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

).( fSn

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

.)(

dffSP n

Example: Find the power when the power spectral density is equal to

.)( || fn efS

.2)]1(0[]01[0

0

0

0)(||

ff

fff

ee

dfedfedfeP

Solution:

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?

.999.1)1(2

)]1([]1[

10

101010

0

0

10

10

0

0

10

)(10

10

||

e

eeee

dfedfedfeP

ff

fff

Solution:

Example: The noise power spectral density is

.2

)( 0NfSn

This noise is passed through a filter whose bandwidth is B. Find the output noise power.

.)2(22 0

00 BNBN

dfN

PB

B

Solution:

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

.1

1)(

2ffSn

.

1

12df

fP

Solution:

This integral can be evaluated using trigonometric substitution. Let f = tan.

2/

2/

2/

2/

22

.

sectan1

1

d

dP

Example: The noise power spectral density is

.2

)( 0NfSn

This noise is passed through a filter whose bandwidth is B and whose voltage gain is G. Find the output noise power.

.)2(22

20

2020 BGNBGN

dfGN

PB

B

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

Exercise: White noise whose power spectral density is N0/2 is input to a low-pass filter whose transfer function (gain versus frequency) is

jffH

1

1)(

Find the output noise power. The gain-squared will be H(f)H*(f). Use the results of previous integrations in this lecture.

Given the spectrum of a signal, the power may be computed using one of two forms of a theorem called Parseval’s theorem.

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.

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:

.)( 0

n

tjnneXtx

The instantaneous power is proportional to x2(t):

.)(2

2 0

n

tjnneXtx

.)( 00 *2

m

tjmm

n

tjnn eXeXtx

This instantaneous power can also be expressed as

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

.)( 0)(*2

n m

tmnjmn eXXtx

To find the average power, we integrate x2(t) and divide by the time interval.

T

avg dttxT

P0

2 )(1

n m

T tmnjTmn

T

n m

tmnjmnTavg

dteXX

dteXXP

.0

)(1*

0

)(*1

0

0

Inserting the expression for x2(t) into the integral for average power, we have

It can be shown that (exercise for the student):

).(0

),(10

)(1 0

mn

mne

T tmnjT

This function acts like a selector function sifting out those terms for which n=m. So,

.

n

n

navg XP

Example: Using Parseval’s theorem for Fourier series, find the power in a simple sinewave

.cos)( ttx c

Solution: First, we convert cosct to complex exponentials:

.cos 21 tjtj

ccc eet

Thus we have, X1 = X-1=½. The power according to Parseval’s theorem is

.212

212

21

1

1

2 n

navg XP

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.

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:

.12

)(

odd

0

nn

tjnenj

tx

Taking the sums of the squares of the Fourier series components, we have

.124

12

4

1412

361

251

91

2

odd1

2

2

odd odd

2

2

22

nn

nn

nn

n

nnP

N

nn

N nP

odd1

2

2

18

Letting

We can calculate PN for various values of N.

N PN

3 0.9015 0.9337 0.9509 0.960

11 0.96613 0.97115 0.97517 0.97819 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.]

1

-1

x(t) 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:

.2coscos)( 21 tttx

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).

.)(2

dttxE

We can manipulate the integral expression by using Fourier transforms:

dffXfX

dfdtetxfX

dtdfefXtxdttxE

ftj

ftj

)()(

)()(

)()()(

2

22

.)(

)()(

2

*

dffX

dffXfX

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

)()( tuetx t

first, using the integral of x2(t), then using the integral of X2(t).

.)01(

)(

21

21

021

0

221

0

22

eee

dtedttxE

t

t

.21

1)(

fjfX

.)2(1

1)(

2

2df

fdffXE

Let 2f = tan.

.tan1

sec

21

21

2/

2/ 2

2

21

dE

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

.t

tsin t sinc)(

tx

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):

.sin)(cos)()( ttnttntn cscc

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 cosct and sinct 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

)()()()(

)()()()()(

21

21

21

21

cscscccc

cscsjcccc

ffNffNjffNffN

ffNffNffNffNfN

Let nc(t) and ns(t) have spectra which look like

f

Nc(f), Ns(f)

The spectrum of n(t) will look like what we see on the following slide.

B

A

f

Re{N(f)}

f

Im{N(f)}

A/2

fc-fc

The magnitude of the spectrum of N(f) is as follows:

f

|N(f)|

B

A/2

(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.

f

Sn(f)

B

A2/2

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:

f

Snc(f), Sns(f)

A2

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.