purplebird said:
How do you do the convolution of
exp (x(n))*u(x(n)) and exp(x(n-1))*u(x(n-1))
where u(x) is the unit step function. Thanks.
Because convolution is commutative,
[tex]y(t) = u(t) * e^{t} = e^{t} * u(t)[/tex]
therefore, using the general result
[tex]h(t)*u(t) = \int_{-\infty} ^{t} *e^{\tau} d\tau= e^{t} - e^{-\infty } = e^{t}[/tex]
So, as an example given that
[tex]h(t) = e^{-t}*u(t)[/tex]
[tex]f(t) = e^{-2t}*u(t)[/tex]
and [tex]y(t) = h(t)* f(t)[/tex] , determine [tex]y(1).[/tex]
Solution,
[tex]y(t) = \int_{-\infty} ^{\infty} h(\tau)f(t-\tau) d\tau = \int_{-\infty} ^{\infty} [e^{-\tau}u(\tau)] [e^{-2(t-\tau)} u(t - \tau)] d\tau[/tex]
for any value of t, it fallows that
[tex]y(1) = \int_{-\infty} ^{\infty} [e^{-\tau}u(\tau)] [e^{-2(1-\tau)} u(1 - \tau)] d\tau = \int_{0} ^{1} e^{-\tau}e^{-2(1-\tau) }d\tau = e^{-2}(e^1 - 1)=e^{-1}-e^{-2} \approx 0.233.[/tex]
I hope this helps a little bit.
[tex]Reference. Convolution examples from Kudeki and Munson[/tex]