Discrete Time Convolution of Sums

[Mr.Tibbs]

Evaluate the following discrete-time convolution:

y[n] = cos($\frac{1}{2}$$\pi$n)*2$^{n}$u[-n+2]

Here is my sloppy attempt:

y[n] = $\sum$cos($\frac{1}{2}$$\pi$k)2$^{n-k}$u[-n-k+2] from k = -∞ to ∞

= $\sum$cos($\frac{1}{2}$$\pi$k)2$^{n-k}$ from k = -∞ to 2

We can re-write the cos as [e$^{0.5j\pi}$-e$^{-0.5j\pi}$]0.5

using the property of summation of geometric series:

0.5$\sum$2$^{n-k}$(e$^{0.5j\pi k}$-e$^{-0.5j \pi k}$)

from k = -∞ to 2


so more or less am I on the right track?

 

[Valkarie]

Yes, you are on the right track. You can simplify the expression further by noting that the sum of the geometric series is equal to 2k+1. So your final expression should look like this: y[n] = 0.5\times(2^{n-2}(e^{0.5j\pi(2)} - e^{-0.5j\pi(2)}) + \sum_{k=-\infty}^{-2}2^{n-k}(e^{0.5j\pi k} - e^{-0.5j \pi k}))