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Digital Signal Processing, © 2011 Robi Polikar, Rowan University
Discrete Time Fourier Transform
(DTFT)
DSPSignals
Sinusoids &
Exponentials
Impulse, step,
rectangular
Phasors
Frequency
Characterization
Time domain
representation
Representation in
frequency domain
Spectrum
Power / Energy Periodicity Cont. / Discrete
Convolution
Regular / Circular
Sampling
Nyquist Thm.
CFT
Transforms
ZPoles & Zeros
ROC
DFTDTFT FFT
(LTI) Systems
Discrete LTI
Systems
Classification
Impulse Resp.
Linearity
Time Inv.
Causality Memory
Stability
Time Domain Rep.
Diff. Equation
Ideal vs. Practical
Freq. Domain Rep.
Filtering
LPF HPF BPF BSF APF Notch
FIR / IIR
Filter Design
FIR IIR
Windows
Linear Phase
Specs
Bilinear. Tran.
Butterworth
Chebychev
Elliptic
Stability
Filter Structure
FIR IIR
Direct
Cascade
Lattice
Transfer Func.
Frequency Res.
Quantization
Finite
Worldlength
A/D D/A
Number Rep.
Fixed/Floating
Quantization
Noise
Overflow
Effects
Advanced Topics
Random Signal Analysis
Multirate Signal Proc.
Time Frequency Analysis
Adaptive Signal Process.
RP
This Week in DSP
The Discrete Time Fourier Transform
Definition
Key theorems
• DTFT of the output of an LTI system
• The frequency response
• Periodicity of DTFT
• Definition of discrete frequency
• Existence of DTFT
DTFTs of some important sequences
DTFT properties
DTFT in Matlab
Fourier Series and
Fourier Transform
Any periodic signal x(t), with fundamental period is T0, can be represented as a finite
and discrete sum of complex exponentials (sines and cosines) that are integer
multiples of Ω0, the fundamental frequency: FOURIER SERIES
A non-periodic continuous time signal can
also be represented as an (infinite and
continuous) sum of complex exponentials: FOURIER TRANSFORM
k
ktjkectx 0)(
00
0
0)(1
0
Tt
t
tjkk dtetx
Tc
1
( ) ( )
1( ) ( ) ( )
2
j t
j t
X x t x t e dt
x t X X e d
k
ck
1 2 3-1-2-3 0
1/2j1/2 1/2
-1/2j
jeX
XXX
( )x t XF
RP
Key Facts
to Remember
All FT pairs provide a transformation between time and frequency domains: The
frequency domain representation provides how much of which frequencies exist in
the signal More specifically, how much ejΩt exists in the signal for each Ω.
In general, the frequency representation is complex (except when the signal is even).
|X(Ω)|: The magnitude spectrum the power of each Ω component
Ang X(Ω): The phase spectrum the amount of phase delay for each Ω component
The FS is discrete in frequency domain, since it is the discrete set of exponentials –
integer multiples of Ω0 – that make up the signal. This is because only a finite
number of frequencies are required to construct a periodic signal.
The FT is continuous in frequency domain, since exponentials of a continuum of
frequencies are required to reconstruct a non-periodic signal.
Both transforms are non-periodic in frequency domain.
the discrete signal, ω, is 2π. While counterintuitive, ω is NOT discrete in frequency! ,
Digital Frequency
Recall that a discrete time signal can be obtained from a continuous time signal through the process of sampling: take a sample every Ts second
When time is discretized, what happens to frequency? Consider the following: (where we use Ω to represent continuous frequency, and ω to represent discrete frequency)
Note that when Ω=Ωs ω = ωs= 2π What does this mean?
,2,1,0,1,2,)()()(
ntxnTxnxsnTts
( ) sin( )
( ) sin( )s s s
y t A t
y nT A T n
sss
F
FfT
2
Sampling frequency (rad/s) Sampling frequency (sam/s)
Spectrum of discrete signals – by definition – is normalized with respect to sampling frequency. If the analog frequency is equal to the sampling frequency, the corresponding frequency of
Digital frequency Analog frequency
2s
F
F
Discrete –Time
Fourier Transform (DTFT)
Similar to continuous time signals, discrete time sequences can also be periodic or
non-periodic, resulting in discrete-time Fourier series or discrete – time Fourier
transform, respectively.
Most signals in engineering applications are non-periodic, and DTFS is really a
special case of DTFT, so we will concentrate on DTFT.
We will represent the discrete signal‟s frequency as ω, measured in radians/sample.
)(][ Xnx
1[ ] ( )
2
( ) [ ]
j n
j n
n
x n X e d
X x n e
Quick facts:
• Since x[n] is discrete, we can only add them, summation in theanalysis equation. ejωn, however, is continuous function for each n
• The sum of x[n], weighted with continuous exponentials ejωn, is
continuous the DTFT X(ω) is continuous (non-discrete)
• Since X(ω) is continuous, x[n] is obtained as a continuous integral
of X(ω), weighed by the same complex exponentials.
• x[n] is obtained as an integral of X(ω), where the integral is over
an interval of 2π. This is our first clue that DTFT is periodic
with 2π in frequency domain.
• X(ω) is sometimes denoted as X(ejω) or X(jω) in some books, including
yours. While X(ejω) is more accurate, we will use X(ω) for brevity.
• The ∞ indicates that the sum is simply over all available samples!
2 sf f
Digital Signal Processing, © 2011 Robi Polikar, Rowan University
Proof
We now show that x[n] and X(ω) are indeed FT pairs, that is one can
be obtained from the other:
Lemma: complex exponentials* are orthogonal, that is
j n
n
X x n e
1
2
j nx n X e d
1
2
j j nx n x e e d
0
0 ( )
0
0
0,[ ]
,
t T
j k m t
t
k me dt T k m
T k m
Integral over
one period
* Remember: Continuous time complex exponentials are ALWAYS periodic with some fundamental frequency 0 –unlike the discrete complex exponentials, which are only periodic for rational values of angular frequency
Analysis Equation Synthesis Equation
( )0,
2 [ ]2 ,
j n ln l
e d n ln l
Time domain case
Frequency domain case
Important Theorems
Theorem 1: The frequency response of the output of any system is
the product of the spectrum of the input signal and that of the
frequency response
Theorem 2: The DTFT of the impulse response is the frequency
response of the system.
Theorem 3: DTFT is periodic with 2π.
Theorem 4: The digital frequency 2π corresponds to the linear
sampling frequency of the signal.
Theorem 5: DTFT only exists for sequences that are absolutely
summable.
System Output
If x[n] is input to an LTI system with an impulse response of h[n],
then the DTFT of the output is the product of X(ω) and H(ω)
x[n] y[n]=x[n]*h[n]h[n]
X(ω) Y(ω) = X(ω) . H(ω)H(ω)
RP
Theorem 2:
If the input to an LTI system with an impulse response of h[n] is a complex
exponential ej0n , then the output is the SAME complex exponential whose
magnitude and phase are given by |H(ω)| and <H(ω), evaluated at ω = ω0.
Frequency Response
h[n]ejω0n 0
0 0
[ ] [ ]
[ ]
j n k
k
j k j n
k
y n h k e
h k e e
H(ω0)
0
0[ ]j n
y n H e
If the system input is a complex exponential at a specific frequency ω0, then the system
output is the same exponential, at the same frequency ω0 but weighted by a complex
amplitude that is a function of this input frequency. This complex amplitude, H(ω0), is
the DTFT of system impulse function h[n], evaluated at ω0 , and it is called the
frequency response of the system.
Quantity independent of n
Frequency Response
This theorem constitutes the fundamental cornerstone for the concept
of frequency response. H(ω), the DTFT of h[n], is called the
frequency response of the system
Why is it important?
If a sinusoidal sequence with frequency 0 is applied to a system whose frequency
response is H(ω), then the output can be obtained simply by evaluating H(ω) at
ω = ω0.
Since all signals can be written as a superposition of sinusoids at different
frequencies, then the output to an arbitrary input can be obtained as the
superposition of H(ω) for each component ω0 that makes up the input signal!
Most importantly, this is cornerstone of filter design: If you want to design a filter
that blocks a certain frequency ωcut, then we design the system such that H(ωcut)=0;
and if we want the system to pass a certain frequency ωpass then we make sure that
H(ωpass)=1
A Simple
Filtering Example
Consider the ideal lowpass filter
We apply an input, , which has
two frequency components in it, ω1 that falls within the passband of our filter, and
ω2 that falls outside of the passband.
According to Theorem 2, the output will be at the same frequencies, but multiplied
by a constant specified by the frequency response of the system, evaluated at those
frequencies.
c
cH
,0
,1
H(ω)
ω-ωc ωc
1
1 1 2 2 1 2cos cos , 0x n C n C n
-π πpassband
1 1 1 2 2 2 1 1 1
0
cos cos cosy n H C n H C n H C n
(why is this an approximation,but not an equality?)
The output does not include the frequency ω2 , i.e., it is filtered out.
Hello Filter(My First Filter Design)
Let‟s design a simple lowpass FIR filter that blocks frequencies above
0.7π, but passes the frequencies below 0.2π perfectly.
We want our filter to be as simple as possible, so let’s assume that we have length
3, symmetric impulse response filter. That is, h[0]=h[2]=1 , and h[1]= 2.
Then our filter should have a frequency response of the form
We want H(0.2π)=1, and H(0.7π)=0. Furthermore, right now, we are primarily
concerned with the magnitude and not phase. Plugging these values into H(ω),
0 1 2
2 2
1 2 1 1 2
1 2 1 2
cos
0 1
22
2
1
2 cos
j j j
j j j j
j j
jj j
H h e h e h e
e e e
e
e
e ee e
1
0.2
11 2 1 2
0.72
1 2 1 2
1
0.3580,0.2 2 cos 0.2 1 2 cos 0.2 1
0.4208.0.7 2 cos 0.7 0 2 cos 0.7 0
magnitude
j
j
magnitude
H e
H e
Hello FilterDid My First Filter Work?
b=[0.3580 0.4208 .3580];a=1;[H w]=freqz(b,a, 1024);plot(w/pi, abs(H))gridxlabel('Normalized Angular Frequency / \pi ')ylabel('Magnitude Frequency Response')
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
1.4
Normalized Angular Frequency /
Ma
gn
itu
de
Fre
qu
en
cy
Re
sp
on
se
H(0.2π)=1
H(0.7π)=0
RP
Periodicity of DTFT
Theorem 3:
The DTFT of a discrete sequence is periodic with the period 2π, that is
X( ) X( 2 ) for any integer k
The periodicity of DTFT can be easily verified from the definition:
j n
n
X x n e
2
2
2 [ ]j k n
n
j k n j nj n
n n
X k x n e
x n e e x n e X
Why…?
Implications of the
Periodicity Property
2H H
Theorem 4 (You-will-flunk-if-you-do-not-understand-this-fact theorem):
The discrete frequency 2π rad. corresponds to the sampling frequency Ωs used to
sample the original continuous signal x(t) to obtain x[n].
Proof: sin sins sx t A t x nT A T n
ω=ΩTs For Ω= Ωs, we have ω=ΩsTs=2πfsTs=2π
Understanding the
Periodicity of the DTFT
Recall the following example
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-10
-5
0
5
10 Sin(220t)+4Cos(250t)+2Sin(2100t)
Time, s
-500 -400 -300 -200 -100 0 100 200 300 400 5000
1000
2000
3000
4000 Frequency domain representation of Sin(220t)+4Cos(250t)+2Sin(2100t)
Frequency, Hz.
t=0:0.001:2; % sampling frequency = 1000Hz
x=sin(2*pi*20*t)+4*cos(2*pi*50*t)+2*sin(2*pi*100*t); subplot(211)
plot(t(1:200),x(1:200))
grid
title('Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)') xlabel('Time, s')
subplot(212)
X=abs(fft(x));
X2=fftshift(X);
f=-499.9:1000/2001:500;
plot(f,X2);
grid
title(' Frequency domain representation of
Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)') xlabel
('Frequency, Hz.')
RP
Understanding the
Periodicity of the DTFT
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-10
-5
0
5
10
Sin(220t)+4Cos(250t)+2Sin(2100t)
Time, s
0 500 1000 1500 2000 25000
1000
2000
3000
4000 Frequency domain representation of Sin(220t)+4Cos(250t)+2Sin(2100t)
Normalized Frequency
t=0:0.001:2; %Sampling frequency = 1000Hz% Note that the length of the signal is 2001 samples.x=sin(2*pi*20*t)+4*cos(2*pi*50*t)+2*sin(2*pi*100*t);subplot(211)plot(t(1:200),x(1:200)) %plot a portion of the signalgridtitle('Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)')xlabel('Time, s')subplot(212)X=abs(fft(x));plot(X)gridtitle(' Frequency domain representation of Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)')xlabel('Normalized Frequency')
What do these mean?
FFT computes the FT at the same N number of points as the length of the originl signal with the assumption that the frequency range is one period of [0 2π]. Then, sample 1 of the FFT corresponds to frequency 0, and sample N corresponds to frequency 2π(which is the sampling frequency).
RP
0 0.05 0.1 0.15 0.2-10
0
10Sin(220t)+4Cos(250t)+2Sin(2100t)
Time, s
0 500 1000 1500 2000 25000
2000
4000 Frequency domain representation of Sin(220t)+4Cos(250t)+2Sin(2100t)
Normalized Frequency
0 500 1000 1500 2000 25000
2000
4000
Normalized Frequency
-500 -400 -300 -200 -100 0 100 200 300 400 5000
2000
4000
Frequency, Hz.
t=0:0.001:2; %Sampling frequency = 1000Hzx=sin(2*pi*20*t)+4*cos(2*pi*50*t)+2*sin(2*pi*100*t);subplot(411)plot(t(1:200),x(1:200)) %plot a portion of the signalgridtitle('Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)')xlabel('Time, s')subplot(412)X=abs(fft(x));%Take the DFT(FFT)plot(X) % Plots X in the default [0 2π] rangegridtitle(' Frequency domain representation of Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)')subplot(413)X2=fftshift(X); % Flip the FT so that 0 frequency is at the center, and the
% frequency range is now [-π π] instead of [0 2π]
plot(X2)gridtitle(' Frequency domain representation of Sin(2\pi20t)+4Cos(2\pi50t)+2Sin(2\pi100t)')subplot(414)f=-499.9:1000/2001:500; %Create frequency axis to plot -500 to 500 Hzplot(f,X2);gridxlabel('Frequency, Hz.') RP
Existence of DTFT
Theorem 5:
The DTFT of a sequence exists if and only if, the sequence x[n] is absolutely
summable, that is, if
because:
Hence, if x[n] is absolutely summable, then |X(ω)|is finite, which means that X(ω)
exists.
We should add that this is sufficient, but not required to have a DTFT. Certain
sequences that do not satisfy this requirement also have DTFTs, if they satisfy
“mean square convergence.” These will be discussed later within z-transform.
n
nx ][
j n j n
n n n
X x n e x n e x n
This quantity is always ≤ 1
Important DTFT Pairs
Impulse Function
The DTFT of the impulse function is “1” over the entire frequency
band. 1][ n
Extend of the frequency band
in discrete frequency domain
01 1j n j n
n
n n e e
Summations terms are all zero,
except for n=0
[n]
n
x[n] X(ω)
ω
1DTFT
-π π RP
Important DTFT Pairs
Constant Function
Note that x[n]=1 (or any other constant) does not satisfy absolute
summability. However, we can show that the DTFT of the constant
function is an impulse at ω=0. (this should make sense!!!)
m
m 221 j n
n
X x n e
Cannot use
We can show that this transformation is correct, by computing the inverse DTFT of the above function
1 012 2 2 2 1
2
j n j n
m m
m m e d e
n
x[n]X(ω)
ω
2π(ω)
DTFT
-2π 2π
2π(ω-2π)2π(ω+4π)
-4π 4π-π π
RP
Matlab Approximation
In class demo!
x=zeros(1000,1);
x(500)=1;
subplot(211)
plot(x); grid
title('Original unit delta sequence \delta[n]')
X=abs(fft(x));
subplot(212)
w=-pi:2*pi/999:pi;
plot(w/pi, fftshift(X)); grid
title('Magnitude spectrum')xlabel('Angular Frequency (x \pi)')
0 100 200 300 400 500 600 700 800 900 10000
0.5
1Original unit delta sequence [n]
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11
1
1
1Magnitude spectrum
Angular Frequency (x )
RP
In class Demo
x=ones(1000,1);
subplot(211)
plot(x); grid
title('Original constant sequence')
X=abs(fft(x));
subplot(212)
w=linspace(-pi, pi, 1000);
plot(w/pi, fftshift(X)); grid
title('Magnitude spectrum')xlabel('Angular Frequency (x \pi)')
0 100 200 300 400 500 600 700 800 900 10000
0.5
1
1.5
2Original constant sequence [n]
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10
200
400
600
800
1000Magnitude spectrum
Angular Frequency (x )
RP
Important DTFT Pairs
The Complex Exponential
The DTFT of the complex exponential:
0
0[ ] 2 2j
k
x n e X k
ω0 ω0+2π ω0+4πω0-4π ω0-2π
. . . . . .
We are only interested in
[- π π] range, where there is
only one spectral component
Hence, the spectrum of a single
complex exponential at a specific
frequency is an impulse at that
frequency.
This can be verified by computing
the inverse DTFT of X(ω) given
above, as in the previous example.
RP
Important DTFT Pairs
Real Exponential
j
n
enunx
1
1][][
0 0
1
1
nn j n j
jn n
X e ee
In Matlab
This is an important function in signal processing. Why?
j
n
enunx
1
1][][
t=0:0.01:10;
x=(0.5).^t;
plot(t,x)
X=fftshift((fft(x)));
subplot(311)
plot(t,x); grid
subplot(312)
plot(abs(X)); grid
f=-50:100/1000:50;
plot(f,abs(X)); grid
subplot(313)
plot(f, unwrap(angle(X))); grid
In Matlab, periodicity of X(ω) is assumed
RP
Important DTFT Pairs
The sinusoid at ω=ω0
By far the most often used DTFT pair (it is less complicated then it looks):
mm
mmnnx 000 22cos][
00[ ] 2 2
j n
m
x n e m
The above expression can also be obtained
from the DTFT of the complex exponential
through the Euler’s formula.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
(ω+ ω0)
ω
X(ω)
(ω- ω0)
ω0- ω0
DTFT
Cos (ω0t)
. . .. . .
-2π 2π-π π
RP
Important DTFT Pairs
Rectangular Pulse
Rectangular pulse train is also very commonly used in DSP (it is the
moving average filter).
What if we change the order?
Whose Fourier transform would
be a rectangular pulse (which
of course is ideal LPF)
0,2sin
21sin
otherwise,0
,1][rect][
Me
MnMnnx
M
Mn
nj
M
RP
Ideal Lowpass Filter
The ideal lowpass filter is defined as
Taking its inverse DTFT, we can obtain the corresponding impulse function h[n]:
c
cH
,0
,1
1 1[ ] 1 1
2 2
sin,
1
2, 0
c
c
c c
j n j n
cj n j n
c
h n e d e d
nn
e e n
jn jnn
Ideal Lowpass Filter
Note that:
The impulse response of an ideal LPF is infinitely long This is an IIR filter. In
fact h[n] is not absolutely summable its DTFT cannot be computed an ideal
h[n] cannot be realized!
One possible solution is to truncate h[n], say with a window function, and then
take its DTFT to obtain the frequency response of a realizable FIR filter.
How does this code work?(Carefully analyze at home)
%This function creates an ideal LPF in frequency domain and then computes
its inverse Fourier transform.
%Robi Polikar, Feb 12 2007
Clear; close all
L=input('Enter the length of the frequency vector: L= ');
wc=input('Enter the LPF corner frequency in rad (times pi): wc= ')
%These values create good looking graphs L=128; wc=0.25*pi;
w=linspace(-pi, pi, L); t=linspace(-0.5,0.5,L);
%Create a LPF with a cutoff frequency of wc
H=zeros(L,1);
H(L/2-round(wc*(L/2)/pi): L/2+round(wc*(L/2)/pi))=1;
subplot(311); plot(w/pi, H); grid
xlabel('Angular frequency, x\pi')
Wc=wc/pi; %Convert to a multiple of pi
title(['Magnitude spectrum of LPF with cutoff frequency \omega_c =', …
num2str(Wc), '\pi rad']);
H=fftshift(H); %This is necessary to reset H into the [0 2pi] range as
%expected by Matlab
h=real(ifft(H)); h=fftshift(h);
subplot(312); plot(t, h); grid
axis([t(1) t(L) 0.8*min(h) 1.2*max(h)])
xlabel('Time, s.')
title('Inverse FT --> impulse response of H(\omega)')
n=-L/2:L/2;
h_alt=sin(wc*n)./(pi*n);
if (rem(L,2)== 0)
h_alt(round(L/2)+1)=wc/pi;
end
subplot(313)
plot(n,h_alt); grid
axis([-L/2 L/2 0.8*min(h_alt) 1.2*max(h_alt)])
xlabel('Time index n')
title('impulse response of H(\omega) computed analytically')
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10
0.5
1
Angular frequency, x
Magnitude spectrum of LPF with cutoff frequency c =0.2 rad
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5
0
0.1
0.2
Time, s.
Inverse FT --> impulse response of H()
-60 -40 -20 0 20 40 60
0
0.1
0.2
Time index n
impulse response of H() computed analytically
Ideal_lpf_by_sinc.m
RP
Ideal LPF defined in frequency domain
Ideal LPF defined in time domain
Some Useful
Matlab Functions
Matlab cannot explicitly calculate the DTFT, since the frequency axis is continuous.
However, it can calculate an approximation of the DTFT using a given number of points.
y=fft(x, N) – Calculates the discrete Fourier transform of the signal x at N points. If N is not
provided, length of y is the same as x. DFT is a sampled version of the DTFT, where the
samples are taken at N equidistant points around the unit circle from 0 to 2π.
[h,w] = freqz(b,a,N,'whole') – Calculates the frequency response of a filter whose CCLDE
coefficients are given as b and a, using N number of points around the unit circle. If „whole‟
is included, it returns a frequency base of w from 0 to 2π, otherwise, from 0 to π.
y=abs(x)- Calculates the absolute value of signal x. For complex values signals, the output is
the magnitude (spectrum) of the complex argument.
y=angle(x) – Calculates the phase (spectrum) of the signal x.
q = unwrap(p) corrects the radian phase angles in a vector p by adding multiples of 2π when
absolute jumps between consecutive elements of p are greater than the default jump tolerance
of π radians.
y = fftshift(x) rearranges the outputs of fft by moving the zero-frequency component to the
center of the array. It is useful for visualizing a Fourier transform with the zero-frequency
component in the middle of the spectrum. It changes the default [0 2π] range to [-π π]
Other Important
Properties of DTFT
We will study the following properties of the DTFT:
Linearity DTFT is a linear operator
Time reversal x[-n] X(-ω)
Time shift x[n-n0] X(ω)e-jωn0
Frequency shift x[n] ejω0n X(ω-ω0)
Convolution in time x[n]*y[n] X(ω).Y(ω)
Convolution in frequency
Differentiation in frequency nx[n] j (dX(ω)/dω)
Parseval’s theorem Conservation of energy in time and frequency domains
Symmetry properties
)(][ Xnx
)(][ Yny
Linearity &
Time Reversal
Given x(t) and X(ω) form a DTFT pair, we can show that
The DTFT is a linear operator
A reversal in of the time domain variable causes a reversal of the frequency variable
ax n by n aX bY
x n X
x n X
,j n
n
j n j n j m
n n mm
X x n e if y n x n
Y y n e x n e x m e X
Proof:
Time & Frequency Shift
A shift in time domain by m samples causes a phase shift of e-jωm in the frequency
domain
Note that the magnitude spectrum is unchanged by time shift.
Similarly, a shift in frequency domain by ω0 causes a time delay of ejω0n
,j Mx n M X e M
00
jx n e X
j k Mj n
k n Mn kn k M
j k j M j M j k j M
k k
y k x n M Y x n M e x k e
x k e e e x k e X e
y k x n M Y X Why not?
Importance of the Linearity &
Time Shift Properties
Frequency Response of FIR Systems
Here is another way to show that the frequency response of a system,
whose impulse response is h[n], is in fact H(ω)=F{h[n]}
Given the impulse response h[n] of an FIR system, the output y[n] to the input
x[n] is of course the convolution sum
To compute the spectrum of the output, Y(ω), let’s take the DTFT of both sides.
Assuming that X(ω) and Y(ω) exist, and using the linearity and time shifting
properties, we have
Now, since we know that in frequency domain, the output Y(ω) = H(ω). X(ω), we
have
proving that the frequency response of the system is indeed the DTFT of the
impulse response h[n].
k
y n h k x n k
j k
k
Y h k X e
j k
k
H h k e
Importance of the Linearity &
Time Shift Properties
Frequency Response of IIR Systems
Consider the CCLDE of a typical LTI system:
The linearity of DTFT allows us to compute the DTFT of the entire
expression, by computing the DTFTs of the individual terms, and then
combine them:
1 1
0
0 0
[ ] [ ], 1N M
i
i
a y n i b x n j a
1 2 1 0 1 1[ ] [ 1] [ 2] 1 [ ] [ 1] 1N My n a y n a y n a y n N b x n b x n b x n M
11 2
1 2 1
11 2
0 1 2 1
1 11 1
1 1 0 1 1
11
0 1 1
11
1 1
1
1
j Nj j
N
j Mj j
M
j N j Mj j
N M
j MjkM
j Nj
N
Y a Y e a Y e a Y e
b X b X e b X e b X e
Y a e a e X b b e b e
bY b b e b eH
X a e a e
1
0
1
0
M j k
k
N j k
kk
e
a e
Differentiation in Frequency
Multiplying the time domain signal with the independent time variable
is equivalent to differentiation in frequency domain.
Example: What is the DTFT of
d
dXjnnx
)(][
1 ny n n a u n
2
1
1
1
1
n
j
j
Let x n a u n Xae
dXy n nx n x n Y j X
d ae
Convolution in Time
Convolution in time domain is equivalent to multiplication in
frequency domain
Proof: Let y[n]=x[n]*h[n], then we need to prove that Y(ω)=X(ω)H(ω)
This is one of the fundamental theorems in DSP. It allows us to compute the
filter response in frequency domain using the frequency response of the filter.
x n h n X H
m
k
j n j n
n n
j
k
j m j
k
m n k
k
H
k m
j k j k
k k
k
m
x k h n
y n x n
m
h n x k h n k
Y y n y n e e
x k h e
x
k
k h m e e
x k e x k e HH X H
Convolution in Frequency
(Modulation)
Multiplication in time domain is equivalent to convolution integral in
frequency domain
This property is also called the modulation theorem, since it involves
the modulation of one signal x[n] with the other h[n].
Recall our discussion on the meaning of signal and system. Here, h[n] can be
considered as another signal.
1
2x n h n X H d
Parseval’s Theorem
The energy of the signal , whether computed in time domain or the frequency
domain, is the same!
Alternatively:
2 21
2n
x n X d
Energy density spectrum
of the signal
1
2n
x n y n X Y d
Your text lists several symmetry properties of DTFT
While all of these properties are important for academic reasons, the following are
important for practical reasons:
• The Fourier transform of a real signal is conjugate symmetric: the magnitude spectrum
is an even function of ω (symmetric), whereas the phase spectrum is an odd function
of ω (antisymmetric). That is, for a real signal x[n]
• The Fourier transform of a symmetric signal is real! More generally, if
then, the following is true:
since the even part of any signal is necessarily symmetric, it follows that the DTFT of
any symmetric signal must also necessarily be real!
Symmetry Properties
of DTFT
* * ,X X X X X X
real imagx n X X jX
,even real odd imagx n X x n X
Resolution of
the DTFT
Recall the definition of the DTFT:
The analysis equation is an infinite sum. In reality, we cannot deal with the
infinitely long signals (particularly using computers), as we do not have infinite
memory. Hence, we usually have a finite length signal, say of N samples, in which
case the DTFT becomes:
…but then, it is obvious that the larger N, the larger number of components are
used to make up X(ω), and hence the more accurate X(ω) becomes. We will see
that the limitation that we can only use a finite number of samples will create some
artifacts in the spectrum
1
2
j n j n
n
X x n e x n X e d
1
0
1, 0 1
2
Nj n j n
n
X x n e x n X e d n N
Resolution of
the DTFT
Consider the following three sinusoids, all at the same (some angular) frequency of
π/2, and their corresponding spectra:
-1 -0.5 0 0.5 10
2
4
6Spectrum of x
1[n]
-1 -0.5 0 0.5 10
5
10Spectrum of x
2[n]
-1 -0.5 0 0.5 10
10
20Spectrum of x
3[n]
-1 -0.5 0 0.5 10
10
30
50Spectrum of x
4[n]
Normalized frequency - to
Sp
ectr
al A
mp
litu
de
%four time bases of length 10, 20 and 40 and 100
t1 = 0:9;
t2 = 0:19;
t3 = 0:39;
t4 = 0:99;
%Create frequency base of 1024 points between +/- pi
w = linspace(-pi, pi, 1024);
omega=pi/2; %Common frequency for all four signals
x1=cos(omega*t1); x2=cos(omega*t2);
x3=cos(omega*t3); x4=cos(omega*t4);
%Compute 1024 point FFTs
X1 = fftshift(abs(fft(x1, 1024)));
X2 = fftshift(abs(fft(x2, 1024)));
X3 = fftshift(abs(fft(x3, 1024)));
X4 = fftshift(abs(fft(x4, 1024)));
subplot(411); plot(w/pi, X1); title('Spectrum of x_1[n]'); grid
subplot(412); plot(w/pi, X2); title('Spectrum of x_2[n]'); grid
subplot(413); plot(w/pi, X3); title('Spectrum of x_3[n]'); grid
subplot(414); plot(w/pi, X4); title('Spectrum of x_4[n]'); grid
xlabel('Normalized frequency -\pi to \pi') RP
Frequency Resolution
of the DTFT
Observe the following:
While the peak spectral component is correctly
located at π/2 in each case, the peak is not
sharp there appears to be other frequencies:
• These are also due to using a finite length
signal, which can be interpreted as windowing
with a rectangular function (more on this later)
• Gibbs’ phenomenon
• These are artifacts – these frequencies do not
actually exist in the signal
The width of the main lobe is inversely
proportional to the length of the signal
The amplitude of the peaks are directly
proportional to the length of the signal
The longer the signal, the larger the number
of side lobes but with narrower width for
each.
-1 -0.5 0 0.5 10
2
4
6Spectrum of x
1[n]
-1 -0.5 0 0.5 10
5
10Spectrum of x
2[n]
-1 -0.5 0 0.5 10
10
20Spectrum of x
3[n]
-1 -0.5 0 0.5 10
10
30
50Spectrum of x
4[n]
Normalized frequency - to
Sp
ectr
al A
mp
litu
de
-0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2
0
2
4
Spectrum of x1[n]
-0.7 -0.65 -0.6 -0.55 -0.5 -0.45 -0.4 -0.35 -0.30
2
4
6
8
Spectrum of x2[n]
-0.6 -0.55 -0.5 -0.45 -0.4 -0.350
5
10
15
Spectrum of x3[n]
-0.58 -0.56 -0.54 -0.52 -0.5 -0.48 -0.46 -0.44 -0.42 -0.40
10
30
Spectrum of x4[n]
Normalized frequency - to
RP
What did we
learn this week?
The Discrete Time Fourier Transform
Definition
Key theorems
• DTFT of the output of an LTI system
• The frequency response my first filter : Hello filter
• Periodicity of DTFT
• Definition of discrete frequency
• Existence of DTFT
DTFTs of some important sequences
DTFT properties
• Linearity and time reversal
• Time and frequency shifting
• Differentiation, convoltion, symmetry properties
• Parseval’s theorem
• The resolution of the DTFT
DTFT in Matlab