anonymous
  • anonymous
Help please: x(T)= -1, t<0 0, t=0 1, t>0 system out: y(t)= u(t)(1-2(e^-t)) -2u(t)e^(-t)-u(-t) a) Fourier transform x(t) b) Fourier transform y(t)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
fourier transform of x(t) is \[\int\limits_{-\infty}^{\infty} x(t) e ^{j \omega t} \delta t\]
anonymous
  • anonymous
so, \[\int\limits_{-\infty}^{\infty}xdx = \int\limits_{-\infty}^{0-} xdx+\int\limits_{0-}^{0+}xdx+\int\limits_{o+}^{\infty}\]
anonymous
  • anonymous
oops the last term is\[\int\limits_{0+}^{\infty}xdx\]

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anonymous
  • anonymous
so, 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
  • anonymous
\[=\int\limits_{0+}^{\infty} -1/e^{j \omega t} \delta t + \int\limits_{0+}^{\infty} e^{j \omega t} \delta t \]
anonymous
  • anonymous
got it?
anonymous
  • anonymous
yes, thanks, I think that I understand. I'm looking for the impulse response can u help me with that?
anonymous
  • anonymous
the second one you mean
anonymous
  • anonymous
i'm looking at that. hold on.
anonymous
  • anonymous
remind me again, u(t) = 1 for t = 0 and u(t) = 0 for all other values. Am i right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so y(t) = u(t)(1-2(e^-t)) -2u(t)e^(-t)-u(-t) = u(t) - 2u(t)e^-t - 2u(t)e^-t-u(-t) = u(t)-u(-t)-4u(t)e^-t
anonymous
  • anonymous
is u(-t) = -u(t)?
anonymous
  • anonymous
no it isn't.
anonymous
  • anonymous
don't bother. it is not -u(t)
anonymous
  • anonymous
so, fourier transform of y(t) is \[\int\limits_{0-}^{0+}u(t)-u(-t)-4e ^{-t} e ^{j \omega t } \delta t\]
anonymous
  • anonymous
notice the limits
anonymous
  • anonymous
nice! agree
anonymous
  • anonymous
now, 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
  • anonymous
what did you get as the fourier transform of y?
anonymous
  • anonymous
yes
anonymous
  • anonymous
impulse response and input x output y , how relate
anonymous
  • anonymous
impulse response is the response of a function at t = 0
anonymous
  • anonymous
or rather, when a brief signal, or impulse is input, the output that you get from the system is called impulse response
anonymous
  • anonymous
ok directly from the previous limit
anonymous
  • anonymous
i am not sure, but I think the fourier transform of the impulse response is 0.
anonymous
  • anonymous
One more question please: If we enter another signal to the system and
anonymous
  • anonymous
Y{2t}=h(t)*X2{t}
anonymous
  • anonymous
\[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t)
anonymous
  • anonymous
\[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t) \]
anonymous
  • anonymous
i am sorry. Post it again tomorrow. I am off to bed :)
anonymous
  • anonymous
ok dont problem thanks 4 erything

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