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SPECTRAL DENSITY, I
OUTLINE
- 7-1 Introduction
- 7-2 Relation of Spectral Density to Fourier Transform
- 7-6 Mean-Square values from Spectral Density
- 7-7 White Noise
Reading: G. R. Cooper & C. D. McGillem 7.1, 7.2, 7.6, 7.
EE/STAT 322, #20 1
INTRODUCTION
- Fourier transform (FT) of a signal y(t) is defined as
FY (ω) =
−∞
y(t)e−jωtdt.
- Since ω = 2πf , we may write FY (f ) =
−∞ y(t)e
−j 2 πf tdt.
Here, we assume
−∞ |y(t)|dt <^ ∞^ (for^ the^ Fourier^ transform^ to converge).
- The inverse Fourier transform is given by
y(t) = F−^1 {FY (f )} =
−∞
Fy(f )ej^2 πf tdf =
2 π
−∞
Fy(f )ejωtdω.
POWER SPECTRAL DENSITY (PSD)
- Since X(t) may have infinite energy for t ∈ (−∞, ∞), we define a finite window:
XT (t) =
X(t) −T < t < T 0 elsewhere , such that^
∫ T
−T |XT^ (t)|
(^2) dt < ∞.
We define FX (ω) =
∫ T
−T XT^ (t)e
−jωtdt, 0 < T < ∞.
- Let us derive PSD and show that X^2 =
−∞ SX(f^ )df^.
- Background: Parseval’s theorem: If f (t) and g(t) have the Fourier transforms F (ω) and G(ω), respectively, then ∫ (^) ∞
−∞
f (t)g(t)dt =
2 π
−∞
F (ω)G(−ω)dω.
EE/STAT 322, #20 3
DERIVATION OF PSD
- Let f (t) = g(t) = XT (t),
⇒
∫ T
−T X
2 T (t)dt^ =^
1 2 π
−∞ |FX^ (ω)|
(^2) dω = ∫^ ∞ −∞ |FX^ (f^ )|
(^2) df. (ω = 2πf .)
Dividing both side by 2 T leads to 1 2 T
∫ T
−T X
2 T (t)dt^ =^
1 2 π
−∞
|FX (ω)|^2 2 T dω. Taking expectation and letting T → ∞ on both sides,
⇒limT →∞ E
1 2 T
∫ T
−T X
2 T (t)dt
= limT →∞ E
1 2 π
−∞
|FX (ω)|^2 2 T dω
⇒limT →∞ (^21) T
∫ T
−T X
(^2) dt =
1 2 π limT^ →∞
−∞
E|FX (ω)|^2 2 T dω
〈X^2 〉 =
2 π
−∞
lim T →∞
E|FX (ω)|^2 2 T
dω.
POWER SPECTRAL DENSITY (CONT.)
Example: (Ex. 7-2.2) A stationary process has a two-sided PSD given by SX (ω) = (^) ω (^224) +16.
(a) Find the mean-square value of the process.
(b) Find the mean-square value of the process in the frequency band of ± 1 Hz centered at the origin.
Solution:
(a) SX (f ) = (^) (2πf^24 ) (^2) +16. X^2 =
−∞
24 (2πf )^2 +16df^ =^
−∞
24 /(2π)^2 f 2 +4^2 /(2π)^2 df^.
Let f = (^) π^2 tan θ, ⇒df = (^) cos^2 /π (^2) θ dθ.
f = (^) π^2 tan θ ∈ (−∞, ∞) ⇒θ ∈ (−π/ 2 , π/2).
X^2 =
∫ (^) π/ 2 −π/ 2
3 2 cos
(^2) θ 2 /π cos^2 θ dθ^ =^ π^ ·^
3 2 ·^
2 π = 3.
EE/STAT 322, #20 7
POWER SPECTRAL DENSITY (CONT.)
(b) X^2 =
− 1
24 /(2π)^2 f 2 +4^2 /(2π)^2 df^.
Let f = (^) π^2 tan θ. f = (^) π^2 tan θ ∈ (− 1 , 1) ⇒tan θ ∈ (−π/ 2 , π/2) ⇒θ ∈ (− tan−^1 (π/2), tan−^1 (π/2)) ⇒θ ∈ (− 1. 004 , 1 .004).
X^2 =
− 1. 004
3 2 cos
(^2) θ 2 /π cos^2 θ dθ^ = 2.^008 ·^
3 2 ·^
2 π =^
6 π = 1.^918.
PSD AND AUTOCORRELATION FUNCTION
The PSD is the Fourier transform of the autocorrelation function RX (τ ).
SX (ω) =
−∞
RX (τ )e−jωτ^ dτ.
Proof: SX (ω) = limT →∞ E[|FX^ (ω)|
(^2) ] 2 T , where^ FX^ (ω) =^
∫ T
−T XT^ (t)e
−jωtdt.
⇒SX (ω) = limT →∞ (^21) T E
[∫ T
−T XT^ (t^1 )e
jωt (^1) dt 1 ∫^ T −T XT^ (t^2 )e
−jωt (^2) dt 2
]
= limT →∞ (^21) T E
[∫ T
−T dt^2
∫ T
−T XT^ (t^1 )XT^ (t^2 )e
−jω(t 2 −t 1 )dt 1
]
EE/STAT 322, #20 9
PSD AND RX (τ ) (CONT.)
Let τ = t 2 − t 1 , d 2 = dτ. We get
SX(ω) = lim T →∞
2 T
∫ (^) T −t 1
−T −t 1
dτ
∫ T
−T
RX (t 1 , t 1 + τ )e−jωτ^ dt 1
−∞
{ lim T →∞
2 T
∫ T
−T
RX (t 1 , t 1 + τ )dt 1 }e−jωτ^ dτ.
When X(t) is stationary and ergodic,
limT →∞ (^21) T
∫ T
−T RX^ (t^1 , t^1 +^ τ^ )dt^1 =^ 〈X(t^1 )X(t^1 +^ τ^ )〉^ =^ RX^ (τ^ ) =^ RX^ (τ^ ).
⇒SX (ω) = F{RX (τ )} =
−∞ RX^ (τ^ )e
−jωτ (^) dτ.
PSD AND RX (τ ) (CONT.)
Example (Ex 7-6.1) A stationary process has an autocorrelation function of the form RX (τ ) = 2e−|τ^ |^ + 4e−^4 |τ^ |. Find the PSD SX (ω).
Solution: F{R 1 (τ )} = S 1 (ω), F{R 1 (τ )} = S 2 (ω), ⇒F{R 1 (τ ) + R 2 (τ )} = S 1 (ω) + S 2 (ω).
SX (ω) = F{ 2 e−|τ^ |}+F{ 4 e−^4 |τ^ |} =
12 + ω^2
42 + ω^2
10 ω^2 + 40 ω^4 + 17ω^2 + 16
EE/STAT 322, #20 13
PSD AND RX (τ ) (CONT.)
Example (Ex 7-6.2) A stationary process has a PSD of the form
SX (ω) = 8 ω
(^2) + ω^4 +20ω^2 +64. Find^ RX^ (τ^ ).
Solution: Using partial fraction expansion leads to
SX (ω) = (^) ω (^216) +4 + (^) ω 2 −+16^8
⇐⇒RX (τ ) = 4e−^2 |τ^ |^ − e−^4 |τ^ |.
WHITE NOISE
Definition: Ideal white noise has a PSD uniformly distributed in (−∞, ∞).
White noise has an infinite bandwidth (and infinite power) RX (τ ) = S 0 δ(τ ) ⇒SX (f ) =
−∞ RX^ (τ^ )e
−j 2 πf τ (^) dτ = ∫^ ∞ −∞ S^0 δ(τ^ )e
−j 2 πf τ (^) dτ = S 0.
S (^) X ( f )= S 0
S 0
f
RX (τ )= S 0 δ(τ )
τ
EE/STAT 322, #20 15
BANDLIMITED WHITE NOISE
Bandlimited white noise: SX (f ) =
S 0 |f | ≤ W 0 elsewhere
RX (τ ) = F−^1 {SX (ω)} = F−^1 {S 0 rect( 2 fW )} =
∫ W
−W S^0 e
j 2 πf τ (^) df =
2 W S 0 sinc(2W τ ), where rect( 2 fW ) is a rectangular window function with
interval (−W, W ), and sinc(X) = sin( πXπX )is the sinc-function.
S (^) X ( f )= S 0 rect ( f /( 2 W ))
− W 0^ W
S 0
f
RX (τ )
2 W 0 1
2 WS 0
τ
W
1 2 W
3 W − 1 2 W −^1 2 W −^3
WHITE NOISE (CONT.)
(b) SX (f ) =
- 01 200 ≤ |ω| < 250 0 elsewhere
Bandwidth is 2 · (250 − 200) = 100 Hz.
(c) X^2 = 2 ·
200 0.^01 df^ = 2^ ·^50 ·^0 .01 = 1.
