Chapter 3Fourier Series Representation of Periodic Signals

If an arbitrary signal x(t) or x[n] is expressed as a linear combination of some basic signals, the response of an LTI system becomes the sum of the individual responses of those basic signals.Such basic signal must:

• be capable of representing a large number of signals,• have system responses simple enough for computational convenience.Complex exponentials, given below, are such basic signals/functions:

est for CT systems,zn for DT systems, where s, z are complex numbers.

Both have the property that:• Response to an LTI system has the same form as the input with a change in the amplitude only. •A function with this property is called Eigen function and the corresponding amplitude ratio is called Eigen value. 

H(s)

x(t) y(t)

est H(s) est

H[z]  

x[n] y[n]

 

zn H[z] zn

1

• est and zn Eigen functions,

• H(s) and H[z] Eigen values.

Consider an input x(t) = est in a CT system. The corresponding output:

... (1)

Similarly, for a DT system:Input: x[n] = zn ,Output:

→

→

d e )(h)s(H

where,e )s(H d e )(he

d e )(h d )t(x )(h)t(h)t(x)t(y

s

stsst

)t(s

∫

∫

∫∫

∞

∞−

τ−

∞

∞−

τ−

∞

∞−

τ−∞

∞−

ττ=

=ττ=

ττ=ττ−τ=∗=

,z ]z[H z].k[hz

z].k[h]kn[x].k[h]n[h]n[x]n[y

n

k

kn

k

kn

k

==

=−=∗=

∑

∑∑∞

−∞=

−

∞

−∞=

−∞

−∞=

2

where

... (2)

CT systems:LTI system. Thus, with an input

the output :

In general: If ... (3)

then ... (4)

DT systems:

If , ... (5)

then, ... (6)

•If the input to an LTI system is a linear combination of complexexponentials, its output is also the linear combination of the same exponentials.

∑∞

−∞=

−=k

kz ]k[h]z[H

,eaeaea)t(x ts3

ts2

ts1

321 ++=

. e )s(Hae )s(Hae )s(Ha)t(y ts33

ts22

ts11

321 ++=

,e a)t(xk

tsk

k∑=

.e )s(Ha)t(yk

tskk

k∑=

∑=k

nkk z a]n[x

∑=k

nkkk z ]z[H a]n[y

3

In Fourier series, such complex exponentials are:s = jω, for CT systems

z = e jω, for DT systems.

Fourier Series of CT Periodic Signals:

A signal x(t) with a fundamental frequency of ω0 is expressed as a linear combination of complex exponential and its harmonics

as:

... (7)

k = 1 , fundamental frequency= 2, second harmonics

= 3, third harmonics, so on.

R.H.S. Is called the ‘Fourier series’ and ak is called the k-th harmonic component.

Fourier Series Coefficient ak :From eqn. (7), we may write:

∑∞

−∞=

ω=k

tjkk

0e a)t(x

∑∑∞

−∞=

ω−∞

−∞=

ω−ωω− ==k

t)nk(j k

k

ntjtjk k

ntj 0000 eae eae )t(x

4

.. ,e .e t3jt2j 00 ω±ω±

tj 0e ω

Integrating over an interval 0 to T (T is the fundamental period):

... (8)

Using Euler’s formula:

Thus,

... (9)

From eqn. (8):

∑ ∫∫ ∑∫∞

−∞=

ω−∞

−∞=

ω−ω− ==k

T

0

t)nk(jk

T

0 k

t)nk(jk

T

0

ntj )dt e( adt e adt e )t(x 000

n .k for ,T nkfor 0,

dt t)nksin(jdt t)nkcos(dt eT

00

T

0

T

00

t)nk(j 0

==≠=

ω−+ω−= ∫∫ ∫ω−

n kfor ,0

nk for ,T dt eT

0

t)nk(j 0

≠=

==∫ ω−

. Tadt e )t(x n

T

0

tj 0 =∫ ω−

5

or

... (10a)

As x(t) is periodic, eqn. (9) is applicable to any time interval T. Hence,

.... (10 b)

•Eqn. (7) is called Fourier synthesis equation and eqn. (10b) as Fourier analysis equation.

If x(t) is a real periodic function, then , consequently

... (11)

Comparing eqn. (7) with eqn. (11):

∫ ω−=T

0

tjnn dt e )t(x

T1a 0

∫ ω−=T

tjnn dt e )t(x

T1a 0

)t(x)t(x ∗=

ktj

kk

ktj

kk

ktj

k kk k

0

00kt0j

e a

e ae a)t(xea)t(x

ω∞

−∞=

∗−

ω−∞

∞=

∗−

ω−∞

−∞=

∞

−∞=

∗∗

∑

∑∑ ∑

=

==== ω

. aa kk −∗ =

6

Problem 3.3 p. 251 of text.Find fundamental frequency and Fourier co-efficient ak of:

Solution:Using Euler’s formula:

Hence, the fundamental frequency is ω0 = π/3. It consists of the fundamental, 2nd and 5th harmonics only.Thus,

Problem 3.22 p. 256 of text.d) Find the Fourier series of:

).t3

5( sin4)t3

2( cos2)t(x π+

π+=

∑∞

∞=

ω

π−

ππ−

π

=

−+++=

-k

ktjk

t3

5jt3

5jt3

2jt3

2j

0e a

ej2 e

j2 e

21 e

212)t(x

.a ,j2a ,

21a ,

21 a ,2a

giving ,e ae ae aeaa)t(x

j2

5-52-20

t5j5

t5j5

t2j2

t2j 20

0000

−=====

++++= ω−−

ωω−−

ω

  x(t)

1 11

-2

t0 1-1-2

-2 -2

2 3

7

Solution:

Here, T =2, ω0 = (2π/T)= π.

Problem 3.5 p. 193 of text. [Very important result]:

Find the Fourier series of:

Solution:

∑∞

−∞=

ω=k

ktjk

0e a)t(x

.)1(21)(cos

21

]) (e21[21] e21[

21 dt] e2. dt1[

21

]dt e )1t( 2dt e )t([21dt e )t(x

21a

kk

k-jπ]jk2

1jk1

0

2

1ktj1

0ktj2

0ktj

k0

−−=π−=

−=−=−=

−δ−+δ==

π−π−

π−π−ω−

∫∫

∫∫∫

 x(t)

-T1 t

1

0 T1 -T/2 T/2 T -T

.k

)kTsin(ω )kTsin(ω.kTω2

]ee [kTjω

1ee.

T1dt e

T1a

1010

0

kTjωkTjω

0

T

Tkj

ktjT

T

ktjk

1010

1

10

01

1

0

π==

−−

=== −

−ω−

ω−

−

ω−∫

8

Special case:

Convergence of the Fourier Series:

For a function to have Fourier series, the series is to be convergent. The convergence conditions for x(t) is given by Dirichlet conditions as follows:

Over any period, x(t):i) must be absolutely integrable over any period, i.e.,

Example:

The function has discontinuities and hence not convergent.

ii) does not have more than a finite number of discontinuities over any period.

iii) may have at most a finite number of finite discontinuities.

.TT2

k

kT.T2

kkTa Lima 1

110

k0k0 =π

π

=π

ω== →

∞<∫ dt)t(x

T

 

x(t)

t1 0

9

Gibbs Phenomenon:• As the number of terms N is increased, Fourier series represents the function more accurately. •At discontinuities, the series has a value ½ the sum of values just before and just after the discontinuity.•The series shows ripples at discontinuity.•As N increases, the ripples get compressed toward the discontinuity, but its peak amplitude remained unchanged.

Properties of Fourier Series:• Linearity:

•Time Shift:kk

k

k

B b A aB y(t) A x(t) Then

b )t(y a )t(x If

+→+

→→

ktjk

0

k

a . e )tt(x

a )t(x00ω−→−

→

10

Time Reversal:

Time Scaling:

Multiplication:

Conjugate:

Conjugate Symmetry:For real x(t):

k

k

a )t(xa )t(x

−→−→

∑∞

−∞=

αω=⇒→α

→

k

t)(jkkk

k

0e a a )t(x

a )t(x

. b . a )t(y).t(x

b )t(ya )t(x

llkl

k

k

∑∞

−∞=−→

→→

∗−

∗ →

→

k

k

a )t(x

a )t(x

kk

kk

a a

a a

−∗

∗−

=

=

11

Derivative:

Parseval’s Relation:

• Consult Table 3.1 p. 206 of text. • Go through example 3.8 p. 208 of text.

Examples:Prob. 3.6 p.251 of text.a)Which of the following are real valued?b) Which are even?

k0ajk dt

)t(dxω→

∑∫∞

−∞=

=k

2kT

2 adt )t(xT1

∑

∑

∑

−=

π

−=

π

=

π

π=

π=

=

100

100k

t502jk

3

100

100k

t502jk

2

100

0k

t502jkk

1

e )2

ksin(j)t(x

e )kcos()t(x

e )21()t(x

12

Solution:a)

. x1(t) is not real-valued.

x

∑∞

∞−

ω= tj k1

0ea)t(x

else. 0,

100 k 0 ,)21( a a

else. 0.

100 k 0 ,)21( a )t(x

kkk

kk1

=

≤≤==

=

≤≤=⇒

−∗

∗−≠ kk a a

real. is (t) x . aa

else 0, 100k100 ),kcos( a

else. 0, 100k100 ),kcos( a :)t(x

2kk

k

k2

∗−

−

=

=≤≤−π=

=≤≤−π=

real. is )t(x . aa

else. 0.

100k100 ,2

ksinja

else. 0.

100k100 ,2

ksinja

else. 0.

100k100 ,2

ksinja :)t(x

3kk

k

k

k3

∗−

∗−

−

=

=

≤≤−π

=

=

≤≤−π

−=

=

≤≤−π

=

13

Problem 3.22(a) p.255 of text,Find the Fourier series of x(t):

Solution:

where

∑∑∞

−∞=

π∞

−∞=

ω ==

π=π

=ω=≤≤−=

k

ktjk

k

ktjk

0

eaeax(t)

hence ,2

2 ,2T and ;1t1 for ,t)t(x

0

.)1( kj)kcos(

kj

]ee [k21]ee [

kπ2j

} dt kj

e .1kj

e .t {21dt e t

21a

k

jkjk22

jkjk

1

1

ktj1

1

ktj1

1

ktjk

−π

=ππ

=

−π

++=

π−−

π−==

ππ−π−π

−

π−

−

π−

−

π− ∫∫

14

-1

-1

1

1 2 3 -2 0 4 t

x(t)

Fourier Series Representation of DT Signals

DT Fourier Series:As with CT signals, a DT signal can be expressed as a series combination of complex exponentials. The series, however, is finite.Let x[n] be periodic with a fundamental period N, i.e.,

x[n]=x[n+N].

Consider a complex function

For an integer r ,

or ... (12)

• φk[n] repeats itself with a period of N. Thus,

• a set of N values of φk[n] is enough to represent a periodic DT signal.

15

n)N2(jknjk

k ee]n[ 0

πω ==φ

].n[ee .e

e .ee]n[

kn)

N2(jkrN2jn)

N2(jk

n)N2.(jrNn)

N2(jkn)

N2)(rNk(j

rNk

φ===

==φπ

ππ

πππ+

+

]n[]n[ rNkk +φ=φ

Therefore, a periodic DT signal may be expressed as

... (13)

where <N> means any arbitrary consecutive N exponentials.

Eqn. (13) is the DT Fourier series, ak is called Fourier coefficient.

Determination of ak:

Consider

Now:

With k = r,

.N2 ,e a]n[ a]n[x 0

Nk

njkk

Nkkk

0 π=ω=φ= ∑∑

>=<

ω

>=<

∑>=<

ω−−ω− =Nk

n)rk(jk

njr 00 eae]n[x

else. 0,

.. 2N. N, 0, r-k if N,

e a

e ae a e]n[x

Nk Nn

n)N2)(rk(j

k

Nk Nn

n)rk(jk

Nn Nk

n)rk(jk

Nn

njr 000

=±±==

=

==

∑ ∑

∑ ∑∑ ∑∑

>=< >=<

π−−

>=< >=<

ω−−

>=< >=<

ω−−

>=<

ω−

.N ae]n[x rNn

njr 0 =∑>=<

ω−

16

Replacing r by k ,

... (14)

Eqn. (13) is called synthesis equation and eqn. (14) the analysis equation.

As ak in an interval N repeats itself in any other periods,

• Since the series is finite having N terms, it is always convergent.

Example:Example 3.12 p.218 of text. Find the Fourier series of:

Here, N = 9.

. N2 ,e ]n[x

N1a 0

njk

Nnk

0 π=ω= ω−

>=<∑

... aaa N 2 kN kk === ±±

 

 

 

 

   -2 -1 0 1 2 n

x[n]

.... -N

.... N

.

2k sin

211N2k sin

N1

e e N1e

N1e

N1a

0

0

1N2

0m

mjk1Njk1N2

0m

)1Nm(jk1N

1Nn

njkk

0000

ω

+ω

=

=== ∑∑∑=

ω−ω

=

−ω−

−=

ω−

17

For k = 0,

With N=9, there are 9 terms, e.g.,a-4, a-3, a-2, a-1, a0, a1, a2, a3, and a4.

If the R.H.S. Of eqn. (14) is truncated to less than 9 terms, distortion occurs. See Fig. 3.18 p.220 of text.

Note that Gibbs phenomenon does not occur in DT signal approxi-mation.

Problem 3.9 p. 252 of text.Find the Fourier series of

Solution:

N12N1 1)(2N1

N1

2k

)11N2(k N1a

0

00

+=+=

ω+ω

=

}4m]-1-[n84m]-[n{4]n[x m

∑∞

∞−=

δ+δ= 

 

 

 

   -4 -3 -2 -1 0 1 2 3 4 5 n

x[n]

. . 4

8 8 8

4 4

. . . . . .

18

Here,

givinga0=3, a1 = 1-j2, a2= -1, a3= 1+j2.a4 = a0 =a-4 = .... = 3;a5 = a1 = a-3 = a9 = ... = 1-j2, etc.

Problem 3.30 p. 258 of text.Find the Fourier coefficients of x[n], y[n] and z[n] given by:

Solution:N=6 for x[n], y[n] and hence for z[n].

e21]e84[41]e8e4[

41

e]}1n[8]n[4{N1e]n[x

N1a

. 24

2 ,4N

k2j

kj1.kj0.knj

3

0n

knj

Nn

knjk

0

000

00

π−ω−ω−ω−

=

ω−

>=<

ω−

+=+=+=

−δ+δ==

π=

π=ω=

∑∑

].n[y . ]n[x]n[z

)4

n6

2sin(]n[y

)n6

2cos(1]n[x

=

π+

π=

π+=

.36

20

π=

π=ω

3620π=π=ω

19

a)

Hence,

b)

(c)

.ea

)ee(211)ee(

211n)

62cos(1]n[x

Nkk

jnjn3

jn3

j

kn0j

n0j0

∑>=<

−ωπ

−π

ω

ω

=

++=++=π

+=

.21a ,

21a ,1a 110 === −

.j2

e - b ,j2

eb ,0b

giving ,e j2

ee j2

e

]ee[j2

1)4

nsin()4

n3

sin(]n[y

4j-

14

j

10

nj4j-

nj4j

)4

n(j)4

n(j0

00

00

π

−

π

ω−

π

ω

π

π+ω−

π+ω

===

−=

−=π

+ω=π

+π

=

.0cc , cj4

ec ,cj2

ec

,4

sin21

j2e

210

j2e

21abababc

giving ,0ababababc

.0bbb and ,0aaa ,a and b ercov to N gsinChoo

.abc ]n[y].n[x

3324

j

214

j

1

4j

4j

1100110

1k1k01k1

4

1lklk

432432

1-1-

Nllklk

======

π=−+=++=

+++==

======><

=→

∗−

∗−

π

∗−

π

π−

π

−−

−+−−

−

>=<−

∑

∑

20

Fourier Series and LTI SystemsFor a system with an impulse response H(jω) or H[ejω] :

For CT: ... (15a)

For DT:... (15b)

CT: With Fourier series of x(t):

... (16)

DT:... (17)

To find the system output y(t) or y[n]: 1.Find H(jω) or H(ejw) using eqn. 15a or 15b,2. Find F. S. of x(t) or x[n].3.Find y(t) or y[n] using eqn. 16 or eqn.17.

 

h(t) h[n]

x(t) y(t)

x[n] y[n] H(jω) H(ejω) ejωn y[n]

ejωτ y(t)

∫∞

∞−

ωτ− ττ=ω d e )(h)j(H j

∑∞

−∞=

ω−ω =k

kjj e ]k[h)e(H

∑∞

−∞=

ω=k

ktjk , ea)t(x 0

tjk

k0k

0e )jk(H a)t(y ω∞

−∞=∑ ω=

∑>=<

ωω=Nk

nkjkjk

00 e )e(H a]n[y

21

Prob. 3.34 p. 260 of text.Given h(t)=e-4|t| , find y(t) with

(a) ,

(c) x(t) given as:

Solution:

(a)

Here, T=1, ω0 =2π.

∑∞

−∞=

−δ=n

)nt()t(x

 

x(t)

-1 -1/4 0 1/4 1 t

1 1 1

.j4

1j-4

1

d e ed e ed e )(h)j(H0

j4-0

j4j

ω++

ω=

τ+τ=ττ=ω ∫∫∫∞

ωτ−τ

∞−

ωτ−τ∞

∞−

ωτ−

 

x(t)

1

-2 -1 0 1 2 t

... ...∑∞

−∞=

ω=k

ktjk . ea)t(x 0

∫

∫ ∑∫ ∑

==

δ=−δ= ω−∞

−∞=

ω−

1

0

T

1

0

0.kjT

n

ktjk

1dt.111

dt e )t(T1dt e )nt(

T1a 00

22

Hence,

(c)

where,

Prob. 3.38 p. 261:Given:

.e ]k2j4

1k2j4

1[

e )k2j(He )kj(H a)t(y

k

kt2j

kt2j

k

ktj

k0k

0

∑

∑∑∞

−∞=

π

π∞

−∞=

ω∞

−∞=

π++

π−=

π=ω=

∑∑

∑∑

∞

−∞=

ππ∞

−∞=

π∞

−∞=

ω∞

−∞=

=π+

+π−π

π

=

π

π

=π

ω=

k

kt2jk

kt2j

k

kt2j

k

ktj

k

10

e be ]k2j4

1k2j4

1[ k

)2ksin(

)t(y

.e k

)2ksin(

e k

)kTsin()t(x 0

odd. k for ],k2j4

1k2j4

1[k

)k2

sin(

0 k for ,41

even k for ,0bk

→π+

+π−π

π

=

==

→=

∑∞

−∞=

−δ=

=−≤≤−−=

≤≤=

k

].k4n[]n[x

.else .0 1n2 ,1

2n0 ,1]n[h

23

Find the Fourier coefficients of output y[n].

Where,

 

h[k]

-2 -1 0 2 k

1

-1

ωωωω=

ω−−

−=

ω−

∞

−∞=

ω−ω

−++−=

+=

=

∑∑

∑

j2j2-jj-

2

0k

kj1

2k

kj

k

kjj

ee1ee

e .1e -1)(

e ]k[h)e(H

.41e

41e ]n[

41e ]n[x

N1a

.24

2 4,N here ,e a]n[x

0.k0j00

0

3

0n

knj

Nn

knjk

0Nk

njk

==δ==

π=

π=ω==

ω−

∑∑

∑

=

ω−

>=<

ω−

>=<

ω

∑

∑

∑

=

ω

ωωω−ω

=

ω−

=

ωω

=

−++−=

=

3

0k

knjk

knjk2jk2jkj3

0k

kj

3

0k

knjkjk

0

00000

00

e b41

e]ee1ee[41

e )e(H a]n[y

].ee1[41

]ee1ee[41b

k2

jk2

j

kjkjk2

jk2

jk

ππ−

ππ−ππ

−

−+=

−++−=

  x[n] 1

-4 -3 -2 -1 0 1 2 3 4 n . . . . . .

24

Frequency-Shaping or Frequency-Selective Filters:Ideal Filters:Low-pass:

High-pass:

Band-pass:

Band-stop:

Non-ideal Filters:CT Filter:

25

0,

,1)j(H

c

c

ω>ω=

ω<ω=ω 

|H(jω)|

-ωc 0 ωc ω

1

1,

,0)j(H

c

c

ω>ω=

ω<ω=ω

 

|H(jω)|

-ω2 −ω1 0 ω1 ω2 ω

1 1 else. 0

,1)j(H 21

=

ω<ω<ω=ω

 

|H(jω)|

-ω2 −ω1 0 ω1 ω2 ω

1 1 else. 1,

,0)j(H 21

=

ω<ω<ω=ω

R

C y(t) = v0(t) x(t) = vi(t)

 

|H(jω)|

-ωc 0 ωc ω

1 1

System differential equation:

The response plot:

This is a low-pass filter.

26

)t(x)t(ydt

)t(dyRC =+

0

1-

)(1

1

0

0

tjtjtj

-tan)H(j ,)H(j

RC1 ,

j1

1RCj1

1)X(j)Y(j) H(j

giving 1, )RC)H(jj(1 or

e)j(Hjdt

)t(dy ,e )H(j y(t),e)t(x

2

0

ωω

=ω∠=ω

=ω

ωω

+=

ω+=

ωω

=ω

=ωω+

ωω=ω==

ωω

+

ωωω

 

-ω0 0 ω0

|H(jω)|

ω

21

)j(H ω∠

ω

4π

4π

−  

−ω0 ω0

DT FiltersConsider a first-order filter given by the difference equation:

y[n] – a y[n-1] = x[n], |a| < 1.

With x[n] = ejωn,

The magnitude |H(ejω)| and the phase angle for a=0.6 and a=-0.6 are shown in Fig, 3,34 p. 246 of the text.

• For 0 < a < 1 , it is low-pass filter.

• For 0> a > -1, it is high-pass filter.

27

. e a1

1)e(H

giving ,ee ]ae1)[e(H

or ,ee )e(H ae ).e(H

jj

njnjjj

nj)1n(jjnjj

ω−ω

ωωω−ω

ω−ωωωω

−=

=−

=−

)e(H jω∠

Problem 3.16 p. 253 of text.With the given impulse response H(ejω) , find the filter output with the following inputs:

Solution:

Here, N=2 and ω0=π.

None is passed.

28

)48

3sin(1]n[x )b

)1(]n[x )a

2

n1

π+

π+=

−=

n1 )1(]n[x )a −=

 

|H(ejω)|

π −5π/12 −π/3 0 π/3 5π/12 π ω

1 1

  x1[n] 1

-2 -1 0 1 2 n

-1

. . . . . .

0). )e(H ce(sin .0

e ).e(H ae )e(H a]n[y

.1a ,0a

giving ),e1(21e)1(

21a

j

njj1

1

0kk

10

1

0n

kjknjnk

knjkj

==

==

==

−=−=

π

ππ

=

=

π−π−

π−π−

∑

∑

b)

Hence,

ω= 1 is blocked, ω= 3π/8 is passed.

29

0. others ,ej2

1a ,ej2

1a ,1a

e )e j2

1(e )e j2

1(1]ee[j2

11]n[x

.816

2 16,N )48

3sin(1]n[x

4j

34

j30

83j

4j

83j

4j)

483(j)

483(j

2

02

=−===⇒

−+=−+=

π=

π=ω=⇒

π+

π+=

π−

−

π

π−π

−πππ

+π

−π

+π

)4

n8

3sin(]ee[j2

1

e.1.e2j1e.1.e

2j1

e )e(H ae )e(H a0

e )e(H ae )e(H ae )e(H a

e )e(H a]n[y

)4

n8

3(j)4

n8

3(j

n83j

4jn

83j

4j

n8

3j8

3j3

n8

3j8

3j3

n3j3j3

n3j3j3

0j0j0

Nk

knjkjk

0000

00

π+

π=−=

−=

++=

++=

=

π+

π−

π+

π

π−

π−

ππ

π−

π−

−

ππ

ω−ω−−

ωω

>=<

ωω∑