(→Part A) |
(→Part B) |
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==Part B== | ==Part B== | ||
I REFERRED TO RONY WIJAYA'S ANSWER | I REFERRED TO RONY WIJAYA'S ANSWER | ||
| + | |||
| + | |||
Signal defined in Question 1: | Signal defined in Question 1: | ||
<math>x(t) = cos(3\pi t+\pi) \!</math> <br> | <math>x(t) = cos(3\pi t+\pi) \!</math> <br> | ||
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From Question 1: | From Question 1: | ||
| − | <math> | + | <math> = -\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> | + | <math>a_3 = -\frac{1}{2}</math> |
| − | <math>a_{- | + | |
| − | <math> | + | <math>a_{-3} = -\frac{1}{2}</math> |
| − | + | ||
| + | |||
| + | <math> = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}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>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> | ||
Revision as of 17:37, 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:
$ = -\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} $
$ = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\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}\, $
