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Signals Energy of 2 signals - Integral limits correct?

  • Thread starter thomas49th
  • Start date
  • #1
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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:

[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
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
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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
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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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