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ECE 302: Lecture A.8 Mean and Autocorrelationthrough LTI Systems
Prof Stanley Chan
School of Electrical and Computer EngineeringPurdue University
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Mean function
Theorem
If X (t) passes through an LTI system to yield Y (t), then the meanfunction of Y (t) is
E[Y (t)] = µX
∫ ∞−∞
h(τ)dτ. (1)
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Proof
Suppose that Y (t) = h(t) ∗ X (t). Then,
µY (t) = E[Y (t)] = E[∫ ∞−∞
h(τ)X (t − τ)dτ
]=
∫ ∞−∞
h(τ)E[X (t − τ)]dτ
=
∫ ∞−∞
h(τ)µXdτ
= µX
∫ ∞−∞
h(τ)dτ,
where the last equality holds because X (t) is WSS so thatE[X (t − τ)] = µX .
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Example
Example 1. Consider a WSS random process X (t) such that each sampleis an i.i.d. Gaussian random variable with zero mean and unit variance. Wesend this process through an LTI system with impulse response h(t), where
h(t) =
{10(1− |t|), −1 ≤ t ≤ 1,
0, otherwise.
Find µY (t).Solution. X (t) is i.i.d. Gaussian, and so
µX (t) = E[X (t)] = 0. (2)
Hence,
µY (t) = E[Y (t)] = µX
∫ ∞−∞
h(τ)dτ = 0. (3)
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Example
-10 -5 0 5 10-4
-2
0
2
4
X(t)
X(t)
Y(t)
Y(t)
Figure: When sending a WSS random process through an LTI system, the meanand the autocorrelation functions are changed.
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Autocorrelation
Theorem
If X (t) passes through an LTI system to yield Y (t), then theauto-correlation function of Y (t) is
RY (τ) =
∫ ∞−∞
∫ ∞−∞
h(s)h(r)RX (τ + s − r)dsdr . (4)
A short hand notation of the above formula is RY (t) = [h ~ (h ∗ RX )](t),where ∗ denotes the convolution and ~ denotes the correlation.
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Proof
We start with the definition of Y (t):
RY (τ) = E[Y (t)Y (t + τ)]
= E[∫ ∞−∞
h(s)X (t − s)ds
∫ ∞−∞
h(r)X (t + τ − r)dr
](a)=
∫ ∞−∞
∫ ∞−∞
h(s)h(r)E [X (t − s)X (t + τ − r)dsdr ]
=
∫ ∞−∞
∫ ∞−∞
h(s)h(r)RX (τ + s − r)dsdr ,
where in (a) we assume that integration and expectation areinterchangeable.
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Example
Example 2. Same as example 1: X (t) is i.i.d. Gaussian, and
h(t) =
{10(1− |t|), −1 ≤ t ≤ 1,
0, otherwise.
Find RY (τ).Solution. X (t) is i.i.d. Gaussian, and so
RX (τ) = δ(τ). (5)
Hence,
RY (τ) =
∫ ∞−∞
∫ ∞−∞
h(s)h(r)RX (τ + s − r)dsdr
=
∫ ∞−∞
∫ ∞−∞
h(s)h(r)δ(τ + s − r)dsdr
=
∫ ∞−∞
h(s)h(τ + s)ds
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Example
-2 -1 0 1 2-0.05
0
0.05
0.1
0.15
0.2
RX(t)
RY(t)
Figure: When sending a WSS random process through an LTI system, the meanand the autocorrelation functions are changed.
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Power spectral density through LTI systems
Theorem
If X (t) passes through an LTI system to yield Y (t), then the powerspectral density of Y (t) is
SY (ω) = |H(ω)|2SX (ω). (6)
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Proof
By definition, power spectral density SY (ω) is the Fourier transform of theautocorrelation function RY (ω). Therefore,
SY (ω) =
∫ ∞−∞
RY (τ)e−jωτdτ
=
∫ ∞−∞
∫ ∞−∞
∫ ∞−∞
h(s)h(r)RX (τ + s − r)dsdr e−jωτdτ
Letting u = τ + s − r , we have
SY (ω) =
∫ ∞−∞
∫ ∞−∞
∫ ∞−∞
h(s)h(r)RX (u)e−jω(u−s+r)dsdrdu
=
∫ ∞−∞
h(s)e jωsds
∫ ∞−∞
h(r)e−jωrdr
∫ ∞−∞
RX (u)e−jωudu
= H(ω)H(ω)SX (ω),
where H(ω) is the complex conjugate of H(ω).11 / 16
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Example
Example 3: A WSS process X (t) has a correlation function
RX (τ) = sinc(πτ).
Suppose that X (t) passes through an LTI system with input/outputrelationship
2d2
dt2Y (t) + 2
d
dtY (t) + 4Y (t) = 3
d2
dt2X (t)− 3
d
dtX (t) + 6X (t).
Find RY (τ).
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Solution: The sinc function has a Fourier transform given by
sinc(Wt)←→F
π
Wrect
( ω
2W
).
Therefore, the auto-correlation function is
RX (τ) = sinc(πτ) ←→F
π
πrect
( ω2π
).
By taking the Fourier transform on both sides, we have
SX (ω) =
{1, −π ≤ ω ≤ π,0, elsewhere.
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The system response can be found from the differential equation as
H(ω) =3(jω)2 − 3(jω) + 6
2(jω)2 + 2(jω) + 4
=3[(2− ω2)− jω
]2 [(2− ω2) + jω]
.
Taking the magnitude square yields
|H(ω)|2 =3[(2− ω2)− jω
]2 [(2− ω2) + jω]
3[(2− ω2) + jω
]2 [(2− ω2)− jω]
=9
4
(2− ω2)2 + ω2
(2− ω2)2 + ω2
=9
4.
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Therefore, the output power spectral density is
SY (ω) = |H(ω)|2SX (ω)
=9
4SX (ω).
Taking the inverse Fourier transform, we have
RY (τ) =9
4sinc(πτ).
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Questions?
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