How to calculate the bandwidth of a continuous signal in an interval?

Question

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

Answer 1

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.