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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

fourier transform of x(t) is $\int\limits_{-\infty}^{\infty} x(t) e ^{j \omega t} \delta t$

2. anonymous

so, $\int\limits_{-\infty}^{\infty}xdx = \int\limits_{-\infty}^{0-} xdx+\int\limits_{0-}^{0+}xdx+\int\limits_{o+}^{\infty}$

3. anonymous

oops the last term is$\int\limits_{0+}^{\infty}xdx$

4. anonymous

so, substituting, we get, $\int\limits_{-\infty}^{+\infty}x(t)e ^{j \omega t} \delta t = \int\limits_{-\infty}^{0-}-1e^{j \omega t} \delta t +0+\int\limits_{0+}^{\infty}e^{j \omega t} \delta t$

5. anonymous

$=\int\limits_{0+}^{\infty} -1/e^{j \omega t} \delta t + \int\limits_{0+}^{\infty} e^{j \omega t} \delta t$

6. anonymous

got it?

7. anonymous

yes, thanks, I think that I understand. I'm looking for the impulse response can u help me with that?

8. anonymous

the second one you mean

9. anonymous

i'm looking at that. hold on.

10. anonymous

remind me again, u(t) = 1 for t = 0 and u(t) = 0 for all other values. Am i right?

11. anonymous

yes

12. 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

13. anonymous

is u(-t) = -u(t)?

14. anonymous

no it isn't.

15. anonymous

don't bother. it is not -u(t)

16. anonymous

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

notice the limits

18. anonymous

nice! agree

19. 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)

20. anonymous

what did you get as the fourier transform of y?

21. anonymous

yes

22. anonymous

impulse response and input x output y , how relate

23. anonymous

impulse response is the response of a function at t = 0

24. anonymous

or rather, when a brief signal, or impulse is input, the output that you get from the system is called impulse response

25. anonymous

ok directly from the previous limit

26. anonymous

i am not sure, but I think the fourier transform of the impulse response is 0.

27. anonymous

One more question please: If we enter another signal to the system and

28. anonymous

Y{2t}=h(t)*X2{t}

29. anonymous

$\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t) 30. anonymous \[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t)$

31. anonymous

i am sorry. Post it again tomorrow. I am off to bed :)

32. anonymous

ok dont problem thanks 4 erything

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