- #1
Mr.Tibbs
- 24
- 0
Evaluate the following discrete-time convolution:
y[n] = cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]n)*2[itex]^{n}[/itex]u[-n+2]
Here is my sloppy attempt:
y[n] = [itex]\sum[/itex]cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]k)2[itex]^{n-k}[/itex]u[-n-k+2] from k = -∞ to ∞
= [itex]\sum[/itex]cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]k)2[itex]^{n-k}[/itex] from k = -∞ to 2
We can re-write the cos as [e[itex]^{0.5j\pi}[/itex]-e[itex]^{-0.5j\pi}[/itex]]0.5
using the property of summation of geometric series:
0.5[itex]\sum[/itex]2[itex]^{n-k}[/itex](e[itex]^{0.5j\pi k}[/itex]-e[itex]^{-0.5j \pi k}[/itex])
from k = -∞ to 2
so more or less am I on the right track?
y[n] = cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]n)*2[itex]^{n}[/itex]u[-n+2]
Here is my sloppy attempt:
y[n] = [itex]\sum[/itex]cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]k)2[itex]^{n-k}[/itex]u[-n-k+2] from k = -∞ to ∞
= [itex]\sum[/itex]cos([itex]\frac{1}{2}[/itex][itex]\pi[/itex]k)2[itex]^{n-k}[/itex] from k = -∞ to 2
We can re-write the cos as [e[itex]^{0.5j\pi}[/itex]-e[itex]^{-0.5j\pi}[/itex]]0.5
using the property of summation of geometric series:
0.5[itex]\sum[/itex]2[itex]^{n-k}[/itex](e[itex]^{0.5j\pi k}[/itex]-e[itex]^{-0.5j \pi k}[/itex])
from k = -∞ to 2
so more or less am I on the right track?