ECE 302: Lecture A.8 Mean and Autocorrelation through LTI ...

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c Stanley Chan 2020. All Rights Reserved. ECE 302: Lecture A.8 Mean and Autocorrelation through LTI Systems Prof Stanley Chan School of Electrical and Computer Engineering Purdue University 1 / 16

Transcript of ECE 302: Lecture A.8 Mean and Autocorrelation through LTI ...

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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),

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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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