0
$\begingroup$

I am struggling with a task I thought would be a good exercise. I am tasked with calculating the bandwidth of the equation below and I'm struggling. I managed to find and understand why $\text{sinc}(t)=1$. However, it's the other function that I can't figure out.

$$x(t) = \left[\frac32 + \frac3{10} \sin(2π t) + \sin\left(\frac{2π}3 t\right) − \sin\left(\frac{2π}{10} t\right)\right] \cdot \operatorname{sinc}(t)\quad \text{ for }-5\le t\le5.$$

I know that bandwidth is defined as the difference between the highest and lowest frequencies of a given signal and according to the book, it should also be equal to 1, but I can't figure out how.

Help is greatly appreciated

$\endgroup$
4
  • $\begingroup$ "sinc(t) is equal to 1" Well, if it was constantly equal to 1, then it would be called "1" not "sinc(t)". I think you mean the bandwidth of it is 1, but that will only help you later on, if at all! $\endgroup$ May 9, 2023 at 10:07
  • $\begingroup$ and your book is wrong, if your definition is really "the distance between highest and lowest nonzero power frequency component", then the bandwidth of $x(t)$ is not 1. $\endgroup$ May 9, 2023 at 10:08
  • $\begingroup$ @MarcusMüller: I think OP is saying "the bandwidth of sinc(t) is one". At least I hope that's what they are saying :-) $\endgroup$
    – Hilmar
    May 9, 2023 at 12:35
  • $\begingroup$ Is $x(t)$ zero for $t$ outside of $[-5, 5]$? If so, edit your question to say so. $\endgroup$
    – TimWescott
    May 9, 2023 at 19:40

1 Answer 1

2
$\begingroup$

Some hints:

  1. The whole thing has 5 different components. Start with determining the lowest and highest frequency of all 5 components.
  2. By multiplying out the bracket you end up with 4 components that are each a multiplication of two components.
  3. Recall that multiplication in time is convolution in frequency.
  4. Using the convolution you can figure out the highest and lowest frequency of all 4 components
  5. Once you have these, you can determine the highest and the lowest frequency of the whole contraption.

$\text{ for }-5\le t\le5$

That's problematic. If that implies that $x(t) = 0 \text{ for } |t| > 5$ then the signal has unlimited bandwidth and this is a poorly defined question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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