UNIT - 1: Discrete Fourier Transforms
(DFT)[1, 2, 3, 4, 5]
Dr. Manjunatha. P
manjup.jnnce@gmail.com
Professor
Dept. of ECE
J.N.N. College of Engineering, Shimoga
September 11, 2014
DSP Syllabus
Introduction
Digital Signal Processing: Introduction [1, 2, 3, 4]
Slides are prepared to use in class room purpose, may be used as a
reference material
All the slides are prepared based on the reference material
Most of the figures/content used in this material are redrawn, some
of the figures/pictures are downloaded from the Internet.
I will greatly acknowledge for copying the some the images from the
Internet.
This material is not for commercial purpose.
This material is prepared based on Digital Signal Processing for
ECE/TCE course as per Visvesvaraya Technological University (VTU)
syllabus (Karnataka State, India).
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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DSP Syllabus
DSP Syllabus
PART - A
UNIT - 1: Discrete Fourier Transforms (DFT)
Frequency domain sampling and reconstruction of discrete time signals.
DFT as a linear transformation, its relationship with other transforms.
Dr. Manjunatha. P
(JNNCE)
7 Hours
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
The concept of frequency in continuous and discrete time signals
The concept of frequency in continuous and discrete time signals
Continuous Time Sinusoidal Signals
The concept of frequency is closely related to specific type of motion called harmonic
oscillation which is directly related to the concept of time.
A simple harmonic oscillation is mathematically described by:
xa (t) = Acos(Ωt + θ),
−∞<t <∞
The subscript a is used with x(t) to denote an analog signal. A is the amplitude, Ω is the
frequency in radians per second(rad/s), and θ is the phase in radians. The Ω is related by
frequency F in cycles per second or hertz by
Ω = 2πF
xa (t) = Acos(2πFt + θ),
−∞<t <∞
Figure 1: Example of an analog sinusoidal signal
Cycle and Period
The completion of one full pattern waveform is called a cycle. A period is defined as the
amount of time required to complete one full cycle.
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
4 / 91
Introduction
The concept of frequency in continuous and discrete time signals
Complex exponential signals
xa (t) = Ae j(Ωt+θ)
where
e ±jφ
=
xa (t) = Acos(Ωt + θ) =
cosφ ± jsinφ
A j(Ωt+θ)
A
e
+ e −j(Ωt+θ)
2
2
As time progress the phasors rotate in opposite directions with angular ±Ω frequencies
radians per second.
A positive frequency corresponds to counterclockwise uniform angular motion, a negative
frequency corresponds to clockwise angular motion.
Figure 2: Representation of cosine function by phasor
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
Discrete Time Sinusoidal Signals
Discrete Time Sinusoidal Signals
A discrete time sinusoidal may expressed as
x(n) = Acos(ωn + θ),
−∞<t <∞
where n is an integer variable called the sample number.
A is the amplitude, ω is the frequency in radians per sample(rad/s), and θ is the phase in
radians.
The ω is related by frequency f cycles per sample by
ω = 2πf
x(n) = Acos(2πfn + θ),
−∞<t <∞
A discrete time signal x(n) is periodic with period N(N > 0) if and only if
x(n + N) = x(n)
for all n
Figure 3: Discrete signal
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
Discrete Time Sinusoidal Signals
Periodic and aperiodic (non-periodic) signals.
A periodic signal consists a continuously repeated pattern. Signal is periodic if it exhibits
periodicity i.e.
x(t + T )
=
x(t) for all t
It has a property that it is unchanged by a time shift of T.
An aperiodic signal changes constantly without exhibiting a pattern or cycle that repeats
over the time.
Figure 4: Periodic signals
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
Discrete Time Sinusoidal Signals
Periodic and aperiodic (non-periodic) signals.
Figure 5: Periodic signal
Figure 6: Periodic discrete time signal
Figure 8: Periodic discrete time
Figure 7: Periodic discrete time signal
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
Discrete Time Sinusoidal Signals
Periodic and aperiodic (non-periodic) signals.
Figure 9: Aperiodic signals
Figure 10: Aperiodic discrete time
signals
Figure 11: Aperiodic discrete time signals
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4, September
5]
11, 2014
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Introduction
Discrete Time Sinusoidal Signals
Periodic and aperiodic (non-periodic) signals.
Figure 12: Aperiodic (random) signal
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
10 / 91
Fourier series
Fourier series
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
11 / 91
Fourier series
Fourier series
Sinusoidal functions are wide applications in Engineering and they are easy to generate.
Fourier has shown that periodic signals can be represented by series of sinusoids with
different frequency.
A signal f (t) is said to be periodic of period T if f (t) = f (t + T ) for all t.
Periodic signals can be represented by the Fourier series and non periodic signals can be
represented by the Fourier transform.
For example square wave pattern can be approximated with a suitable sum of a
fundamental sine wave plus a combination of harmonics of this fundamental frequency.
Several waveforms that are represented by sinusoids are as shown in Figure 14. This sum
is called a Fourier series.
The major difference with respect to the line spectra of periodic signals is that the spectra
of aperiodic signals are defined for all real values of the frequency variable ω.
Figure 13: Square Wave from
Fourier Series
Dr. Manjunatha. P
(JNNCE)
Figure 14: Waveforms from Fourier Series
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
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Fourier series
Fourier series
Fourier analysis: Every composite periodic signal can be represented with a series of sine
and cosine functions with different frequencies, phases, and amplitudes.
The functions are integral harmonics of the fundamental frequency f of the composite
signal.
Using the series we can decompose any periodic signal into its harmonics.
f (θ)
=
a0 +
∞
X
an cos(nθ) +
n=1
∞
X
bn sin(nθ)
n=1
where
a0
=
1
2π
Z2π
f (θ)dθ
0
an
=
1
π
Z2π
f (θ)cos(nθ)dθ
0
bn
=
1
π
Z2π
f (θ)sin(nθ)dθ
0
Line spectra, harmonics
The fundamental frequency f0 = 1/T . The Fourier series coefficients plotted as a function
of n or nf0 is called a Fourier spectrum.
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
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Fourier series
Dr. Manjunatha. P
(JNNCE)
Fourier series
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
14 / 91
Fourier series
Fourier series
Figure 15: Square Wave
an
f (θ) =
A when 0 < θ < π
−A when π < θ < 2π
Z
=
1
π
Z
=
1
sin nθ π
1
sin nθ 2π
−A
+
A
=0
π
n
π
n
0
π
f (θ + 2π) = f (θ)
a0
f (θ) cos nθdθ
0
π
Z
2π
A cos nθdθ +
0
(−A) cos nθdθ
π
2π
1
2π
Z
=
1
2π
Z
Z
=
1
2π
=
2π
1
π
=
f (θ)dθ
0
π
Z
2π
f (θ)dθ +
0
π
Z
π
2π
Adθ +
0
Dr. Manjunatha. P
f (θ)dθ
(−A)dθ = 0
π
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
15 / 91
Fourier series
bn
=
=
=
=
=
=
Fourier series
2π
1
π
Z
1
π
Z
f (θ) sin nθdθ
0
π
Z
2π
0
(−A) sin nθdθ
A sin nθdθ +
π
1
cos nθ π
cos nθ 2π
1
−A
A
+
π
n
π
n
0
π
A
[− cos nπ + cos 0 + cos 2nπ − cos nπ]
nπ
A
[1 + 1 + 1 + 1]
nπ
4A
when n is odd
nπ
A
[− cos nπ + cos 0 + cos 2nπ − cos nπ]
nπ
A
=
[−1 + 1 + 1 − 1]
nπ
= 0 when n is even
4A
1
1
1
sin θ + sin 3θ + sin 5θ + sin 7θ + · · ·
π
3
5
7
bn
Dr. Manjunatha. P
(JNNCE)
=
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
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Fourier series
4A
π
1
1
1
sin θ + sin 3θ + sin 5θ + sin 7θ + · · ·
3
5
7
Figure 16: Square Wave from Fourier
Series
Dr. Manjunatha. P
Fourier series
Figure 17: Square Wave from Fourier
Series
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
17 / 91
Fourier series
Fourier series
Fundamental sinusoidal signal
sin(θ)
1
0.5
0
−0.5
clc;clear all; close all;
f=100;%Fundamental frequency 100 Hz
t=0:.00001:.05;
xsin = sin(2*pi*f*t);
x1 = sin(2*pi*f*t);
x3 = (1/3)*sin(3*2*pi*f*t);
x5 = (1/5)*sin(5*2*pi*f*t);
Figure
x7 = (1/7)*sin(7*2*pi*f*t);
x=x1+x3+x5+x7;
subplot(2,1,1)
plot(t,xsin,’linewidth’,2);
xlabel(’\theta’,’fontsize’,16)
ylabel(’sin(\theta)’,’fontsize’,16)
title(’Fundamental sinusoidal signal’)
subplot(2,1,2)
plot(t,x,’linewidth’,2);
xlabel(’\theta’,’fontsize’,16)
ylabel(’f (\theta)’,’fontsize’,16)
title(’Reconstructed square wave by Fourier ’)
−1
0
0.005
0.01
0.015
0.02
0.025
θ
0.03 0.035
0.04 0.045
0.05
0.04 0.045
0.05
Reconstructed square wave by Fourier
1
f (θ)
0.5
0
−0.5
−1
0
Dr. Manjunatha. P
(JNNCE)
0.005
0.01
0.015
0.02
0.025
θ
0.03 0.035
18: Square Wave
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier series
Fourier series
f (t) =
t when − T4 ≤ t ≤ T4
−t + T2 when T4 ≤ t ≤
3T
4
Figure 19: Triangular Wave
bn
T
2
T
Z
=
4
T
Z
Z
=
4
T
=
4
T
" 2
T
2
sin
2πn
=
=
=
2πn
t
T
f (t) sin
0
T /2
f (t) sin
0
T /4
t sin
0
dt
2πn
t
T
2πn
t
T
dt
4
dt +
T
#
πn
Z
T /4
−t +
0
T
2
sin
2πn
t
T
dt
2
2T
sin πn
2
2π 2 n2
0 when n is even
2T
2π
1
6π
1
10π
sin
t
−
sin
t
+
sin
t
−
·
·
·
π2
T
32
T
52
T
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
19 / 91
Fourier series
Figure 20: Square Wave[6]
Dr. Manjunatha. P
(JNNCE)
Fourier series
Figure 21: Sawtooth Signal[6]
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
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Fourier series
Fourier series
The Exponential (Complex) Form of Fourier Series
f (θ) = a0 +
∞
X
an cos nθ +
n=1
cosθ =
∞
X
bn sin nθ
n=1
e jθ + e −jθ
2
sinθ =
e jθ − e −jθ
2j
an cos nθ + bn sin nθ =
=
an
e jnθ − e −jnθ
e jnθ + e −jnθ
+ bn
2
2j
=
an
e jnθ − e −jnθ
e jnθ + e −jnθ
− jbn
2
2
an − jbn
an + jbn
jnθ
e +
e −jnθ
2
2
=
let
cn
=
an − jbn
2
Dr. Manjunatha. P
(JNNCE)
c−n =
an + jbn
2
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier series
an cos nθ + bn sin nθ
f (θ)
=
c0 +
=
Fourier series
cn e jnθ + c−n e −jnθ
∞ X
cn e jnθ + c−n e −jnθ
In exponential Fourier series only one
integral has to be calculated and it is
simpler integration.
n=1
∞
X
=
cn e jnθ
n=−∞
where
cn
=
an − jbn
2
The coefficient cn can be evaluated as.
cn
=
=
=
Z π
Z π
1
j
f (θ) cos nθdθ −
f (θ) sin nθdθ
2π −π
2π −π
Z π
1
f (θ) (cos nθ − j sin nθ)dθ
2π −π
Z π
1
f (θ)e −jnθ dθ
2π −π
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier series
Figure 22: Square Wave[6]
Fourier series
Figure 23: Sawtooth Signal[6]
Find the
e frequency
y componen
nts of a sig
gnal buried
d in noise. Consider data
d
sample
ed at
1000 Hzz. Form a signal consissting of 50 Hz and 120
0 Hz sinuso
oids and co
orrupt the signal
s
with random noise.
Example
Find the spectrum of the following signal:
f=0.25+2sin(2π5k)+sin(2π12.5k)+1.5sin(2π20k)+0.5sin(2π35k)
>>
>>
>>
>>
>>
>>
>>
>>
>>
N=256; % number of samples
T=1/128; % sampling frequency=128Hz
k=0:N-1; time=k*T;
f=0.25+2*sin(2*pi*5*k*T)+1*sin(2*pi*12.5*k*T)+…
+1.5*sin(2*pi*20*k*T)+0.5*sin(2*pi*35*k*T);
plot(time,f); title('Signal sampled at 128Hz');
F=fft(f);
magF=abs([F(1)/N,F(2:N/2)/(N/2)]);
hertz=k(1:N/2)*(1/(N*T));
stem(hertz,magF), title('Frequency components');
Figure 24: Square Wave[6]
Dr. Manjunatha. P
(JNNCE)
Figure 25: Sawtooth Signal[6]
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
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Fourier series
Fourier series
5
4
3
2
1
0
-1 0
200
400
600
0
800
1000
1200
14
400
-2
-3
-4
-5
Figure 27: Sawtooth Signal[6]
Figure 26: Square Wave[6]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
24 / 91
Fourier Transform
Fourier Transform
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
25 / 91
Fourier Transform
Fourier Transform
The Fourier transform is a generalization of the Fourier series representation of functions.
The Fourier series is limited to periodic functions, while the Fourier transform can be used
for a larger class of functions which are not necessarily periodic. S
Sinusoidal functions are wide applications in Engineering and they are easy to generate.
Fourier has shown that periodic signals can be represented by series of sinusoids with
different frequency.
A signal f (t) is said to be periodic of period T if f (t) = f (t + T ) for all t.
Periodic signals can be represented by the Fourier series and non periodic signals can be
represented by the Fourier transform.
For example square wave pattern can be approximated with a suitable sum of a
fundamental sine wave plus a combination of harmonics of this fundamental frequency.
Several waveforms that are represented by sinusoids are as shown in Figure 14. This sum
is called a Fourier series.
The major difference with respect to the line spectra of periodic signals is that the spectra
of aperiodic signals are defined for all real values of the frequency variable ω.
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
26 / 91
Fourier Transform
f (θ)
∞
X
=
cn e jnθ
Fourier Transform
when θ = π
n=−∞
π=
where
cn
1
2π
=
π
Z
2πt
⇒
T
f (θ)e −jnθ dθ
−π
f (t)
t=
∞
X
=
T
2
cn e jnωt
n=−∞
θ = ωt
ω is the angular velocity in radians per
second.
ω = 2πf
θ=
2π
t
T
and
and
cn
=
1
T
Z
T /2
f (t)e −jnωt dt
−T /2
θ = 2πft
dθ =
2π
dt
T
when θ = −π
−π =
2πt
⇒
T
Dr. Manjunatha. P
t=−
(JNNCE)
T
2
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Relationship from Fourier series to Fourier Transform
f (t)
∞
X
=
cn e jnωt
f (t)


Z T /2
∞
1  X
−jωt  jωt
f (t)e
dt e ∆ω
2π ω=−∞ −T /2
=
n=−∞
T ⇒ ∞ ∆ω ⇒ dω and
cn
=
1
T
Z
T /2
f (t)e −jnωt dt
−T /2
f (t) =
As T approaches infinity
ω approaches zero
and n becomes meaningless
nω ⇒ ω ω ⇒ ∆ω
2π
T ⇒ ∆ω
f (t)
∞
Z
=
cω e jωt
∞
Z
⇒
R
f (t)e(−jωt) dt e(jωt) dω
−∞
−∞
Z
ω
X
=
f (t)
1
2
P
1
2
∞
F (ω) =
Z
∞
F (ω)e(jωt) dω
−∞
f (t)e(−jωt) dt
−∞
n=−ω
cω
=
∆ω
2π
Dr. Manjunatha. P
Z
T /2
f (t)e −jωt dt
−T /2
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Figure 28: Rectangular Pulse
Fourier Transform
Figure 29: Sinc Function
Z1/2
F (ω)
=
exp(−jωt)dt =
1
1/2
[exp(−jωt)]−1/2
−jω
−1/2
=
=
=
=
Dr. Manjunatha. P
(JNNCE)
1
[exp(−jω/2) − exp(jω/2)]
−jω
1 exp(jω/2) − exp(−jω/2)
(ω/2)
2j
sin(ω/2)
(ω/2)
sinx
sinc(ω/2)
since it is
form
x
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Figure 30: Exponential
Figure 31: Gaussian
Z∞
F (ω)
exp( − at) exp(−jωt)dt
=
0
Z∞
exp(−at − jωt)dt =
=
0
=
=
=
Dr. Manjunatha. P
Z∞
exp(−[a + jω]t)dt
0
−1
−1
exp(−[a + jω]t)|+∞
=
[exp(−∞) − exp(0)]
0
a + jω
a + jω
−1
[0 − 1]
a + jω
1
a + jω
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Z∞
δ(t) exp(−iω t) dt = exp(−iω [0]) = 1
−∞
Z∞
1 exp(−iω t) dt = 2π δ(ω)
−∞
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Z∞
F {exp(iω0 t)}
=
exp(iω0 t) exp(−i ω t) dt
−∞
Z∞
exp(−i [ω − ω0 ] t) dt
=
−∞
= 2π δ(ω − ω0 )
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Z∞
F {cos(ω0 t)}
=
cos(ω0 t) exp(−j ω t) dt
−∞
=
1
2
Z∞
[exp(j ω0 t) + exp(−j ω0 t)] exp(−j ω t) dt
−∞
=
1
2
Z∞
exp(−j [ω − ω0 ] t) dt
+
−∞
=
Dr. Manjunatha. P
(JNNCE)
π δ(ω − ω0 )
1
2
Z∞
exp(−j [ω + ω0 ] t) dt
−∞
+
π δ(ω + ω0 )
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Figure 32: A signal with four different frequencies
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Fourier Transform
Fourier Transform
Figure 33: A signal with four different frequency components at four different time intervals
Figure 34: Each peak corresponds to a frequency of a periodic component
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(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
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Discrete Fourier Transform (DFT)
Many applications demand the processing of signals in frequency domain.
The analysis of signal frequency, periodicity, energy and power spectrums can be analyzed
in frequency domain.
Frequency analysis of discrete time signals is usually and most conveniently performed on
a digital signal processor.
Applications of DFT:
Spectral analysis
Convolution of signals
Partial differential equations
Multiplication of large integers
Data compression
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
37 / 91
Discrete Fourier Transform (DFT)
Summary of FS and FT
Fourier Series is
x(θ)
=
a0 +
∞
X
an cos(nθ) +
n=1
where
a0 =
1
2π
2π
R
∞
X
bn sin(nθ)
n=1
x(θ)dθ
0
an
=
1
π
Z2π
x(θ)cos(nθ)dθ
bn =
Z2π
1
π
x(θ)sin(nθ)dθ
0
0
The Exponential (Complex) Form
x (θ)
=
∞
X
cn e jnθ
where cn =
n=−∞
x(t)
=
∞
X
cn e jnωt
where cn =
n=−∞
1
2π
Z
x(θ)e −jnθ dθ
−π
T /2
Z
1
T
π
x(t)e −jnωt dt
−T /2
Fourier Transform pair is
Z
∞
X (ω) =
−∞
Dr. Manjunatha. P
(JNNCE)
f (t)e (−ωt) dt
and x(t) =
1
2
Z
∞
X (ω)e (jωt) dω
−∞
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
38 / 91
Discrete Fourier Transform (DFT)
Summary of FS and FT
Time Domain
Frequency Domain
Transform
Continuous Periodic
Continuous nonperiodic
Discrete nonperiodic
Discrete periodic
Discrete nonperiodic
Continuous nonperiodic
Continuous nonperiodic
Discrete periodic
Fourier series
Fourier Transform
Sequences Fourier Transform
Discrete Fourier Transform
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
39 / 91
Discrete Fourier Transform (DFT)
Summary of FS and FT
The Fourier Series for Continuous time Periodic Signals
Synthesis Equation
∞
P
x(t) =
Analysis Equation
cn =
1
T
ck e j2πnft
k=−∞
∞
R
x(t)e −j2πnft dt
−∞
The Fourier Transform for Continuous Time Aperiodic Signals
∞
R
Synthesis Equation (Inverse transform)
x(t) =
X (F )e j2πFt dF
Analysis Equation (Direct transform)
−∞
∞
R
x(t)e −j2πFt dt
X (F ) =
−∞
The Fourier Series for Discrete time Periodic Signals
Synthesis Equation
x(n) =
Analysis Equation
ck =
N−1
P
1
N
k=0
N−1
P
ck e j2πkn/N
x(n)e −j2πkn/N
n=0
The Fourier Transform of Discrete Time Aperiodic Signals
Synthesis Equation (Inverse transform)
Analysis Equation (Direct transform)
x(n) =
X (ω) =
1
2π
∞
R
−∞
∞
P
X (F )e j2πF1 t dF
x(n)e −j2πkn/N
n=−∞
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
40 / 91
Discrete Fourier Transform (DFT)
Summary of FS and FT
Signal
Time
Amplitude
Amplitude
DFT transforms the time domain signal samples to the frequency domain components.
Signal
Spectrum
Frequency
Figure 35: Discrete Fourier Transform
Signal
Continuous and periodic
Continuous and aperiodic
Discrete and periodic
Discrete and aperiodic
Types of Transforms
Fourier Series
Fourier Series
Fourier Series
Fourier Series
Dr. Manjunatha. P
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
(JNNCE)
Example Waveform
sine wave
41 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Need For Frequency Domain Sampling
In practical application, signal processed by computer has two main characteristics: It
should be Discrete and Finite length
But nonperiodic sequences Fourier Transform is a continuous function of ω, and it is a
periodic function in ω with a period 2π.
So it is not suitable to solve practical digital signal processing.
Frequency analysis on a discrete-time signal x(n) is achieved by converting time domain
sequence to an equivalent frequency domain representation, which is represented by the
Fourier transform X (ω) of the sequence x(n).
Consider an aperiodic discrete time signal x(n) and its Fourier transform is
X (ω) =
∞
X
x(n)e −jωn
n=−∞
The Fourier transform X (ω) is a continuous function of frequency and it is not a
computationally convenient representation of the sequence.
To overcome the processing, the spectrum of the signal X (ω) is sampled periodically in
frequency at a spacing of δω radians between successive samples.
The signal X (ω) is periodic with period 2π and take N equidistant samples in the interval
0 ≤ ω ≤ 2π with spacing δ = 2π/N .
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
42 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Figure 36: Frequency domain
sampling
Figure 37: Frequency domain
sampling
To Determine The Value Of N
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
43 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Now consider ω = 2πk/N
X
X
2π
k
N
2π
k
N
=
∞
X
x(n)e −j2πkn/N
k = 0, 1, 2, . . . N − 1
n=−∞
−1
X
=
··· +
x(n)e −j2πkn/N +
2N−1
X
x(n)e −j2πkn/N
n=0
n=−N
+
N−1
X
x(n)e −j2πkn/N + · · ·
n=N
=
∞
X
lN+N−1
X
n=−∞
n=lN
x(n)e −j2πkn/N
By changing the index in the inner summation from n to n − lN and interchanging the
order of summation


N−1
∞
X
X
2π
2π

k
=
X
x(n − lN) e −j N k(n−lN)
N
n=0 n=−∞


N−1
∞
X
X
2π

=
x(n − lN) e −j N kn e −j2πkl
n=0
n=−∞
e −j2πkl = 1 ∵ both k and l integers
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
44 / 91
Discrete Fourier Transform (DFT)
X
2π
k
N
=
N−1
X


∞
X
x(n − lN) e −j2πkn/N

n=0
Discrete Fourier Transform (DFT)
k = 0, 1, 2, . . . N − 1
n=−∞
Let
xp (n)
∞
X
=
x(n − lN)
n=−∞
The term xp (n) is obtained by the periodic repetition of x(n) every N samples hence it is
a periodic signal. This can be expanded by Fourier series as
xp (n)
=
N−1
X
ck e j2πkn/N
n = 0, 1, · · · N − 1
k=0
where ck is the fourier coefficients expressed as
ck
=
N−1
1 X
xp (n)e −j2πkn/N
N n=0
ck
=
k = 0, 1, · · · N − 1
Upon comparing
Dr. Manjunatha. P
(JNNCE)
1
X
N
2π
k
N
k = 0, 1, · · · N − 1
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
45 / 91
Discrete Fourier Transform (DFT)
xp (n)
=
Discrete Fourier Transform (DFT)
N−1
1 X
2π
X
k e j2πkn/N
N k=0
N
n = 0, 1, · · · N − 1
xp (n) is the reconstruction of the periodic signal from the spectrum X (ω) (IDFT).
The equally spaced frequency samples X 2π
k = 0, 1, · · · N − 1 do not uniquely
N
represent the original sequence when x(n) has infinite duration. When x(n) has a finite
duration then xp (n) is a periodic repetition of x(n) and xp (n) over a single period is
x(n)
0≤n ≤L−1
xp (n) =
0
L≤n ≤N −1
For the finite duration sequence of length L the Fourier transform is:
X (ω)
=
L−1
X
x(n)e −jωn
0 ≤ ω ≤ 2π
n=0
When X (ω) is sampled at frequencies ωk = 2πk/N
X (k)
=
X
2πk
N
=
L−1
X
n=0
k = 0, 1, 2, . . . N − 1 then
x(n)e −j2πkn/N =
N−1
X
x(n)e −j2πkn/N
n=0
The upper index in the sum has been increased from L − 1 to N − 1 since x(n)=0 for n ≥ L
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
46 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
DFT and IDFT expressions are
DFT expressions is
X (k)
=
N−1
X
x(n)e −j2πkn/N
k = 0, 1, . . . N − 1
n=0
IDFT expressions is
xp (n)
N−1
1 X
2π
X
k e j2πkn/N
N k=0
N
=
n = 0, 1, · · · N − 1
If xp (n) is evaluated for n = 0, 1, 2 . . . N − 1 then xp (n) = x(n)
x(n)
Dr. Manjunatha. P
(JNNCE)
=
N−1
1 X
X (k) e j2πkn/N
N k=0
n = 0, 1, · · · N − 1
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
47 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
DFT as a Linear Transformation
X (k)
N−1
X
=
x(n)e −j
2π kn
N
k = 0, 1, . . . N − 1
n=0
Let
WN = e −j
X (k)
=
2π
N
is called twiddle factor
N−1
X
x(n)WNnk
for k = 0, 1.., N − 1
x=0








X (0)
X (1)
X (2)
.
..
X (3)


 
 
 
=
 
 


Dr. Manjunatha. P
1
1
1
1
.
..
1
(JNNCE)
1
W
W2
W3
1
W2
W2
W6
1
W3
W4
W9
...
...
...
...
1
W N−1
W 2(N−1)
W 3(N−1)
W N−1
W N−2
W N−3
...
W (N−1)(N−1)



 
 
 
.
 
 


x(0)
x(1)
x(2)
..
.
x(N − 1)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014







48 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Periodicity property of WN
WN = e −j
2π
N
Let us consider for N=8
W8 = e −j
2π
8
1 = e −j
π
4
π
kn
W8kn = e − 4 kn
Result
0
1
2
3
4
5
6
7
8
9
10
W80 = e 0
π
π
W81 = e −j 4 1 = e −j 4
π
−j π
2
−j
2
W8 = e 4 = e 2
π
3
−j3
−j π
3
4
W8 = e 4 = e
−j π
4
4
−jπ
W8 = e 4 = e
π
π
W85 = e −j 4 5 = e −j3 5
π
π
W86 = e −j 4 6 = e −j3 2
π
−j π
7
−j7
7
4
W8 = e 4 = e
−j π
8
8
−j2π
W8 = e 4 = e
π
π
W89 = e −j 4 9 = e −j(2π+ 4 )
π
π
W810 = e −j 4 10 = e −j(2π 2 )
π
3π
W811 = e −j 4 11 = e −j2π+ 4
Magnitude 1 Phase 0
Magnitude 1 Phase −π/4
Magnitude 1 Phase −π/2
Magnitude 1 Phase −3 π4
Magnitude 1 Phase −π
Magnitude 1 Phase −5π/4
Magnitude 1 Phase −3π/2
Magnitude 1 Phase −7π/4
Magnitude 1 Phase −2π W88 = W80
Magnitude 1 Phase (−2π + π/4) W89 = W81
Magnitude 1 Phase (−2π +π/2) W810 = W82
11
Dr. Manjunatha. P
(JNNCE)
W811 = W83
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
49 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Imaginary
part of WN
W85 = W813 = ..
=−
W86 = W814 = .. = j
1
1
+j
2
2
W87 = W815 = ..
=
= −1
W83 = W811 = ..
1
1
−j
2
2
2
+j
1
2
Real part
of WN
W80 = W88 = .. = 1
W84 = W812 = ..
=−
1
W81 = W89 = ..
W82 = W810 = .. = − j
=
1
2
−j
1
2
Figure 38: Periodicity of WN and its values
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
50 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Imaginary
part of WN
W43 = j = W47 = W411
W42 = −1 = W46
W40 = 1 = W44
= W410
= W48
Real part
of WN
W41 = − j = W45 = W49
Figure 39: Periodicity of WN and its values
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
51 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find Discrete Fourier Transform (DFT) of x(n) = [2 3 4 4]
Solution:
X (k)
=
N−1
X
x(n)e −j
2π nk
N
for k = 0, 1.., N − 1
n=0
π
2
=
3π
2
=
e −j
e −j
π
π
− jsin = −j
2
2
3π
3π
− jsin
=j
cos
2
2
cos
e −jπ = cos(π) − jsin(π) = −1
e −j2π = cos(2π) − jsin2(π) = 1
for k=0,1,2,3
3
P
X (0) =
x(n)e 0 = 2e 0 + 3e 0 + 4e 0 + 4e 0 = [2 + 3 + 4 + 4] = 13
X (1) =
n=0
3
P
x(n)e −j
X (2) =
n=0
3
P
x(n)e
−j4πn
4
= 2e 0 + 3e −jπ + 4e −j2π + 4e −j3π = [2 − 3 + 4 − 4] = [−1 − 0j] = −1
X (3) =
n=0
3
P
x(n)e
−j6πn
4
= 2e 0 + 3e −j3π/2 + 4e −j3π + 4e −j9π/2 = [2 + 3j − 4 − 4j][−2 − j]
2πn
4
= 2e 0 + 3e −jπ/2 + 4e −jπ + 4e −j3π/2 = [2 − 3j − 4 + 4j] = [−2 + j]
n=0
The DFT of the sequence x(n) = [2 3 4 4] is [13, -2+j, -1, -2-j]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
52 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find Discrete Fourier Transform (DFT) of x(n) = [2 3 4 4]
X (k)
=
N−1
X
x(n)e −j
2π nk
N
for k = 0, 1.., N − 1
n=0
Matlab code for the DFT equation is:
clc; clear all; close all
xn=[2 3 4 4]
N=length(xn);
n=0:N-1;
k=0:N-1;
WN=exp(-1j*2*pi/N);
nk=n’*k;
WNnk=WN.^nk;
Xk=xn*WNnk
Matlab code using FFT command
clc; clear all; close all
xn=[2 3 4 4]
y=fft(xn)
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
53 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find DFT for a given a sequence x(n) for 0 ≤ n ≤ 3 where
x(0) = 1, x(1) = 2, x(2) = 3, x(3) = 4
Solution:
x(n) = [1 2 3 4]
for k=0,1,2,3
3
P
X (0) =
x(n)e 0 = 4e 0 + 2e 0 + 3e 0 + 4e 0 = [1 + 2 + 3 + 4] = 10
X (1) =
n=0
3
P
x(n)e −j
X (2) =
n=0
3
P
X (3) =
n=0
3
P
2πn
4
= 1e 0 + 2e −jπ/2 + 2e −jπ + 4e −j3π/2 = [1 − j2 − 3 + j4] = [−2 + j2]
x(n)e
−j4πn
4
= 1e 0 + 2e −jπ + 3e −j2π + 4e −j3π = [1 − 2 + 3 − 4] = [−1 − 0j] = −2
x(n)e
−j6πn
4
= 1e 0 + 2e −j3π/2 + 3e −j3π + 4e −j9π/2 = [1 + 2j − 3 − 4j][−2 − j2]
n=0
The DFT of the sequence x(n) = [1 2 3 4] is [10, − 2 + j2, − 2, − 2 − j2]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
54 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find 8 point DFT for a given a sequence x(n) = [1, 1, 1, 1] assume imaginary part is zero. Also
calculate magnitude and phase
Solution:
x(n) = [1 1 1 1]
The 8 point DFT is of length 8. Append zeros at the end of the sequence. x(n) = [1 1 1 1 0 0 0
0]











X (0)
X (1)
X (2)
X (3)
X (4)
X (5)
X (6)
X (7)
 1
  1
 
  1
 
  1
=
  1
 
  1
  1
1

Dr. Manjunatha. P
(JNNCE)
1
W81
W82
W83
W84
W85
W86
W87
1
W82
W84
W86
W88
W810
W812
W814
1
W83
W86
W89
W812
W815
W818
W821
1
W84
W88
W812
W816
W820
W824
W828
1
W85
W810
W815
W820
W825
W830
W835
1
W86
W812
W818
W824
W830
W836
W842
1
W87
W814
W821
W828
W835
W842
W849


 
 
 
 
 
=
 
 
 
 
x(0)
x(1)
x(2)
x(3)
x(4)
x(5)
x(6)
x(7)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014











55 / 91
Discrete Fourier Transform (DFT)
W80
W81
W82
W83
W84
W85
W86
W87
=
=
=
=
=
=
=
=












Discrete Fourier Transform (DFT)
W88 = W816 = W824 = W840 ... = 1
W89 = W817 = W825 = W833 ... = √1 − j √1
2
2
W810 = W818 = W826 = W834 ... = −j
11
19
27
35
1
W8 = W8 = W8 = W8 ... = − √ − j √1
2
2
W812 = W820 = W828 = W836 ... = −1
13
21
29
37
1
1
W8 = W8 = W8 = W8 ... = − √ + j √
2
2
W814 = W822 = W830 = W838 ... = j
15
23
31
39
1
1
√
√
W8 = W8 = W8 = W8 ... =
+j
2
2
X (0)
X (1)
X (2)
X (3)
X (4)
X (5)
X (6)
X (7)












=










1
1
1
√
1
− j √1
2
1
1
−j
− √1 − j √1
1
1
−1
− √1 + j √1
2
2
1
1
1
√
1
−j
2
2
2
2
j
+ j √1
2
 4
√
 1 − j(1 + 2)
 0

√

 1 + j(1 − 2)

 0
√

 1 − j(1 − 2)

0
√
1 + j(1 + 2)
Dr. Manjunatha. P
(JNNCE)
1
− √1 − j √1
2
−1
j
1
√
1
−j
1
√
−1
j
2
2
j
− j √1
−1
+ j √1
2
1
−1
2
2











1
− √1 + j √1
2
1
1
2
−j
− √1 + j √1

XR (0)
 XR (1)




 XR (2)



 XR (3)
=

 XR (4)

 X (5)


R

 X (6)
R
XR (7)

1
−1
2
4
1
0
1
0
1
0
1
1
√
2
1
√
2
−j
+ j √1
2
2
−1
− j √1
2
1
−1
j
− √1 − j √1







 and














2
XI (0)
XI (1)
XI (2)
XI (3)
XI (4)
XI (5)
XI (6)
XI (7)
2
1
j
1
+ j √1
1
√
2

2
−1
−j
j
− √1 + j √1
1
j
−1
− √1 − j √1
−1
−j
1
√
2
 0
√
  −(1 + 2)
 0

√

  (1 − 2)

 0
√

  −(1 − 2)

0
√
(1 + 2)
2
2
2
2
−j
− j √1


1


 1 




 1 




1 
=



 0 

 0 




 0 


0
2











UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
56 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find DFT for a given a sequence x(n) = [2 3 4 4]
Solution:
WN = e −

 
1
X (0)
 X (1)   1

=
 X (2)   1
X (3)
1
2π
N
= e−
2π
4
1
W
W2
W3
1
W2
W4
W6
 
1
2
3
 3
W 


=
W6   4
4
W9




π
= e − 2 = −j
W 2 = −j 2 = −1, W 3 = −j 3 = j
Using the property of periodicity of W W p = W P+r .N = j with basic period N = 4
W 4 = W 4−4 = W 0 = 1, W 6 = W 6−4 = W 2 = −1, W 9 = W 9−2.4 = W 1 = −j

 
X (0)
 X (1)  

 
 X (2)  = 
X (3)
1
1
1
1
1
−j
−1
j
1
−1
1
−1

1
2

j 
 3
−1   4
−j
4



13
  −2 + j 
=

  −1

−2 − j
The DFT of the sequence x(n) = [2 3 4 4] is [13, − 2 + j, − 1, − 2 − j]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
57 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find DFT for a given a sequence x(n) = [1 2 3 4]
Solution:
WN = e −

 
1
X (0)
 X (1)   1

=
 X (2)   1
X (3)
1
2π
N
= e−
2π
4
1
W
W2
W3
1
W2
W4
W6
 
1
1
3
 2
W 


=
W6   3
4
W9




π
= e − 2 = −j
W 2 = −j 2 = −1, W 3 = −j 3 = j
Using the property of periodicity of W W p = W P+r .N = j with basic period N = 4
W 4 = W 4−4 = W 0 = 1, W 6 = W 6−4 = W 2 = −1, W 9 = W 9−2.4 = W 1 = −j

 
X (0)
 X (1)  

 
 X (2)  = 
X (3)
1
1
1
1
1
−j
−1
j
1
−1
1
−1

1
1

j 
 2
−1   3
−j
4



10
  −2 + 2j 
=

  −2

−2 − 2j
The DFT of the sequence x(n) = [1 2 3 4] is [10, − 2 + j2, − 2, − 2 − j2]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
58 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Inverse DFT: Find the IDFT for X (k) = [10, − 2 + j2, − 2, − 2 − j2]
x(n)
x(n)
x(1)
=
=
=
Dr. Manjunatha. P
N−1
2π
1 X
X (k)e N kn
N k=0
N−1
1 X
X (k)W ∗kn
N k=0
=
x(0)
=
for n = 0, 1.., N − 1
for n = 0, 1.., N − 1 where
W∗ = e
2π
N
=
N−1
1 X
X (k)e j0 = X (0)e j0 + X (1)e j0 + X (2)e j0 + X (3)e j0
4 k=0
=
1
(10 + (−2 + j2) − 2 + (−2 − j2)) = 1
4
N−1
kπ
π
3π
1 X
X (k)e j 2 = X (0)e j0 + X (1)e j 2 + X (2)e jπ + X (3)e j 2
4 k=0
1
(X (0) + jX (1) − X (2) − jX (3)
4
1
(10 + j(−2 + j2) − (−2) − j(−2 − j2)) = 2
4
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
59 / 91
Discrete Fourier Transform (DFT)
x(2)
=
=
=
x(1)
=
=
=
Discrete Fourier Transform (DFT)
N−1
kπ
1 X
X (k)e j 2 = X (0)e j0 + X (1)e jπ + X (2)e j2π + X (3)e j3π
4 k=0
1
(X (0) − X (1) + X (2) − X (3)
4
1
(10 − (−2 + j2) + (−2) − (−2 − j2)) = 3
4
N−1
kπ3
3π
9π
1 X
X (k)e j 2 = X (0)e j0 + X (1)e j 2 + X (2)e j3π + X (3)e j 2
4 k=0
1
(X (0) − jX (1) − X (2) + jX (3)
4
1
(10 − j(−2 + j2) − (−2) + j(−2 − j2)) = 4
4
Matlab command used to calculate the Inverse DFT is ifft
x=[10 -2+j2 -2 -2-j2]
y=ifft(x)
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
60 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find the Discrete Fourier Transform of the following signal: x(n), n = 0,1,2,3 = [1, 1, -1, -1].
Solution:
N=4 The matrix notation is
X
where T is matrix of the

X (0)
 X (1)

 X (2)
X (3)
=
T .f
transform with elements Tkn = WNkn
 

1
1
1
1
1
  1 −j −1

j 
=
 1
  1 −1
1
−1   −1
1
j
−1 −j
−1
k, n = 0, 1.., N
 
0
  2 − 2j
=
  0
2 + 2j
−1




The DFT of the sequence x(n) = [1 1 -1 -1] is [0, 2 − j2, 0, 2 + j2]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
61 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find the Inverse Discrete Fourier Transform of the following signal: x(n), n = 0,1,2,3 = [0, 2-2j,
0, 2+2j].
Solution:
The IDFT of the discrete signal X(k)is x(n):
N = 4 and W4 = e −π/2
x(n)
=
N−1
1 X
X (k)WN−kn
N k=0
N=4 The matrix notation is

1
1
 1 −j
[WN ] = 
 1 −1
1
j



x(0)
 x(1) 
1 



 x(2)  = N 
x(3)
1
−1
1
−1
1
1
1
1
for n = 0, 1.., N − 1 where

1
j 

−1 
−j
1
j
−1
−j
1
−1
1
−1
W = e−
2π
N


1
1
1
1
 1
j
−1 −j 

[WN∗ ] = 
 1 −1
1
−1 
1 −j −1
j

 

1
0
1




−j   2 − 2j   1 

 =  −1 
−1   0
j
2 + 2j
−1
The IDFT of the sequence X (k) = [0, 2 − j2, 0, 2 + j2] is [1 1 -1 -1]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
62 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find DFT for a given a sequence x[0]=1, x[1]=2, x[2]=2, x[3]=1, x[n]=0 otherwise: x =
[1,2,2,1]
Solution:
x(n) = [1 2 2 1]
for k=0,1,2,3
3
P
X (0) =
x(n)e 0 = 1e 0 + 2e 0 + 2e 0 + 1e 0 = [1 + 2 + 2 + 1] = 6
X (1) =
n=0
3
P
x(n)e −j
X (2) =
n=0
3
P
x(n)e
−j4πn
4
= 1e 0 + 2e −jπ + 2e −j2π + 1e −j3π = [1 − 2 + 2 − 1] = [0] = 0
X (3) =
n=0
3
P
x(n)e
−j6πn
4
= 1e 0 + 2e −j3π/2 + 2e −j3π + 1e −j9π/2 = [1 + 2j − 2 − 1j][−1 + j1]
2πn
4
= 1e 0 + 2e −jπ/2 + 2e −jπ + 1e −j3π/2 = [1 − j2 − 2 + j1] = [−1 − j1]
n=0
The DFT of the sequence x(n) = [1 2 2 1] is [6, − 1 − j1, 0, − 1 + j1]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
63 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find IDFT for a given a sequence X[0]=6, X[1]=-1-j1, X[2]=0, X[3]=-1+j1, X[n]=0 otherwise:
x = [6, − 1 − j1, 0, − 1 + j1]
Solution:
x(n) = [6, − 1 − j1, 0, − 1 + j1]
for k=0,1,2,3
3
P
X (0) = 14
x(n)e 0 = 6e 0 + (−1 − j1)e 0 + 0e 0 + (−1 + j1)e 0 = 14 [6−1−j1+0−1+j1] = 1
X (1) =
1
4
n=0
3
P
x(n)e j
2πn
4
= 6e 0 + (−1 − j1)e jπ/2 + 0e jπ + (−1 + j1)e j3π/2 =
n=0
1
[6
4
− j + 1 + j + 1] = [2]
3
j4πn
P
X (2) = 14
x(n)e 4 = 6e 0 + (−1 − j1)e jπ + 0e j2π + (−1 + j1)e j3π =
n=0
1
[6
4
+ (−1 − j1)(j) + 0 + (−1 + j)(−j)] = 14 [6 − j1 + 1 + 0 + 1 + j] = [2]
3
−j6πn
P
X (3) = 14
x(n)e 4 = 6e 0 + (−1 − j1)e j3π/2 + 0e j3π + (−1 + j1)e j9π/2 =
n=0
1
[6
4
+ (−1 − j1)(−j) + 0 + (−1 + j)(j)] = 14 [6 + j1 − 1 + 0 − 1 − j] = [1]
The IDFT of the sequence [6, − 1 − j1, 0, − 1 + j1] is x(n) = [1 2 2 1]
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
64 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Continuous Time Fourier Transform (CTFT)
Z∞
X (jω) =
x(t)e −jωt dt
−∞
x(t) =
Z∞
1
2π
X (jω)e jωt dω
−∞
Discrete Time Fourier Transform (DTFT)
X (e jω ) =
∞
X
x(n)e −jωn
−∞
x(n) =
1
2π
Z
X (e jω )e jωn dω
2π
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
65 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Unit sample δ(n)
x(n) =
X (k)
=
1 for n = 0
0 for n 6= 0
N−1
X
x(n)e −j
2π nk
N
x=0
=
N−1
X
x(0)e 0 = 1 × 1 = 1
x=0
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
66 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find the N Point DFT of x(n) = an for 0 ≤ n ≤ N − 1
X (k)
=
N−1
X
x(n)e −j
2π nk
N
x=0
=
N−1
X
an e −j
x=0
X (k)
e −j2πk
=
1 − aN e −j2πk
1 − ae −j2πk /N
2π nk
N
=
N−1
X
(ae −j
2πk
N
)n
x=0
Using series expansion
N−1
X
ak =
k=0
aN1 − aN2 +1
1−a
!
=1
X (k)
=
1 − aN
1 − ae −j2πk /N
x[n] = (0.5)n u[n] 0 ≤ n ≤ 3
X (k)
Dr. Manjunatha. P
(JNNCE)
=
1 − (0.5)4
0.9375
=
1 − 0.5e −j2πk /4
1 − 0.5e −jπ/2k
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
67 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find the 4 Point DFT of x(n) = cos( nπ
)
4
Solution:
x(0) = cos(0) = 1
x(1) = cos( 1π
) = 0.707
4
x(2) = cos( 2π
)=0
4
) = −0.707
x(3) = cos( 3π
4


x(0)
 x(1) 
1


 x(2)  = N
x(3)
Dr. Manjunatha. P
(JNNCE)

1
 1

 1
1
1
−j
−1
j
1
−1
1
−1

1
1

j 
  0.707
−1   0
−j
−0.707

1
  1 − j1.414 
=

  1

1 + j1.414


UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
68 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
If the length of x[n]
√ is N=4, and
√ if its 8-point
√ DFT is:
√ X8 [k]
k = 0..7 = [5, 3 − 2j, 3, 1 − 2j, 1, 1 + 2j, 3, 3 + 2j] , find the 4 point DFT of the signal
x[n].
Solution:
The samples of Xs [k] are eight equally spaced samples from the frequency spectrum of
the signal x[n]:
More precisely,
they are samplesfrom the spectrum for the following frequencies:
ωT = 0, π4 , π2 , 3π
, π 5π
, 3π
, 7π
4
4
2
4
With 4-point DFT of x[n] we get 4 samples from the spectrum of x[n]: ωT = 0, π2 , π,
3π
2
By comparing the two sets of frequencies:
X4 [0] = X (e j0 ) = Xs [0] = 5
X4 [1] = X (e jπ/2 ) = Xs [2] = 3
X4 [2] = X (e jπ ) = Xs [4] = 1
X4 [3] = X (e j3π/2 ) = Xs [6] = 3
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
69 / 91
Discrete Fourier Transform (DFT)
Discrete Fourier Transform (DFT)
Find the Fourier Transform of the sequence
x[n] =
1
0
0≤n≤4
else
∞
X
x [n] e −jwn
X e jw =
n=−∞
sin (5ω/2)
X e jω = e −j2ω
sin (ω/2)
Consider a causal sequence x[n] where;
x[n] = (0.5)n u[n]
Its DTFT X e jw can be obtained as
∞
∞
X
X
X e jw =
(0.5)n u[n]e −jwn =
(0.5)n (1)e −jwn
n=−∞
=
∞ n
X
0.5e jw
=
n=0
Dr. Manjunatha. P
(JNNCE)
n=0
1
1 − 0.5e −jw
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
70 / 91
Discrete Fourier Transform (DFT)
Find the N Point DFT of x(n) = 4 +
Discrete Fourier Transform (DFT)
cos 2 ( 2πn
)
N
Solution:
x(0) = 4 + cos 2 (0) = 5
x(1) = 4 + cos 2 ( 2π1
) = 4.6545
10
x(2) = 4 + cos 2 ( 2π2
) = 4.09549
10
x(3) = 4 + cos 2 ( 2π3
) = 4.09549
10
) = 4.09549
x(4) = 4 + cos 2 ( 2π4
10
x(5) = 4 + cos 2 ( 2π5
)=5
10
x(6) = 4 + cos 2 ( 2π6
) = 4.6545
10
x(7) = 4 + cos 2 ( 2π7
) = 4.09549
10
x(8) = 4 + cos 2 ( 2π8
) = 4.09549
10
x(9) = 4 + cos 2 ( 2π9
) = 4.6545
10
x(n) = x(N − n)
Cosine function is even function
x(n) = x(−n)
X (k) =
N−1
X
x(n)cos
n=0
X (k) =
N−1
X
n=0
Dr. Manjunatha. P
(JNNCE)
4 + cos 2 (
2πkn
N
0≤k ≤N −1
2πn
2πkn
) cos
N
N
0≤k ≤N −1
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
71 / 91
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
72 / 91
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Relationship to the Fourier series coefficients of periodic sequence
DFT expression is
X (k) =
N−1
X
x(n)e −j2πkn/N
k = 0, 1, . . . N − 1
(1)
n=0
IDFT expression is
x(n) =
N−1
1 X
X (k)e j2πkn/N
N k=0
n = 0, 1, · · · N − 1
(2)
Fourier series is
xp (n) =
N−1
X
ck e j
2π nk
N
−∞≤n ≤∞
(3)
k=0
Fourier series coefficients are expressed as:
ck =
N−1
2π
1 X
xp (n)e −j N nk
N k=0
k = 0, 1.., N − 1
(4)
By comparing X (k) and ck fourier series coefficients has the form of a DFT.
x(n) = xp (n) 0 ≤ n ≤ N − 1
X (k) = Nck
0≤n ≤N −1
Fourier series has the form of an IDFT
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
73 / 91
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Relationship to the Fourier transform of an aperiodic sequence (DFT and DTFT)
Fourier transform X (ω)
X (ω) =
∞
X
x(n)−jωn
−∞≤n ≤∞
(5)
n=−∞
∞
X
X (k) = X (ω|ω=2πk/N ) =
x(n)−j
2π nk
N
−∞≤n ≤∞
(6)
n=−∞
DFT coefficients are expressed as:
xp (n) =
∞
X
x(n − lN)
(7)
n=−∞
xp (n) is determined by aliasing x(n) over the interval 0 ≤ n ≤ N − 1. The finite duration
sequence
xp (n) 0 ≤ n ≤ N − 1
x̂(n) =
0 Otherwise
The relation between x̂(n) and x(n) exist when x(n) is of finite duration
x(n) = x̂(n)
Dr. Manjunatha. P
0≤n ≤N −1
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
74 / 91
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Relationship to the Z Transform
Z transform of the sequence x(n) is
X (z) =
∞
X
x(n)z −n
(8)
n=−∞
Sample X(z) at N equally spaced points on the unit circle. These points will be
Zk = e j2πk/N
k = 0, 1, · · · N − 1
∞
X
X (z)|zk = e j2πk/N =
x(n)e −j2πkn/N
(9)
(10)
n=−∞
If x(n) is causal and has N number of samples then
X (z)|zk = e j2πk/N =
∞
X
x(n)e −j2πkn/N
(11)
n=0
This is equivalent to DFT X(k)
X (k) = X (z)|zk = e j2πk/N
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
(12)
75 / 91
Relationship of the DFT to other Transforms
Relationship of the DFT to other Transforms
Parseval’s Theorem
Consider a sequence x(n) and y(n)
DFT
x(n) ↔ X (k)
DFT
y (n) ↔ Y (k)
N−1
X
x(n)y ∗ (n) =
n=0
N−1
1 X
X (k)Y ∗ (k)
N k=0
(13)
N−1
1 X
|X (k)|2
N k=0
(14)
When x(n)=y(n)
N−1
X
n=0
|x(n)|2 =
This equation give the energy of finite duration sequence it terms of its frequency
components
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
76 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
Determine the DFT of the sequence for N=8, h(n) =
1
2
0
−2≤n ≤2
otherwise
Plot the magnitude and phase response for N=8
Solution:



1 1 1 1 1

h(n) =
, , , ,

2 2 2 2 2

↑
Consider a sequence x(n) and its DFT is
DFT
x(n) ↔ X (k)
DFT
xp (n) ↔ X (k)
where xp (n) is the periodic sequence of x(n) in this example x(n) is of h(n) and is of



1 1 1 1 1

h(n) =
, , , ,


2
2
2
2
2


↑
There are 5 samples in h(n) append 3 zeros to the right side of the sequence h(n)




1 1 1 1 1

h(n) =
, , , , , 0, 0, 0


2 2 2 2 2

↑
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
77 / 91
Problems and Solutions on DFT
The DFT of h(n) and hp (n) is
Problems and Solutions on DFT
DFT
DFT
h(n) ↔ H(k)
hp (n) ↔ H(k)
h( n )
1
2
-2 -1 0
1 2 3
h p (n )
-10 -9 -8
-7 -6
-5 -4 -3
-2 -1
0
4 5
n
These are new
sequences x(n)
1
2
3
4
5
6
7
8
9
10 11 12 13
n
Figure 40: Plot of h(n) and hp (n)
The value of h(n) from the Figure is represented as
hp (n) 0 ≤ n ≤ N − 1
h(n) =
0 Otherwise
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
78 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
The new sequence h(n) from hp (n) is




1 1 1
1 1
h(n) =
, , , 0, 0, 0, ,

2 2
2 2 2

↑











X (0)
X (1)
X (2)
X (3)
X (4)
X (5)
X (6)
X (7)
 1
  1
 
  1
 
  1
=
  1
 
  1
  1
1

Dr. Manjunatha. P
(JNNCE)
1
W81
W82
W83
W84
W85
W86
W87
1
W82
W84
W86
W88
W810
W812
W814
1
W83
W86
W89
W812
W815
W818
W821
1
W84
W88
W812
W816
W820
W824
W828
1
W85
W810
W815
W820
W825
W830
W835
1
W86
W812
W818
W824
W830
W836
W842
1
W87
W814
W821
W828
W835
W842
W849


 
 
 
 
 
=
 
 
 
 
x(0)
x(1)
x(2)
x(3)
x(4)
x(5)
x(6)
x(7)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014











79 / 91
Problems and Solutions on DFT
|H(k)|
Magnitude
1 1.5
2
2
W84 = W812 = W820 = W828 = W836 ... = −1
W85 = W813 = W821 = W829 = W837 ... = − √1 + j √1
2
W86 = W814 = W822 = W830 = W838 ... = j
W87 = W815 = W823 = W831 = W839 ... = √1 + j √1
2
.5
2
W82 = W810 = W818 = W826 = W834 ... = −j
W83 = W811 = W819 = W827 = W835 ... = − √1 − j √1
2
0
W80 = W88 = W816 = W824 = W840 ... = 1
W81 = W89 = W817 = W825 = W833 ... = √1 − j √1
2
2.5
Problems and Solutions on DFT
0
180
2
1 2
3 4
5
6 7
k
6 7
k
Phase of H(k)
90
0
1 2
3 4
5
2
-90
-180












X (0)
X (1)
X (2)
X (3)
X (4)
X (5)
X (6)
X (7)












=










1
1
1
1
1
1
1
1
1
√
1
− j √1
2
2
−j
− √1 − j √1
2
−1
+ j √1
− √1
2
1
√
1
−j
2
2
2
j
+ j √1
2
1
√
1
−j
1
√
−1
j
− √1










(JNNCE)
2
−1
j

Dr. Manjunatha. P
1
− √1 − j √1
2
2
2
j
− j √1
−1
+ j √1
2
2
2.5
1.207
−0.5
−0.207
0.5
−0.207
−0.5
1.207
1
1
1
−1
2
−j
+ j √1
1
−1
2
1
−1
1
− √1 + j √1
2
1
√
2
1
√
2
− √1
−j
+ j √1

2.5∠0

 1.207∠0



 0.5∠ − 180



 0.207∠ − 180
=

 0.5∠0

 0.207∠ − 180



 0.5∠ − 180
1.207∠0

2
−1
− j √1
2
2
−1
−j
1
j
2
j
− j √1
1
j
2
−1
−j
1
+ j √1
1
√
2

2
j
− √1 + j √1
2
2
− √1
2
2
−1
− j √1
1
√
2
−j
− j √1
 1
2


 1
 2


 1
 2



= 0

 0


 0




 1

2
2












1
2











UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
80 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
The unit sample response of the first order recursive filter is given as h(n) = an u(n)
i) Determine the Fourier transform H(ω)
ii) DFT H(k) of h(n)
iii) Relationship between H(ω) and H(k)
H(ω)
∞
X
=
h(n)e −jωn
n=−∞
∞
X
=
an u(n)e −jωn
n=−∞
=
∞
X
(ae −jω)
n
∵ u(n) = 0 for n < 0
n=0
N−1
X
k=0
H(ω)
Dr. Manjunatha. P
(JNNCE)
=
(
ak =
N
1−aN
1−a
for a = 1
for a 6= 1
(ae −jω )0 − (ae −jω )∞+1
1
=
1 − ae −jω
1 − ae −jω
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
81 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
The DFT of h(n) and hp (n) is
DFT
DFT
h(n) ↔ H(k)
hp (n) ↔ H(k)
where hp (n) is related as
∞
X
hp (n) =
h(n − lN)
l=−∞
Consider l=-p
hp (n) =
−∞
X
h(n + pN)
hp (n) =
l=∞
∞
X
h(n + pN)
l=−∞
N point DFT H(k) in terms of hp (n) is
H(k) =
N−1
X
h(n)e −j
2π kn
N
n=0
H(k) =
N−1
X
(JNNCE)
∞
X

n=0
Dr. Manjunatha. P


h(n + pN) e −j
2π kn
N
n=−∞
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
82 / 91
Problems and Solutions on DFT
H(k)
=
N−1
X

=
a
N−1
X
n=0
(n+pN)
u(n + pN) e −j
2π kn
N
n=−∞
"
n=0
=

∞
X

n=0
N−1
X
Problems and Solutions on DFT
∞
X
#
a
(n+pN)
e −j
2π kn
N
∵ u(n) = 0 for n < 0
n=0
"
∞
X
#
an apN e −j
2π kn
N
n=0
Interchanging the summations
H(k) =
∞
X
apN
n=0
N
X
p=0
Dr. Manjunatha. P
(JNNCE)
apN =
∞
X
p=0
p
aN =
an e −j
2π kn
N
n=0
ak =
k=0
∞
X
N−1
X
1 − aN
1−a
1 − (aN )∞+1
1
=
1 − aN
1 − aN
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
83 / 91
Problems and Solutions on DFT
N−1
X
an e −j
2π kn
N
=
N−1
X
n=0
N−1
X
(ae −j
Problems and Solutions on DFT
2π k)n
N
=
n=0
(ae −j
2π k)n
N
=
n=0
1 − aN e −j2πk
1 − aN
=
−j2πk/N
1 − (ae
)
1 − (ae −j2πk/N )
H(k)
=
=
H(ω)
=
(ae −j2πk/N )0 − (ae −j2πk/N )N
1 − (ae −j2πk/N )
=
∵ e −j2πk = 1
1
1 − aN
1 − aN 1 − (ae −j2πk/N )
1
1 − (ae −j2πk/N )
1
1 − ae −jω
and
H(k) =
1
1 − (ae −j2πk/N )
H(k) = H(ω)|ω=2πk/N
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
84 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
Compute the DFT of the following finite length sequence of length N x(n) = u(n) − u(n − N)
u(n)
Unit step sequence u(n)
0
1
2
3
4
N-3 N-2 N-1 N
N+1 N+2 … n
u(n-N)
Unit step sequence delayed by N samples
0
1
2
3
4
N-3 N-2 N-1 N
N+1 N+2 … n
x(n)=u(n)-u(n-N)
Subtraction of u(n-N) from u(n)
0
1
2
3
4
N-3 N-2 N-1 N
N+1 N+2 … n
Figure 41: Generation of x(n) = u(n) − u(n − N)
The value of x(n) as shown in Figure is represented as
1 0≤n ≤N −1
x(n) =
0 Otherwise
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
85 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
DFT expression is
X (k)
=
N−1
X
x(n)e −j2πkn/N
k = 0, 1, . . . N − 1
n=0
=
N−1
X
1e −j2πkn/N
n=0
=
N−1
X
(e −j2πk/N )n
(1)
"N−1
X
n=0
X (k) =
ak =
k=0
1 − aN
1−a
#
1−1
1 − e −j2πk
= −j2πk/N = 0
e −j2πk/N
e
When k=0 From the expression (1)
X (k) =
N−1
X
(1)n = N
n=0
X (k) =
0 when k 6= 0
N when k = 0
X (k) = Nδ(k)
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
86 / 91
Problems and Solutions on DFT
Problems and Solutions on DFT
If x(n)=[1,2,0,3,-2, 4,7,5] evaluate the following i) X(0) ii) X(4) iii)
7
P
X (k) iv)
k=0
7
P
|X (k)|2
k=0
X(0) is
X (k) =
N−1
X
x(n)e −j2πkn/N
n=0
with k=0 and N=8
X (0) =
N−1
X
x(n) = 1 + 2 + 0 + 3 − 2 + 4 + 7 + 5 = 20
n=0
X(4) is
X (k) =
N−1
X
x(n)e −j2πkn/N
n=0
with k=4 and N=8
X (4) =
N−1
X
n=0
x(n)e −j2π4n/8 =
N−1
X
n=0
x(n)e −jπn =
N−1
X
x(n)(−1)n
n=0
X (4) == 1 − 2 + 0 − 3 − 2 − 4 + 7 − 5 = −8
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
87 / 91
Problems and Solutions on DFT
iii)
7
P
Problems and Solutions on DFT
X (k)
k=0
We Know the IDFT expression as
x(n) =
N−1
1 X
X (k)e j2πkn/N
N n=0
With n=0 and N=8 it becomes
x(0) =
∴
N−1
X
N−1
1 X
X (k)
8 n=0
X (k) = 8x(0) = 8 × 1 = 8
n=0
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
88 / 91
Problems and Solutions on DFT
The value of
7
P
Problems and Solutions on DFT
|X (k)|2 is
k=0
The expression for Parseval’s theorem is
N−1
X
|x(n)|2 =
N−1
1 X
|X (k)|2
N k=0
|x(n)|2 =
N−1
1 X
|X (k)|2
N k=0
|x(n)|2 =
7
1X
|X (k)|2
8 k=0
n=0
7
P
|X (k)|2 is related as
k=0
N−1
X
n=0
N=8 Then
7
X
n=0
7
X
|X (k)|2 = 8
k=0
Dr. Manjunatha. P
(JNNCE)
7
X
|x(n)|2 = 8[1 + 4 + 0 + 9 − 4 + 4 + 49 + 25] = 864
n=0
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
89 / 91
Thank You
Dr. Manjunatha. P
(JNNCE)
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
90 / 91
References
J. G. Proakis and D. G. Monalakis, Digital signal processing Principles Algorithms &
Applications, 4th ed. Pearson education, 2007.
Oppenheim and Schaffer, Discrete Time Signal Processing.
Hall, 2003.
S. K. Mitra, Digital Signal Processing.
L. Tan, Digital Signal Processing.
Tata Mc-Graw Hill, 2004.
Elsivier publications, 2007.
J. S. Chitode, Digital signal processing.
Technical Pulications.
B. Forouzan, Data Communication and Networking, 4th ed.
Dr. Manjunatha. P
(JNNCE)
Pearson education, Prentice
McGraw-Hill, 2006.
UNIT - 1: Discrete Fourier Transforms (DFT)[1, 2, 3, 4,September
5]
11, 2014
91 / 91