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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)(1-2(e^-t)) -2u(t)e^(-t)-u(-t) a) Fourier transform x(t) b) Fourier transform y(t)

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  1. anonymous
    • 5 years ago
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    fourier transform of x(t) is \[\int\limits_{-\infty}^{\infty} x(t) e ^{j \omega t} \delta t\]

  2. anonymous
    • 5 years ago
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    so, \[\int\limits_{-\infty}^{\infty}xdx = \int\limits_{-\infty}^{0-} xdx+\int\limits_{0-}^{0+}xdx+\int\limits_{o+}^{\infty}\]

  3. anonymous
    • 5 years ago
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    oops the last term is\[\int\limits_{0+}^{\infty}xdx\]

  4. anonymous
    • 5 years ago
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    so, substituting, we get, \[\int\limits_{-\infty}^{+\infty}x(t)e ^{j \omega t} \delta t = \int\limits_{-\infty}^{0-}-1e^{j \omega t} \delta t +0+\int\limits_{0+}^{\infty}e^{j \omega t} \delta t \]

  5. anonymous
    • 5 years ago
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    \[=\int\limits_{0+}^{\infty} -1/e^{j \omega t} \delta t + \int\limits_{0+}^{\infty} e^{j \omega t} \delta t \]

  6. anonymous
    • 5 years ago
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    got it?

  7. anonymous
    • 5 years ago
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    yes, thanks, I think that I understand. I'm looking for the impulse response can u help me with that?

  8. anonymous
    • 5 years ago
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    the second one you mean

  9. anonymous
    • 5 years ago
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    i'm looking at that. hold on.

  10. anonymous
    • 5 years ago
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    remind me again, u(t) = 1 for t = 0 and u(t) = 0 for all other values. Am i right?

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    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

  13. anonymous
    • 5 years ago
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    is u(-t) = -u(t)?

  14. anonymous
    • 5 years ago
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    no it isn't.

  15. anonymous
    • 5 years ago
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    don't bother. it is not -u(t)

  16. anonymous
    • 5 years ago
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    so, fourier transform of y(t) is \[\int\limits_{0-}^{0+}u(t)-u(-t)-4e ^{-t} e ^{j \omega t } \delta t\]

  17. anonymous
    • 5 years ago
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    notice the limits

  18. anonymous
    • 5 years ago
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    nice! agree

  19. anonymous
    • 5 years ago
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    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)

  20. anonymous
    • 5 years ago
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    what did you get as the fourier transform of y?

  21. anonymous
    • 5 years ago
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    yes

  22. anonymous
    • 5 years ago
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    impulse response and input x output y , how relate

  23. anonymous
    • 5 years ago
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    impulse response is the response of a function at t = 0

  24. anonymous
    • 5 years ago
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    or rather, when a brief signal, or impulse is input, the output that you get from the system is called impulse response

  25. anonymous
    • 5 years ago
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    ok directly from the previous limit

  26. anonymous
    • 5 years ago
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    i am not sure, but I think the fourier transform of the impulse response is 0.

  27. anonymous
    • 5 years ago
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    One more question please: If we enter another signal to the system and

  28. anonymous
    • 5 years ago
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    Y{2t}=h(t)*X2{t}

  29. anonymous
    • 5 years ago
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    \[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t)

  30. anonymous
    • 5 years ago
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    \[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t) \]

  31. anonymous
    • 5 years ago
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    i am sorry. Post it again tomorrow. I am off to bed :)

  32. anonymous
    • 5 years ago
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    ok dont problem thanks 4 erything

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