t) = e^{-at}u(t)$$ to get $h(t-T)$ replace all instances of $t$ with $t-T$
 P: 185 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.
 P: 113 yes now you can substitute the expresions for $x(T)$ and $h(t-T)$ in the convolution integral.