Noise
Noise is like a weed. Just as a weed is a plant
where you do not wish it to be, noise is a signal
where you do not wish it to be. The noise signal
interferes with the signals of interest.
If we knew exactly what the interfering noise signal
was, we could simply subtract it from the original
signal. (Assuming that the noise is additive.)
Since we generally do not know what the noise
signal is, we must know something about its nature
in order to attenuate its interfering effect on other
signals.
Power Spectral Density
One thing that we typically know about noise is
something about its spectrum.
If the noise signal is n(t), the noise spectrum is N(f).
Instead of working with the noise spectrum [N(f)], we
typically work with the noise power spectral
density. The notation for the power spectral density
of the noise n(t) is
Sn ( f ).
The power spectral density of the noise is the power
per Hz of the noise. Dimensionally,
Power = (Power Spectral Density) x (Bandwidth).
In general, when the power spectral density is not
constant, we can find the power from the power
spectral density using

P   Sn ( f ) df .

Example: Find the power when the power spectral
density is equal to
| f |
Sn ( f )  e .
Solution:

P  e
| f |

e
f 0

0
df   e

e
f 
0
(  f )

df   e df
f
0
 [1  0]  [0  (1)]  2.
Example: Suppose the noise in the previous
example is passed through a low-pass filter with
cutoff frequency of 10 Hz. What is the output noise
power?
Solution:
10
P
e

| f |
10
e
f 0
10
0
df   e
10
e
 f 10
0
(  f )
10
df   e df
f
0
 [1  e
 2(1  e 10 )  1.999.
10
]  [e
10
 (1)]
Example: The noise power spectral density is
N0
Sn ( f ) 
.
2
This noise is passed through a filter whose
bandwidth is B. Find the output noise power.
Solution:
B
N0
N0
P 
df 
(2B)  N0 B.
2
2
B
This kind of noise has a (power) spectrum which is
constant over all frequencies. For this reason, the
noise is called white noise.
Example: Find the power when the power spectral
density is equal to
Solution:
1
Sn ( f ) 
.
2
1 f

1
P 
df
.
2
1

f

This integral can be evaluated using trigonometric
substitution. Let f = tanq.
 /2
1
2
P 
sec
q
d
q
2
1  t an q
 / 2
 /2

d
q


.

 / 2
Example: The noise power spectral density is
N0
Sn ( f ) 
.
2
This noise is passed through a filter whose
bandwidth is B and whose voltage gain is G. Find
the output noise power.
Solution: We proceed as before, but we introduce a
factor proportional to the power gain of the signal.
Power Gain = [Voltage Gain]2
For this example, the power gain is G2. The power
spectral density at the output of the filter is N0G2/2.
Thus the power is
B
N0 2
N0 2
2
P 
G df 
G (2B)  N0G B.
2
2
B
Exercise: White noise whose power spectral density
is N0/2 is input to a low-pass filter whose transfer
function (gain versus frequency) is
1
H( f ) 
1  jf
Find the output noise power. The gain-squared will
be H(f)H*(f). Use the results of previous integrations
in this lecture.
In future calculations, the spectrum of a signal or the
spectrum of noise will be given, an the powers will
need to be computed from these spectra.
Given the spectrum of a signal, the power may be
computed using one of two forms of a theorem
called Parseval’s theorem.
Parseval’s Theorem for
Fourier Series
Given a signal described by a Fourier series, the
power can be found from its Fourier series
coefficients by using something called Parseval’s
Theorem for Fourier series.
Let x(t) be a periodic signal described by a Fourier
series:
x(t ) 

X e
n  
n
jn 0t
.
The instantaneous power is proportional to x2(t):
2

jn 0t 
x (t )    X n e
 .
n 


2
This instantaneous power can also be expressed as
x (t ) 
2

X e
n  
n
jn 0t

X
m  
*  jm 0t
m
e
The term Xm* is the complex conjugate of Xm.
(The magnitude-squared of any complex number is
given by |x|2=xx*.)
.
This instantaneous power can now be expressed as
x (t ) 
2


 X
n   m  
n
*
m
X e
j ( n  m ) 0t
To find the average power, we integrate x2(t) and
divide by the time interval.
Pavg
1 T 2
  x (t )dt
T 0
.
Inserting the expression for x2(t) into the integral for
average power, we have
Pavg 



  X
T
1
T 0

n   m  

X
n   m  
n
X
n
*
m
X e
T

* 1
m T
0
j ( n  m ) 0t
e
dt
j ( n  m ) 0t
dt.
It can be shown that (exercise for the student):
T

1
T 0
e
j ( n  m ) 0t
1 (n  m),

0 (n  m).
This function acts like a selector function sifting out
those terms for which n=m. So,
Pavg 

X
n  
n
n
.
Example: Using Parseval’s theorem for Fourier
series, find the power in a simple sinewave
x(t )  cosct.
Solution: First, we convert cosct to complex
exponentials:
cosct 
1
2
e
jct
e
 jct
.
Thus we have, X1 = X-1=½. The power according to
Parseval’s theorem is
Pavg 
1
X
n  1
2
n

  
1 2
2
1 2
2
 .
This result should come as no surprise since the
average power is equal to the square of the rms
voltage, and the rms voltage is equal to 1/2.
1
2
Example: Find the average power in a squarewave
using Parseval’s theorem. Compare the result to
finding the average power by finding the average of
the square of the squarewave.
Solution: The squarewave has the following Fourier
series expansion:
2
x(t ) 
j

1 jn0t
e
.

n   n
n odd
Taking the sums of the squares of the Fourier series
components, we have
2
P 
 
2

4
1
   2


n    n 
1
 

n    n 
n odd
n odd

1
 2 2 
 n 1  n 
4
2

2
n odd

1
1
1


2
1





.
9
25
36
2
4

2
Letting
1
PN  2   
 n1  n 
8
N
2
n odd
We can calculate PN for various values of N.
PN
N
3
5
7
9
11
13
15
17
19
0.901
0.933
0.950
0.960
0.966
0.971
0.975
0.978
0.980
The series seems to converge to one.
Now we compute the average of the square of the
squarewave. In the graph below, the unsquared
squarewave is denoted in green; the squared
squarewave is denoted in blue (dashed). [(-1)2 = 1.]
x(t)
1
-1
x2(t)
We see that the value of the square of the
squarewave is one, and that the average value is
one.
Exercise: Find the average power in the following
signal:
x(t )  cost  12 cos2t.
Use Parseval’s theorem for Fourier series.
Parseval’s Theorem for
Fourier Transforms
Parseval’s theorem for Fourier transforms uses
Fourier transforms to calculate the energy in a
signal.
Energy is equal to power times time. If the power
is not constant, energy is equal to the integral of the
power over time.
Energy = (Power) x (Time).

E
x
(
t
)
dt
.

2

We can manipulate the integral expression by using
Fourier transforms:

E
x

2
(t ) dt 






j 2ft
X
(
f
)
x
(
t
)
e
dt
df




 x(t )  X ( f )e



 X ( f ) X ( f ) df

j 2ft
df dt


*
X
(
f
)
X
( f ) df




 X(f )
2
df .

So, generally speaking, the integral of the square of
the time-domain signal is equal to the integral of the
square of the corresponding frequency-domain
signal.
Example: Find the energy in the signal
t
x(t )  e u(t )
first, using the integral of x2(t), then using the integral
of X2(t).


E   x (t ) dt   e dt
2

 e
1
2
0
 2t 
0
 (1  0)  .
1
2

 e
1
2
1
2
 2t

e
0

1
X(f ) 
.
1  j 2f
E



1
X ( f ) df  
df
.
 1  ( 2f ) 2
Let 2f = tanq.
2
sec
q
E  21 
d
q
 / 2 1  tan2 q
1
1
 2   2 .
 /2
2
We see that Parseval’s theorem gives consistent
results in the computation of energy for both the time
and the frequency domain.
Exercise: Find the energy in the function
sint
x(t )  sinc t 
.
t
Use Parseval’s theorem for Fourier transforms: it is
impossible to integrate sinc t analytically.
The Quadrature
Decomposition of Noise
Very often noise is added to a modulated signal, that
is a signal whose frequency components are
centered about some high frequency fc.
The corresponding noise is usually passed through
a band-pass filter in the demodulation process.
The noise of interest is a band-pass signal.
A bandpass signal can be represented as a
modulated form of lowpass signals. The lowpass
signals are, effectively, upconverted to bandpass
signals.
If we have a bandpass noise signal n(t), it can be
represented in terms of lowpass signals nc(t) and
ns(t):
n(t )  nc (t ) cosct  ns (t ) sin ct.
The component nc(t) is called the inphase
component, and the component ns(t) is called the
quadrature component. Both are lowpass signals
that are upconverted by cosct and sinct
respectively.
Now, let us relate the lowpass signals with the
quadrature representation.
Taking the Fourier transform of the quadrature
representation of n(t), we have
N ( f )  12 N c ( f  f c )  N c ( f  f c )  21j N s ( f  f c )  N s ( f  f c )
 12 N c ( f  f c )  N c ( f  f c )  j 12 N s ( f  f c )  N s ( f  f c )
Let nc(t) and ns(t) have spectra which look like
Nc(f), Ns(f)
A
f
B
The spectrum of n(t) will look like what we see on
the following slide.
Re{N(f)}
A/2
f
-fc
Im{N(f)}
fc
f
The magnitude of the spectrum of N(f) is as follows:
|N(f)|
A/2
f
B
(The magnitude or height can be found by taking the
square-root of the sum of the squares of the real and
the imaginary parts of N(f).)
The power spectral density of n(t) is proportional to
the square of its spectrum.
Sn(f)
A2/2
f
B
The power in n(t) can be calculated by integrating
the power spectral density.
P = 2 [ B•A2/2] = BA2.
If we compute the power spectral density and the
power of nc(t) and ns(t) we get the same value:
Snc(f), Sns(f)
A2
f
B
Thus, we can conclude that the power in each the
in-phase and quadrature components, nc(t), ns(t),
are the same as that of the noise itself, n(t).
Exercise: Suppose we had narrow-band noise
whose power spectral density is N0/2. Find the
power spectral density magnitude of the in-phase
and quadrature-phase components.