Can someone please explain how to compute the bandwidth of a signal? Like why the bandwidth of $5\sin(2t)$ is $0$ and of $\sin(2t) + \sin(3t)$ is $1$? Thanks.
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2 Answers
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I suppose the correct expressions you meant are:
a) $5\sin(2t)$
b) $\sin(2t) + \sin(3t)$
Bandwidth is defined as the difference between the highest and lowest frequencies of a given signal ou system.
With this in mind, signal a) has one single frequency of 2 rad/s and so its bandwidth is 2-2=0 rad/s. Similarly, signal b) has 2 frequencies: 2 rad/s and 3 rad/s. So, bandwidth is 3-2=1 rad/s.
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In the frequency domain a sinusoid is a "spike", an impulse function at the sinusoid's frequency. Theoretically at least, a single sinusoid has 0 bandwidth. In practice they actually have a small but non-zero bandwidth because in order to have zero bandwidth the sinusoid would have to be perfect and started at time $-\infty$ and end at time $+\infty$. But nevermind that, let's stick to the ideal theoretical world.
The reason that the bandwidth of the two sinusoids together is one is because the first sinusoid is at 2 Hz (assuming that there is an implied $2\pi$ in the frequency constant) and the second sinusoid is at 3 Hz. Individually they have a bandwidth of 0 Hz, but since they are 1 Hz apart together they have a bandwidth of 1 Hz.
