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Discrete-Time Fourier Transform

The document discusses the discrete-time Fourier transform (DTFT). It provides the definition of the DTFT as the sum from negative infinity to positive infinity of the discrete signal multiplied by an exponential term. An example problem calculates the DTFT of a rectangular pulse and draws its magnitude spectrum, showing it is a sinc function.

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114 views5 pages

Discrete-Time Fourier Transform

The document discusses the discrete-time Fourier transform (DTFT). It provides the definition of the DTFT as the sum from negative infinity to positive infinity of the discrete signal multiplied by an exponential term. An example problem calculates the DTFT of a rectangular pulse and draws its magnitude spectrum, showing it is a sinc function.

Uploaded by

anil kadle
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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DISCRETE-TIME FOURIER TRANSFORM

Fourier Series  Periodic signals


Fourier Transform  Non-Periodic Signals.
DTFT and CTFT

Discrete-time Fourier transform is applied on discrete non-periodic signals.

The DTFT of the signal x [n] is given as


∞
X ( e j Ω )= ∑ x [ n ] e− j Ωn
n=−∞

e− j Ωn=cosΩn− jsinΩn

Inverse DTFT is given as,


❑
1
x [ n ] = ∫ X ( e j Ω ) e j Ωn d Ω
2 π 2π
Problems on DTFT
Prob 1: Find the DTFT of the signal x [ n ] =α n u [ n ] ,|α|<1. Draw the
magnitude spectrum

Sol:
∞
X ( e j Ω )= ∑ x [ n ] e− j Ωn
n=−∞

∞
X ( e j Ω )= ∑ α n u [ n ] e− j Ωn
n=−∞

u[n]

-2 -1 0 1 2 3 4 5 6

∞
X ( e j Ω )=∑ α n e− j Ωn
n=0

∞
n
X ( e j Ω )=∑ ( α e− j Ω )
n=0

1
X ( e− j Ω )=
1−α e− j Ω

Magnitude spectrum |α |<1  -1<α <1


i. 0< α <1 ii. −1<α <0

-2π -π 0 π 2π
-2π -π 0 π 2π

Prob 2: Find the DTFT of the signalδ [ n ]. Draw the magnitude


spectrum
Sol:
∞
X ( e j Ω )= ∑ x [ n ] e− j Ωn
n=−∞

∞
jΩ
X ( e )= ∑ δ [ n ] e− j Ωn
n=−∞

δ[n]

-2 -1 0 1 2

X ( e j Ω )=δ [ 0 ] e− j Ω 0

X ( e− j Ω )=1

Magnitude spectrum

-2π -π 0 π 2π

Prob 3: Find the DTFT of the signal x [ n ] ={1 ,3 ,5 , 3 ,1 } and evaluate


X ( e j Ω )at Ω=0

Sol:
∞
X ( e j Ω )= ∑ x [ n ] e− j Ωn
n=−∞

X ( e j Ω )=…+ x [ −2 ] e− j Ω (−2)+ x [ −1 ] e− j Ω (−1 )+ x [ 0 ] e− j Ω 0 + x [ 1 ] e− j Ω 1 + x [ 2 ] e− j Ω 2 +…


X ( e j Ω )=x [−2 ] e j Ω 2 + x [−1 ] e j Ω 1 + x [ 0 ] e− j Ω 0+ x [ 1 ] e− j Ω 1+ x [ 2 ] e− j Ω 2

X ( e j Ω )=1 e j Ω 2 +3 e j Ω 1 +5+3 e− j Ω 1+1 e− j Ω 2

X ( e j Ω )=( e j 2 Ω+ e− j 2 Ω ) +3 ( e j Ω 1+ e− j Ω 1 ) +5

X ( e j Ω )=5+2 cos ( 2 Ω ) +6 cos ⁡(Ω)

They have also asked to evaluate X ( e j Ω )at Ω=0


X ( e j Ω )=5+2 cos ( 0 )+ 6 cos ⁡(0)

X ( e j Ω )=5+2+6

X ( e j Ω ) |Ω=0=13

Prob 4: Find the DTFT of a rectangular pulse defined as

x [ n ] = 1 ;∨n∨≤ m
{ 0 ;|n|>m

Draw the magnitude spectrum

Sol:
∞
X ( e j Ω )= ∑ x [ n ] e− j Ωn
n=−∞

x[n]

-m m
m
jΩ
X ( e )= ∑ 1 e− j Ωn
n=−m

n n−m +1
∑ ak=am 1−a
1−a
k=m

1−e− j Ω (m +m +1)
X ( e j Ω )=e− j Ω (−m) −jΩ
1−e

1−e− j Ω (2 m+ 1)
X ( e j Ω )=e j Ωm
1−e− j Ω

e x =e x/ 2 . e x /2

1−e− j Ω (2 m+ 1)/ 2 e− j Ω (2 m +1)/ 2


X ( e j Ω )=e j Ωm
1−e− j Ω/ 2 e− j Ω /2

e− j Ω ( 2 m+1 )/2 e j Ω ( 2 m+1 )/2 −e− j Ω ( 2m +1) /2


X ( e j Ω )=e j Ωm −jΩ/2 j Ω /2 − j Ω /2
e e −e

2 jsin Ω ( 2m+1 ) /2
X ( e j Ω )=
2 jsin Ω/2

sin Ω ( 2m+1 ) /2
X ( e j Ω )=
sin Ω/2

Magnitude spectrum

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