(Part B)
(Part B)
 
(3 intermediate revisions by the same user not shown)
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From Question 1:
 
From Question 1:
<math>     = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t}</math><br>
+
<math>   x(t) = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t}</math><br>
 
With this expression we can conclude:<br>
 
With this expression we can conclude:<br>
 
<math>a_3 = -\frac{1}{2}</math>
 
<math>a_3 = -\frac{1}{2}</math>
Line 38: Line 38:
  
  
<math>     = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t}</math><br>
+
<math> y(t)  = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t}</math><br>
 
+
 
+
<math>x(t) = 3j\omega_0e^{j2\pi t}+3j\omega_0e^{-j2\pi t} + 4j\omega_0e^{j4\pi t}-4j\omega_0e^{-j4\pi t}\,</math><br>
+
 
+
<math>\omega_0\,</math> value as the base frequency is 2
+
 
+
<math>x(t) = 6je^{j2\pi t}+6je^{-j2\pi t} + 8je^{j4\pi t}-8je^{-j4\pi t}\,</math><br>
+

Latest revision as of 18:40, 26 September 2008

Part A

$ y(t) = K x(t-a) $

if $ x(t)=e^{jwt} $ was inputed to the system

$ y(t) = K e^{jw(t-a)} $

$ = K e^{-jwa}e^{jwt} $


eigen function is $ e^{-jwa} $


$ H(jw)=Ke^{-jwa} $

$ h(t)=K\delta (t-a) $

$ H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as} $

Part B

I REFERRED TO RONY WIJAYA'S ANSWER


Signal defined in Question 1: $ x(t) = cos(3\pi t+\pi) \! $

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\, $

From Question 1: $ x(t) = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t} $
With this expression we can conclude:
$ a_3 = -\frac{1}{2} $

$ a_{-3} = -\frac{1}{2} $


$ y(t) = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t} $

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