Engineering and PhysicsUniversity of Central Oklahoma

Dr. Mohamed Bingabr

Ch6Continuous-Time Signal Analysis

ENGR 3323: Signals and Systems

Outline

• Introduction• Fourier Series (FS) Representation of

Periodic Signals.• Trigonometric and Exponential Form of FS.• Gibbs Phenomenon.• Parseval’s Theorem.• Simplifications Through Signal Symmetry.• LTIC System Response to Periodic Inputs.

Sinusoidal Wave and phasex(t) = Asin(ωt) = Asin(2π50t)

A

t

T0 = 20 msec

x(t)

A

ttd = 2.5 msec

x(t-0.0025)= Asin(2π50[t-0.0025])= Asin(2π50t-0.25π)= Asin(2π50t-45o)

Time delay td = 2.5 msec correspond to phase shift θ = 45o

Representation of Quantity using Basis

• Any number can be represented as a linear sum of the basis number {1, 10, 100, 1000}

Ex: 10437 =10(1000) + 4(100) + 3(10) +7(1)

• Any 3-D vector can be represented as a linear sum of the basis vectors {[1 0 0], [0 1 0], [0 0 1]}

Ex: [2 4 5]= 2 [1 0 0] + 4[0 1 0]+ 5[0 0 1]

Basis Functions for Time Signal• Any periodic signal x(t) with fundamental frequency ω0 can be represented by a linear sum of the basis functions {1, cos(ω0t), cos(2ω0t),…, cos(nω0t), sin(ω0t), sin(2ω0t),…, sin(nω0t)}

Ex:x(t) =1+ cos(2πt)+ 2cos(2 π2t)+ 0.5sin(2π3t)+ 3sin(2πt)

x(t) =1+ cos(2πt)+ 2cos(2 π2t)+ 3sin(2πt)+ 0.5sin(2π3t)

+ +

+ =

Purpose of the Fourier Series (FS)

FS is used to find the frequency components and their strengths for a given periodic signal x(t).

The Three forms of Fourier Series

• Trigonometric Form

• Compact Trigonometric (Polar) Form.

• Complex Exponential Form.

Trigonometric Form

• It is simply a linear combination of sines and cosines at multiples of its fundamental frequency, f0=1/T.

• a0 counts for any dc offset in x(t).• a0, an, and bn are called the trigonometric Fourier

Series Coefficients. • The nth harmonic frequency is nf0.

( ) ( ) ( )∑ ∑∞

=

∞

=

++=1 1

000 2sin2cosn n

nn ntfbntfaatx ππ

Trigonometric Form• How to evaluate the Fourier Series Coefficients

(FSC) of x(t)?

( )∫=00

01

T

dttxT

a

To find a0 integrate both side of the equation over a full period

( ) ( ) ( )∑ ∑∞

=

∞

=

++=1 1

000 2sin2cosn n

nn ntfbntfaatx ππ

Trigonometric Form

( ) ( ) ( )∑ ∑∞

=

∞

=

++=1 1

000 2sin2cosn n

nn ntfbntfaatx ππ

( ) ( )∫=0

00

2cos2

Tn dtntftx

Ta π

To find an multiply both side by cos(2πmf0t) and then integrate over a full period, m =1,2,…,n,…∞

To find bn multiply both side by sin(2πmf0t) and then integrate over a full period, m =1,2,…,n,…∞

( ) ( )∫=0

00

2sin2

Tn dtntftx

Tb π

Example

• Fundamental periodT0 = π

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

( ) ( ) ( )

( )

( )

. as amplitudein decrease and 1618 504.0 2sin21612 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)

( ) ( ) ( )( )

+

++= ∑

∞

=12 2sin42cos

16121504.0

nntnnt

ntf

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

Simplifications Through Signal Symmetry

• If x (t) is EVEN: It must contain Cosine Terms and it may contain DC. Hence bn = 0.

• If x(t) is ODD: It must contain ONLY Sines Terms. Hence a0 = an = 0.

A periodic signal x(t), has a Fourier series if it satisfies the following conditions:1. x(t) is absolutely integrable over any period,

namely

2. x(t) has only a finite number of maxima and minima over any period

3. x(t) has only a finite number of discontinuitiesover any period

Dirichlet Conditions

∫ ∞<0

)(T

dttx

• Using single sinusoid,

• are related to the trigonometric coefficients anand bn as:

( )

( )∑∞

=

++=1 harmonicnth

0component dc

0 2cosn

nn ntfCCtx θπ

00 aC =

nnC θ and ,

and22nnn baC +=

−= −

n

nn a

b1tanθ

Compact Trigonometric Form

The above relationships are obtained from the trigonometric identity

a cos(x) + b sin(x) = c cos(x + θ)

Role of Amplitude in Shaping Waveform

( ) ( )∑∞

=

++=1

00 2cosn

nn ntfCCtx θπ

Role of the Phase in Shaping a Periodic Signal

( ) ( )∑∞

=

++=1

00 2cosn

nn ntfCCtx 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

0 2 4 6

• x(t) can be expressed as

( ) ∑∞

−∞=

=n

ntfjneDtx 02π

D-n = Dn*

( ) ,....2,1,0 , 102 ±±== ∫ − ndtetx

TD

oT

ntfj

on

π

Exponential Form

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 of x(t) in the Exponential Form

Cn = 2|Dn| Cn = Dn

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

...||||

||||...)(

2021010

2221

012

2

001

0102

+++++=

++

++++= −−−

−−−

θωθω

ωθωθ

ωθωθ

tCtCCtx

eeDeeD

DeeDeeDtxtjjtjj

tjjtjj

C0 = D0

Example

τ/2−τ/2

1f(t)

−Το Το

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

Το/2−Το/2

𝐷𝐷𝑛𝑛 =1𝑇𝑇0

�−𝜏𝜏/2

𝜏𝜏/2

𝑓𝑓(𝑡𝑡)𝑒𝑒−𝑗𝑗𝜔𝜔𝑜𝑜𝑛𝑛𝑛𝑛𝑑𝑑𝑡𝑡

𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2

If To = 2π and τ = π then

τ/2−τ/2

1f(t)

−Το ΤοΤο/2−Το/2

𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2

𝜏𝜏 = 0.2𝜋𝜋

𝑇𝑇0 = 𝜋𝜋

𝜏𝜏 = 0.1𝜋𝜋

𝑇𝑇0 = 𝜋𝜋

𝜏𝜏 = 0.1𝜋𝜋

𝑇𝑇0 = 0.5𝜋𝜋

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

ππ/2−π/2

1f(t)

−π π 2π−2π

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

[ ]∑∞

=

−

−−++=

oddn

nntn

tx1

2/)1(

21)1(cos2

21)( π

π

Does the Fourier series converge to x(t) at every point?

𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2

3( )x t 9 ( )x t

Gibbs Phenomenon

21( )x t 45 ( )x t

overshoot: about 9 % of the signal magnitude (present even if )N →∞

Gibbs Phenomenon – Cont’d

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.

2T0T0-T0-2T0

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

)(0

tTδ

( )( )

000

5.05.0

5.05.0

CaDeCCD

jbaDDjbaD

njnnnn

nnnn

nnn

===∠=

+==

−=∗

−

θθ

Relationships between the Coefficients of the Different Forms

{ }( ) { }

( )( )

000

sincos

Im2Re2

cDaCb

CaDDDjb

DDDa

nnn

nnn

nnnk

nnnn

==−=

=−=−=

=+=

−

−

θθ

000

1

22

2

tan

DaCDDC

ab

baC

nn

nn

n

nn

nnn

==∠=

=

−=

+=

−

θ

θ

• 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

LTIC System Response to Periodic Inputs

H(s)H(jω)

tje 0ω tjejH 0)( 0ωω

A periodic signal x(t) with period T0 can be expressed astjn

nneDtx 0)( ω∑

∞

−∞=

=

For a linear system

H(s)H(jω)

tjn

nneDtx 0)( ω∑

∞

−∞=

= ∑∞

−∞=

=n

tjnneDjnHty 0)()( 0

ωω

Fourier Series Analysis of DC Power Supply

A full-wave rectifier is used to obtain a dc signal from a sinusoid sin(t). The rectified signal x(t) is applied to the input of a lowpass RC filter, which suppress the time-varying component and yields a dc component with some residual ripple. Find the filter output y(t). Find also the dc output and the rms value of the ripple voltage.

05.0rms ripple

0025.0136)41(

2

||2

22

1

2

==

=+−

=

= ∑∞

=

ripple

ripple

n

nnripple

P

Pnn

D

DP

π

Ripple rms is only 5% of the input amplitude

20 /4 /2 ππ == DCPD

∑

∑

∑

∞

−∞=

∞

−∞=

∞

−∞=

+−=

=

+=

−=

−=

n

ntj

n

tjnn

n

ntj

n

enjn

ty

ejnHDty

jjH

en

tx

nD

22

0

22

2

)16)(41(2)(

)()(

131)(

)41(2)(

)41(2

0

π

ω

ωω

π

π

ω

Fourier Series Analysis of DC Power Supply

clear allt=0:1/1000:3*pi;for i=1:100

n=i;yp=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));n=-i;yn=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));y(i,:)=yp+yn;

endyf = 2/pi + sum(y);plot(t,yf, t, (2/pi)*ones(1,length(yf)))axis([0 3*pi 0 1]);

Power=0;for n=1:50

Power(n) = abs(2/(pi*(1-4*n^2)*(j*6*n+1)));endTotalPower = 2*sum((Power.^2));figure; stem( Power(1,1:20));

This Matlab code will plot y(t) for -100 ≤ n ≤100 and find the ripple power according to the equations below

0025.0||2

)16)(41(2)(

1

2

22

==

+−=

∑

∑∞

=

∞

−∞=

nnripple

n

ntj

DP

enjn

tyπ

Fourier Series Analysis of DC Power Supply