| Line 12: | Line 12: | ||
:<math>H(z)=10 e^{0 z} + e^{-1 z} \, </math> | :<math>H(z)=10 e^{0 z} + e^{-1 z} \, </math> | ||
:<math>H(z)=10 + e^{- z} \, </math>, where z =jw | :<math>H(z)=10 + e^{- z} \, </math>, where z =jw | ||
| + | |||
| + | ==Part B== | ||
| + | :<math>x[n]= -5(e^{j \dfrac{\pi}{2} n}) \,</math> | ||
Revision as of 07:16, 25 September 2008
DT LTI system
The system is:
- $ y(n)=4x(n)+x(n-3) $
unit impulse response
Obtain the unit impulse response h(t) and the system function H(s) of your system. :
- $ d (n) => System =>4 d (n) + d(n-3)\, $
- $ h(t)=4d(n) +d(n-3)\, $
- $ H(z)=\sum_{-\infty}^{\infty} h(n)e^{-s n} $
- $ H(z)=\sum_{-\infty}^{\infty} (4d(n) +d(n-3))e^{-z n} $
Using the shifting property,
- $ H(z)=10 e^{0 z} + e^{-1 z} \, $
- $ H(z)=10 + e^{- z} \, $, where z =jw
Part B
- $ x[n]= -5(e^{j \dfrac{\pi}{2} n}) \, $
