King Fahd University of Petroleum & Minerals Comppg gputer ... · 2 Signals • A signal is a function representing information • Voice signal – microphone output • Video signal

1

King Fahd University of Petroleum & MineralsComputer Engineering Deptp g g p

COE 241 – Data and Computer CommunicationsTerm 122

2/18/2013 Dr. Ashraf S. Hasan Mahmoud 1

Dr. Ashraf S. Hasan MahmoudRm 22-420Ext. 1724 Email: [email protected]

Lecture Contents1. Fourier Analysis

a. Fourier Series Expansion b. Fourier Transform c. Ideal Low/band/high pass filters


2

Signals• A signal is a function representing

information• Voice signal – microphone output• Video signal – camera output• Etc.

• Types of SignalsA l ti l ti ti


• Analog – continuous-value continuous-time• Discrete – discrete-value continuous-time

• Digital – predetermined discrete levels – much easier to reproduce at receiver with no errors

• Binary – only two predetermined levels: e.g. 0 and 1

Example of Continuous-Value Continuous-time signal• s1(t) and s2(t) are two example of analog

signalsg

s (t)


t

s1(t)

t

s2(t)

3

Example of Discrete-Value Continuous-time signal• d1(t) and d2(t) are two example of discrete

signalsg• d1(t) – takes more than two levels • d2(t) – takes only two levels - binary

d (t)


t

d1(t)

t

d2(t)

Note d2(t) extends to ±∞

Time Domain Representation• Time domain representation – we plot

value (voltage, current, electric field

Applies to BOTH analog and digital signals

( g , ,intensity, etc.) versus time

• Can infer rate of change (speed or frequency) information – e.g. s2(t) seems faster than s1(t)• Using calculus terms: rate of change for s2(t) > rate of

change for s1(2)s1(t) d1(t)


t

t

s2(t)

t

1

t

d2(t)

4

Frequency - Bandwidth• s2(t) faster than s1(t)

• s2(t) contains higher frequencies than those


2contained in s1(t)

• s1(t) and s2(t) contain more than one frequency

• Minimum frequency = fmin

• Maximum frequency = fmax


• Bandwidth = Range of frequencies contained in signal

= fmax – fmin

Frequency – Bandwidth (2)• For our example signals, assume:

• S1(t): fmin = 10 Hz, fmax = 500 HzS2(t) f i 5 H f 1000 H


• S2(t): fmin = 5 Hz, fmax = 1000 Hz

• This means:• BW for s1(t) = 500 – 10 = 490 Hz• BW for s2(t) = 1000 – 5 = 995 Hz

• Note that: because s2(t) is “faster than” s1(t) it


• Note that: because s2(t) is faster than s1(t) it should contain frequencies higher than those in s1(t)

• E.g. s2(t) contains frequencies (500,1000] which do not exist in s1(t)

5

Frequency – Bandwidth (3)• Consider the discrete signals d1(t) and

d2(t)


• The function plots have points of infinite slope

• rate of change = ∞ frequency = ∞

• Therefore for signals that look like d1(t) and d2(t), fmax = ∞


2( ),• Furthermore, BW = ∞• Example:

• d2(t) contains frequencies from some minimum fmin Hz to fmax = ∞ Hz

Example of Signal BW• Consider the human speech

• Typically fmin ~ 100Hz• fmax ~ 3500 Hz

• BW of the human speech signal = 3100 Hz


6

Bandwidth for Systems• For a system to respond (amplify, process,

Tx, Rx, etc.) for a particular signal with all , , ) p gits details, the system should have an equal or greater bandwidth compared to that of the signal

• Example:


Example:• The system required to process s2(t) should

have a greater bandwidth than the system required to process s1(t)

Bandwidth for Systems (2)• Example 2: consider the human ear system

• Responds to a range of frequencies only• fmin = 20 Hz fmax = 20,000 Hz BW = 19,980 Hz• It does not respond to sounds with frequencies outside

this range

• Example 3: consider the copper wire • It passes (electric) signals only between a certain fmin

and a certain fmax


and a certain fmax• The higher the quality of the wire – the wider the BW

• More on Systems BW later!

7

Frequency Representation• How to represent signals and indicate

their frequency content?


• The X-axis: frequency (in Hertz or Hz)• What is the Y-axis then? – the answer will be

postponed!

BW f ( ) f f i f

fmax = 500 Hz


BW for s2(t): range of frequencies ffmin = 5 Hz fmax = 1000 Hz

BW for s1(t): range of frequencies ffmin = 10 Hz

Periodic Signals• A periodic signal repeats itself every T

seconds


• Period T seconds

• In calculus terms:• s(t) is periodic if s(t) = s(t+T) for any - ∞ < t < ∞

• For previous examples: s1(t), s2(t), and


p p 1( ), 2( ),d1(t) are not periodic – however, d2(t) is periodic

8

Periodic Signals (2)• A periodic signal has a FUNDAMENTAL

FREQUENCY – f0


Q 0• f0 = 1 / T – where T is the period

• A periodic signal may also has frequencies other than the fundamental frequency f0


BW for periodic s(t) ffmin fmax

fmin <= f0 <= fmax

x x x x x x

Periodic Signals (3)• Examples of other periodic signals:


s4(t)

t

s3(t)

T

t

s5(t)

~ t2


T

t

4( ) t

T

9

Energy/Power of Signals• Energy for any signal is defined as


where the integral is carried over ALL range of t

• In other words, Es is the area under the

= dttsEs

2)(


,absolute squared of the signal

• The unit of energy is Joules

Energy/Power of Signals (2)• Note that for periodic signal Es is equal to

infinity since it is defined on (-∞, ∞)


• However power is FINITE for these type of signals

• Power is defined as the average of the absolute squared of the signal, i.e.


• The unit of power is Joules/sec or Watt

=T

s dttsT

P0

2)(

1Note the integral can be performed on [0,T], [-T/2, T/2], or any continuous interval of length T

10

A VERY SPECIAL Analog Signal• A function of the form

s(t) = A cos(2πft + θ)

s (t)

AAcos(θ)


t

A

-APeriod T

Characteristics of COSINE • Completely specified by:

• Amplitude – AAmplitude A• Phase - θ• Frequency – f

• s(t = 0) = A cos(θ)• Periodic signal – repeats itself every T

seconds


seconds• T = 1 / f

• Time to review your trigonometry !!• E.g. sin(x) = cos(x-π/2)

11

Characteristics of COSINE (2)• Energy for this signal, Es = infinity• Power for this signal, Pg = A2/2g , g

• Note Pg is dependent only on the amplitude A• Exercise: Verify the above results using the power

formula

• It contains ONLY ONE frequency f • The “purest” form of analog signals


p g g

• Frequency representation:BW for s (t)

ffmin = fmax = f Hz

x

Characteristics of COSINE (3)• Very Useful Properties (f = 1/T)

=+T

dtft0

0)2cos( θπ

=+T

dtnft 0)2cos( θπ

=+T

dtftT 0

2 2/1)2(cos1 θπ

=+T

dtnftT

2 2/1)2(cos1 θπ


0 fT 0

)(

=≠

=T

mn

mndtmftnft

T 0 2/1

0)2cos()2cos(

1 ππ

12

Example of Cosine Functions• Y1(t) – has

• a frequency f of 2 ( )Hz (T = ½ sec)

• An amplitude of 3• PY1 = 32/2 = 4.5

Watts

• Y2(t) - has • a frequency f of 1

Hz (T =1/1 = 1 0

1

2

3Examples of analog contineous-time functions

valu

e

y1 = 3 cos(2*pi*2*t)

y2 = cos(2*pi*1*t)


Hz (T =1/1 = 1 sec)

• An amplitude of 1• PY2 = 12/2 = 0.5

Watts-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-3

-2

-1

time - t

v

Fourier Series Expansion• Can we use the basic cosine functions to

represent periodic signals?

ONLY FOR PERIODIC SIGNALS

p p g

• YES – Fourier Series Expansion

s3(t)∑

t


t

3( )

T

∑t

t

factorization

13

Fourier Series Expansion (2)• For a periodic signal s(t) can be

represented as a sum of sinusoidal psignals as in

[ ]∞

=

++=1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Ats ππ

where the coefficients are computed using:T2


= dttsT

A0

0 )(2

=T

n dttnftsT

A0

0 )2cos()(2 π =

T

n dttnftsT

B0

0 )2sin()(2 π

f0 is the fundamental frequencyof s(t) and is equal to 1/T

Fourier Series Expansion (3)• Another form for the series:

∞C =

++=1

00 )2cos(

2)(

nnn tnfC

Cts θπ

where the coefficients are computed using:

00 AC =


22nnn BAC +=

−= −

n

nn A

B1tanθ An

Bn

Cn

θn

14

Notes on Fourier Series Expansion• The representation (the sum of

sinusoids) is completely identical and equivalent to the original specification of s(t)

• It is applies to any periodic signal –analog or digital!


Very powerful tool – it reveals all frequencies contained in the original periodic signal s(t)

Notes on Fourier Series Expansion (2)• In general, s(t) contains

• DC term – the zero frequency term = A0/2DC term the zero frequency term A0/2• A (possibly infinite) number of harmonics (or

sinusoids) at multiples of the fundamental frequency, f0

• The contribution of a harmonic with frequency nf0 is proportional to |A 2 B 2| C 2


|An2+Bn

2| or Cn2

• E.g. if Cn2 ~ 0, then we say the harmonic at nf0 (or

higher does not contribute significantly towards building s(t) – more on this when we discuss total power!

15

Notes on Fourier Series Expansion (3)• A harmonic with frequency equal to nf0

(n>0), has a period of 1/(nT)( ), p /( )

• In general the series expansion of s(t) contains INFINITE number of terms (harmonics)

• However for less than 100% accurate


• However for less than 100% accurate representation one can ignore higher terms – terms with frequencies greater than certain n*f0

Notes on Fourier Series Expansion (4)• Lets define the following function:

s e(n=k)s_e(n=k)

To be the series expansion of s(t) up to and including the n = k term

It should be noted that s_e(n=k) is periodic with period T

• Examples:


2/)0(_ 0Anes ==

)2sin()2cos(2/)1(_ 01010 tfBtfAAnes ππ ++==)2cos(2/ 1010 θπ ++= tfCA

16

Notes on Fourier Series Expansion (5)• Examples – cont’d:

)2sin()2cos(2/)2( 01010 tfBtfAAnes ++== ππ)22sin()22cos(

)2sin()2cos(2/)2(_

0202

01010

tfBtfA

tfBtfAAnes

×+×+++

ππππ

)22cos()2cos(2/ 2021010 θπθπ +×+++= tfCtfCA

∞A


[ ]=

++=∞=1

000 )2sin()2cos(

2)(_

nnn tnfBtnfA

Anes ππ

∞

=

++=1

00 )2cos(2/n

nn tnfCA θπ

Notes on Fourier Series Expansion (6)• It is obvious that s(t) is 100%

represented by s_e(n=∞)

• s_e(n = n* < ∞) produces a less than 100% accurate representation of the original s(t)

i l i di i l


• For most practical periodic signals s_e(n=10) provides a more than enough accuracy in representing s(t)

• No need for infinite number of terms

17

Example 1:• Consider the following s(t)

s (t)

t

A

-T/4 T/4 3T/4 5T/4T

• Over one period, the signal is defined as

s(t) = A -T/4 < t <= T/4= 0 T/4 < t <= 3T/4

Fi di th S i E i


• Finding the Series Expansion:• The DC term A0

A

AT

Tdtts

TA

T

T

=

××== − 2

2)(

2 4/

4/

0

Example 1: cont’d• The term An:

4/4/ TT

−−

==4/

4/

0

4/

4/

0 )2cos(2

)2cos()(2 T

T

T

T

n dttnfT

Adttnfts

TA ππ

)2

sin(24/

4/)2sin(

2

20

0

ππ

ππ

n

n

A

Tt

Tttnf

Tnf

A ××=−=

==

= 6420 n


=−

=

=

=

,...11,7,32

,...9,5,12

,...6,4,20

nn

A

nn

An

π

πRemember1. f0 = 1/T2. ∫ cos(ax) dx = -1/a sin(ax)3. sin(nπ) = 0 for integer n4. sin(nπ/2) = 1 for n=1,5,9, …5. sin(nπ/2) = -1 for n=3,7,11, …

18

Example 1: cont’d• Therefore An is given by:

( )

=×−

== − ,...7,5,3,1

2)1(

,...6,4,202/1 n

n

An

n

π


Remember(-1)(n-1)/2 = 1 for n = 1,5,9, …

= -1 for n = 3,7,11, …

Example 1: cont’d• The term Bn:

4/4/ TT

−−

==4/

4/

0

4/

4/

0 )2sin(2

)2sin()(2 T

T

T

T

n dttnfT

Adttnfts

TB ππ

−−×−=

−==−= )

2cos()

2cos(

2

4/

4/)2cos(

2

20

0

πππ

ππ

nn

n

A

Tt

Tttnf

Tnf

A

0=


Remember1. ∫ cos(ax)dx = -1/a sin(ax)2. cos(x) = cos(-x)

19

Example 1: cont’d• Therefore, the overall series expansion is

given byg y( )

)2cos()1(2

2)( 0

5,3,1

2/1

tnfn

AAts

n

n

ππ

×−+= ∞

=

−

)32cos(3

2)2cos(

2

2)( 00 ×−×+= tf

Atf

AAts ππ


...)72cos(7

2)52cos(

5

232

00

00

+×−××+ tfA

tfA π

ππ

π

ππ

Example 1: cont’d• Original s(t) and

the series up to and including n = 0

1.2

i i l ( )= 0

• i.e. Comparing:

s(t)

vs.

s_e(n=0) = A/20.4

0.6

0.8

1

original s(t)up to n = 0


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2

20


the series up to and including n = 1.2

original s(t)g1

• i.e. Comparing:

s(t)

vs.0.4

0.6

0.8

1



s_e(n=1) =A/2 +

2A/π cos(2πf0t) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2


the series up to and including n = 1.2

original s(t)g3

• i.e. Comparing:

s(t)

vs.0.4

0.6

0.8

1



s_e(n=3) = A/2 + 2A/πcos(2πf0t) –2A/(3π)cos(2π3f0t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2

21


the series up to and including n = 11

1.2

original s(t)11

• i.e. Comparing:

s(t)

vs.

s_e(n=11) =0.4

0.6

0.8

1



A/2 + 2A/πcos(2πf0t) – 2A/(3π)cos(2π3f0t) + 2A/(5π)cos(2π5f0t) – 2A/(7π)cos(2π7f0t)

+2A/(11π)cos(2π11f0t)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.2

0

0.2

Example: cont’dclear all

T = 1;A = 1;t = 1:0 01:1;

•The matlab code for plotting and t = -1:0.01:1;n_max = 11;

s = (A*square(2*pi/T*(t+T/4))+A)/2;

figure(1)plot(t, s);gridaxis([0 1 -0.2 1.2]);

s_e = A/2*ones(size(t));

for n=1:2:n_max

evaluating the Fourier Series Expansion•This code builds the series incrementallyusing the “for” loop• Make sure you study this code!!


s_e = s_e + (-1)^((n-1)/2) * 2*A/(n*pi) * cos(2*pi*n/T*t);end

figure(2)plot(t, s,'b-', t, s_e,'r--');axis([0 1 -0.2 1.2]);legend('original s(t)', 'up to n = 11');grid

22

Notes Previous Example• The more terms included in the series

expansion the closer the representation to the original s(t)

• i.e. comparing s(t) with s_e(n=n*), the greater the n* the closer the representation is

• How to measure “closeness”?


• How to measure closeness ?

• Answer: Let’s use power!!

Power Calculation Using Fourier Series Expansion• Rule: if s(t) is represented using Fourier

Series expansion, then its power can be p , pcalculated using:

[ ]∞

=

++==1

222

0

0

2

2

1

4)(

1

nnn

T

s BAA

dttsT

P


∞

=

+=1

22

0

2

1

4 nnC

A

23

Power Calculation Using Fourier Series Expansion (2)• The previous result is based on the

following two facts:

• (1) For f(t) = constant power of f(t) = constant2

Proof:


Proof:

power = 1/T X ∫0Tconstant2 dt

= 1/T X constant2 X T= constant2 Watts

Power Calculation Using Fourier Series Expansion (3)• The previous result is based on the following

facts (continued):

• (2) For f(t) = A cos(2πnf0t+θ) power of f(t) = A2/2

Proof:

+==TT

f dttnfA

dttfP 02

22

)2(cos)(1 θπ


f fT

fT 0

0

0

)()(

++=T

dttnfT

A

0

0

2

)]24cos(2

1

2

1[ θπ

2]0

2[

22 AT

T

A =+=

24

Example 2:• Problem: What is the power of the signal

s(t) used in previous example? And find ( ) p pn* such that the power contained in s_e(n=n*) is 95% of that existing in s(t)?

• Solution:


Solution:Let the power of s(t) be given by Ps

22

2

0

25.0

22

1)(

1A

ATA

Tdtts

TP

T

s ==××==

Example 2: cont’d• Now it is desired to compute the power

using the Fourier Series Expansiong p

• What is the power in s_e(n=0) = A/2?• Ans: we apply the power formula:

T

21


===nes dtnesT

P0

2

)0(_ )0(_1

222

25.044

1A

AT

A

T==××=

25

Example 2: cont’d• What is the power in

s e(n=1) = A/2 + 2A/π cos(2πf0t)s_e(n=1) = A/2 + 2A/π cos(2πf0t)• Ans: we can use the result on slide Power

Calculation Using Fourier Series Expansion:

222 2

)1(1 AA

dtnesPT

+


20

)1(_ 4)1(_

πdtnes

TP nes +=== =

222

4526.02

4

1AA =

+=

π


s e(n=3) = A/2 + 2A/π cos(2πf0t) –s_e(n=3) = A/2 + 2A/π cos(2πf0t) 2A/(3π) cos(2π3f0t)

• Ans: we can use the result on slide Power Calculation Using Fourier Series Expansion:

2222 22

)3(1 AAA

dtnesPT

++


220

)3(_ 94)3(_

ππdtnes

TP nes ++=== =

2222

4752.09

22

4

1AA =

++=

ππ

26


s e(n=5) = A/2 + 2A/π cos(2πf0t) –s_e(n=5) = A/2 + 2A/π cos(2πf0t) 2A/(3π) cos(2π3f0t) + 2A/(5π) cos(2π5f0t)


22222 222

)5(1 AAAA

dtnesPT

+++


2220

)5(_ 2594)5(_

πππdtnes

TP nes +++=== =

22222

4833.025

2

9

22

4

1AA =

+++=

πππ


( ))(

)1(2)(

2/1

fAA n−

∞ −


∞

∞= +=∞==22

222

)(_

12

4)(_

1 T

nes n

AAdtnes

TP

π

)2cos()1(2

2)(_ 0

5,3,1

tnfn

AAnes

n

ππ

×+=∞= =


=10 4 n nT π22

122

5.012

4

1AA

nn

=

+=

∞

=π

This the EXACT SAME power contained in s(t) –This is expected since s(t) is 100% represented by s_e(n=∞)

27

Example 2: cont’ds_e(n=k) Expression Power % Power+

k = 0 A/2 0.25 A2 (0.25A2)/(0.5A2)

= 50%

k = 1 A/2 + 2A/πcos(2πf0t) 0.4526 A2 (0.4526A2)/(0.5A2)

= 90.5%

k = 2* A/2 + 2A/πcos(2πf0t) 0.4526 A2 90.5%

k = 3 A/2 + 2A/πcos(2πf0t) – 0 4752 A2 95 0%


k 3 A/2 + 2A/πcos(2πf0t) 2A/(3π)cos(2π3f0t)

0.4752 A 95.0%

k = 5 A/2 + 2A/πcos(2πf0t) –2A/(3π)cos(2π3f0t) + 2A/(5π)cos(2π5f0t)

0.4833 A2 96.7%

+ % power = power of s_e(n=k) relative to original power in s(t) which is equal to 0.5A2

* For k = 2, the expression s_e(n=k) is the same as that for s_e(k=1). Why?

Example 2: cont’d

• Therefore, s_e(n=n*) such that 95% of power is contained n* = 3


28

Power Spectral Density Function• Fourier Series Expansion:

• Specifies all the basic harmonics contained inSpecifies all the basic harmonics contained in the original function s(t)

• Cn2/2 = (An

2+Bn2) /2 determines the power

contribution of the nth harmonic with frequency nf0

• The power Spectral Density function is a function specifying: how much power is


function specifying: how much power is contributed by a given frequency

Power Spectral Density Function (2)• Typical PSD function for

periodic signals:

Periodic s(t)

(1) Fourier Series p g

(A0/2)2

C12/2

C 2/2

PSD(f)

∞

=

++=1

00 )2cos(

2)(

nnn tnfC

Ats θπ

∞

=

+==1

22

0

2

0 2

1

4)(

1

nn

T

s CA

dttsT

P

(2) Power calculation

Expansion


frequency0 f0 2f0 3f0 4f0

C22/2

C32/2

C42/2f0 f0

=10 n

(3) Plotting power versus frequency

29

Power Spectral Density Function (3)• A mathematical expression for PSD(f) can be

written as fA 04/2

×==

=otherwise

fnfC

fA

fPSD n

0

2/

04/

)( 02

0

• Another way (more compact) of writing PSD(f) is as follows:

∞

22

0 1A


=

−×+×=1

020 )(

2

1)(

4)(

nn nffCf

AfPSD δδ

where δ(t) is defined by

≠=

=00

01)(

f

ffδ

Power Spectral Density Function (4)

• δ(f) is referred to as the dirac function• δ(f) is referred to as the dirac function or unit impulse function

δ(f)

1


f0

30

Note on the PSD Function• PSD function has units of Watts/Hz

• For periodic signals PSD is a discrete function - defined for integer multiples of the fundamental frequency

• Specifies the power contribution of every harmonic component Cn

2/2 nf0

Th ti b t th di t


• The separation between the discrete components is at least f0

• It is exactly f0 if all Cn’s are not zeros• E.g. for the previous s(t) example, Cn=0 for

even n separation = 2f0

Note on the PSD Function (2)• To calculate the total power of signal

Integrate PSD over all contained gfrequencies

• For discrete PSD: integration = summation

• Therefore total power of s(t),


Ps = (An/2)2 +∑ Cn2/2 in Watts

31

Example 3:

• Find the PSD function of the periodic signal s(t) considered in Example 1.considered in Example 1.

• From Example 1, s(t) is given by ( )

)2cos()1(2

2)( 0

5,3,1

2/1

tnfn

AAts

n

n

ππ

×−+= ∞

=

−

• Using Example 2:• Power at the zero frequency = (A/2)2 = A2/4


• Power at the nth harmonic (n odd) is equal to 2A2/(nπ)2

• Power at the nth harmonic (n even) is zero• Therefore the PSD function is given by

∞

=

−×+×=5,3,1

022

22

)(12

)(4

)(n

nffn

Af

AfPSD δ

πδ

Example 3: cont’d

• The PSD is plotted as shown (A = 1, T = 1)

0 1

0.15

0.2

0.25Power Spectral Density function for s(t) - A = 1, T = 1

∞

=

−×+×=5,3,1

022

22

)(12

)(4

)(n

nffn

Af

AfPSD δ

πδ


0 2 4 6 8 10 120

0.05

0.1

frequency - Hz

32

Example 3: cont’d• Matlab Code to plot PSD

clear allclear all

T = 1;A = 1;t = -1:0.01:1;n_max = 11;

Frequency = [0:1:n_max];PwrSepctralD = zeros(size(Frequency));

% Record the DC term power at f = 0PwrSepctralD(1) = (A/2)^2;


% Record the nth harmonic power at f = nf0for n=1:2:n_max

PwrSepctralD(n+1) = (2*A/(n*pi))^2 / 2;end

figure(1)stem(Frequency, PwrSepctralD,'rx');title('Power Spectral Density function for s(t) - A = 1, T = 1');xlabel('frequency - Hz');grid

The “stem” function is typically used to plot discrete functions

Example 4:• Problem: Consider the periodic half-wave

rectified signal s(t) depicted in figure.

This is a typical exam question

• Write a mathematical expression for s(t)• Calculate the Fourier Series Expansion for s(t)• Calculate the total power for s(t)• Find n* such that s_e(n*) has 95% of the total power• Determine the PSD function for s(t)• Plot the PSD function for s(t)

s(t)

2/18/2013 Dr. Ashraf S. Hasan Mahmoud 64t

s(t)

A

T-T

33

Example 4: cont’d• Answer:(a) To write a mathematical expression for ( ) p

s(t), remember that the general form of a sinusoidal function is given by

A cos(2π X Freq X t), orA cos(2π /Period X t)


Therefore s(t) is given by

s(t) = A cos(2π/T t) -T/4 < t ≤ T/4= 0 T/4 < t ≤ 3T/4

Example 4: cont’d• Answer:(b) The F S E of s(t): [ ]

∞

++= 0 )2sin()2cos()( tnfBtnfAA

ts ππ(b) The F.S.E of s(t):The DC term is given by

−−

×==4/

4/

4/

4/

0 )/2cos(2

)(2 T

T

T

T

dtTtT

Adtts

TA π

[ ])2/sin()2/sin()/2sin(4/

4/πππ −−=×= = A

TtA Tt

Tt

[ ]=

++=1

00 )2sin()2cos(2

)(n

nn tnfBtnfAts ππ


[ ])()()(4/ ππ −= Tt

πA2=

34

Example 4: cont’d• Answer:The An term is given by (remember 1/T = f0)

Remember:

sin(ax+bx) sin(ax-bx)∫[cos(ax)cos(bx)] dx = ------------- + -------------

2(a+b) 2(a-b)for a ≠ b

sin(a+/-b) = sin(a)cos(b) +/- cos(a)sin(b)

The An term is given by (remember 1/T = f0)

−−

×==4/

4/

0

4/

4/

0 )2cos()/2cos(2

)2cos()(2 T

T

T

T

n dttnfTtT

Adttnfts

TA πππ

( )( )( )

( )( )( )

4/

00

14

12sin

14

12sin2Tt

f

tfn

f

tfn

T

A=

−+++×= ππ For n ≠ 1


( ) ( )4/00 1414

TtfnfnT

−=

−+ ππ

( )( )

( )( )

−

−++

×=1

2/cos

1

2/cos

n

n

n

nA πππ

For n ≠ 1

This means: the n=1 should be special!

Example 4: cont’dBut

cos(nπ/2) = 0 n = odd, n≠1( ) ,= (-1)(1+n/2) n = even

Therefore

++ )2/1()2/1(


( ) ( )

−

−−++

−×=++

1

)1)(1(

1

)1( )2/1()2/1(

nn

AA

nn

n πFor n even

For n odd, n≠10=

35

Example 4: cont’dThe expression for An (for even n) can be

further simplified top

( ) ( )

−

−−++

−×=++

1

)1)(1(

1

)1( )2/1()2/1(

nn

AA

nn

n π

( ) ( )( )( )

−+

+−−+−−×=++

11

1)1)(1(1)1( )2/1()2/1(

nn

nnA nn

π


( ) ( ) ( )[ ]1)1(1)1(1

)2/1()2/1(2

+−−−−×−

= ++ nnn

A nn

π

( )1

)1(22

)2/1(

−−=

+

n

A n

π For n even

Example 4: cont’dAn is still not completely specified – we still need to

calculate it for n=1; in other words we need to calculate A1:calculate A1:

−−

= ×=××=4/

4/

0

4/

4/

01 )2cos()/2cos(2

)12cos()(2 T

T

T

T

n dttfTtT

Adttfts

TA πππ

×=4/

02

1 )2(cos2 T

dttfT

AA π

Therefore:


− 4/

01 )(T

fT

( ) ( ) ( )

−−+×=

×

+×==

−= 0

4/

4/

00 8

sinsin

4

24sin

24

1

2

2

f

T

T

Atf

f

t

T

ATt

Ttπ

ππππ

2

A=

36

Example 4: cont’dThis mean An is equal to the following:

A = 2A/π n = 0An = 2A/π n = 0

0 n odd, n≠1

A/2 n = 1

2A(-1)(1+n/2)

2 4 6


------------- n = 2, 4, 6, …π(n2-1)

The above expression specifies An for ALL POSSIBLE values of n specification is complete

Example 4: cont’dWe still need to compute Bn:

4/4/ TT

Remember:- cos(ax+bx) cos(ax-bx)

∫[sin(ax)cos(bx)] dx = ------------- - -------------2(a+b) 2(a-b)

for a ≠ b

−−

×==4/

4/

0

4/

4/

0 )2sin()/2cos(2

)2sin()(2 T

T

T

T

n dttnfTtT

Adttnfts

TB πππ

( )( )( )

( )( )( )

4/

4/0

0

0

0

14

12cos

14

12cos2Tt

Ttfn

tfn

fn

tfn

T

A=

−=

−

−−+

+×=π

ππ

πFor n ≠ 1

( )( ) ( )( ) ( )( ) ( )( )

2/18/2013 Dr. Ashraf S. Hasan Mahmoud 72This means: the n=1 should be special!

( )( ) ( )( )( )

( )( ) ( )( )( )

−

−−−−−+

+−++−×=1

12/cos12/cos

1

12/cos12/cos

2 n

nn

n

nnA πππππ

For n ≠ 10=

37

Example 4: cont’dBn is still NOT completely specified – we still need to

calculate it for n=1; in other words we need to calculate B :

Remember:

sin(2ax) =2 cos(ax) sin(ax)

calculate B1:

−−

= ×=××=4/

4/

0

4/

4/

01 )2sin()/2cos(2

)12sin()(2 T

T

T

T

n dttfTtT

Adttfts

TB πππ

×=×=4/

0

4/

001 )4sin()2sin()2cos(2 TT

dttfT

Adttftf

T

AB πππ

Therefore:


−− 4/

0

4/

001 )()()(TT

fT

ffT

[ ])cos()cos(4

)4cos(4

4/

4/0 πππ

ππ

−−×−=×−= =

−=

Atf

A Tt

Tt

0= This means Bn = 0 for all n

Example 4: cont’d• To summarize:

An = 2A/π n = 0

0 n odd, n≠1

A/2 n = 1

2A(-1)(1+n/2)

------------- n = 2, 4, 6, …π(n2-1)

And


Bn = 0 for all n

• Having computed An and Bn we are now in a position to write the Fourier Series Expansion for s(t)

38

Example 4: cont’d• The Fourier Series Expansion for s(t) is

given byg y

[ ]∞

=

++=1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Ats ππ

∞ +

−−++=

64202

)2/1(

0 )2cos(1

)1(2)2cos(

2

n

tnfn

Atf

AA ππ

ππ


= 6,4,2 12 n nππThe Cn terms (there is a typo in the textbook) are as follows:

C0 = A/π 2A(-1)(1+n/2)

C1 = A/2 Cn = --------------, n = 2, 4, 6, …π(n2 – 1)

0 n odd, n≠1

Example 4: cont’dPlot for s(t) and

the Fourier Series E i f

Remember:

sin(2ax) = ½ cos(ax) sin(ax)

1.2Original signal and its Fourier Series Expansion

orignal s(t)s

e(k=0)

s (k=1)Expansion for k=0, 1, 2, and 4

Note: As k increases s_e(n=k) approaches the original s(t)

0.4

0.6

0.8

1

Am

plitu

de

se(k 1)

se(k=2)

se(k=4)


-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2

0

0.2

Time (sec)

39

Example 4: cont’d• The total power of s(t) is given by:

−−

×==4/

4/

224/3

4/

2)/2(cos)(

1 T

T

T

T

s TtT

Adtts

TP π

4/

4/

2

/8

)/4sin(

2

Tt

TtTt

Ttt

T

A=

−=

+×=

ππ


4

2A=

Therefore total power of s(t) = 0.25 A2

Example 4: cont’d• To find n* such that power of s_e(n=n*) = 95%

of total power:

s_e(n=k) Expression Power % Power+

k = 0 A/π 0.1013 A2 (0.1013A2)/(0.25

A2) = 40.5%

k = 1 A/π + A/2 cos(2πf0t) 0.2263 A2 (0.2262A2)/(0.25A2)

= 90.5%

k 2 A/ A/2 (2 f t) 0 2488 A2 (0 2488A2)/(0 25A2)


k = 2 A/π + A/2 cos(2πf0t) + 2A/(3π) cos(2π2f0t)

0.2488 A2 (0.2488A2)/(0.25A2)

99.5%

Therefore n* = 2 power of s_e(n=2) = 0.2488 A2

which is 99.5% of total power of s(t)

40

Example 4: cont’d• The PSD function for s(t) is as follows:

• Power for DC term = (A/π)2Power for DC term (A/π)• Power for harmonic at f = f0: (A/2)2/2 = A2/8• Power for harmonic at f = nf0 (n=2,4,6, …):

[2A/(π(n2-1))]2/2 = 2A2/(π(n2-1))2

• Therefore PSD function equals to


q

( )∞

= −−+−+

=

6,4,222

02

2

0

22

1

)(2)(

8)()(

n n

nffAff

Af

AfPSD

δπ

δδπ

Example 4: cont’d• Plot of

The PSD 0.14Power spectral density for s(t)

function for s(t)

0.06

0.08

0.1

0.12

Pow

er -

Wat

ts


0 2 4 6 8 10 120

0.02

0.04

Multiples of fundamental freuqncy f (Hz)

41

Fourier Transform• Fourier Series Expansion analysis is

applicable for PERIODIC signals ONLYpp g• There are important signals that are not

periodic such as• Your voice waveform• Pulse signal p(t) – used for modulation and

transmission


• Examples: p1(t) and p2(t)

t

p1(t)

τ/2−τ/2 t

p2(t)

A

τ/2−τ/2

Fourier Transform (2)• How to find the frequency content of

such signals?g

• Use FOURIER TRANSFORM

∞

dfX jftπ2)()(


∞−

−= dtetxfX jftπ2)()(

∞

∞−

= dfefXtx jftπ2)()(

42

Notes on Fourier Transform• F.T describes a two-way transformation

x(t) X(f)

where x(t) is the time representation of the signal, while X(f) is the frequency representation of the signal


• X(f) is defined on a continuous range of frequencies

• All frequencies within the range of X(f) where X(f) is not zero contribute towards building x(t)

Notes on Fourier Transform (2)• The magnitude of the contribution of a

particular frequency f* in x(t) is proportional to |X(f*)|2|X(f*)|2

• Example: Consider the F.T. pair shown below –clearly frequencies belonging to (-1/τ,1/τ) contribute significantly more compared to frequencies belonging to (1/τ,∞) or (-∞,-1/τ)


t

x(t)

τ/2−τ/2

F.T.

f0

X(f)

1/τ-1/τ

43

Properties of Fourier Transform• If x(t) is time-limited X(f) is not

frequency-limitedfrequency limited• i.e. the range of X(f) = (- ∞, ∞)

• X(f) is complex-valued (has magnitude and phase) in general, i.e. ( ) ∈ ℂ


i.e.

• If x(t) is a real-valued symmetric X(f) is real-valued, i.e.

( ) ∈ ℂ( ) ∈ ℝ

Relation between Fourier Series Expansion and Fourier Transform

• Consider the following two signals:(t)

t

x(t)

τ/2−τ/2

X(f)

t

s (t)

A

-T/4 T/4 3T/4 5T/4T

S(f)

x(t) is the same as s(t) except that its period T = ∞


f0

X(f)

1/τ-1/τf01/T-1/T

2/T

3/T4/T-2/T

-3/T

-4/T

For S(f) Separation between the discrete frequencies equal to f0 or 1/T

Separation become zero as T ∞ and the discrete function S(f) continuous function X(f)

44

Relation between Fourier Series Expansion and Fourier Transform (2)

• The separation between spectral lines for a periodic signal is 1/Tlines for a periodic signal is 1/T

• As T infinity and s(t) becomes non periodic the separation between spectral lines zero (i.e. it becomes continuous)


Example 5:• Problem: Consider the square pulse

function shown in figure:function shown in figure:• Write a mathematical expression for

p(t)• Find the Fourier transform for p(t)• Plot P(f)

p(t)


t

A

-τ/2 τ/2

45

Example 5: cont’d• Answer: p(t) can be expressed as

p(t) = A |t| ≤ τ/2= 0 otherwise

Th F T f (t) P(f) i i b


The F.T. for p(t), P(f) is given by

∞

∞−

−= dtetpfP jftπ2)()(

Example 5: cont’d• Which is equal to

τ2∞

P(f) is real-valued – for this specific example of p(t); why?

dteAdtetpfP jftjft π

τ

π 2

2

2)()( −

−∞−

− ==

( )τπτπ

τ

τ

π

ππjfjfjft ee

jf

Adte

jf

A −×−=−

= −

−

− 22

2

2

2

( )eeA jfjf τπτπ −


( )j

ee

f

A jfjf

2π−×=

τπτπτ

f

fA

)sin(=

Remember: Euler identity :ejx = cos(x) + j sin(x), ORcos(x) = (ejx+e-jx)/2sin(x) = (ejx-e-jx)/(2j)

46

Example 5: cont’d• P(f) plot for A = 1 and τ

= 1

• Note: • P(f) is define on (-∞, ∞)• P(f) is continuous

(except at f = 0 Hz)

• P(f) = ZERO for f = n/τ

• For practical pulses 0.2

0.4

0.6

0.8

1Fourier Transform of p(t) - A = 1, Taw = 1


• For practical pulses –P(f) approaches zero as f ±∞

• Most of the energy of p(t) is contained in the period of (-1/τ, 1/τ)

-5 -4 -3 -2 -1 0 1 2 3 4 5-0.4

-0.2

0

frequency - Hz

Energy Spectral Density Function (ESDF)• ESDF is defined as ( ) = 1 ( ) ∗( )

where P(f) is the F.T. of the pulse p(t). P*(f) is the complex conjugate of P(f).

• ESDF is a measure of how much energy is contained at a particular frequency f

( ) = 2 ( ) ( )


p q y

• Units of ESDF is Joules per Hz

• How would you compare ESDF with PSDF?

47

Example 5 (For A = 1 Volts and Taw = 1 sec) - cont’d

0.8

1

ts)

0.8

1

(a) – pulse (time domain) (b) – Magnitude of F.T (freq. domain)

-1 -0.5 0 0.5 10

0.2

0.4

0.6

time - (sec)

am

plid

e (

volt

-4 -2 0 2 40

0.2

0.4

0.6

frequency - (Hz)

ab

s(P

(f))

0.14

0.16

(c) – ESDF (freq. domain)( ) = | | < 2⁄0 th i (a)

2/18/2013 Dr. Ashraf S. Hasan Mahmoud 93-4 -2 0 2 40

0.02

0.04

0.06

0.08

0.1

0.12

frequency - (Hz)E

SD

P (

Jou

les/

Hz)

( ) = 12 ( ) ∗( ) = 12 sin( ) 2

( ) 0 otherwise( ) = ( ) −2∞−∞ = sin( )

(b)

(c)

( )

Example 5 (For A = 1 Volts and Taw = 1 sec) – Matlab Code

clear all; LineWidth = 3; FontSize = 14;

• Code for producing plots on previous slide

clear all; LineWidth = 3; FontSize = 14;%Example of rectangular pulseA = 1; Taw = 1; % parameters for the rectangle pulset_step = 0.01; f_step = 0.01; Nmax = 4;t = -Taw:t_step:Taw; % define the time axisf = -Nmax/Taw:f_step:Nmax/Taw; % define the frequency axis

p_t = A*rectpuls(t/Taw); % The rectangle pulse of height A and width TawP_f = A*Taw*sin(pi*Taw*f)./(pi*Taw*f); % The corresponding F.T P(f)ESDF_f = P_f.*conj(P_f)/(2*pi); % The ESDF

figure(1); clf; set(gca, 'FontSize', FontSize);h = plot(t, p_t, '-r', 'LineWidth', LineWidth);


xlabel('time - (sec)'); ylabel('amplide (volts)'); grid on;

figure(2); clf; set(gca, 'FontSize', FontSize);h = plot(f, abs(P_f), '-r', 'LineWidth', LineWidth);xlabel('frequency - (Hz)'); ylabel('abs(P(f))'); grid on;

figure(3); clf; set(gca, 'FontSize', FontSize);h = plot(f, ESDF_f, '-r', 'LineWidth', LineWidth);xlabel('frequency - (Hz)'); ylabel('ESDP (Joules/Hz)'); grid on;

48

Example 6:• Repeat example 5 for triangular pulse shown in

textbook Figure A.2 page 839. Let be equal to 2 seconds and be equal to 3 volts.


Example 7:• Problem: If the FSE expansion for the function

g(t) is as given below. Compute the FSE for the function s(t) without computing the FSE coefficients for s(t).

t

g(t)

A

-Τ/2 Τ/2 t

s(t)

A

-Τ/2 Τ/2

2/18/2013 Dr. Ashraf S. Hasan Mahmoud

49

Z-Transform• Digital signals

• Discrete-time signals – sampled continues-time signals• E.g. I/O signal used by micro-controllers

• Systems described by difference equations• E.g. CD player contains digital signal processing

system (digital filter) to manipulate the digital audio signal

• For such systems input and output are


• For such systems input and output are related by difference equations and the Z-transform is used to solve these systems

Z-Transform - Introduction• For Modern systems

processing – filtering noise, equalizing music, ddi ff di / id

A/DSignal

processorD/A

analogsignal

analogoutput

samples samples


adding effects to audio/video, etc.

50

Z-Transform - Definition• Let y(t) be a continuous-time signal (function)• Define T – sampling period

• Then – y(k) = yk = y(kT) for k = 0, 1, 2, 3, …defines uniformly spaced samples of the original signal y(t)

• The Z-Transform for yk is defined as


if the summation converges

= ( ) = −∞=−∞

Z-Transform – Example 1• Compute the Z-transform for the

sequence ( ) = 0 0.8

1

y0 = 1 and a = 0.5

for k = 0, 1, 2, … and 0<a<1• Compute the Z-transform for the

sequence• A plot for y(k) is shown for 0< a <

1• Note the y(k) is defined only on

specific indices (time samples) 0 1 2 3 4 50

0.2

0.4

0.6

sample index (k)

sig

na

l y(k

)

A l i th d fi iti


• Applying the definition

for |a/z| < 1.

( ) = −∞=−∞ = 0 −∞

=0 = 0 ( ⁄ )∞=0 = 01 − ⁄ = 0−

0 ↔ 0− pair (1)

for k=0, 1, 2, …

51

1.5

2


sequence ( ) = 2(0.9) y0 = 2 and a = 0.9

0 1 2 3 4 50

0.5

1

sample index (k)

sig

na

l y(k

)for k = 0, 1, 2, … and 0<a<1• Compute the Z-transform for the

sequence• A plot for y(k) is shown for 0< a <

1

U i th i lt


• Using the previous result

for |0.9/z| < 1.

( ) = 21 − 0.9⁄ = 2− 0.9

Z-Transform – Example 3• Compute the Z-transform for

the sequence( ) = 1 = 0 0.8

1

Discrete-time Delta Function

• Compute the Z-transform for the sequence

( ) = 0 ≠ 0

U i th d fi iti

-5 0 50

0.2

0.4

0.6

sample index (k)

sig

na

l δ(k

)


• Using the definition

for all z.

Δ( ) = ( ) −∞=−∞ = (0) 0 = 1

0 ( ) ↔ 0 pair (2)

52

Z-Transform – Example 3• Compute the Z-transform for

the sequence( ) = 1 = 0, 1, 2, …Discrete-time Unit-Step Function

0.8

1


( ) = 0 otherwise


-5 0 50

0.2

0.4

0.6

sample index (k)

sig

na

l δ(k

)



for |1/z| < 1.

( ) = ( ) −∞=−∞ = −∞

=0 = 0 + −1 + −2 + … = 11 − 1⁄ = − 1

0 ( ) ↔ 0− 1 pair (3)


sequence

y(k) = [2 4 6 4 2 0]

Discrete-time arbitrary function

4

5

6

y(k) = [2, 4, 6, 4, 2, 0]

for k = 0, 1, 2, 3, 4, and 5, respectively



0 1 2 3 4 50

1

2

3

4

sample index (k)

sig

na

l δ(k

)



for |1/z| < 1.

( ) = ( ) −∞=−∞ = 2 0 + 4 −1 + 6 −2 + 4 −3 + 2 −4 + 0 −5

53

Inverse Z-Transform – Example 5• Compute the Inverse Z-Transform for( ) = 2• We use pair (1) y0 = 2 and a = 3

Or y(k) is given by

for k = 0, 1, 2, …

( ) = − 0.3= 2(0.3)


y(k) may be rewritten as= 2(0.3) ( )

Inverse Z-Transform – Example 6• Compute the Inverse Z-Transform for( ) = 10• Solution?

( ) = ( − 0.3)( − 0.9)


Documents

King Fahd University of Petroleum & Minerals Comppg gputer ... · 2 Signals • A signal is a function representing information • Voice signal – microphone output • Video signal