Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

got it?

Answer 7

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

Answer 8

the second one you mean

Answer 9

i'm looking at that. hold on.

Answer 10

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

Answer 11

yes

Answer 12

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

Answer 13

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

Answer 14

no it isn't.

Answer 15

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

Answer 16

so, fourier transform of y(t) is \[\int\limits_{0-}^{0+}u(t)-u(-t)-4e ^{-t} e ^{j \omega t } \delta t\]

Answer 17

notice the limits

Answer 18

nice! agree

Answer 19

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)

Answer 20

what did you get as the fourier transform of y?

Answer 21

yes

Answer 22

impulse response and input x output y , how relate

Answer 23

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

Answer 24

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

Answer 25

ok directly from the previous limit

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

\[\sum_{-\infty}^{+\infty} C1e ^{jl2t} * h(t) \]

Answer 31

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

Answer 32

ok dont problem thanks 4 erything