How can the inner product of two signals be calculated?

[Jncik]

Homework Statement

Hi

suppose

you're given two signals for example

$$x_{1}(t) = cos(3 \omega_{0} t)$$
$$x_{2}(t) = cos(7 \omega_{0} t)$$

and you want to find out the inner product

Homework Equations

The Attempt at a Solution

I mean, it's an integral right? But what will the boundaries be? from -oo to +oo or are we interested only in 1 period, hence from 0 to T?

thanks in advance

 

[mathfeel]

Homework Statement

Hi

suppose

you're given two signals for example

$$x_{1}(t) = cos(3 \omega_{0} t)$$
$$x_{2}(t) = cos(7 \omega_{0} t)$$

and you want to find out the inner product

Homework Equations

The Attempt at a Solution

I mean, it's an integral right? But what will the boundaries be? from -oo to +oo or are we interested only in 1 period, hence from 0 to T?

thanks in advance

Both signals have fundamental frequency $\omega_0$. So integration over any integer multiple of the fundamental period gives the same result.

 

[Jncik]

thanks for your reply

so, if I took from -T to T it would be an interval 2 times larger than the interval of a period, which is correct right?

but if I integrate from 0 to T I will get a different result

do you mean by "the same result" that the answer to a question that wants an inner product can include any interval of integration no matter what the final arithmetic result will be?

thanks in advance

 

[mathfeel]

thanks for your reply

so, if I took from -T to T it would be an interval 2 times larger than the interval of a period, which is correct right?

but if I integrate from 0 to T I will get a different result

do you mean by "the same result" that the answer to a question that wants an inner product can include any interval of integration no matter what the final arithmetic result will be?

thanks in advance

Product of periodic function is still periodic and any integration over integer multiple of period, once or twice should give the same result, or you are doing the integration wrong.

If this helps, try this identity:
$\cos a \cos b = \frac{\cos (a+b) + \cos(a-b)}{2}$