Frequency domain description can be derived for periodic and pulse-like aperiodic phenomena.
- Noise is not repetitive and infinite in time. So not a periodic or aperiodic pulse type signal.
- As an approximation take a sample of noise of interest from -T/2 to T/2 consider it repetitive.
- Purely random so safely assume no DC exists. Now Fourier description applicable.
n
T
(
s
)
t
=
∑
k
=
1
∞
a
k
cos
2πkΔ
ft
+
b
k
sin
2πkΔ
ft
=
∑
k
=
1
∞
c
k
cos
2πkΔ
ft
+
θ
k
n
T
(
s
)
t
=
∑
k
=
1
∞
a
k
cos
2πkΔ
ft
+
b
k
sin
2πkΔ
ft
=
∑
k
=
1
∞
c
k
cos
2πkΔ
ft
+
θ
k
size 12{n rSub { size 8{T} } rSup { size 8{ \( s \) } } left (t right )= Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } { left (a rSub { size 8{k} } "cos"2πkΔ ital "ft"+b rSub { size 8{k} } "sin"2πkΔ ital "ft" right )= Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } {c rSub { size 8{k} } "cos" left (2πkΔ ital "ft"+θ rSub { size 8{k} } right )} } } {}
Also,
ck2=ak2+bk2ck2=ak2+bk2 size 12{c rSub { size 8{k} } rSup { size 8{2} } =a rSub { size 8{k} } rSup { size 8{2} } +b rSub { size 8{k} } rSup { size 8{2} } } {} and
θk=−tan−1bkakθk=−tan−1bkak size 12{θ rSub { size 8{k} } = - "tan" rSup { size 8{ - 1} } { {b rSub { size 8{k} } } over {a rSub { size 8{k} } } } } {}
- Two sided power spectrum and mean power spectral density may be derived
Power associated with each spectral term is
c
k
2
2
=
a
k
2
2
+
b
k
2
2
c
k
2
2
=
a
k
2
2
+
b
k
2
2
size 12{ { {c rSub { size 8{k} } rSup { size 8{2} } } over {2} } = { {a rSub { size 8{k} } rSup { size 8{2} } } over {2} } + { {b rSub { size 8{k} } rSup { size 8{2} } } over {2} } } {}
(1)
The power spectral density at k∆f is
G
n
kΔf
≡
G
n
−
kΔf
≡
c
k
2
4Δf
=
a
k
2
+
b
k
2
4Δf
G
n
kΔf
≡
G
n
−
kΔf
≡
c
k
2
4Δf
=
a
k
2
+
b
k
2
4Δf
size 12{G rSub { size 8{n} } left (kΔf right ) equiv G rSub { size 8{n} } left ( - kΔf right ) equiv { {c rSub { size 8{k} } rSup { size 8{2} } } over {4Δf} } = { {a rSub { size 8{k} } rSup { size 8{2} } +b rSub { size 8{k} } rSup { size 8{2} } } over {4Δf} } } {}
(2)
And total power in ∆f at k∆f is
P
k
=
2G
n
kΔf
Δf
P
k
=
2G
n
kΔf
Δf
size 12{P rSub { size 8{k} } =2G rSub { size 8{n} } left (kΔf right )Δf} {}
(3)
Half of this power is associated with k∆f and other half with -k∆f
- The power spectrum above is deterministic in the sense that it has been derived for a specific waveform, and hence the a,b,c values are specific calculable values. In the general case we can treat these as random variables, replace them by the ensemble average values.
- Let T tend to infinity - and ∆f tend to 0. then the actual noise waveform results.
n
t
=
lim
Δf
→
0
∑
k
=
1
∞
a
k
cos
2πkΔ
ft
+
b
k
sin
2πkΔ
ft
=
lim
Δf
→
0
∑
k
=
1
∞
c
k
cos
2πkΔ
ft
+
θ
k
n
t
=
lim
Δf
→
0
∑
k
=
1
∞
a
k
cos
2πkΔ
ft
+
b
k
sin
2πkΔ
ft
=
lim
Δf
→
0
∑
k
=
1
∞
c
k
cos
2πkΔ
ft
+
θ
k
size 12{n left (t right )= {"lim"} cSub { size 8{Δf rightarrow 0} } Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } { left (a rSub { size 8{k} } "cos"2πkΔ ital "ft"+b rSub { size 8{k} } "sin"2πkΔ ital "ft" right )= {"lim"} cSub { size 8{Δf rightarrow 0} } Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } {c rSub { size 8{k} } "cos" left (2πkΔ ital "ft"+θ rSub { size 8{k} } right )} } } {}
(4)
- Power contribution from coefficients is now replaced be mean square of the random coefficients which vary with chosen T or ensemble member.
Power spectral density can now be defined from mean square coefficients
- The Power spectral density then becomes
G
n
f
=
lim
Δf
→
0
c
k
2
¯
4Δf
=
lim
Δf
→
0
a
k
2
¯
+
b
k
2
¯
4Δf
G
n
f
=
lim
Δf
→
0
c
k
2
¯
4Δf
=
lim
Δf
→
0
a
k
2
¯
+
b
k
2
¯
4Δf
size 12{G rSub { size 8{n} } left (f right )= {"lim"} cSub { size 8{Δf rightarrow 0} } { { {overline {c rSub { size 8{k} } rSup { size 8{2} } }} } over {4Δf} } = {"lim"} cSub { size 8{Δf rightarrow 0} } { { {overline {a rSub { size 8{k} } rSup { size 8{2} } }} + {overline {b rSub { size 8{k} } rSup { size 8{2} } }} } over {4Δf} } } {}
(5)
- The power in the frequency range f1 to f2 is given by
P
f
1
→
f
2
=
∫
−
f2
−
f1
G
n
f
df
+
∫
f1
f2
G
n
f
df
=
2
∫
f1
f2
G
n
f
df
P
f
1
→
f
2
=
∫
−
f2
−
f1
G
n
f
df
+
∫
f1
f2
G
n
f
df
=
2
∫
f1
f2
G
n
f
df
size 12{P left (f rSub { size 8{1} } rightarrow f rSub { size 8{2} } right )= Int rSub { size 8{ - f2} } rSup { size 8{ - f1} } {G rSub { size 8{n} } left (f right ) ital "df"} + Int rSub { size 8{f1} } rSup { size 8{f2} } {G rSub { size 8{n} } left (f right ) ital "df"} =2 Int rSub { size 8{f1} } rSup { size 8{f2} } {G rSub { size 8{n} } left (f right ) ital "df"} } {}
(6)
P
T
=
∫
−
∞
∞
G
n
f
df
=
2
∫
0
∞
G
n
f
df
P
T
=
∫
−
∞
∞
G
n
f
df
=
2
∫
0
∞
G
n
f
df
size 12{P rSub { size 8{T} } = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {G rSub { size 8{n} } left (f right ) ital "df"} =2 Int rSub { size 8{0} } rSup { size 8{ infinity } } {G rSub { size 8{n} } left (f right ) ital "df"} } {}
(7)