Solving Sampling with Aliasing Homework

[Bromio]

Homework Statement

Let the signal $x(t) = \left(\displaystyle\frac{\sin(50\pi t)}{\pi t}\right)^2$, which we want to sample with sampling frequency $\omega_s = 150\pi$ in order to obtain a signal, $g(t)$ whose Fourier transform is $G(\omega)$. Determine the maximun value for $\omega_0$ which guarantees that $G(\omega) = 75X(\omega)$ for $\left|\omega\right| \leq{\omega_0}$

Homework Equations

Sampling Nyquist theoreme: $\omega_s > 2B$, where $B$ is the signal band-with.

The Attempt at a Solution

$X(\omega) = FT\{x(t)\}$ is a triangular signal with $B = 100\pi$ and amplitude $X(0) = 25$.

From sampling Nyquist theoreme, $150\pi > 200\pi$ is false, so there is aliasing.

I don't know how to finish the problem.

Thank you.

 

[marcusl]

There is no aliasing. You should not define signal bandwidth to include both positive and negative frequencies if you plan to use a relation with 2B. (That is, 2B assumes that B was defined as one-sided, whereupon the factor of 2 accounts for the negative frequency portion.) With this definition, you should write
$$B=50\pi$$ and 2B<ws, so you are sampled properly.

 

[Bromio]

I've defined bandwith to include only positive frequencies, so my problem must be an error in Fourier transform.

I think $X(\omega) = FT\{x(t)\}$ is a triangular signal whose bandwith is $B = 100\pi$ and $X(0) = 25$

Is this wrong?

Thank you.

 

[marcusl]

Hmm. Now that I work it out, I also get B=100pi but X(0)=50 (although the value of X(0) is not needed to solve the problem). Aliasing changes the shape of G so that the spectrum becomes horizontal between w=50pi and 75pi. You can compare the shape of G to that of X; I don't see that G exceeds X by 75, however.

 

[Bromio]

You're right: B=100pi and X(0)=50.

Intuitively I understand that a horizontal line appears in the spectrum, but how can you know that this occurs between w=50pi and 75pi?

Regarding the problem question (which I've checked that it is correctly copied), I know $G(\omega) = \displaystyle\frac{1}{T_s}\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)$. As $\omega_s = 150\pi$, then $T_s = 2\pi/150\pi = 1/75$, this expression can be written as $G(\omega) = 75\displaystyle\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)$, where 75 appears, but I don't know how to continue.

Thank you.

 

[marcusl]

You're right: B=100pi and X(0)=50.

Intuitively I understand that a horizontal line appears in the spectrum, but how can you know that this occurs between w=50pi and 75pi?

You can do it graphically. I'll discuss only the positive frequencies, but everything to follow is mirrored for w<0. The spectrum X is a triangle X(0)=50 and X(100*pi)=0, and G is X replicated at every integer multiple of ws. The first folding frequency is 75*pi, so fold the bit of triangle from 75 to 100 pi back and add it to X to give G. The flat part of G extends from 50 to 75 pi.

Regarding the problem question (which I've checked that it is correctly copied), I know $G(\omega) = \displaystyle\frac{1}{T_s}\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)$. As $\omega_s = 150\pi$, then $T_s = 2\pi/150\pi = 1/75$, this expression can be written as $G(\omega) = 75\displaystyle\sum_{k=-\infty}^{\infty}X(\omega-k\omega_s)$, where 75 appears, but I don't know how to continue.

Thank you.

Excellent, you've shown that G=75X in the absence of aliasing. Aliasing raises the value of G between 50 and 75 pi, so 50pi is the maximum frequency where G=75X.

 

[Bromio]

Thank you.

I've just understood this problem.