# Signals Energy of 2 signals - Integral limits correct?

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:

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

because $$\int x(t)y(t)dt$$ = 0 because of integration by parts:

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

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

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

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

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Dick
Homework Helper
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.

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