# Bandwidth and Taylor Series

## Homework Statement

Consider the PM (phase modulated) signal, s(t) = Acos(wt+x(t)) where x(t) is the information bearing signal. Assume that |x(t)|< y, which is not necessarily small. Using Taylor's series expansion, derive an estimate for the bandwidth of the PM signal s(t).

## The Attempt at a Solution

I was able to get the Taylor series expansion of s(t). I'm guessing the relevant part of the expansion is (wt+x(t))^2k. At this point I would guess that you'd want to substitute the maximum value of x(t) in - which is y - to get, (wt+y)^2k. If you couldn't tell, this required a lot of guess work, but I'm lost as to what to do at this point.

I am not sure where your k came from but the Taylor series expansion up to fourth order is
$$1 - \frac{(wt + x(t))^2}{2} + \frac{(wt + x(t))^4}{24} - \cdots$$
Is ##\lvert x(t)\rvert## absolute value or the modulus?

k is the index for the summation. I guess I should have explained that. |x(t)|, I presume is absolute.

k is the index for the summation. I guess I should have explained that. |x(t)|, I presume is absolute.

The we could say an estimate of ##s(t)## is then
$$s(t) \approx A\Bigg(1 - \frac{(wt + x(t))^2}{2}\Bigg) < A - A\frac{(wt + y)^2}{2}$$
where I took only up to second order terms. You wouldn't have a k in your solution.

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What can we infer about the bandwidth of that?

rude man
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Gold Member
cos(wt + x(t)) = cos(x) cos(wt) - sin(x) sin (wt).
Expand cos(x) and sin(x) in power series. You know |x| is limited to y and I would think y = pi is the biggest it can get. So keep terms in n until xn/n! << 1. These are the coefficients of cos(wt) and sin(wt). When you multiply cos (nwT) by cos(wt) you get what? keep the sum term.

That should get you where you want to go.