Estimating Bandwidth of Phase Modulated Signal Using Taylor Series

Click For Summary

Discussion Overview

The discussion revolves around estimating the bandwidth of a phase modulated (PM) signal represented by the equation s(t) = Acos(wt + x(t)), where x(t) is the information-bearing signal. Participants explore the use of Taylor series expansion to derive this estimate, considering the condition |x(t)| < y, which is not necessarily small.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the PM signal and attempts to derive the bandwidth using Taylor series expansion, expressing uncertainty about the next steps after obtaining the expansion.
  • Another participant questions the notation used in the Taylor series expansion and clarifies the meaning of |x(t)|, suggesting it refers to absolute value.
  • A participant explains that the index k in the Taylor series is for summation and confirms the interpretation of |x(t)| as absolute.
  • Further, a participant proposes an estimate for s(t) using second-order terms from the Taylor expansion, indicating that the maximum value of x(t) should be substituted.
  • Another participant suggests expanding cos(wt + x(t)) using power series for cos(x) and sin(x), and discusses the implications of the limits on |x| for estimating bandwidth.

Areas of Agreement / Disagreement

Participants express differing views on the steps to take in deriving the bandwidth estimate, with no consensus on the final approach or solution. The discussion remains unresolved as participants explore various methods and interpretations.

Contextual Notes

Participants have not reached a consensus on the specific steps to derive the bandwidth, and there are uncertainties regarding the assumptions made about the behavior of x(t) and its maximum value.

ace1719
Messages
23
Reaction score
2

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).


Homework Equations





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'll thank you for your help in advance.
 
Physics news on Phys.org
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.
 
ace1719 said:
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.
 
Last edited:
  • Like
Likes   Reactions: 1 person
What can we infer about the bandwidth of that?
 
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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
4K
Replies
1
Views
8K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 1 ·
Replies
1
Views
14K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
17
Views
9K