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anonymous
 5 years ago
Help please:
x(T)= 1, t<0
0, t=0
1, t>0
system out:
y(t)= u(t)(12(e^t)) 2u(t)e^(t)u(t)
a) Fourier transform x(t)
b) Fourier transform y(t)
anonymous
 5 years ago
Help please: x(T)= 1, t<0 0, t=0 1, t>0 system out: y(t)= u(t)(12(e^t)) 2u(t)e^(t)u(t) a) Fourier transform x(t) b) Fourier transform y(t)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fourier transform of x(t) is \[\int\limits_{\infty}^{\infty} x(t) e ^{j \omega t} \delta t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, \[\int\limits_{\infty}^{\infty}xdx = \int\limits_{\infty}^{0} xdx+\int\limits_{0}^{0+}xdx+\int\limits_{o+}^{\infty}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops the last term is\[\int\limits_{0+}^{\infty}xdx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, substituting, we get, \[\int\limits_{\infty}^{+\infty}x(t)e ^{j \omega t} \delta t = \int\limits_{\infty}^{0}1e^{j \omega t} \delta t +0+\int\limits_{0+}^{\infty}e^{j \omega t} \delta t \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\int\limits_{0+}^{\infty} 1/e^{j \omega t} \delta t + \int\limits_{0+}^{\infty} e^{j \omega t} \delta t \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, thanks, I think that I understand. I'm looking for the impulse response can u help me with that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second one you mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm looking at that. hold on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remind me again, u(t) = 1 for t = 0 and u(t) = 0 for all other values. Am i right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so y(t) = u(t)(12(e^t)) 2u(t)e^(t)u(t) = u(t)  2u(t)e^t  2u(t)e^tu(t) = u(t)u(t)4u(t)e^t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't bother. it is not u(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, fourier transform of y(t) is \[\int\limits_{0}^{0+}u(t)u(t)4e ^{t} e ^{j \omega t } \delta t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, u(t) = 1 for t = 0+ or t = 0 so the u(t) terms cancel out. (i am not sure about this. ask some one else about this)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what did you get as the fourier transform of y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0impulse response and input x output y , how relate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0impulse response is the response of a function at t = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or rather, when a brief signal, or impulse is input, the output that you get from the system is called impulse response

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok directly from the previous limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am not sure, but I think the fourier transform of the impulse response is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One more question please: If we enter another signal to the system and

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{\infty}^{+\infty} C1e ^{jl2t} * h(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{\infty}^{+\infty} C1e ^{jl2t} * h(t) \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am sorry. Post it again tomorrow. I am off to bed :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok dont problem thanks 4 erything
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