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

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

• "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! May 9, 2023 at 10:07
• 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. May 9, 2023 at 10:08
• @MarcusMüller: I think OP is saying "the bandwidth of sinc(t) is one". At least I hope that's what they are saying :-) May 9, 2023 at 12:35
• Is $x(t)$ zero for $t$ outside of $[-5, 5]$? If so, edit your question to say so. May 9, 2023 at 19:40

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