So I have an AM signal, given by: $$v(t)=50\cos(\pi 10^{6} t )+ 20\sin(\pi 10^3t)\cos(\pi 10^6 t)$$ I was asked to sketch the spectrum of the signal obtained, so I found the fourier transform of it, and that's what I got,

$$V(f)=25 \left[ \delta(f - 500k)+ \delta(f+500k)\right]-5i\left[\delta(f-999.5k)-\delta(f+999.5k) \right]-5i\left[\delta(f-500.5k)-\delta(f+500.5k) \right]$$

And so far I only knew how to sketch the first term, as shown below,

enter image description here

But I don't know how to sketch the terms with the $-5i$.


1 Answer 1


Since a signal's spectrum is complex, it can't be drawn in a single two-dimensional plot. What is usually done is to do two plots; one is the magnitude spectrum and the other is the phase spectrum.

When looking at signals, most of the time, the magnitude spectrum contains all the information that is needed and the phase spectrum is ignored. When looking at filters, the phase spectrum is important because one usually wants the phase response to be linear in the filter's passband.

So, what you need is the magnitude of $V(f)$:

$$|V(f)| = 25 \left[ \delta(f - 500k)+ \delta(f+500k)\right] + 5\left[\delta(f-999.5k) + \delta(f+999.5k) \right] + 5\left[\delta(f-500.5k)+\delta(f+500.5k) \right].$$

The magnitude spectrum is real, so it can be drawn in a regular 2-D plot. Note that all values of the magnitude spectrum are positive.


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