Signals Energy of 2 signals - Integral limits correct?

thomas49th

If signals x(t) and y(t) are orthogonal and if z(t) = x(t) + y(t) then
E_{z} = E_{x} + E_{y}:


Proof:

[tex] E_{z} => \int^{\infty}_{-\infty} {(x(t) + y(t))^{2}} dt
=> \int {(x(t) + y(t))^{2}}^{2} dt
=> \int (x^{2}(t)) + \int(y^{2}(t))dt + \int x(t)y(t)dt
=> E_{x} + E_{y}
[/tex]

because [tex]\int x(t)y(t)dt[/tex] = 0 because of integration by parts:

u = x(t) dv/dt = y(t)
u' = dx/dt, v = [tex]frac{y^{2}(t)}{2}[/tex]

so [tex]x(t)\frac{y^{2}(t)}{2} - \int {\frac{y^{2}(t)}{2}\frac{dx}{dt}}dt[/tex]
[tex]x(t)\frac{y^{2}(t)}{2} - \int {\frac{y^{2}(t)}{2}}dx[/tex]
we can treat y^2(t) as a constant so:

[tex]x(t)\frac{y^{2}(t)}{2} - \int^{\infty}_{-\infty} {\frac{y^{2}(t)}{2}}dx[/tex]
[tex]x(t)\frac{y^{2}(t)}{2} - } [{\frac{y^{2}(t)x}{2}}]^{\infty t}_{-\infty t}[/tex]

but the problem is that the limits were destined for integrating with respect to time. I'm not integrating with respect to x.

Any suggestions?
Thanks
Thomas

Dick

The integral of x(t)y(t) isn't zero because of some bogus 'integration by parts' argument. It's zero because that's what 'orthogonal' means.

thomas49th

Ofcourse! Execellent. May I ask, out of interest alone what the integral of x(t)y(t) with respect to t should be?

Thanks
Thomas

Dick

There's really nothing in particular you can say about it without knowing more about x(t) and y(t). y(t)dt can't be integrated to y(t)^2/2. That's y(t)dy(t). So integration by parts isn't useful.