Need Help On Signals & Systems Questions

i need some help on this past year exam paper.
How do i get the graphical method for it in w form?
(1) x(t) = sin(2 pi t)e^-tu(t)]
(b) x(t) = [2sin(3 pi t) / (pi t)] [sin(2 pi t)/(pi t)]
(c) x(t) d/dt (te^-2tsin(t)u(t))
(d) x(t) = integrate from t to - infinity sin(2 pi t)/(pi t) dt

anyone can advise me?

are you trying to convert x(t) to x(w) domain?

are you trying to convert x(t) to x(w) domain?

Yes.

a) x(t) = sin(2 pi t)e^-tu(t)]

You know that sin(2 pi t) <-> -j*pi*[b(w-2*pi)-b(w+2*pi)]
where b=delta dirac function

You also know that e^-tu(t) <-> 1/(j*w+1)

Now draw these two functions in frequency domain.

Next multiplication in time domain is convolution in frequency domain with 1/(2*pi) amplitude.

So it becomes easy to convolve the delta diracs with 1/(j*w+1) at w=2*pi


b) x(t) = [2sin(3 pi t) / (pi t)] [sin(2 pi t)/(pi t)]

The two functions here look like sinc functions. So for example this seems to be a triangle or a trapezoid in frequency domain (e.g. a convolution of 2 square pulses). One square pulse is limited by +/-3pi and has an amplitude of 2, while the other is limited by +/-2pi and has an amplitude of 1.
If you're not sure how to do it graphically, then you can always represent each pulse as an addition of two step responses in frequency domain. That can sometimes make work alot simpler.

c) x(t)=d/dt (te^-2tsin(t)u(t))

Question: is there any more brackets in this expression?

d) x(t) = integrate from t to - infinity sin(2 pi t)/(pi t) dt

here convert the sin(2 pi t)/(pi t) (or sinc function as it is sometimes called) into a pulse in its frequency domain. This conversion should be somewhere in your FT tables. Then it becomes a very easy graphical addition of a signal shaped by dirac delta and the pulse signal scaled by 1/jw amplitude.


Anyway I hope that this helps.

But if you need the graphical method, you should just draw the graph for each function, and then choose one to be the (tau) function, and the other the (t - tau) function. Then choose accordingly the ranges of tau, and your function would be the product of the intersection of the functions.

Here's some helpful (I hope) pics...

Alternatively,
You can take the Laplace Transform of the given signals & put s=jw to get the answer.
Taking Laplace Transform is always easier than to get Headon with Fourier Transform.

yeah, but sometimes, we're just stuck on Fourier. We weren't able 2 use Laplace in some exams. Maybe shermaine has a similar case?