Signal Analysis - Using Convolution

[cathode-ray]

Homework Statement

Consider the signal x(t)=cos(4t)+cos(5t)+cos(6t), and the SLIT with impulse response:


h(t)=\begin{cases} 1, & \mbox{if } |t|<T \\ 0, & \mbox{if } |t|>T \end{cases}

For what value of T is the output of the system y(t) equal to Acos(4t)+Bcos(5t), when x(t) is the input?

The Attempt at a Solution

I know that this can be solved through Fourier Transform and the solution is T=\pi /3. My problem is that I tried to do this using the convolution, but it gaves me a sum of sin and i don't know how to progress to solve by that way, or if it is possible to do it.

 

[ideasrule]

How were you planning to solve it using convolution directly? You'd have to compute the integral of h(s)*x(t-s)*ds, but first you'd have to express h(s)*x(t-s) in terms of analytical functions that can be easily integrated. One way to do this would be to use the Fourier transform, but if you're going to do that, you might as well use the convolution theorem.

 

[cathode-ray]

I was thinking to do it with the convolution directly:

\intop_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau=\intop_{-T+t}^{T+t}x(\tau)h(t-\tau)d\tau=\intop_{-T+t}^{T+t}[cos(4\tau)+cos(5\tau)+cos(6\tau)]d\tau.

Integrating I get a very complicated expression with a sum of sin. I don't know how to progress further at this point.

 

[ideasrule]

I was thinking to do it with the convolution directly:

\intop_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau=\intop_{-T+t}^{T+t}x(\tau)h(t-\tau)d\tau=\intop_{-T+t}^{T+t}[cos(4\tau)+cos(5\tau)+cos(6\tau)]d\tau.

Integrating I get a very complicated expression with a sum of sin. I don't know how to progress further at this point.

Yup, that should work. (Ignore what I said in the previous post; I was just being stupid.)

Integrating that should give you 1/4*(sin(4*(T+t)) - sin(4*t-T)) plus 2 other terms. You can use the sum-to-product identity:

sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))

to simplify each term into a cosine factor that depends on t, and a sine factor that depends on T.

Using both this method and the convolution theorem, I got T=pi/6. Are you sure it's pi/3? (Not a rhetorical question; it wouldn't be the first time I made an algebra mistake!)

 

[cathode-ray]

Thanks for your help! I rellay didn't have any idea how to progress. That trigonometric identity really helps a lot.

Yes its a multiple choice question and the correct option according to the solution is \pi/3.