Engineering and PhysicsUniversity of Central Oklahoma

Dr. Mohamed Bingabr

Ch2Signals and Signal Space

ENGR 4323/5323Digital and Analog Communication

Outline

• Size of a Signal• Classification of Signals• Useful Signals and Signal Operations• Signals Versus Vectors• Correlation of Signals• Orthogonal Signal Sets• Trigonometric and Exponential Fourier Series

Signal Energy and Power

Energy Signal

Power Signal

Energy

Power

𝐸𝐸𝑔𝑔 = �−∞

∞𝑔𝑔(𝑡𝑡) 2𝑑𝑑𝑡𝑡

𝑃𝑃𝑔𝑔 = lim𝑇𝑇→∞

1𝑇𝑇�−𝑇𝑇/2

𝑇𝑇/2|𝑔𝑔 𝑡𝑡 |2𝑑𝑑𝑡𝑡

Signal Classification

1. Continuous-time and discrete-time signals2. Analog and digital signals3. Periodic and aperiodic signals4. Energy and power signals5. Deterministic and probabilistic signals6. Causal and non-causal7. Even and Odd signals

Analog continuous Digital continuous

Analog Discrete Digital Discrete

PeriodicDeterministic Aperiodic Probabilistic

Useful Signal Operation

Time Shifting Time Scaling Time Inversion

1-1

12

t

x(t)

Useful Signals

Unit impulse Signal

Unit step function u(t)𝑑𝑑𝑑𝑑(𝑡𝑡)𝑑𝑑𝑡𝑡

= 𝛿𝛿(𝑡𝑡)

Signals Versus Vectors

By sampling, a continuous signal g(t) can be represented as vector g.

g = [ g(t1) g(t2) … g(tn)]

Vector ApproximationTo approximate vector g using another vector x then we need to choose c that will minimize the error e.

g = cx + e

Remember: Dot product: <g, x> = ||g||.||x|| cos θ

θ θ θ

Signals Versus Vectors

Value of c that minimizes the error

Signal Approximationg(t) = cx(t) + e(t)

x(t)

Correlation of Signals

Two vectors are similar if the angle between them is small.

Correlation coefficient

Note: Similarity between vectors or signals does not depend on the length of the vectors or the strength of the signals.

Example

ρ1=1ρ2=1ρ3=-1

ρ4=0.961ρ5=0.628ρ6=0

Which of the signals g1(t), g2(t), …, g6(t) are similar to x(t)?

Cross-Correlation Function

2

g(t-τ)

13 4

Autocorrelation Function

g(t)

g(t-τ)

Orthogonal Signal Sets

g = c1x1 + c2x2 + c3x3

Orthogonal Vector Space

g(t) = c1x1(t)+ c2x2(t) + … + cNxN(t)Orthogonal Signal Space

Parseval’s Theorem

Trigonometric Fourier Series

( ) ( ) ( )∑ ∑∞

=

∞

=

++=1 1

000 2sin2cosn n

nn tnfbtnfaatx ππ

( ) ( )∫=0

00

2cos2

Tn dttnftx

Ta π

( ) ( )∫=0

00

2sin2

Tn dttnftx

Tb π

( )∫=00

01

T

dttxT

a

Example

• Fundamental periodT0 = π

• Fundamental frequencyf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s

( ) ( ) ( )

( )

( )

. as amplitudein decrease and 1618 504.0 2sin2

1612 504.0 2cos2

504.0121

2sin2cos

202

202

20

20

10

∞→

+

==

+

==

≈

−−==

++=

∫

∫

∫

∑

−

−

−−

∞

=

nban

ndtnteb

ndtntea

edtea

ntbntaatf

nn

t

n

t

n

t

nnn

π

π

ππ

π

π

ππ0 π−π

1e-t/2

f(t)

( ) ( ) ( )

+

++

+= ∑∞

=122 2sin

16182cos

16121504.0

nnt

nnnt

ntf

To what value does the FS converge at the point of discontinuity?

00 aC =22

nnn baC +=

−= −

n

nn a

b1tanθ

Compact Trigonometric Fourier Series

We can use the trigonometric identitya cos(x) + b sin(x) = c cos(x + θ)

to find the compact trigonometric Fourier series

C0, Cn, and θn are related to the trigonometric coefficients an and bn as:

( ) ( )∑∞

=

++=1

00 2cosn

nn tnfCCtx θπdc component nth harmonics

Role of Amplitude in Shaping Waveform

( ) ( )∑∞

=

++=1

00 2cosn

nn tnfCCtx θπ

Role of the Phase in Shaping a Periodic Signal

( ) ( )∑∞

=

++=1

00 2cosn

nn tnfCCtx θπ

Compact Trigonometric

• Fundamental periodT0 = π

• Fundamental frequencyf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s

( ) ( )

nab

nbaC

aCn

nb

na

a

ntCCtf

n

nn

nnn

o

n

n

nnn

4tantan

1612504.0

504.01618 504.0

1612 504.0

504.0

2cos

11

2

22

0

2

2

0

10

−−

∞

=

−=

−=

+=+=

==

+

=

+

=

≈

−+= ∑

θ

θ

0 π−π

1e-t/2

f(t)

( ) ( )∑∞

=

−−+

+=1

1

24tan2cos

1612504.0504.0

nnnt

ntf

• The amplitude spectrum of x(t) is defined as the plot of the magnitudes |Cn| versus ω

• The phase spectrum of x(t) is defined as the plot of the angles versus ω

• This results in line spectra• Bandwidth the difference between the

highest and lowest frequencies of the spectral components of a signal.

Line Spectra of x(t)

)( nn CphaseC =∠

Line Spectra

nn

CC

n

n

4tan1612504.0 504.0

1

20

−−=

+==

θ0 π−π

1e-t/2

f(t)

( ) ( )∑∞

=

−−+

+=1

1

24tan2cos

1612504.0504.0

nnnt

ntf

f(t)=0.504 + 0.244 cos(2t-75.96o) + 0.125 cos(4t-82.87o) +0.084 cos(6t-85.24o) + 0.063 cos(8t-86.24o) + …

0.504

0.244

0.1250.084

0.063

Cn

ω0 2 4 6 8 10

ω

θn

-π/2

( ) ∑∞

−∞=

=n

ntfjneDtx 02π

D-n = Dn*

( ) ,....2,1,0 , 102 ±±== ∫ − ndtetx

TD

oT

ntfj

on

π

Exponential Fourier Series

To find Dn multiply both side by and then integrate over a full period, m =1,2,…,n,…∞

ntfje 02π−

Dn is a complex quantity in general Dn=|Dn|ejθ

Even Odd

|Dn|=|D-n| Dn = - D-n

D0 is called the constant or dc component of x(t)

• The line spectra for the exponential form has negative frequencies because of the mathematical nature of the complex exponent.

Line Spectra in the Exponential Form

Cn = 2 |Dn| n ≠ 0

Dn = Cn

...)2cos()cos()(

...||||

||||...)(

2021010

2221

012

2

001

0102

+++++=

++

++++= −−−

−−−

θωθω

ωθωθ

ωθωθ

tCtCCtx

eeDeeD

DeeDeeDtxtjjtjj

tjjtjj

D0= C0

D-n = - Cn

Example

• Fundamental periodT0 = 2π

• Fundamental frequencyf0 = 1/T0 = 1/2π Hzω0 = 2π/T0 = 1 rad/s

π/2−π/2

1f(t)

−π π 2π−2π

Find the exponential Fourier Series for the square-pulse periodic signal.

=−≠

=

=

=

==

= ∫−

−

,15,11,7,3,15,11,7,3 allfor 0

odd /1even 0

21

)2/sinc(5.02/sin

21

0

2/

2/

nn

nnn

D

D

nn

n

dteD

n

n

jntn

πθ

π

πππ

π

π

π

|Dn|

Dn

1 1

1 1

Exponential Line Spectra

Example

=−≠

=

=

=

,15,11,7,3,15,11,7,3 allfor 0

odd 2

even 021

0

nn

nn

nC

C

n

n

πθ

π

π/2−π/2

1f(t)

−π π 2π−2π

The compact trigonometric Fourier Series coefficients for the square-pulse periodic signal.

[ ]∑∞

=

−

−−++=

1

2/)1(

21)1(cos2

21)(

n

nntn

tf ππ

• Fundamental frequencyf0 = 1/T0 = 1/2π Hzω0 = 2π/T0 = 1 rad/s

{ }( ) { }

( )( )

000

sincos

Im2Re2

cDaCb

CaDDDjb

DDDa

nnn

nnn

nnnk

nnnn

==−=

=−=−=

=+=

−

−

θθ

Relationships between the Coefficients of the Different Forms

000

1

22

2

tan

DaCDDC

ab

baC

nn

nn

n

nn

nnn

==∠=

=

−=

+=

−

θ

θ

( )( )

000

5.05.0

5.0

5.0

CaDeCCD

jbaDDjbaD

njnnnn

nnnn

nnn

===∠=

+==

−=∗

−

θθ

ExampleFind the exponential Fourier Series and sketch the corresponding spectra for the impulse train shown below. From this result sketch the trigonometric spectrum and write the trigonometric Fourier Series.Solution

+=

====

=

=

∑

∑

∞

=

∞

−∞=

10

0

000

0

0

0

)cos(211)(

/1||/2||2

1)(

/1

0

0

0

nT

nn

n

tjnT

n

tnT

t

TDCTDC

eT

t

TD

ωδ

δ ω

2T0T0-T0-2T0

)(0

tTδ

1- The function g(t) is absolutely integrable over one period.

∫𝑇𝑇0 𝑔𝑔 𝑡𝑡 𝑑𝑑𝑡𝑡 < ∞

2- The function g(t) can have only a finite number of maxima, minima, and discontinuities in one period.

Dirichlet Conditions for FS Convergence

• Let x(t) be a periodic signal with period T• The average power P of the signal is defined as

• Expressing the signal as

it is also

( ) ∑∞

=

++=1

00 )cos(n

nn tnCCtx θω

∑∞

=

+=1

220 2

nnDDP∑

∞

=

+=1

220 5.0

nnCCP

Parseval’s Theorem

∫−=2/

2/

2)(1 T

Tdttx

TP

Numerical Calculation of Fourier Series

( ) ,....2,1,0 , 10 ±±== ∫ − ndtetg

TD

oT

tjn

on

ω

𝐷𝐷𝑛𝑛 = lim𝑇𝑇𝑠𝑠→0

1𝑇𝑇0

�𝑘𝑘=0

𝑁𝑁0−1

𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛𝜔𝜔0𝑘𝑘𝑇𝑇𝑠𝑠 𝑇𝑇𝑠𝑠

𝐷𝐷𝑛𝑛 = lim𝑇𝑇𝑠𝑠→0

1𝑁𝑁0

�𝑘𝑘=0

𝑁𝑁0−1

𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛Ω0𝑘𝑘

𝐷𝐷𝑛𝑛 =1𝑁𝑁0

�𝑘𝑘=0

𝑁𝑁0−1

𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛Ω0𝑘𝑘

Ω0 = 𝜔𝜔0𝑇𝑇𝑠𝑠

𝑁𝑁0 =𝑇𝑇0𝑇𝑇𝑠𝑠

Matlab Exercise Page 80

T0 = pi; N0 = 256; Ts = T0/N0;t = 0: Ts : Ts*(N0-1); g = exp(-t/2); g(1) = 0.604;Dn = fft(g)/N0;[Dnangle, Dnmag] = cart2pol( real(Dn), imag(Dn));k =0:length(Dn)-1; k = 2*k;subplot(211), stem(k, Dnmag)subplot(212), stem(k,Dnangle)

0 π−π

1e-t/2

g(t)

T0 = πf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s