Suppose I have a waveform $x(t)$
where $x(t) = 5 + 30\mathrm{cos}(2000\pi t) + 10\mathrm{cos}(6000\pi t)$ that is to be uniformly sampled for digital transmission, how do I calculate the bandwidth of this signal?

I reckon that it would $4000 \textrm{Hz}$ because we're subtracting the highest and lowest frequencies $(6000-2000)$. Is this right?

However, the answer in my book says it's $3000 \textrm{Hz}$. I do not understand how did they get this answer.

Thank you for reading


1 Answer 1


A sinusoidal signal is represented as $$x(t) = \mathrm{cos}(\omega t) = \mathrm{cos}(2\pi f t)$$

$\omega$ is the angular frequency and $f$ is the frequency. See Frequency definition.

Your signal \begin{align} x(t) &= 5 + 30\mathrm{cos}(2000\pi t) + 10\mathrm{cos}(6000\pi t)\\ &= 5\mathrm{cos}(2\pi \times 0 t) + 30\mathrm{cos}(2\pi \times 1000 t) + 10\mathrm{cos}(2\pi \times 3000 t)\end{align}

There are three frequencies $0, 1000, \textrm{ and } 3000\textrm{Hz}$ thus the bandwidth is $3000\textrm{Hz}$.


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