Fourier Series

Fourier Representations : Part-I

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Fourier Representations of Signals & LTI Systems

A signal can be represented as a weighted superposition of complex sinusoids.

LTI system:

Output = A weighted superposition of the system response to each complex sinusoid.

Complex Sinusoids and Frequency Response of LTI Systems

Frequency response The response of an

Discrete-time LTI system:

Impulse response of discrete-time

Output:

The output of a complex sinusoidal input to an LTIinput, multiplied by the frequency response of the system.

Frequency response:

Continuous-time LTI system

Impulse response of continuous-time

Output:

Frequency response:

Polar form complex number c = a

[ ] [ ] [

=

=

knxkhny

[ ] j n j k j j ny n e h k e H e e = =

( ) [ ]j jH e h k e

( )( ) ( ) ( ) ( )

j t j t j

j t

y t h e d e h e d

H j e

= =

=

( ) ( )H j h e d

arg{ }j cc c e=


A signal can be represented as a weighted superposition of complex sinusoids.



The response of an LTI system to a sinusoidal input.

time LTI system = h[n], input = x[n] = e j n

The output of a complex sinusoidal input to an LTI system is a complex sinusoid of the same frequency as the input, multiplied by the frequency response of the system.

time LTI system = h(t), input = x(t) = e j t

+ jb: where

] [ ] ( )

=

=

k

knjekhk

[ ] ( )j n j k j j nk

y n e h k e H e e

=

= =

( ) [ ]j jk

H e h k e

=

=

( )( ) ( ) ( )j t j t jy t h e d e h e d

= =

( ) ( ) jH j h e d

=

2 2 1and arg{ } tanc a b c= + =

Dept of E&CE, NMIT




system is a complex sinusoid of the same frequency as the

( )2 2 1and arg{ } tan ba


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Polar form for H (j):

where

RC Circuit: Frequency response

The impulse response of the system relating to the input voltage to the voltage across the capacitor

Find an expression for the frequency response, and plot the magnitude and phase response.

Frequency response:

Frequency response of the

( ) ( )H j H j e =

( ) Magnitude response and arg ( ) Phase respnseH j H j = =( ) ( ) {( argj t H jy t H j e +=

1( ) ( )h t e u tRC

=

( )RC

jH

=1

1 1RC j

=

1

=

RC

jRC

1

+=



Frequency response of the RC circuit (a) Magnitude response. (b) Phase response.

arg{ ( )}( ) ( ) j H jH j H j e =

{ }( ) Magnitude response and arg ( ) Phase respnseH j H j = =( )})j t H j

/( ) ( )t RCh t e u t

( )

deue jRC

eRC

RCj

+

=0

11

1

0

1 11

jRCe

jRC

+

+

( )101

1

+

RCj

RC

RC1

Dept of E&CE, NMIT



circuit (a) Magnitude response. (b) Phase response.

( ) Magnitude response and arg ( ) Phase respnse

dRC

1


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Orthogonality of the Complex exponentials

Definition : Two signals are orthogonal if their inner product is zero. The inner product is defined using complex conjugation when the signals are complex valued. For continuousproduct is defined in terms of an integral as

For discrete-time signals with period N, their inner product is defined as


Fourier Representations for Four classes of Signals

Periodic Signals: Fourier Series Representations

x[n] = discrete-time signal with fundamental period

. x(t) = continuous-time signal with fundamental period

^ denotes approximate value. A

The complex sinusoids exp(jkon) are 0 0is the th harmonic of .jk t j te k e


Two signals are orthogonal if their inner product is zero. The inner product is defined using complex conjugation when the signals are complex valued. For continuous-time signals with period T, the inner product is defined in terms of an integral as

, .

time signals with period N, their inner product is defined as

,


Fourier Representations for Four classes of Signals

Series Representations

time signal with fundamental period N. DTFS of x[n] is

time signal with fundamental period T. FS of x(t) is

A[k] = the weight applied to the kth harmonic.

) are N-periodics in the frequency index k. 0 0is the th harmonic of .jk t j te k e

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Two signals are orthogonal if their inner product is zero. The inner product is defined using time signals with period T, the inner


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There are only N distinct complex sinusoids of the form exp(

If x[n] is symmetric in odd or even, we can choose

k = (N 1)/2 to (N 1)/2 if N is odd

Continuous-time complex sinusoid exp(

FS of continuous-time periodic signal

Mean-square error (MSE) between the signal and its series representation:

Discrete-time case:

Continuous-time case:

Nonperiodic Signals: Fourier-Transform RepresentationsFT of continuous-time signal:

DTFT of discrete-time signal:

( ) njNnKNjeee 00

+=

=

=

1

0][

N

knx

$( ) [ ]k

x t A k e

=

=

MSE x n x n dt=

MSE x t x t dt=

( ) ( )12

j tx t X j e d pi

=

[ ] ( )12 j j nx n X e e d pi

=

distinct complex sinusoids of the form exp(jk0n) should be used in Eq

] is symmetric in odd or even, we can choose k as:

1)/2 if N is odd

time complex sinusoid exp(jk0t) with distinct frequencies k0 are always distinct.

time periodic signal x(t) becomes

square error (MSE) between the signal and its series representation:

Transform Representations

njk 0 njNnj ee 02 = pi jke =

0][ njkekA

0( ) [ ] jk tx t A k e

$1 2

0

1 [ ] [ ]N

n

MSE x n x n dtN

=

=

$2

0

1 ( ) ( )TMSE x t x t dtT

=

x n X e e d

Dept of E&CE, NMIT

) should be used in Eq

are always distinct.

n0


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Example 3.2 Determining DTFS Coefficients

Find the frequency domain representation of the signal depicted in Fig

Period: N = 5 ; o = 2pipipipi/5

Odd symmetry: n = 2 to n = 2

One period of the DTFS coefficients

[ ] [ ]2

2 / 5

2

15

jk n

n

X k x n e pi=

= [{51

x =

2 / 5 2 /51 1 1[ ] {1 }5 2 21

{1 sin( 2 / 5)}5

jk jkX k e e

j k

pi pi

pi

= +

= +

Determining DTFS Coefficients

Find the frequency domain representation of the signal depicted in Fig

One period of the DTFS coefficients X[k], k = 2 to k = 2:

] [ ] [ ] [ ]05/25/4 1012 pipi jjkjk exexexe +++

2 / 5 2 /5[ ] {1 }pi pi

Dept of E&CE, NMIT

[ ] }5/45/2 2 pipi jkjk ex +


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Calculate X[k] using n = 0 to n = 4:

Magnitude and phase of the DTFS coefficients for the signal x(n)

Example 3.3 Computation of DTFS by Inspection

Determine the DTFS coefficients of

Period: N = 6 : o = 2pipipipi/6 = pipipipi/3

Using Eulers formula, x[n] can be expressed as

Compare above eqn with the DTFS equation with

[ ] (5

5/4sin512 jX = pi

[ ] 02.0510 jeX == [ ]

511X +=

[ ] [ ] [ ]{ /20 1051 pijkj

exexkX +=

+= 5/221

211

51 pi jkjk ee

( ) ( )3 3

[ ]2 2 2

j n j ne e

x n e e e e

pi pi + ++

= = +

[ ] 3 / 32

2 / 3 / 3 / 3 2 / 3

[ ]

[ 2] [ 1] [0] [1] [2] [3]

jk n

kj n j n j n j n j n

x n X k e

X e X e X X e X e X e

pi

pi pi pi pi pi

=

=

= + + + + +

= 4:

;


Computation of DTFS by Inspection

Determine the DTFS coefficients of x[n] = cos (npipipipi/3 + ), using the method of inspection.

] can be expressed as

------

Compare above eqn with the DTFS equation with o = pipipipi/3, written by summing from

8 /5 2 2 /5 2 /5jk jk jk jke e e epi pi pi pi = =

) 531.0232.05 je= [ ] (5

5/2sin511 jX = pi

( ) 760.0276.05

5/2sin jej =+ pi [ ] ( )5

5/4sin512 jX += pi

[ ] [ ] [ ] }5/85/65/45 432 pipipi jkjkjk exexex +++

5/8pijk

[ ] M agnitude spectrum of [ ]X k x n

{ }arg [ ] Phase spectrum of [ ]X k x n

( ) ( )3 31 1

2 2 2j n j nj jx n e e e e

pi pi

+ +

= = +

2 / 3 / 3 / 3 2 / 3[ 2] [ 1] [0] [1] [2] [3]j n j n j n j n j nX e X e X X e X e X epi pi pi pi pi= + + + + +

Dept of E&CE, NMIT


), using the method of inspection.

------1

/3, written by summing from k = 2 to k = 3:

---------2

8 /5 2 2 /5 2 /5jk jk jk jke e e epi pi pi pi

) 760.0276.05 je=

531.0232.0 je=

}

[ ] M agnitude spectrum of [ ]X k x n

arg [ ] Phase spectrum of [ ]X k x n

j n j n j n j n j npi pi pi pi pi


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Equating terms in Eq1 with those in Eq2 having equal frequencies,

Magnitude spectrum and phase spectrum of

Example 3.4 DTFS Representation of An Impulse TrainFind the DTFS coefficients of the N

It is convenient to evaluate Eq. over the interval

Example 3.5 The Inverse DTFS

Determine the time-domain signal

[ ] [; 3DTFSx n X k e kpi

= =

[ ] [ ]10

1 1N jkn Nn

X k n eN N

=

= =

Equating terms in Eq1 with those in Eq2 having equal frequencies, kpipipipi/3, gives

Magnitude spectrum and phase spectrum of X[k]:

DTFS Representation of An Impulse Train N-periodic impulse train

It is convenient to evaluate Eq. over the interval n = 0 to n = N 1 to obtain

domain signal x[n] from the DTFS coefficients depicted in Fig

[ ]/ 2, 1/ 2, 1

0, otherwise on 2 3

j

je k

x n X k e kk

=

= =

[ ] [l

x n n lN

=

=

2 /1 1jkn NN N

pi= =

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] from the DTFS coefficients depicted in Fig

]x n n lN=


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Period of DTFS coefficients = 9 ;

It is convenient to evaluate x(n) over the interval

Example 3.6 DTFS Representation of A Square Wave Find the DTFS coefficients for the

That is, each period contains 2M + 1 consecutive ones and the remaining depicted in Fig. Note that this definition requires that

Period of DTFS coefficients = N ;

It is convenient to evaluate Eq.(3.11) over the interval

For k = 0, N, 2N, , we have

[ ] [ ]

(

42 / 9

42 / 3 6 / 9 / 3 4 / 9 / 3 4 / 9 2 / 3 6 / 92 1 2

2cos 6 / 9 2 / 3 4cos 4 / 9 / 3 1

jk n

kj j n j j n j j n j j n

x n X k e

e e e e e e e e

n n

pi

pi pi pi pi pi pi pi pi

pi pi pi pi

=

=

= + + +

= +

[ ]

2M +1.

N ; o = 2pipipipi/N

It is convenient to evaluate Eq.(3.11) over the interval n = M to n = N M 1.

, , we have

) ( )2 / 3 6 / 9 / 3 4 / 9 / 3 4 / 9 2 / 3 6 / 92 1 2

2cos 6 / 9 2 / 3 4cos 4 / 9 / 3 1

j j n j j n j j n j j ne e e e e e e e

n n

pi pi pi pi pi pi pi pi

pi pi pi pi

= + + +

= +

MM

0

0

1

[ ] [ ]

1

N Mjk n

n M

Mjk n

n M

X k x n e

e

=

=

0

0 0

2( )

02

0

1

1

Mj k m M

m

Mjk M j k m

m

X k eN

e eN

=

=

1o ojk jke e = =

0

1 2 11 , 0, , 2 ,M MX k k N N

N N=

+= = = K

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M +1) values are zero, as

1.

2 / 3 6 / 9 / 3 4 / 9 / 3 4 / 9 2 / 3 6 / 9j j n j j n j j n j j ne e e e e e e epi pi pi pi pi pi pi pi


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For k 0, N, 2N, , we may sum the geometric series in Eq to obtain

Substituting o = 2pipipipi/N, yields

For this reason, the expression for

The DTFS coefficients for the square wave shown in Fig., assuming a period

0 01[ ] , 0 , , 2 , . . . . . .1

j k M j k M Me eX k k N N

N e

+ =

[ ]( )0 0 0 0

0 0 0 0

2 1 / 2 2 1 2 1 / 2 2 1 / 2

/ 2 / 2 / 21 1 1jk M jk M jk M jk M

jk jk jk jke e e eX k

N e e N e e

+ + + +

= =

[ ] (( (sin2sin1 0

=

kk

NkX

[ ]((

sin 2 1 /1sin /

2 1 / , 0, , 2 ,

k M NX k N k N

=

(0, , 2 ,

sin 2 1 /1 2 1limsin /k N Nk M N

N k N Npi

K

[ ] (( ( NkMk

NkX

/sin12sin1

pi

pi +=

, , we may sum the geometric series in Eq to obtain

For this reason, the expression for X[k] is commonly written as

The DTFS coefficients for the square wave shown in Fig., assuming a period N = 50: (a)

k N N

0 0

0

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

1

j k M j k M M

jke eX k k N N

N e

+

=

( ) ( )0 0 0 00 0 0 0

2 1 / 2 2 1 2 1 / 2 2 1 / 2

/ 2 / 2 / 21 1 1

1

jk M jk M jk M jk M

jk jk jk jke e e e

N e e N e e

+ + + +

= =

) )) ,2/

2/120

+M

0 , , 2 ,k N N=

( ) )( )

)

sin 2 1 /, 0, , 2 ,

sin /2 1 / , 0, , 2 ,

k M Nk N N

N k NM N k N N

pi

pi

+

+ =

K

K

( ) )( )

sin 2 1 /1 2 1sin /k M N M

N k N Npi

pi

+ +=

) ))

N/1

Dept of E&CE, NMIT

= 50: (a) M = 4. (b) M = 12.

0 , , 2 , ... ...k N N

[ ] , 0 , , 2 , . . . . . .X k k N N

( )0 0 0 00 0 0 0

2 1 / 2 2 1 2 1 / 2 2 1 / 2

/ 2 / 2 / 2

jk M jk M jk M jk M

jk jk jk jke e e e

N e e N e e

+ + + +

0 , , 2 ,k N N= K


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Symmetry property of DTFS coefficient:

1. DTFS of Eq. given below can be written as a series involving harmonically related cosines.

2. Assume that N is even and let k

3. Use X[m] = X[ m] and the identity

Define the new set of coefficients

and write the DTFS for the square wave in terms of a series of harmonically related cosines as

A similar expression may be derived for

Example 3.7 Building a Square Wave From

The contribution of each term to the square wave may be illustrated by defining the partialapproximation to x[n] in above Eq. as

where J N/2. This approximation contains the first 2a square wave has period N = 50 and + 1 term approximation for

1

0[ ] [ ]

Njk n

kx n X k e

=

=

[ ] [ ]+=

=

2/

12/

N

Nk

jkekXnx

[ ] [ ] [+= /0 NXXnx

[ ] [ ] [ ]0 / 2 2j nx n X X N e X mpi= + +

[ ] [ ] (0 / 2 cos 2 cosX X N n X m m n= + +

[ ]2 , 1, 2 , , / 2 1X k k N

B k

=

/ 2

[ ] [ ]cos( )N

kx n B k k n

=

=

$

0[ ] [ ]cos( )

J

Jk

x n B k k n=

=

Symmetry property of DTFS coefficient: X[k] = X[ k].

can be written as a series involving harmonically related cosines.

---------------1

range from ( N/2 ) + 1 to N/2. Eq. 1 becomes

] and the identity No = 2pipipipi to obtain


A similar expression may be derived for N odd.

Building a Square Wave From DTFS Coefficients The contribution of each term to the square wave may be illustrated by defining the partial

] in above Eq. as

/2. This approximation contains the first 2J + 1 terms centered on k = 50 and M = 12. Evaluate one period of the Jth term in Eq. (3.18) and the 2

for J = 1, 3, 5, 23, and 25.

0[ ] [ ] jk nx n X k e

0 n

] [ ] [ ](=

++

12/

1

2/ 002N

m

njknjkemXemXe

[ ] 0 0/ 2 11

0 / 2 22

jm n jm nNj n

m

e ex n X X N e X m

=

+= + +

( ) [ ] ( )/ 2 1 01

0 / 2 cos 2 cosN

m

X X N n X m m npi

=

= + +

[ ][ ]

, 0 , / 22 , 1, 2 , , / 2 1X k k NX k k N

=

= K

/ 2

00

[ ] [ ]cos( )N

kx n B k k n

=

=

0[ ] [ ]cos( )x n B k k n=

Dept of E&CE, NMIT

can be written as a series involving harmonically related cosines.

becomes


The contribution of each term to the square wave may be illustrated by defining the partial-sum

= 0 in Eq. (3.10). Assume th term in Eq. (3.18) and the 2J

] ) 0njke


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Individual terms in the DTFS expansion of a square wave (left panel) and the corresponding partial(right panel). The J = 0 term is 0[

1. Fig. 3.14 depicts the J th term in the sum, values of J.

2. Only odd values for J are considered, because the evenand M = 12.

3. The approximation improves as

4. The coefficients B[k] associated with values of varying features in the signal, while the coefficients associated with values of high frequency or rapidly varying features in the signal.

expansion of a square wave (left panel) and the corresponding partial= 0 term is 0[n] = and is not shown. (a) J = 1. (b) J = 3. (c) J = 5. (d)

term in the sum, B[J] cos(Jon), and one period of

are considered, because the even-indexed coefficients B

3. The approximation improves as J increases, with x[n] represented exactly when

] associated with values of k near zero represent the low- varying features in the signal, while the coefficients associated with values of k near high frequency or rapidly varying features in the signal.

$x n

Dept of E&CE, NMIT

expansion of a square wave (left panel) and the corresponding partial-sum approximations J[n] = 5. (d) J = 23. (e) J = 25.

for the specified

B[k] are zero when N = 25

when J = N/2 = 25.

frequency or slowly near N/2 represent the

$ [ ]Jx n


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Continuous-Time Periodic Signals

1. Continuous-time periodic signals are represented by the Fourier series (FS).

2. A signal with fundamental period

3. Notation:

Mean-square error (MSE)

1. Suppose we define

and choose the coefficients X[k] according to Eq.2

2. If x(t) is square integrable,

then the MSE between x(t) and

at all values of t; it simply implies that there is zero power in their difference.

3. Pointwise convergence of

except those corresponding to discontinuities if the Dirichlet conditions are satisfied:

x(t) is bounded.

x(t) has a finite number of maximum

x(t) has a finite number of discontinuities in one period.

If a signal x(t) satisfies the Dirichletmidpoint of the left and right limits of

x t

( )( )x t x t=

0( ) [ ] jk tk

x t X k e

=

=

0

0

1[ ] ( )T jk tX k x t e dtT

=

( ) [0;FSx t X k

( )k

x t X k e=

( )x t x t=

Time Periodic Signals: The Fourier Series

time periodic signals are represented by the Fourier series (FS).

2. A signal with fundamental period T and fundamental frequency o = 2pipipipi/T,

--1

--2

] according to Eq.2

is zero, or, mathematically,

; it simply implies that there is zero power in their difference.

is guaranteed at all values of t

except those corresponding to discontinuities if the Dirichlet conditions are satisfied:

) has a finite number of maximum and minima in one period.

) has a finite number of discontinuities in one period.

Dirichlet conditions and is not continuous, then midpoint of the left and right limits of x(t) at each discontinuity.

( )x t

x t

jk tX k x t e dt

]x t X k

[ ] ojk tk

x t X k e

=

=

( )


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Example 3.9 Direct Calculation of FS Coefficients Determine the FS coefficients for the signal

1. The period of x(t) is T = 2, so

2. One period of x(t): x(t) = e 2t, 0

3. FS of x(t):

The Magnitude of X[k] the magnitude spectrum of x(t).

[ ] 2 220 0

1 12 2

t jk tX k e e dt e dt = =

[ ] ( )(21 1 1

2 2 4 2 4 2jk tX k e e ejk jk jk

pi

pi pi pi +

= = =

+ + +

Direct Calculation of FS Coefficients Determine the FS coefficients for the signal x(t) depicted in Fig.

= 2, so o = 2pipipipi/2 = pipipipi.

, 0 t 2.

the magnitude spectrum of x(t); the phase of X[k]

( )2 2 20 0

1 12 2

jk tt jk tX k e e dt e dtpipi + = =

) ( )2

4 2

0

1 1 112 2 4 2 4 2

jk t jk eX k e e ejk jk jkpi pi

pi pi pi

= = =

+ + +

Dept of E&CE, NMIT

] the phase spectrum of

4

2 2 4 2 4 2e

jk jk jkpi pi pi


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Example 3.10 Determine the FS coefficients for the signal x(t) defined by

1. Fundamental period of x(t) is T = 4, each period contains an impulse.

2. By integrating over a period that is symmetric about the origin 2 < t 2, to obtain X[k]:

3. The magnitude spectrum is constant and the phase spectrum is zero.

Inspection method for finding X[k]: Whenever x(t) is expressed in terms of sinusoid, it is easier to obtain X[k] by inspection. The method of inspection is based on expanding all real sinusoids in terms of complex sinusoids and comparing each term in the resulting expansion to the corresponding terms of x(t).

Example 3.11 Calculation of FS Coefficients By Inspection Determine the FS representation of the signal

1. Fundamental period of x(t) is T = 4, and o = 2pipipipi/4 = pipipipi/2. Eq. is written as

2. Equating each term in this expression to the terms in Eq. gives the FS coefficients:

( ) ( )4l

x t t l

=

=

[ ] ( ) ( )2 / 22

1 14 4

jk tX k t e dtpi

= =

( ) ( )4/2/cos3 pipi += ttx

/ 2( ) [ ] jk tk

x t X k e pi

=

=

( )( ) ( )/ 2 / 4 / 2 / 4

/ 4 / 2 / 4 / 23 332 2 2

j t j tj j t j j te e

x t e e e epi pi pi pi

pi pi pi pi+ +

+= = +

/ 4

/ 4

3, 1

23[ ] , 120 , o th e rw ise

j

j

e k

X k e k

pi

pi

=

= =


RN

Time-domain representation obtained from FS coefficients

Example 3.12 Inverse FS . Find the time

Assume that the fundamental period is

1. Fundamental frequency: o = 2pipipipi

2. Putting the fraction over a common denominator results in

Example 3.13 FS for A Square WaveDetermine the FS representation of the square wave depicted in Fig. 3.21.

1. Fundamental frequency: o = 2pipipipi

2. X[k]: by integrating over T/2

[ ] ( ) 20/2/1 pijkk ekX =

( ) ( ) / 20 / 200 1

1/ 2 1/ 2k kjk jk t jk jk tk k


= =

= +

( )

=

+=0

20/2/1k

tjkjkk ee pipi

( ) ( ) (2/111

20/

=+ pipi tje

tx

( ) (cos453

pipi +=

ttx

Magnitude and phase spectra

domain representation obtained from FS coefficients

Find the time-domain signal x(t) corresponding to the FS coefficients

Assume that the fundamental period is T = 2.

pipipipi/T = pipipipi. From Eq. (3.19), we obtain

2. Putting the fraction over a common denominator results in

FS for A Square Wave Determine the FS representation of the square wave depicted in Fig. 3.21.

pipipipi/T.

t T/2

( )/ 20 / 200 1

1/ 2 1/ 2k kjk jk t jk jk tk k


= =

= +

( )

=

+1

20/2/1l

tjljll ee pipi

) ( ) ( ) 12/111

20/20

++ pipi tje

)20/

Dept of E&CE, NMIT

) corresponding to the FS coefficients


RN

For k = 0, we have

3. By means of LHpitals rule, we have

4. X[k] is real valued. Using o = 2

5. Fig. (a)-(c) depict X[k], 50 k

[ ] ( )

( )

0

0

0

0 0 0 0

/ 2

/ 2

0

0

0 0

0

1 1

1

22

2 sin, 0

T Tjk t jk tT T

Tjk t

T

jk T jk T

X k x t e dt e dtT T

e kTjk

e e

Tk jk T

Tk

= =

=

=

=

[ ]

=

10T

X

[ ] ( )0

00sin2

TkTkkX =

02sin( 2 / )[ ]2

k T TX kk

pi

pi=

3. By means of LHpitals rule, we have

= 2pipipipi/T gives X[k] as a function of the ratio To/T:

50, for To/T =1/4, To/T =1/16, and To/T = 1/64, respectively.

00 0

0

0

0 0 0 0

1 1

, 0

, 02

, 0

T Tjk t jk tT T

T

jk T jk T

X k x t e dt e dtT T

e k

e e kTk j

k

= =

=

=

=

=0

0

02TT T

Tdt

( )TT

TkTk

k0

0

000

2sin2lim =

Dept of E&CE, NMIT

1/64, respectively.


RN

Definition of Sinc function:

1) Maximum of sinc function is unity at amplitude dies off as 1/u.

2) The portion of this function between the zero crossings at mainlobe are termed sidelobe.

4) The FS coefficients in Eq. are expressed as

Trigonometric FS

1. Trigonometric FS of real-valued signal

FS coefficients:

2. Relation between exponential FS and trigonometric FS coefficients:

Ex. Find trigonometric FS coefficients of the square wave studied in

1. Substituting Eq.

2. Trigonometric FS expression of

0

1[0] ( )TB x t dtT

=

0

2[ ] ( ) cos( )TB k x t k t dtT

=

[ ] [ ] [ ]B k X k X k= +

0[0] 2 /B T T=02sin( 2 / )[ ] , 0k T TB k k

kpi

pi=

0

1[ ] ( )TX k x t e dtT

=

1) Maximum of sinc function is unity at u = 0, the zero crossing occur at integer values of

2) The portion of this function between the zero crossings at u = 1 is 3) The smaller ripples outside the

4) The FS coefficients in Eq. are expressed as main lobe of the sinc function.

valued signal x(t):

2. Relation between exponential FS and trigonometric FS coefficients:

Ex. Find trigonometric FS coefficients of the square wave studied in Example 3.13.

1. Substituting Eq. into above expressions, gives

2. Trigonometric FS expression of x(t):

sin( )sinc( ) uu

u

pi

pi=

0 01

( ) [0] [ ]cos( ) [ ]sin( )k

x t B B k k t A k k t

=

= + +

0[ ] ( ) cos( )B k x t k t dt 02[ ] ( )sin( )TA k x t k t dtT

=

[ ] ( [ ] [ ])A k j X k X k=

[ ] , 0B k k= [ ] 0A k =

0[ ] ( ) jk tX k x t e dt

00

( ) [ ] cos( )k

x t B k k t

=

=

Dept of E&CE, NMIT

= 0, the zero crossing occur at integer values of u, and the

3) The smaller ripples outside the

Example 3.13.

[ ] 0 02 2sincT TX k kT T

=

0 0( ) [0] [ ]cos( ) [ ]sin( )x t B B k k t A k k t = + +

0[ ] ( )sin( )A k x t k t dt


RN

Example 3.14 Square Wave Partial

Let the partial-sum approximation to the FS in Eq. (3.29), be given by

This approximation involves the exponential FS coefficients with indices wave with T = 1 and To/T = . Depict one period of the

1. The individual terms and partial

2. Each partial-sum approximation passes through the average value (1/2) of the discontinuity, the approximation exhibits ripple.

3. This ripple near discontinuities in partial

4. As J increase, the ripple in the partialthe discontinuities.

Individual terms (left panel) in the FS expansion of a square wave and the corresponding partialapproximations J(t) (right panel). The square wave has period = and is not shown. (a) J = 1.

( ) [ ] (=

=

J

kJ kkBtx

0cos

( ) for 1, 3, 7, 29, and 99.Jx t J =

Square Wave Partial-Sum Approximation

approximation to the FS in Eq. (3.29), be given by

This approximation involves the exponential FS coefficients with indices J k = . Depict one period of the Jth term in this sum, and find

1. The individual terms and partial-sum approximation are depicted in Fig.

sum approximation passes through the average value (1/2) of the discontinuity, the

3. This ripple near discontinuities in partial-sum FS approximations is termed the Gibbs phenomenon.

increase, the ripple in the partial-sum approximations becomes more and more concentrated near

Individual terms (left panel) in the FS expansion of a square wave and the corresponding partial) (right panel). The square wave has period T = 1 and Ts/T = . The

)tk 0

for 1, 3, 7, 29, and 99.

Dept of E&CE, NMIT

J. Consider a square th term in this sum, and find

sum approximation passes through the average value (1/2) of the discontinuity, the

sum FS approximations is termed the Gibbs phenomenon.

sum approximations becomes more and more concentrated near

Individual terms (left panel) in the FS expansion of a square wave and the corresponding partial-sum = . The J = 0 term is 0(t)


RN

Example 3.15 RC Circuit: Calculating The Output By Means of FSLet us find the FS representation for the output the square-wave input depicted in Fig. 3.21, assuming that

1. If the input to an LTI system is expressed as a weighted sum of sinusoid, then the output is also a weighted sum of sinusoids.

2. Input:

3. Output:

4. Frequency response of the system

5. Frequency response of the RC circuit:

6. Substituting for H(jko) with RC

7. We determine y(t) using the approximation

( ) [ ]

=

=

k

tjkekXtx 0

( ) (

=

=

kHty

( )RCj

RCjH/1

/1+

=

[ ] (pi

pi

pi kk

kjkY2/sin

10210

+=

( ) [ ]y t Y k e

(b) J = 3. (c) J = 7. (d) J = 29.

(e) J = 99.

RC Circuit: Calculating The Output By Means of FS Let us find the FS representation for the output y(t) of the RC circuit depicted in Fig. 3.2 in response to

wave input depicted in Fig. 3.21, assuming that To/T = , T = 1 s, and

Figure 3.2 : RC circuit

system is expressed as a weighted sum of sinusoid, then the output is also a

4. Frequency response of the system H(j):

circuit:

RC = 0.1 s and o = 2pipipipi, and To/T = , gives

) using the approximation

1

( ) [ ] tjkekXjk 00

( ) [ ] (0;FSy t Y k H jk X k =

)2

0100

100( ) [ ] jk t

ky t Y k e

=

Dept of E&CE, NMIT

circuit depicted in Fig. 3.2 in response to = 1 s, and RC = 0.1 s.

system is expressed as a weighted sum of sinusoid, then the output is also a

) [ ]0y t Y k H jk X k


RN

8. The circuit attenuates the amplitude of frequency ko increases. The circuit also introduces a frequency

Figure 3.26 The FS coefficients Y[input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal

signal y(t) (solid line). The output signal

8. The circuit attenuates the amplitude of X[k] when k 1. The degree of attenuation increases as the increases. The circuit also introduces a frequency-dependent phase dependent shift.

The FS coefficients Y[k], 25 k 25, for the RC circuit output in response to a squareinput. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal

) (solid line). The output signal y(t) is computed from the partial-sum approximation g

Dept of E&CE, NMIT

1. The degree of attenuation increases as the dependent phase dependent shift.

circuit output in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x(t) dashed line) and output

sum approximation given in Eq. 1

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