integration over entire period of the function divided by that period is called average(sorry i dont know to use math symbol here)
but in case of sinusoidal function the average is zero

 

integration over entire period of the function divided by that period is called average(sorry i dont know to use math symbol here)
but in case of sinusoidal function the average is zero

The best way to display math symbols and expressions is with TEX. Look at the sticky at the top of the forum.

\(
X_{avg} \; = \; \int_{-\infty}^{+\infty} x(t) dt
\)

Now, break that into the sum of two integrals, one from -∞ to zero and the other from zero to +∞, and then apply the definition of an odd signal to the first integral. What is the result.

 

definition of odd signal is
\(x_e=((x(t)+x(-t))/2
\)
if we apply to negative integral
we get

\(x(t)=(\int ^0_{-\infty} x(t)dt +\int ^0_{-\infty} x(-t)dt)/2 \)

 

definition of odd signal is
\(x_e=((x(t)+x(-t))/2
\)
if we apply to negative integral
we get

\(x(t)=(\int ^0_{-\infty} x(t)dt +\int ^0_{-\infty} x(-t)dt)/2 \)

That is NOT the definition of an odd signal. That is, in fact, how you extract the even component of a generic signal. Notice the 'e' subscript -- that does NOT stand for 'odd'!

You gave the defining property of an odd signal just a few posts ago.

 

sorry

A signal x(t) is anti-symmetric about about vertical axis is said to be odd signal

this is what i said and applying mathematical definition


x(t)=-x(-t)

\((\int ^0_{-\infty} x(t)dt=\int ^0_{-\infty} -x(-t)dt \)

 

sorry

this is what i said and applying mathematical definition


x(t)=-x(-t)

\((\int ^0_{-\infty} x(t)dt=\int ^0_{-\infty} -x(-t)dt \)

Let's start with the given integral and work in steps to see what happens.

\(
X_{avg} \; = \; \int_{-\infty}^{+\infty} x(t) dt
\)

We can split this into two integrals.

\(
X_{avg} \; = \; \int_{-\infty}^{0} x(t) dt \; + \; \int_{0}^{+\infty} x(t) dt
\)

If we do a change of variable on the first integral using u = -t we get

\(
X_{avg} \; = \; -\int_{+\infty}^{0} x(-u) du \; + \; \int_{0}^{+\infty} x(t) dt
\)

We can interchange the integration limits by multiplying the integral by -1, and we can change u to t since it is just a dummy variable of integration.

\(
X_{avg} \; = \; \int_{0}^{+\infty} x(-t) dt \; + \; \int_{0}^{+\infty} x(t) dt
\)

We can now combine the two integrals together again since they are over the same limits.

\(
X_{avg} \; = \; \int_{0}^{+\infty} \[ x(t) + x(-t) \]dt
\)

Note that everything up to this point has been completely general and has not involved anything specific to even or odd signals.

Now, what happens to this integral x(t) is an odd signal?

 

area of odd signal is zero
if x(t) is odd then x(t) can be written as -x(-t) since x(t)=-x(-t)
if we do so we get zero

 

So we know that if we have an odd signal, that the average value of that signal is zero. We don't have to do any more math. If you tell me (or if I can show) that a signal is odd, then I can tell you that the average value of that signal is zero. Doesn't that seem like a piece of knowledge that might be useful from time to time?

 

okay but my question is why we going even odd decomposition and it will be more correct to ask like where we neccessarily do even odd decmposition.?

 

If you have a signal that is

x(t) = A·xo(t) + B·xe(t)

where xo(t) and xe(t) are each normalized so that they are comparable -- perhaps they are each divided by their respective peak magnitude -- then if A >> B, can you tell me something, at least approximately, about the average value of x(t)?

The point is the you know some things about a signal just by knowing that it is odd or just by knowing it is even. By extension, that means that if you know how an arbitrary signal is composed of an even and odd signal then you know some things, at least in general terms, about the composite signal.

In addition, when you work with the signal, such as finding its spectral components, breaking it into even and odd components can make the analysis and computations easier and more meaningful.

 

finally got your point ,thanks for patient reply