Signals Energy of 2 signals - Integral limits correct?

thomas49th
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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} =&gt; \int^{\infty}_{-\infty} {(x(t) + y(t))^{2}} dt<br /> =&gt; \int {(x(t) + y(t))^{2}}^{2} dt<br /> =&gt; \int (x^{2}(t)) + \int(y^{2}(t))dt + \int x(t)y(t)dt<br /> =&gt; E_{x} + E_{y}<br />

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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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
 
thomas49th said:
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

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.
 
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