Prove the sum of 2 signals is periodic?

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Discussion Overview

The discussion revolves around the periodicity of the sum of two discrete signals, x1[n] and x2[n], both of which are known to be periodic individually. Participants explore how to prove that their sum is also periodic.

Discussion Character

  • Technical explanation

Main Points Raised

  • John asks how to prove that the sum of two periodic discrete signals is also periodic, referencing the requirement for periodicity.
  • One participant suggests using the lowest common multiple (lcm) of the periods of the two signals as a potential solution.
  • Another participant clarifies that the lcm refers to the lowest common multiple of the periods n1 and n2 of the signals.
  • A further explanation is provided that if k is the lcm of n1 and n2, then the periodicity condition can be satisfied for both signals, leading to the conclusion that the sum is periodic.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the proof, as participants are still exploring the implications of using the lcm and how it applies to the periodicity of the sum.

Contextual Notes

There are assumptions about the definitions of periodicity and the nature of the signals that are not fully detailed, such as the specific values of n1 and n2.

LM741
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hey guys - just a question regarding a systems and signals course:

if I've got two disctrete signals: x1[n] and x2[n]

if i know both signals are periodic (individually) how do i prove the sum of these signals is periodic?

requirement for periodicty (one signal): x[n+N] = x[n]

where n and N are integers - remeber - discrete signals!

thanks!

John
 
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Lcm (n1,n2)
 
Last edited:
EvLer >> Please elaborate...lowest common multiple of n1 and n2? not entirely sure what you mean ..thanks...
 
let k = lcm(n1, n2) which is the lowest common multiplier

x(n+k) = x1(n+k) + x2(n+k)

Since k is the lowest common multiplier, that means that it will satisfy the requirement for both signals (for example: x1(n+2) = x1(n+4) = x1(n+6) = ... ). So you have

x(n+k) = x1(n) + x2(n)
 
Thanks Mohammad!

John
 

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