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Using the shifting property, | Using the shifting property, | ||
| − | :<math>H(z)= | + | :<math>H(z)=4 e^{0 * 3 z} + e^{-3 z} \, </math> |
| − | :<math>H(z)= | + | :<math>H(z)=4 + e^{- 3 z} \, </math>, where z =jw |
| + | |||
| + | ==Part B== | ||
| + | :<math>x[n]= -5(e^{j \dfrac{\pi}{2} n}) \,</math> | ||
| + | :<math>Response = H(s) x(n) \,</math> | ||
| + | :<math>x(t)=-5(e^{j \dfrac{\pi}{2} n})(4+e^{-3j/2}) \, </math> | ||
| + | :<math>x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} n}e^{-3j\pi /2} \, </math> | ||
| + | :<math>x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} n-3j\pi /2} \, </math> | ||
| + | :<math>x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} (n-3) } \, </math> | ||
Latest revision as of 06:27, 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)=4 e^{0 * 3 z} + e^{-3 z} \, $
- $ H(z)=4 + e^{- 3 z} \, $, where z =jw
Part B
- $ x[n]= -5(e^{j \dfrac{\pi}{2} n}) \, $
- $ Response = H(s) x(n) \, $
- $ x(t)=-5(e^{j \dfrac{\pi}{2} n})(4+e^{-3j/2}) \, $
- $ x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} n}e^{-3j\pi /2} \, $
- $ x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} n-3j\pi /2} \, $
- $ x(t)-20e^{j \dfrac{\pi}{2} n}-5e^{j \dfrac{\pi}{2} (n-3) } \, $
