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, 7.7
EE/STAT 322, #20 1
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Power Spectral Density: Definition, Derivation, and Relation to Autocorrelation Function, Study notes of Statistics
An in-depth understanding of power spectral density (psd), its relation to the fourier transform, and its derivation. It also explains how psd is related to the autocorrelation function. Examples and solutions to exercises for better comprehension.
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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, #
INTRODUCTION
Fourier transform (FT) of a signal
y ( t )
is defined as
F
Y
( ω
) =
∞
−∞
y ( t ) e −
jωt
dt.
Since
ω
πf
(^) , we may write
F
Y
( f (^) ) =
∞
−∞
y ( t ) e − j 2
πf t
dt
Here,
we
assume
∞ −∞
| y ( t ) |
dt <
(for
the
Fourier
transform
to
converge).
The inverse Fourier transform is given by
y ( t ) =
F − 1 { F Y ( f
∞
−∞
F
y ( f (^) ) e j 2 πf t
df
π
∞
−∞
F
y ( f (^) ) e jωt
dω.
EE/STAT 322, #
DERIVATION OF PSD
Let
f (^) ( t ) =
g ( t ) =
X
T (^) ( t ) ,
T −
T
X
T 2 (^) ( t ) dt
1
2 π
∫
∞ −∞
F
X
(^) ( ω
) | 2 dω
∞ −∞
F
X
(^) ( f (^) ) | 2 df
ω
πf
Dividing both side by
T
leads to
1
2 T
∫
T −
T
X
T 2 (^) ( t ) dt
1
2 π
∫
∞
−∞
| F X (^) ( ω ) | 2
2 T
dω
Taking expectation and letting
T
on both sides,
lim
T (^) →∞
E
1
2 T
T − T
X
T 2 (^) ( t ) dt
= lim
T (^) →∞
E
1
2 π
∫
∞ −∞
| F X (^) ( ω ) | 2
2 T
dω
lim
T (^) →∞
1
2 T
∫
T −
T
X
2 dt
1
2 π
lim
T (^) →∞
∞ −∞
E
| F X (^) ( ω ) | 2
2 T
dω
〈 X 2 〉 = 1
π
∞
−∞
lim
T (^) →∞
E
F
X
(^) ( ω
) | 2
T
dω.
EE/STAT 322, #
DERIVATION OF PSD (CONT.)
Let
S
X
(^) ( ω
) = lim
T (^) →∞
E
| F X (^) ( ω ) | 2
2 T
, which is called the (power) spectral
density (PSD).
For a stationary process
〈 X 2 〉 = X 2.
X
2
=
1
2 π
∫
∞ −∞
S
X
(^) ( ω
) dω
∞ −∞
S
X
(^) ( f (^) ) df
Answer: Integrating PSD over frequencyWhy the name power spectral density (PSD)?
f
(or
ω
) gives the power.
X
2
is the average power measured in time domain, and
∞ −∞
S
X
(^) ( f (^) ) df
is
the total power measured in frequency domain.
EE/STAT 322, #
POWER SPECTRAL DENSITY (CONT.)
Example:
(Ex. 7-2.2) A stationary process has a two-sided PSD given by
S
X
(^) ( ω
) =
24
ω 2
(b) Find the mean-square value of the process in the frequency band of(a) Find the mean-square value of the process.
(a) Solution: Hz centered at the origin.
S
X
(^) ( f (^) ) =
24
(
πf
(^) ) 2
. X 2 = ∫ ∞
−∞
24
(
πf
(^) ) 2
df
∞ −∞
24
/ (
π ) 2
f (^2)
2 / (
π ) 2 (^) df
Let
f
π 2
tan
(^) θ
,
⇒
df
2 /π
cos
2 θ^ (^) dθ
f
π 2
tan
(^) θ
, ∞ ) ⇒ θ ∈ ( −
π/
, π/
X
2
=
π/
2
−
π/
2
2 3
cos
2 θ^
2 /π
cos
2 θ^ (^) dθ
π
2 3
· π 2
EE/STAT 322, #
POWER SPECTRAL DENSITY (CONT.)
(b)
X
2
=
1
−
1
24
/ (
π ) 2
f (^2)
2 / (
π ) 2 (^) df
Let
f
π 2
tan
(^) θ
.
f
π 2
tan
(^) θ
tan
θ
∈
π/
, π/
θ
∈
tan
−
1 ( π/
tan
− 1 ( π/
⇒ θ ∈ ( − 1.
X 2 = ∫ 1.
004
−
1 . 004
2 3
cos
2 θ^
2 /π
cos
2 θ^ (^) dθ
2 3
· π 2
π 6
EE/STAT 322, #
PSD AND
R
X
(CONT.)
Let
τ = t 2 − t 1 , d 2 =
dτ
(^). We get
S
X
( ω
) =
lim
T (^) →∞
T
T (^) −
t 1
−
T (^) −
t 1 dτ
T
−
T
R
X
(^) ( t 1 , t
1
τ (^) ) e −
jωτ
dt^
1
∞
−∞
lim
T (^) →∞
T
T
− T
R
X
(^) ( t 1 , t
1
τ (^) ) dt
1 } e −
jωτ
dτ.^
When
X
t )
is stationary and ergodic,
lim
T (^) →∞
1
2 T
T −
T
R
X
(^) ( t 1 , t
1 (^) +
(^) τ (^) ) dt
1 = 〈 X ( t 1 ) X ( t 1
(^) τ (^) ) 〉
=
R
X
(^) ( τ (^) ) =
R
X
(^) ( τ (^) ) .
S
X
(^) ( ω
) =
F{
R
X
(^) ( τ (^) ) }
=
∞ −∞
R
X
(^) ( τ (^) ) e −
jωτ
dτ^
EE/STAT 322, #
10
PSD AND
R
X
(CONT.)
Example:
R
X
(^) ( τ (^) ) =
Ae
−
β | τ (^) | ,
A >
, β >
. Find
S
X
(^) ( ω
) .
S Solution:
X
(^) ( ω
) =
∞
−∞
R
X
(^) ( τ (^) ) e −
jωτ
dτ^
∞
0
Ae
−
βτ
e^ −
jωτ
dτ^
0
−∞
Ae
βτ
e^ −
jωτ
dτ^
A
[
β
jω
β
jω
]
Aβ
ω
2
β
2 (^).
The
R
X
(^) ( τ (^) )
can be obtained using the inverse Fourier transform of
S
X
(^) ( ω
) .
R
X
(^) ( τ (^) ) =
F − 1 { S X
ω
) }
=
1
2 π
∫
∞ −∞
S
X
(^) ( ω
) e jωτ
dω^
EE/STAT 322, #
11
PSD AND
R
X
(CONT.)
Example
(Ex 7-6.1) A stationary process has an autocorrelation function
of the form
R
X
(^) ( τ (^) ) = 2
e −|
τ (^) |
- 4
e −
4 | τ (^) |
. Find the PSD
S
X
(^) ( 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 ).
S
X
(^) ( ω
) =
F{
e −|
τ (^) | }
F{
e −
4 | τ (^) | } = 2 · 2 · 1
2
ω
2 (^) +
4 2 + ω 2 =
ω
2
- 40
ω
4
- 17
ω
2
- 16
EE/STAT 322, #
13
PSD AND
R
X
(CONT.)
Example
(Ex 7-6.2) A stationary process has a PSD of the form
S
X
(^) ( ω
) =
8 ω 2
ω 4
ω 2
. Find
R
X
(^) ( τ (^) ) .
Solution:
Using partial fraction expansion leads to
S
X
(^) ( ω
) =
16
ω 2
−
8
ω 2
R
X
(^) ( τ (^) ) = 4
e −
2 | τ (^) | −
e −
4 | τ (^) | .
EE/STAT 322, #
BANDLIMITED WHITE NOISE
Bandlimited white noise:
S
X
(^) ( f (^) ) =
S
0
f (^) | ≤
W
elsewhere
R
X
(^) ( τ (^) )
= F − 1 { S X
ω ) } = F − 1 { S 0
rect
f
2 W ) } = ∫ W
− W S 0 e j 2
πf τ
df^
W S
0 sinc(
W τ
, where rect
f
2 W
(^) )
is a rectangular window function with
interval
W, W
, and
sinc(
X
sin(
πX
)
πX
is the
sinc
-function.
))
2
/(
(
)
(
0
W
f
rect
S
f
S X^
=
W
0
W
−
0
S
f
)
(τ
X
R
W 2 1
0
0
2 WS
τ
W 1
W 2 3
W 1
−
W 2 1
−
W
2 3
−
EE/STAT 322, #
WHITE NOISE (CONT.)
Example:
(Ex 7-7.1) A stationary process has a bandlimited PSD
S
X
( f (^) ) =
{ 0. 1 | f
Hz
elsewhere
(a) Find the mean square value of
X
; (b) Find the smallest value of
τ
for
which
R
X
(^) ( τ (^) ) = 0
; (c) What is the bandwidth of this process?
Solution:
(a)
X
2
=
1000 − 1000
S
X
(^) ( f (^) ) df
(b)
R
X
(^) ( τ (^) ) = 2
W S
0 (^) sinc(
W τ
where
S
0
and
W
Hz.
sinc(
W τ
W τ π
= π ⇒ τ = 1
2 W
(c) Bandwidth is
W
EE/STAT 322, #
WHITE NOISE (CONT.)
(b)
S
X
(^) ( f (^) ) =
ω
| <
elsewhere
Bandwidth is
Hz.
(c)
X
2
= 2
250 200
df
EE/STAT 322, #