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Message started by greatqs on Sep 15th, 2006, 6:50pm

Title: Problem about random process and phase noise
Post by greatqs on Sep 15th, 2006, 6:50pm

I've read in a document as follow:

if v(t)=S(t)+n(t) (eq.1) , where S(t)=A•cos(wc•t)

S(t) is the signal, n(t) is a gaussian noise process.

The document said
v(t) can be expressed by

v(t)=A•cos(wc•t) + x(t)•cos(wc•t) -y(t)•sin(wc•t) (eq.2)
where x(t) and y(t) are real, low pass, Gaussian processes with zero mean and equal variance.

However, it never gives the explanation.
Can anyone explain to me how can we convert eq.1 to eq.2?
Thanks a lot!!


BTW: the document also call x(t) as inpahse noise or amplitude noise and
y(t) as quadrature or "phase noise"

Title: Re: Problem about random process and phase noise
Post by gitarrelieber on Oct 1st, 2006, 1:41pm

It is equivalent to prove random process n(t) is equal to z(t)=x(t)cos(ωct)-y(t)sin(ωct).

To prove random processes n(t) and z(t) are equal, we need to prove the first and second order moments of n(t) and z(t) are equal.

The first order moment is the mean of n(t) and z(t). It is straightforward to show E{n(t)}=E{z(t)}.

Since random process n(t) is Gaussian and real,
Rn(τ) = E{n(t)n(t-τ)} = δ(τ), where δ(τ) is Dirac function and E{} denotes expectation.
Assume x(t) and y(t) are also real and Gaussian, or
Rx(τ) = E{x(t)x(t-τ)}  = Ry(τ) = E{y(t)y(t-τ)}  = c δ(τ),
and indepedent to each other, Rxy(τ)=E{x(t)y(t-τ)}=0;

Rz(τ) = E{z(t)z(t-τ)} = E{(x(t)cos(ωct)-y(t)sin(ωct))(x(t)cos(ωc(t-τ)-y(t)sin(ωc(t-τ))}
= E{x(t)x(t-τ)cos(ωct) cos(ωc(t-τ)) + y(t)y(t-τ)sin(ωct) sin(ωc(t-τ)) }
= E{x(t)x(t-τ)}cos(ωct) cos(ωc(t-τ)) + E{y(t)y(t-τ)}sin(ωct) sin(ωc(t-τ))
= Rx(τ) cos(ωct) cos(ωc(t-τ)) + Ry(τ) sin(ωct) sin(ωc(t-τ))
= c δ(τ) cos2ct) + c δ(τ) sin2ct)
(let c=1)
= δ(τ) = Rn(τ)

Therefore, Rx(τ) = Ry(τ) = Rn(τ)

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