Take the 2-minute tour ×
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.

I have been stuck on this question for a while now. It has to do with vestigial sideband.

enter image description here

I wasn't sure if I should be dividing $H(\omega)$ graph values by 2 because only the positive side of the filter is given....

The answer that I am getting is:

We know that $VSB(t) = A_c f(t)\cos(\omega_c t) \ast h(t)$

then in frequency form I get:

$$VSB(\omega) = A_c\pi 0.5 (\delta(\omega + (\omega_c + \omega_m)) + \delta(\omega + (\omega_c - \omega_m)) + \delta(\omega - (\omega_c + \omega_m)) + \delta(\omega - (\omega_c - \omega_m))) H(\omega)$$

Now using the graph I am multiplying the appropriate $H(\omega)$ values:

$$VSB(\omega) = A_c \pi 0.5 (0.25 \delta(\omega + (\omega_c + \omega_m)) + 0.75 \delta(\omega + (\omega_c - \omega_m)) + 0.25 \delta(\omega - (\omega_c + \omega_m)) + 0.75 \delta(\omega - (\omega_c - \omega_m)))$$

where carrier wave is given by $A_c \cos(\omega_c t)$

I don't have the solution to this, so I have no idea if my answer or reasoning is correct... It would be sooooo helpful if someone could confirm/clarify

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

Your reasoning seems reasonable and correct to me. Assuming that the filter's impulse response $h(t)$ is real, then its magnitude response $\left|H(\omega)\right|$ will be symmetric about $\omega=0$, so the negative-frequency terms will be scaled by the same values as those on the positive side of the frequency axis. I don't see any reason why you would need to multiply or divide by 2 anywhere.

The only thing that might differ if you see a solution somewhere else might be the scaling factor at the front of the expression. The typical transform using angular frequency would be:

$$ \mathcal{F}\{\cos(ax)\} = \pi\left(\delta(\omega - a) + \delta(\omega + a)\right) $$

which assumes a Fourier transform definition of:

$$ \mathcal{F}\{x(t)\} = X(\omega) = \int_{-\infty}^{\infty}x(t) e^{-j\omega t}dt $$

It looks like you have an extra factor of $0.5$ at the front of your expression that you might have gotten from the transform. Note that different implementations or expressions of Fourier transforms often have arbitrary scaling factors included; what's most important is that if you apply the forward and inverse transforms in sequence, you get the original signal back:

$$ \mathcal{F}^{-1}\left\{\mathcal{F}\{x(t)\}\right\} = \mathcal{F}^{-1}\left\{X(\omega)\right\} = x(t) $$

or at least to some known scaling factor.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.