A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
what is the general proof to the theorem in signals. i.e "for any signal  discrete or continuous.. the condition for it to be periodic requires it to have a rational frequency"..?
a particular case of this proof could be arrived by assuming sinusoidal curve.
but that isn't general..
anonymous
 3 years ago
what is the general proof to the theorem in signals. i.e "for any signal  discrete or continuous.. the condition for it to be periodic requires it to have a rational frequency"..? a particular case of this proof could be arrived by assuming sinusoidal curve. but that isn't general..

This Question is Open

KenLJW
 3 years ago
Best ResponseYou've already chosen the best response.0Using the assumption of a linear circuit superposition applies. All signals can be expressed in a Fourier's Series therefore if it applies to one it applies to all. For none linear circuits as large baseemitter signal of a common emitter BJT circuit you must use another analysis.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Actually the proof of this comes from sinusoidal signals itself. Any PERIODIC signal can be expressed as a sum of sine and cosine terms terms right? Yes, there are other condition(Drcihlet's conditions), but lets not go into that. So, if its true for one sinusoid, it must be true for other sinusoid since each one has a frequency which is an integer multiple of the harmonics frequency. So its must be true for that periodic signal since its a sum of all those periodic signals. So it is true for any periodic signal.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.