Lab Manual EKT230/4
Signals & Systems
LABORATORY 5
FOURIER REPRESENTATION
1.0
OBJECTIVES:
1.1
To determine the DTFS coefficients of the signal.
1.2
Building a square wave from DTFS coefficients and partial-sum
approximation.
1.3
Use the fft, fftshift and ifft MATLAB commands.
1.4
Do Fourier Series problems without integration or indeterminate.
1.5
To study the relation in the time domain between a continuous time
signal and discrete time signal.
1.6
To study relation between the continuous time Fourier transform (CTFT)
and DTFT In frequency domain.
2.0
THE DTFS ( DISCRETE-TIME FOURIER SERIES )
Commands covered to evaluate DTFS are,
fft
ifft
The DTFS representation of a periodic signal x[n] with fundamental period N and
fundamental frequency Ω0 = 2N is given by
x[n] 
N 1
 X [k ]e
jk 0 n
______(equation 2.1(a ))
K 0
X [k ] 
Where,
1
N
N 1
 x[n]e
 jk 0 n
______(equation2.1(b))
n 0
are the DTFS coefficients of the signal x [n]. We say that x [n] and x [k] are a
DTFS pair and denote this relationship as;
DTFS ; Ω0
x [n]
X [k] = frequency domain representation for x[n]
Any periodic discrete-time signal maybe describe in terms of equation 2.1 (a)
Given a length-N vector x representing one period of an N periodic signal x [n],
the command
>> x = fft (x) / N
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Produces a length-N vector x containing the DTFS Coefficient X[K]. MATLAB
assumes that the summation in both equations above run from 0 to N-1, so the
first elements of x and X correspond to x [0] and X [0], respectively, while the last
elements correspond to x [N-1] and X [N-1].
Note that division by N is necessary because fft evaluates the sum in equation
2.1(b) without dividing by N. Similarly, given DTFS coefficient in a vector X, the
command
>> x = ifft(x)*N
Produces a vector x that represents one period of the time domain waveform.
Note that ifft must multiplied by N to evaluate equation 2.1(a)
Example 2.1.1
Determine the DTFS coefficients of the signal using MATLAB.
x [n] = 1 + Sin (
x [n]
n 3

)
12
8
DTFS ; 2π/24
X [K]
The signal has period 24, so we define one period and evaluate the DTFS
coefficients using the following commands.
>> x = ones (1,24) + sin ( [0:23]*pi/12 + 3*pi/8 );
>> X1= fft (x)/24
>> X1 = ?
( Note that MATLAB uses i to denote the square roots of -1), your answer will be
X [K] =
1
0.4619-j0.1913
0.4619+j0.1913
0
K=0
K=1
K = 23
otherwise on 0 ≤ K ≤ 23
Using ifft, we may reconstruct the time-domain signal and evaluate the first form
values of the reconstructed signal with the commands.
>> xrecon = ifft (X1)*24;
>> xrecon (1:4);
ans = 1.9239
1.9914 - 0.0000i
1.9914
1.9239 – 0.0000i
Note that the reconstructed signal has an imaginary component, even through
the original signal was purely real.
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3.0
Signals & Systems
THE FOURIER SERIES (FS)
The FS coefficients are known as frequency-domain representation of x(t)
because each FS coefficient is associated with a complex sinusoid of a different
frequency. Continuous-time periodic signals are represented by the FS.

 x[k ]e
x[n] 
jk 0 n
______(equation6.2(a ))
K  
T
X [k ] 
1
x[t ]e  jk0t dt ______(equation6.2(b))

T0
FS ;
x(t)
ω0
X[k]
x(t) and X[k] are an FS pair;
The infinite series in equation 6.2(a) is not guaranteed to converge for all
possible signals
^
x[n] 

 x[k ]e
jk0 n
______    t  
K  
A periodic signal x(t) has a FS if it satisfies the Dirichlet condition given by;
1.
x (t) is absolutely integrable over any period; that is
a T
 x[t ] dt  
for any a
0
2.
x (t) has a finite number of maxima and minima over any period.
3.
x (t) has only a finite number of discontinuities over any period.
Example 3.0.1
Building a square wave from DTFS coefficients
The contribution of each term to the square wave maybe illustrated by defining
the partial-sum approximation to x[n]in equation below:

x[t ] 
 B[k ] cos(k t )
K  
xˆ J [t ] 
0
__________
(equation 6.3(c))

 B[k ] cos(k t )
K  
0
_________
xJhat (equation 6.3(d))
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where J ≤
N
. This approximation contains the first 2J+1 terms centered on k =
2
0 in Equation 6.3(d). Assume a square wave has period N = 50 and M = 12.
Evaluate one period of the Jth term in equation 6.3(d) and the 2J+1 term
x̂ J[n] for J=1,3,5,23 and 25.
approximation
by using the following commands:
>> clear
>> k=1:24;
>> n=-24:25;
>> B(1)=25/50;
>> B(2:25)=2*sin(k*pi*25/50)./(50*sin(k*pi/50));
>> B(26)=sin(25*pi*25/50)/(50*sin(25*pi/50));
>> %coefficient for k=N/2
>> xJhat(1,:)=B(1)*cos(n*0*pi/25);
>> %term in sum for k=0
>> %accumulate partial sums
>> for k=2:26
xJhat(k,:)=xJhat(k-1,:)+B(k)*cos(n*(k-1)*pi/25);
end
>> stem(n,xJhat(6,:))
Example 3.0.2
R
x(t)
+
+
-
C
-
RC Circuit
This example exploits linearity and the FS representation of the square wave in
order to determine the output of an LTI System.
If
x(t ) 

 X [k ]e
jk0t
,
k  
y (t ) 
Then the output is

 H ( jkw ) X [k ]e
k  
0
jk0t
,
Where H(j) is the frequency response of the system. Thus,
FS ;
ω0
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y(t)
Y[k] = H(jk0) X[k]
The frequency response of the RC Circuit was determined to be
H ( j ) 
1
RC
j  1
RC
X [k ] 
&
2Sin (k 2 (T0 T ))
k 2
Now, substituting for H(jk0) with RC = 0.01, 0 = 2 and using T0 T = ¼ ,
givens,

 sin( k / 2) 
100

Y [k ]  

k

 j 2k  100 
The Y[k] go to zero in proportion to 1/k2 as k increases, so a reasonably accurate
representation of y(t) maybe determined with the use of a modest number of
terms in the FS. We plot the magnitude and phase spectra of X[k] and Y[k], and
we determine y(t) using the approximation
y (t ) 
100
 Y [k ]e
jk0t
k  100
From the above explanation, the approximation involves the exponential FS
Coefficient with indices -J≤ k ≤ J. Let using 20 samples per period, we obtain
TS 
T
. Note that the total number of sample in one period is then 20
(20 J max)
Jmax. Assuming Jmax =200 and T=1, we may compute the partial-sum Y[K].
>> clc
>> clear
>> % Set KV into 2 portion and at K = 101 Y = 1/2 k = 0
>> KV = [-100:100]; KV1 = [-100:-1];
KV2 = [1:100]
>> Y(101) = 0.5;
>> % Compute Y[K] and set RC = 0.01 or 1/RC = 100
>> RC = 0.01;
>> Y([1:100 102:201])=(1/RC)*Sin([KV1 KV2].*Pi/2)/((j*2*pi[KV1 KV2]+1/RC).*
[KV1 KV2]*Pi);
>> % Set, Jmax and sampling time delta.
>> Jmax=200; T=1; delta=T/(20*Jmax);
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>> % Set time t for the process
>> t=[-(10*Jmax-1):10*Jmax]*delta;
y=zeros(1,length(t));
>> % initial y
>> % Create loop to find y(t) for K=-100 to 100
for k=1: Jmax+1
y=y+Y(k)*exp(j*t*KV(k)*2*pi);
end
>> % Plot figure for the response
>> figure (1)
>> Plot (t,abs(y)), xlabel(‘t),ylabel(‘y(t)’)
>> axis tight
4.0
RELATING THE DTFS TO THE DTFT
Equation 7.1 states that the DTFS coefficient of a finite-duration signal
correspond to samples of the DTFT, divided by the number of DTFS coefficients,
N.
X [k] =
1
X ( e j )
N
  k 0
…
( equation 7.1)
As discussed in the previous LAB, the MATLAB routine ‘fft’. Calculates N times
the DTFS coefficients, hence ‘fft’ directly evaluates samples of the DTFT of a
finite duration signal. The zero padding process involves appending zeros to the
finite-duration signal before computing the DTFS and the results in a denser
sampling of the underlying DTFT. Zero padding is easy accomplished command
‘fft’ by adding and argument that specifies the number of coefficient to compute.
If ‘x’ is a length-M vector representing a finite duration time signal and ‘n’ is
greater than ‘M’, then the command x = fft(x,n) evaluates ‘n’ samples of the
DTFT of x by first padding x with training zeros to length ‘n’. If n is less than M,
then fft(x,n) first truncates x to length n.
The frequency values corresponding to the samples in x are represented by a
vector n point long command the first element zero and the remaining entries
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spaced at intervals
2
. For example, the command w = [0:( n-1 )]*2*pi/n
n
generates the appropriate vector of frequencies. Note that this describes the
DTFT for 0 ≤   2. It is sometimes more convenient to view the DTFT over a
period centered on zero. That is - <  ≤ . The command y = fftshift (x) swaps
the left and right values of x in order to put the zero-frequency value in the
center. The vector of frequency value corresponding t . The values in y maybe
generated by using :
 = [-n/2 : (n/2-1)]*2*pi/n
Example 4.0.1
Sampling the DTFT of a Cosine Pulse:
Consider the signal
 3
 8
cos 
x [n] =

n

0
, 0  n  31
, otherwise
Using MATLAB to evaluate X (ej)at intervals in frequency of
(a)
2
32
(b)
2
60
and
(c)
2
120
Solution 4.0.1;
For case (a) we use a 32 point DTFS computed from the 32 nonzero values of
the signal. For the cases (b) and (c), we zero pad to the length 60 & 120
respectively.
To sample the DTFT at the specified interval, we evaluate and display the results
on -π  n  π . Using the following commands and observe the result.
>> n = [0:31];
>> x = cos (3*pi*n/8):
>> X32 = abs(fftshift (fft (x))); % Magnitude for 32 points DTFS
>> X60 = abs(fftshift (fft (x,60)));
% Magnitude for 60 points DTFS
>> X120 = abs(fftshift (fft (x,120)));
% Magnitude for 120 points DTFS
>> w32 = [-16:15]*2*pi/32;
w60 = [-30:29]*2*pi/60;
>> w120 = [-60:59]*2*pi/120;
>> stem (w32,X32);
stem (w60,X60);
stem (w120,X120);
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5.0
Signals & Systems
COMPUTATIONAL APPLICATIONS OF THE DTFS
As previously noted, MATLAB’s fft command maybe used to evaluate the
DTFS and thus is used for approximating the FT. In particular, the fft is
used to generate the DTFS approximation.
Example 5.0.1
DTFS Approximation Of Sinusoids:
Use the DTFS to approximate the FT of the periodic signal;
x(t) = cos (2(0.4)t) + ½ cos (2(0.45)t,
Assume that the frequency band of interest is -10π    10π and the desired
sampling interval is ∆ω = 20/M. Evaluate the DTFS approximation for
resolutions of
(a) ωr =

rad/s
2
and
(b) ωr =

100
rad/s
Solution 5.0.1:
First note that the FT of x(t) is given by
X(jω) = πδ( + 0.8) + πδ( - 0.8) +


δ( + 0.9) +
δ( + 0.9)
2
2
The maximum frequency of interest is -10 rad/s, and this is much large than the
highest frequency in X(jω), so aliasing is not a concern and we choose ωs = 2
ωa. This gives Ts = 0.1s. The number of samples, M is determined by
substituting ωs into Equation M 
i.e
M
s
r
20
r
To obtain the resolution specified in case (a), we require that M  40 samples,
while in case (b) we need M  2000 samples, we shall choose M = 40 for case
(a) and M = 2000 for case (b). We substitute ∆ω =

20
.Equation N  s , with
M

equality to obtain N = M, and thus no. zero padding is required
N
s
M
20
@
s = 20 ,
N=μ
The signal is weighted sum of complex sinusoids, so the underlying FT is a
weighted sum of shifted window frequency responses and is given by
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yδ(jω) = ½ W δ( j ( + 0.8)) + ½ W δ( j ( - 0.8)) + ¼ W δ( j ( + 0.9)) +
¼ W δ( j ( - 0.9))
Use the following command to observe the approximating the FT.
>> ta = 0:0.1:3.9;
% Time samples for case (a)
>> xa = cos (0.8*pi*ta) + 0.5* cos (0.9*pi*ta);
>> ya = abs (fft (xa)/40);
>> ydela = abs (fft (xa,8192)/40); % Evaluate 1/M Y- delta (jomega) for case (a)
>> fa = [0:19]*5/20;
>> fdela = [0:4095]*5/4096;
>> plot (fdela , ydela (1:4096));
>> hold on
>> stem (fa , ya (1:20));
>> xlabel (‘Frequency (Hz)’);
ylabel (‘Amplitude’)
>> tb = 0:0.1:199.9; % Time samples for case (b)
>> xb = cos (0.8*pi*tb) + 0.5* cos (0.9*pi*tb);
>> yb = abs (fft (xb)/2000);
>> ydelb = abs (fft (xb,16000)/2000); % Evaluate 1/M Y- delta (jomega) for case (b)
>> fb = [0:999]*5/1000;
>> fdelb = [0:7999]*5/8000;
>> plot (fdelb (560:800), ydelb (560:800));
>> hold on
>> stem (fb (71:100), yb (71:100));
>> xlabel (‘Frequency (Hz)’);
ylabel (‘Amplitude’)
Note that here we evaluated 1/M yδ(jω) by using fft, and zero padding with a
large number of zeros relative to the length of x[n]. Recall that zero padding
decreases the spacing between samples of DTFT that are obtained by the
DTFT.
Hence, by padding with a large number of zeros, we capture sufficient detail
such that plot provides a smooth approximation to the underlying DTFT. If plot
is used to display the DTFS coefficient continued zero padding, then a much
coarser approximation to the underlying DTFT is obtained.
For case (b), using plot and stem, the coefficient is obtained via the following
commands:
>> plot (fb(71:100), yb(71:100))
>> hold on
>> stem(fb(71:100), yb(71:100))
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TASKS OF THE DAY
Q 5.1 (a) Use the defining equation for the DTFS coefficients to evaluate the
DTFS representation of the following signals:
(a) x[ n]  cos(
6n 
 )
17
3
(b) x[n]  2 sin(
14n
10n
)  cos(
) 1
19
19
(b) Repeat by using MATLAB’s fft command
Q 5.2 Using MATLAB, solve the following problem
Let the partial-sum approximation to the FS in equation

x[t ]   B[k ] cos( k 0 t ) , be given by
K 0
J
xˆ[t ]   B[k ] cos( k 0 t )
K 0
This approximation involves the exponential FS coefficient with indices
–J ≤ k ≤ J. Consider a square wave with T = 1 and To/T = ¼. Depict one period
of the Jth term in this sum, and find x J[t] for J = 19, 29, and 59.
coefficient of the Square Wave is X [k] =
2 sin( k 2 T0 T )
k 2
Q 5.3 Use the MATLAB command fft to derive both the DTFT, X(ej), and DTFS,
X[k],of x[n], assuming a period N>31. Evaluate and plot X(ej) and NX[k]
for N = 32, 60 and 120.
x [n] =
cos (5/8 n) , 0  n  31
0
, otherwise
Q 5.4 Use the MATLAB command fft to approximate the FT of the signal
x(t) = e-t/10 u(t) (cos (10t) + cos (12t))
Assume that the frequency band of interest is -20 <  < 20 and the desired
sampling interval is ∆ω = /20 rad/s. Compare the approximation with the
underlying FT for resolutions of ωr = 2 rad/s.
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Q 5.5 Using zero padding and the MATLAB commands fft and fftshift to sample and
plot Y(ej) at 512 points - <    . Consider the signal
x[n] = cos ( 7π/16 n) + cos ( 9π/16 n )
Evaluate the effect of computing the DTFT, using only the 2M + 1 values x[n],
n≤ M.
Hint :
x[n] 
N 1
 X [k ]e jk0n
K 0
DTFT
X (e j )  2

 X [k ] (  k)
k  
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