Additions and Verifications from my Library Bike Voyage
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Dated 5.11.13, these are Phil's notes after a visit to a library, where he checked books related to his spectral document, including Xiong. He shows that for uncorrelated pulse amplitudes α = μ² and β = σ² + μ², so β − α = σ², and rewrites his averaged power spectrum result (35.11) accordingly. He also notes that his pulse train is a PAM digital signal and that the standard term is power spectral density. The notes end with an unresolved attempt to reconcile ensemble averages with sums over discrete values.
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Additions and Verifications from my Library Bike Voyage PhL 5.11.13
On May 11 I rode to Marriott and perused a few books related to spectral doc. The Xiong book, which I later downloaded, revealed that α and β are related to μ and σ2. This is in now in my Appendix D and is incorporated into spectral doc in a reasonable way. I was too stupid to even realize that <ymyn> = <ym><yn> , and I also had the meaning of correlation wrong. It is all cleared up now I think.
1. Everybody uses the phrase "power spectral density" PSD, so probably I should change my phrase to match or at least add a comment on same.
2. My "general" pulse train is really just a PAM digital signal, I should mention that somewhere. I might note that B&D allow for two different pulse shapes, whereas I only allow for one with two different amplitudes. We treated different problems, perhaps a comment on this.
2. Xiong's appendix gets my α and β formula, but interprets the parameters in a way that seems might be good to add, since I already have a correlation theory section especially. Here is my formula
<P(ω)> = Ppulse(ω) { (β-α) + α!Syntax Error, I2π δ(ωT1- 2πm) } (35.11)
n ≠ m α ≡ <ymyn> = [pp] AA + [p(1-p)] AB + [(1-p)p]BA + [(1-p)(1-p)]BB
n = m β ≡ <ym2> = [p]AA + [(1-p)] BB . (35.6)
I am confused by the X appendix A. He gets this result
where I like E{anam}. He then has
and yes, this is a sort of digital autocorrelation function. But then I think basically he is claiming that
<anam> = <an><am>
in the uncorrelated case. Well I guess that is right, I overlooked the obvious. Suppose I add to my list
μ = <ym> = [p]A + [p-1]B
α ≡ <ymyn> = <ym><yn> = μ2 when m ≠ n
This never appeared in my Ng notes because I guess it was not relevant. I might add this
corr(X,Y) = ∫∫ x y p(x,y) dx dy // general case
When things are uncorrelated, the idea is that
p(x,y) = p(x)p(y)
so we get
corr(X,Y) = ∫∫ x y p(x,y) dx dy = (∫ x p(x) dx) (∫ y p(y) dy) = μxμy
In the discrete world this would be
p(x1,x2) = Σi=1m P(x(i)) δ(x1-x(i)1) δ(x2-x(i)2) = Σi=1m P(x(i))δ(2)(x - x(i))
If the two variables are independent, then [ i labels my little clot points ]
P(x(i)) = P(x1(i)) P(x2(i))
then
p(x1,x2) = Σi=1m P(x1(i)) P(x2(i)) δ(x1-x(i)1) δ(x2-x(i)2)
= Σi=1m [P(x1(i)) δ(x1-x(i)1)][P(x2(i)) δ(x2-x(i)2)]
or
p(x,y) = Σi=1m [P(x(i)) δ(x-x(i))][P(y(i)) δ(y-y(i))]
Then for example,
E(XY) = !Syntax Error, I!Syntax Error, I dx dy x y p(x,y) .
= !Syntax Error, I!Syntax Error, I dx dy x y Σi=1m [P(x(i)) δ(x-x(i))][P(y(i)) δ(y-y(i))]
= Σi=1m!Syntax Error, I dx x[P(x(i)) δ(x-x(i))] !Syntax Error, I dy y[P(y(i)) δ(y-y(i))]
= Σi=1m { P(x(i)) x(i) }{ P(y(i) y(i)}
Maybe do it instead this way
E(XY) = ΣiΣj xiyj p(xi,yj) = ΣiΣj xiyj p(xi)q(yj) = [Σixi p(xi) ] [Σjyj q(yj) ] = E(X)E(Y)
OK, but then how do I get something with a single sum
E(XY) = (1/I) Σi xiyi
How can I reconcile these two expressions for E(XY) ?
Idea: I think I am using label i to mean two different things:
i is a label of a pulse train in the ensemble = label of training point in Ng world
i is used to label discrete values of a variable as in xi = {1, 2, 3 }
I am confusing these two meanings and getting meaningless results.
The ensemble pulse trains and the training points all have the same weight in whatever we compute, so they are all equally as likely in some vague sense.
OK, let's not dwell on this Ng connection right now. Go back to
<P(ω)> = T1 Ppulse(ω) (1/T) { α!Syntax Error, I !Syntax Error, I [ eiω(m-n)T] +β!Syntax Error, I !Syntax Error, I [1] } . (35.7)
I can write this setting α = μ2, I now accept that fact, so then
<P(ω)> = T1 Ppulse(ω) (1/T) { μ2!Syntax Error, I !Syntax Error, I [ eiω(m-n)T] +β!Syntax Error, I !Syntax Error, I [1] } . (35.7)
Now what about the other one?
β ≡ <ym2>
Recall that
variance ≡ < [ym - <ym>]2> = !Syntax Error, I (x-μx)2 p(x) dx = σ2
So I guess I can write
σ2 = <ym2> - 2 <ym>2 + <ym>2 = <ym2> – <ym>2 = variance = β - μ2.
Then I can write
β = σ2 + μ2 and then
<P(ω)> = T1 Ppulse(ω) (1/T) { μ2!Syntax Error, I !Syntax Error, I [ eiω(m-n)T] +(σ2+μ2)!Syntax Error, I !Syntax Error, I [1] } . (35.7)
OK, now jump to my endgame where I have for uncorrelated case
<P(ω)> = Ppulse(ω) { (β-α) + α!Syntax Error, I2π δ(ωT1- 2πm) } (35.11)
Then since I just showed that
α ≡ <ymyn> = <ym><yn> = μ2
β = σ2 + μ2
β - α = σ2
I could write my result as
<P(ω)> = Ppulse(ω) { (σ2 + μ2!Syntax Error, I2π δ(ωT1- 2πm) } (35.11)
But using these new names does not really simplify any of my calculations.
OK, I have my little probability digression sitting in exactly the right place to add something on this subject with minimal disruption. But I need a way to talk about independence in the discrete world.
<ymyn> = E(YmYn) = (1/I) !Syntax Error, I ym(i)yn(i) ≡ corr(YmYn)
I think I have to stop and restart this subject. I got a lot done but did not solve the problem.