(Part A)
(Part A)
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<math>H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as}</math>
 
<math>H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as}</math>
 +
 +
==Part B==
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I REFERRED TO RONY WIJAYA'S ANSWER
 +
Signal defined in Question 1:
 +
<math>x(t) = cos(3\pi t+\pi) \!</math> <br>
 +
<br>
 +
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math>
 +
 +
<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math>
 +
 +
From Question 1:
 +
<math>x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br>
 +
With this expression we can conclude:<br>
 +
<math>a_1 = 3\,</math><br>
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<math>a_{-1} = 3\,</math><br>
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<math>a_2 = 4\,</math><br>
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<math>a_{-2} = -4\,</math><br>
 +
 +
<math>x(t) = 3H(s)e^{j2\pi t}+3H(s)e^{-j2\pi t} + 4H(s)e^{j4\pi t}-4H(s)e^{-j4\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>

Revision as of 18:32, 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) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $

$ x(t) = 3H(s)e^{j2\pi t}+3H(s)e^{-j2\pi t} + 4H(s)e^{j4\pi t}-4H(s)e^{-j4\pi t}\, $

$ 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}\, $

$ \omega_0\, $ value as the base frequency is 2

$ x(t) = 6je^{j2\pi t}+6je^{-j2\pi t} + 8je^{j4\pi t}-8je^{-j4\pi t}\, $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood