Difference between revisions of "HW4.4 Scott Erdbruegger ECE301Fall2008mboutin" - Rhea

Latest revision as of 06:27, 25 September 2008

The system is:

$ y(n)=4x(n)+x(n-3) $

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

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) } \,

@@ Line 1: / Line 1: @@
 ==DT LTI system ==
 The system is:
-:<math>y(n)=10x(n)+x(n-1)</math>
+:<math>y(n)=4x(n)+x(n-3)</math>
+==unit impulse response==
+Obtain the unit impulse response h(t) and the system function H(s) of your system. :
+:<math>d (n) => System =>4 d (n) + d(n-3)\,</math>
+:<math>h(t)=4d(n) +d(n-3)\,</math>
+:<math>H(z)=\sum_{-\infty}^{\infty} h(n)e^{-s n}</math>
+:<math>H(z)=\sum_{-\infty}^{\infty} (4d(n) +d(n-3))e^{-z n}</math>
+Using the shifting property,
+:<math>H(z)=4 e^{0 * 3 z} + e^{-3 z} \, </math>
+:<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>