Fourier Series

EEL205 : Signals and Systems

Kushal K. ShahAsst. Prof. @ EE, IIT Delhi

Email : [email protected] : http://web.iitd.ac.in/~kkshah

Fourier Series

Response of LTI systems to complex exponentials

continuous time : est → H (s)est

discrete time : zn → H (z)zn

y (t) =

ˆ∞

−∞

x (τ)h (t− τ)dτ

=

ˆ∞

−∞

h (τ)x (t− τ)dτ

=

ˆ∞

−∞

h (τ)es(t−τ)dτ

= estˆ

∞

−∞

h (τ)e−sτdτ

= H (s)est




y (t) =

ˆ∞

−∞


=

ˆ∞

−∞


=

ˆ∞

−∞

h (τ)es(t−τ)dτ

= estˆ

∞

−∞

h (τ)e−sτdτ

= H (s)est




y (t) =

ˆ∞

−∞


=

ˆ∞

−∞


=

ˆ∞

−∞

h (τ)es(t−τ)dτ

= estˆ

∞

−∞

h (τ)e−sτdτ

= H (s)est




y (t) =

ˆ∞

−∞


=

ˆ∞

−∞


=

ˆ∞

−∞

h (τ)es(t−τ)dτ

= estˆ

∞

−∞

h (τ)e−sτdτ

= H (s)est




y (t) =

ˆ∞

−∞


=

ˆ∞

−∞


=

ˆ∞

−∞

h (τ)es(t−τ)dτ

= estˆ

∞

−∞

h (τ)e−sτdτ

= H (s)est

x (t) = a1es1t +a2e

s2t +a3es3t

⇒ y (t) = a1H (s1)es1t +a2H (s2)e

s2t +a3H (s3)es3t

x (t) = ∑k

akesk t

⇒ y (t) = ∑k

akH (sk)esk t : Continuous Time

x [n] = ∑k

akznk

⇒ y [n] = ∑k

akH (zk)znk : Discrete Time

If the input to an LTI system is represented

as a linear combination of complex exponentials,

then the output can also be represented

as a linear combination of the same complex exponential signals.

Fourier series of periodic continuous time signals

Any periodic continuous time signal with period, T = 2π

/ω0, can be

written as a sum of complex exponentials

x (t) =∞

∑k=−∞

akesk t

wheresk = jkω0 = jk

2π

T

How to determine ak?

Determination of spectral coefficients, ak

x (t) =∞

∑k=−∞

akejkω0t

Multiplying both sides by e−jnω0t and integrating from 0 to T , we get

ˆ T

0

x (t)e−jnω0tdt =

ˆ T

0

∞

∑k=−∞

akejkω0te−jnω0tdt

=∞

∑k=−∞

ak

ˆ T

0

e j(k−n)ω0tdt

=∞

∑k=−∞

akTδ [k−n] = Tan

an =1

T

ˆ T

0

x (t)e−jnω0tdt


x (t) =∞

∑k=−∞

akejkω0t


ˆ T

0

x (t)e−jnω0tdt =

ˆ T

0

∞

∑k=−∞


=∞

∑k=−∞

ak

ˆ T

0

e j(k−n)ω0tdt

=∞

∑k=−∞

akTδ [k−n] = Tan

an =1

T

ˆ T

0

x (t)e−jnω0tdt


x (t) =∞

∑k=−∞

akejkω0t


ˆ T

0

x (t)e−jnω0tdt =

ˆ T

0

∞

∑k=−∞


=∞

∑k=−∞

ak

ˆ T

0

e j(k−n)ω0tdt

=∞

∑k=−∞

akTδ [k−n] = Tan

an =1

T

ˆ T

0

x (t)e−jnω0tdt


x (t) =∞

∑k=−∞

akejkω0t


ˆ T

0

x (t)e−jnω0tdt =

ˆ T

0

∞

∑k=−∞


=∞

∑k=−∞

ak

ˆ T

0

e j(k−n)ω0tdt

=∞

∑k=−∞

akTδ [k−n] = Tan

an =1

T

ˆ T

0

x (t)e−jnω0tdt


x (t) =∞

∑k=−∞

akejkω0t


ˆ T

0

x (t)e−jnω0tdt =

ˆ T

0

∞

∑k=−∞


=∞

∑k=−∞

ak

ˆ T

0

e j(k−n)ω0tdt

=∞

∑k=−∞

akTδ [k−n] = Tan

��

� an =

1

T

ˆ T

0

x (t)e−jnω0tdt

Continuous-time FS : Example 1

x (t) = sinω0t

=e jω0t − e−jω0t

2j

=1

2je jω0t − 1

2je−jω0t

x (t) =∞

∑k=−∞

akejkω0t

⇒ ak =

{1

2j, k =±1

0 , k 6=±1


x (t) = sinω0t


2j

=1

2je jω0t − 1

2je−jω0t

x (t) =∞

∑k=−∞

akejkω0t

⇒ ak =

{1

2j, k =±1

0 , k 6=±1


x (t) = sinω0t


2j

=1

2je jω0t − 1

2je−jω0t

x (t) =∞

∑k=−∞

akejkω0t

a1 =1

2j

a−1 =1

2j


x (t) = sinω0t


2j

=1

2je jω0t − 1

2je−jω0t

x (t) =∞

∑k=−∞

akejkω0t

⇒ ak =

{1

2j, k =±1

0 , k 6=±1


x (t) = 1+ sinω0t +2cosω0t + cos(2ω0t +

π

4

)= 1+

1

2j

[e jω0t − e−jω0t

]+[e jω0t + e−jω0t

]+

1

2

e j(2ω0t+π

/4

)+ e−j(2ω0t+π

/4

)= 1+

(1+

1

2j

)e jω0t +

(1− 1

2j

)e−jω0t

+1

2ejπ

/4

e j2ω0t +1

2e−jπ/

4

e−j2ω0t

= a0 +a1ejω0t +a−1e

−jω0t +a2e2jω0t +a−2e

−2jω0t



π

4

)= 1+

1

2j



]+

1

2

e j(2ω0t+π

/4

)+ e−j(2ω0t+π

/4

)= 1+

(1+

1

2j

)e jω0t +

(1− 1

2j

)e−jω0t

+1

2ejπ

/4

e j2ω0t +1

2e−jπ/

4

e−j2ω0t



−2jω0t



π

4

)= 1+

1

2j



]+

1

2

e j(2ω0t+π

/4

)+ e−j(2ω0t+π

/4

)= 1+

(1+

1

2j

)e jω0t +

(1− 1

2j

)e−jω0t

+1

2ejπ

/4

e j2ω0t +1

2e−jπ/

4

e−j2ω0t



−2jω0t



π

4

)= 1+

1

2j



]+

1

2

e j(2ω0t+π

/4

)+ e−j(2ω0t+π

/4

)= 1+

(1+

1

2j

)e jω0t +

(1− 1

2j

)e−jω0t

+1

2ejπ

/4

e j2ω0t +1

2e−jπ/

4

e−j2ω0t



−2jω0t

Dirichlet conditions

Every continuous periodic signal has a Fourier series representation

For discontinuous signals :

1. Over any period, x (t) must be absolutely integrable

ˆ T

0

|x (t)|dt < ∞

Ex : x (t) = 1/t, 0 < t ≤ 1 violates 1

ˆ1

0

1

tdt = log t

∣∣∣10

= ∞


2. In any finite interval of time, x (t) is of bounded variation;(finite number of maxima and minima in that period)Ex : x (t) = sin

(2π

/t)

, 0 < t ≤ 1 satisfies 1 but violates 2


3. In any finite interval of time, there are only a finite number ofdiscontinuities and each of these discontinuities is finite.


Every continuous periodic signal has a Fourier series representation

For discontinuous signals :

1. Over any period, x (t) must be absolutely integrable

ˆ T

0

|x (t)|dt < ∞

Ex : x (t) = 1/t, 0 < t ≤ 1 violates 1

2. In any finite interval of time, x (t) is of bounded variation;(finite number of maxima and minima in that period)Ex : x (t) = sin

(2π

/t)

, 0 < t ≤ 1 satisfies 1 but violates 2

3. In any finite interval of time, there are only a finite number ofdiscontinuities and each of these discontinuities is finite.

Fourier series of a square wave

x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =1

T

ˆT

x (t)e−jkω0tdt

=1

T

ˆ T1

−T1

e−jkω0tdt

=1

jkω0T

[e jkω0T1− e−jkω0T1

]=

2sin(kω0T1)

kω0T

=sin(kω0T1)

πksince ω0 = 2π

/T


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =1

T

ˆT

x (t)e−jkω0tdt

=1

T

ˆ T1

−T1

e−jkω0tdt

=1

jkω0T


]=

2sin(kω0T1)

kω0T

=sin(kω0T1)

πksince ω0 = 2π

/T


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =1

T

ˆT

x (t)e−jkω0tdt

=1

T

ˆ T1

−T1

e−jkω0tdt

=1

jkω0T


]=

2sin(kω0T1)

kω0T

=sin(kω0T1)

πksince ω0 = 2π

/T


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =1

T

ˆT

x (t)e−jkω0tdt

=1

T

ˆ T1

−T1

e−jkω0tdt

=1

jkω0T


]=

2sin(kω0T1)

kω0T

=sin(kω0T1)

πksince ω0 = 2π

/T


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =1

T

ˆT

x (t)e−jkω0tdt

=1

T

ˆ T1

−T1

e−jkω0tdt

=1

jkω0T


]=

2sin(kω0T1)

kω0T

=sin(kω0T1)

πksince ω0 = 2π

/T


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=0

∑k=−∞

sin(kω0T1)

πke jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=∞

∑k=0

sin(kω0T1)

πke−jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

sin(kω0T1)

πk

(e jkω0t + e−jkω0t

)=

ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=0

∑k=−∞

sin(kω0T1)

πke jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=∞

∑k=0

sin(kω0T1)

πke−jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

sin(kω0T1)

πk


)=

ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=0

∑k=−∞

sin(kω0T1)

πke jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=∞

∑k=0

sin(kω0T1)

πke−jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

sin(kω0T1)

πk


)=

ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=0

∑k=−∞

sin(kω0T1)

πke jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=∞

∑k=0

sin(kω0T1)

πke−jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

sin(kω0T1)

πk


)=

ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=0

∑k=−∞

sin(kω0T1)

πke jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=∞

∑k=0

sin(kω0T1)

πke−jkω0t +

ω0T1

π+

∞

∑k=0

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

sin(kω0T1)

πk


)=

ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

=∞

∑k=−∞

akejkω0t =

∞

∑k=−∞

sin(kω0T1)

πke jkω0t

=ω0T1

π+

∞

∑k=0

2sin(kω0T1)

πkcos(kω0t)

On a computer:

x (t) =ω0T1

π+

N

∑k=0

2sin(kω0T1)

πkcos(kω0t)

Gibbs PhenomenonGibbs phenomenon is the peculiar way in which the Fourier series

of a piecewise continuously differentiable periodic functionbehaves at a jump discontinuity.

x (t) =ω0T1

π+

N

∑k=0

2sin(kω0T1)

πkcos(kω0t)

Properties of Fourier Series : Linearity

x (t)FS←→ ak

y (t)FS←→ bk

If x (t) and y (t) have the same period, T ,

Ax (t)+By (t)FS←→ Aak +Bbk

Properties of Fourier Series : Time Shifting

x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT

x (t− t0)e−jkω0tdt

=1

T

ˆT

x (τ)e−jkω0(τ+t0)dτ

=1

T

ˆT

x (τ)e−jkω0τe−jkω0t0dτ

= e−jkω0t01

T

ˆT

x (τ)e−jkω0τdτ = e−jkω0t0ak

x (t− t0)FS←→ e

−jk(2π

/T

)t0ak

No change in magnitude of FS coefficients due to time-delay


x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT


=1

T

ˆT


=1

T

ˆT


= e−jkω0t01

T

ˆT



−jk(2π

/T

)t0ak



x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT


=1

T

ˆT


=1

T

ˆT


= e−jkω0t01

T

ˆT



−jk(2π

/T

)t0ak



x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT


=1

T

ˆT


=1

T

ˆT


= e−jkω0t01

T

ˆT



−jk(2π

/T

)t0ak



x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT


=1

T

ˆT


=1

T

ˆT


= e−jkω0t01

T

ˆT



−jk(2π

/T

)t0ak



x (t)FS←→ ak

y (t) = x (t− t0)FS←→ bk

bk =1

T

ˆT


=1

T

ˆT


=1

T

ˆT


= e−jkω0t01

T

ˆT



−jk(2π

/T

)t0ak


Properties of Fourier Series : Time Reversal

x (t)FS←→ ak=

1

T

ˆT

x (τ)e−jkω0τdτ

y (t) = x (−t) FS←→ bk

bk =1

T

ˆT

x (−t)e−jkω0tdt

=1

T

ˆT

x (τ)e jkω0τdτ

= a−k

x (−t) FS←→ a−k

If x (t) is even, ak = a−k (FS coefficients are also even)

If x (t) is odd, ak =−a−k (FS coefficients are also odd)


x (t)FS←→ ak=

1

T

ˆT


y (t) = x (−t) FS←→ bk

bk =1

T

ˆT


=1

T

ˆT

x (τ)e jkω0τdτ

= a−k





x (t)FS←→ ak=

1

T

ˆT


y (t) = x (−t) FS←→ bk

bk =1

T

ˆT


=1

T

ˆT

x (τ)e jkω0τdτ

= a−k





x (t)FS←→ ak =

1

T

ˆT


y (t) = x (−t) FS←→ bk

bk =1

T

ˆT


=1

T

ˆT

x (τ)e jkω0τdτ

= a−k





x (t)FS←→ ak=

1

T

ˆT


y (t) = x (−t) FS←→ bk

bk =1

T

ˆT


=1

T

ˆT

x (τ)e jkω0τdτ

= a−k





x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =sin(kω0T1)

πkω0 = 2π

/T

x (t)is even,

x (t) = x (−t)⇒ ak = a−k

Properties of Fourier Series : Time Scaling

x (t)FS←→ ak

y (t) = x (αt)FS←→ bk

bk =α

T

ˆT

/α

x (αt)e−jk(αω0)tdt

=α

T

ˆT

x (τ)e−jkω0τ dτ

α, τ = αt

=1

T

ˆT


= ak


x (t)FS←→ ak


bk =α

T

ˆT

/α


=α

T

ˆT


α, τ = αt

=1

T

ˆT


= ak


x (t)FS←→ ak


bk =α

T

ˆT

/α


=α

T

ˆT


α, τ = αt

=1

T

ˆT

x (τ)e−jkω0τdτ = ak


x (t) =∞

∑k=−∞

akejkω0t

⇒ y (t) = x (αt) =∞

∑k=−∞

akejkω0(αt)

=∞

∑k=−∞

akejk(αω0)t

x (t)FS←→ ak

y (t) = x (αt)FS←→ ak

FS coefficients remain same butFS representation is different


x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =sin(2πkT1

/T)

πkω0 = 2π

/T

x (αt) ←→ ak

Properties of Fourier Series : Multiplication

x (t)FS←→ ak

y (t)FS←→ bk

x (t)y (t)FS←→ hk=

∞

∑l=−∞

ak−lbl = ak ∗bk

x (t)y (t) =∞

∑k=−∞

akejkω0t

∞

∑l=−∞

blejlω0t

=∞

∑k=−∞

∞

∑l=−∞

akblej(k+l)ω0t

=∞

∑m=−∞

(∞

∑l=−∞

am−lbl

)e jmω0t

Valid only if x (t) and y (t) have same time-period


x (t)FS←→ ak

y (t)FS←→ bk


∞

∑l=−∞

ak−lbl = ak ∗bk

x (t)y (t) =∞

∑k=−∞

akejkω0t

∞

∑l=−∞

blejlω0t

=∞

∑k=−∞

∞

∑l=−∞

akblej(k+l)ω0t

=∞

∑m=−∞

(∞

∑l=−∞

am−lbl

)e jmω0t



x (t)FS←→ ak

y (t)FS←→ bk


∞

∑l=−∞

ak−lbl = ak ∗bk

x (t)y (t) =∞

∑k=−∞

akejkω0t

∞

∑l=−∞

blejlω0t

=∞

∑k=−∞

∞

∑l=−∞

akblej(k+l)ω0t

=∞

∑m=−∞

(∞

∑l=−∞

am−lbl

)e jmω0t



x (t)FS←→ ak

y (t)FS←→ bk

x (t)y (t)FS←→ hk =

∞

∑l=−∞

ak−lbl = ak ∗bk

x (t)y (t) =∞

∑k=−∞

akejkω0t

∞

∑l=−∞

blejlω0t

=∞

∑k=−∞

∞

∑l=−∞

akblej(k+l)ω0t

=∞

∑m=−∞

(∞

∑l=−∞

am−lbl

)e jmω0t



x (t) =

{1 , |t|< T1

0 ,T1 < |t|< T/2

ak =sin(kω0T1)

πkω0 = 2π

/T

x (t)x (t) = x (t)

⇒ ak ∗ak = ak

Properties of Fourier Series : Conjugation

x (t)FS←→ ak

x∗ (t)FS←→ a∗−k

x (t) =∞

∑k=−∞

akejkω0t

⇒ x∗ (t) =∞

∑k=−∞

a∗ke−jkω0t =

∞

∑k=−∞

a∗−kejkω0t


x (t)FS←→ ak


x (t) =∞

∑k=−∞

akejkω0t

⇒ x∗ (t) =∞

∑k=−∞

a∗ke−jkω0t =

∞

∑k=−∞

a∗−kejkω0t


x (t)FS←→ ak


If x (t) is real, then

x (t) = x∗ (t)

⇒ ak = a∗−k

FS coefficients are conjugate symmetric


x (t)FS←→ ak


If x (t) is real(ak = a∗−k

)and even (ak = a−k), then

ak = a∗k

FS coefficients are also real and even


x (t)FS←→ ak


If x (t) is real(ak = a∗−k

)and odd (ak =−a−k), then

ak =−a∗k

FS coefficients are purely imaginary and odd with a0 = 0


x (t)FS←→ ak


x(t) FS coefficientsreal conjugate symmetric

real and even real and evenreal and odd purely imaginary and odd with a0 = 0

Properties of Fourier Series : Parseval’s Relation

1

T

ˆT

|x (t)|2 dt =∞

∑k=−∞

|ak |2

Total average power in a periodic signal

equals

the sum of the average powers in all of its harmonic components

Fourier Seriesof

Discrete-time Periodic Signals

Fourier series representation

of a discrete-time periodic signal is a finite series,

as opposed to the infinite series representation

required for continuous-time periodic signals.

x [n] = x [n+N]

ω0 = 2π

/N

Set of all discrete-time complex exponential signals with the sameperiod, N:

φk [n] = e jkω0n = ejk

(2π

/N

)n

,k = 0,±1,±2, ...

In the above, there are only N distinct signals,

φk+N [n] = e j(k+N)ω0n

= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

x [n] = x [n+N]

ω0 = 2π

/N



(2π

/N

)n

,k = 0,±1,±2, ...



= e jkω0ne jNω0n

= e jkω0nejN

(2π

/N

)n

= e jkω0ne j2πn

= e jkω0n

= φk [n]

ar : spectral coefficients of x [n]

x [n] =N−1

∑k=0

akφk [n] =N−1

∑k=0

akejkω0n

Multiplying both sides by e−jrω0n and summing, we get

N−1

∑n=0

x [n]e−jrω0n =N−1

∑n=0

N−1

∑k=0

akejkω0ne−jrω0n

=N−1

∑k=0

ak

N−1

∑n=0

e j(k−r)ω0n

= Nar

ar =1

N

N−1

∑n=0

x [n]e−jrω0n


x [n] =N−1

∑k=0

akφk [n] =N−1

∑k=0

akejkω0n


N−1

∑n=0


∑n=0

N−1

∑k=0

akejkω0ne−jrω0n

=N−1

∑k=0

ak

N−1

∑n=0

e j(k−r)ω0n

= Nar

ar =1

N

N−1

∑n=0

x [n]e−jrω0n


x [n] =N−1

∑k=0

akφk [n] =N−1

∑k=0

akejkω0n


N−1

∑n=0


∑n=0

N−1

∑k=0

akejkω0ne−jrω0n

=N−1

∑k=0

ak

N−1

∑n=0

e j(k−r)ω0n

= Nar

ar =1

N

N−1

∑n=0

x [n]e−jrω0n


x [n] =N−1

∑k=0

akφk [n] =N−1

∑k=0

akejkω0n


N−1

∑n=0


∑n=0

N−1

∑k=0

akejkω0ne−jrω0n

=N−1

∑k=0

ak

N−1

∑n=0

e j(k−r)ω0n

= Nar

ar =1

N

N−1

∑n=0

x [n]e−jrω0n


x [n] =N−1

∑k=0

akφk [n] =N−1

∑k=0

akejkω0n


N−1

∑n=0


∑n=0

N−1

∑k=0

akejkω0ne−jrω0n

=N−1

∑k=0

ak

N−1

∑n=0

e j(k−r)ω0n

= Nar��

��ar =

1

N

N−1

∑n=0

x [n]e−jrω0n

Range of the summation for discrete-time FS

ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑

n=〈N〉x [n]e−jkω0n

ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n+N]e−jkω0(n+N)

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n


ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑


ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0


=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n


ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑


ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0


=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n


ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑


ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0


=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n


ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑


ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0


=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n


ak =1

N

N−1

∑n=0

x [n]e−jkω0n =1

N∑


ak =1

N

N+M−1

∑n=M

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

N+M−1

∑n=N

x [n]e−jkω0n

=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0


=1

N

N−1

∑n=M

x [n]e−jkω0n +1

N

M−1

∑n=0

x [n]e−jkω0n

=1

N

N−1

∑n=0

x [n]e−jkω0n

FS of discrete-time periodic square wave

x [n] =

{1 , |n| ≤ N1

0 , N1 < |n| ≤ N/2

ak =1

N∑

n=〈N〉x [n]e−jkω0n ω0 = 2π

/N

=1

N

N1

∑n=−N1

e−jkω0n =1

N

2N1

∑n=0

e−jkω0(n−N1)

=1

Ne jkω0N1

2N1

∑n=0

e−jkω0n =1

Ne jkω0N1

1− e−jkω0(2N1+1)

1− e−jkω0

=sin[kω0

(N1 +1

/2)]

N sin[kω0

/2]


x [n] =

{1 , |n| ≤ N1

0 , N1 < |n| ≤ N/2

ak =1

N∑


/N

=1

N

N1

∑n=−N1

e−jkω0n =1

N

2N1

∑n=0

e−jkω0(n−N1)

=1

Ne jkω0N1

2N1

∑n=0

e−jkω0n =1

Ne jkω0N1

1− e−jkω0(2N1+1)

1− e−jkω0

=sin[kω0

(N1 +1

/2)]

N sin[kω0

/2]


x [n] =

{1 , |n| ≤ N1

0 , N1 < |n| ≤ N/2

ak =1

N∑


/N

=1

N

N1

∑n=−N1

e−jkω0n =1

N

2N1

∑n=0

e−jkω0(n−N1)

=1

Ne jkω0N1

2N1

∑n=0

e−jkω0n =1

Ne jkω0N1

1− e−jkω0(2N1+1)

1− e−jkω0

=sin[kω0

(N1 +1

/2)]

N sin[kω0

/2]


x [n] =

{1 , |n| ≤ N1

0 , N1 < |n| ≤ N/2

ak =1

N∑


/N

=1

N

N1

∑n=−N1

e−jkω0n =1

N

2N1

∑n=0

e−jkω0(n−N1)

=1

Ne jkω0N1

2N1

∑n=0

e−jkω0n =1

Ne jkω0N1

1− e−jkω0(2N1+1)

1− e−jkω0

=sin[kω0

(N1 +1

/2)]

N sin[kω0

/2]


x [n] =

{1 , |n| ≤ N1

0 , N1 < |n| ≤ N/2

ak =1

N∑


/N

=1

N

N1

∑n=−N1

e−jkω0n =1

N

2N1

∑n=0

e−jkω0(n−N1)

=1

Ne jkω0N1

2N1

∑n=0

e−jkω0n =1

Ne jkω0N1

1− e−jkω0(2N1+1)

1− e−jkω0

=sin[kω0

(N1 +1

/2)]

N sin[kω0

/2]

There are no convergence issues or Gibbs phenomenon

for discrete-time periodic signals

due to the presence of only a finite number of terms

in their Fourier series representation

Date post:	26-Jan-2016
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Fourier Series

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