attenuate the signal

[drolex]

A continuous-time LTI systems has an impulse response
h(t)=piecewise(-A<=t<=A, A-|t|, otherwise, 0)
Find the smallest value of A (A>0) so that the signal x(t)=cos(10^6*2Pi*t) is attenuated to zero by the system.

h(t) is a triangle from -A to A with height A at t=0. The slopes are 1 and -1 on the left and right sides, respectively.

Since we have the impluse response of the system the output of any input can be calculated using a convolution (or Fourier transforms).
Can anyone see an easy way to do this? I don't think it is supposed to be extremely difficult.

thanks

 

[3iMaJ]

This sounds like homework to me.

How do you think we should start? I'll help you but I wont' tell you the answer. By the way a function that is time limited is band unlimited and therefore your signal cannot be attenuated to zero exactly.

 

[drolex]

I was already on my way before I posted here, but I didn't know how to get the fourier transform of h(t). I then remembered stuff and figured out how to do it. The fourier transform of a convolution is the multiplication of the fourier transforms of the signals in the convolution. It's easier to do multiplication in the frequency domain than it is to do convolutions. The signal is attenuated when the product of the FTs is zero. The FT of h(t) is a sinc(t)^2 type function. The FT of x(t) is 2 delta functions at +/- the frequency 2Pi*10^6. The product is zero when the sinc(t)^2 type function is zero at the location of the spikes. My work is attached. Is there still some reason why the signal could not be attenuated to zero?

 

[3iMaJ]

My first thought is (with out doing the math) is that the fourier transform of a triangle function is a sinc squared function with has unlimited bandwidth. Which means its always going to pass a little energy when filtering a signal with a bandwidth B.

However, its possible that your sinc^2 function has a zero at the exact position of the delta function in frequency, which would allow for this to happen. And my guess is that the value A of the triangle function affects the arguement of the sinc function which would allow for this to happen.

Your answer is correct.

P.S. I just checked a fourier transform table and the triangle function does indeed have zeros in frequency at multiples of 1/A.