Power of a Sawtooth Function

  • Thread starter salman213
  • Start date
  • #1
302
1
1. Hi, I'm confused on the method of calculating the power of the sawtooth function.
In my textbook the the general formula for claculating the power is

definition 1. http://img210.imageshack.us/img210/5091/75697758.jpg [Broken]

where the limit approaches infinity and the limits are -T/2 to T/2.

For periodic function they say this can be simplfiied to just

definition 2. http://img156.imageshack.us/img156/5962/54601684.jpg [Broken]

so now the T value is actually the period (finite), and the limits are also the finite numbers where the period is from.



For the SAWTOOTH function I know it is periodic so doing with the function equal to f(x) = Ax from for example -1 to 1 (period) and so T = 2

gets the answer that the power is

Px = A^2/3



If i just use the general formula I get infinity!!!

so for periodic functions is it NECESSARY that you HAVE to use definition 2, and you cannot use definition 1?





example when I use defintion 1 for the saw tooth

i get

lim 1/T A^2 [t^3/3)] limits are from T/2 to - T/2
T-> infinity

so you get

lim 1/T A^2 (T^3/4)
T-> infinity

=

lim A^2 (T^2/4)
T-> infinity


= infinity!!
 
Last edited by a moderator:

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
Those two formulas give you two different quantities. The first one gives you the average power delivered over all time. If the sawtooth source is switched on for an infinitely long time then it delivers infinite power to the load. Hence, the average power is infinite. But no source is ever switched on for an infinitely long time, so that quantity isn't of much interest.

What you are interested in is the quantity given by the second formula, which is the average power delivered to the load in just one period. That is finite, as you found.
 
  • #3
302
1
But if you find the power of a sinusoid using the first formula

you get a finite number even though that sinusoid is define on all t as well.....

ex. Acos(wt)


P = A^2/2 using the first formula
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
Why would you expect the average value for all t of the sinusoid and the sawtooth function to behave the same way? They're completely different functions.
 
  • #5
CEL
656
0
But if you find the power of a sinusoid using the first formula

you get a finite number even though that sinusoid is define on all t as well.....

ex. Acos(wt)


P = A^2/2 using the first formula
When you take the limit of T to infinity, this means that your period is infinite.
For a sinusoid, the amplitude is limited, so the power in one period is a finite number, no matter how long the period is.
For a sawtooth wave, the amplitude grows with time. So, if you take an infinite time, you have an infinite amplitude and an infinite power.
 

Related Threads on Power of a Sawtooth Function

  • Last Post
Replies
0
Views
5K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
9
Views
768
Top