CHAPTER 3

Fourier Representation of
Signals and LTI Systems.

1
3.0 Fourier Representation of
Signals and LTI Systems.
Introduction.

Complex Sinusoids and Frequency Response

of LTI Systems.

Fourier Representation of Four Classes of Signals:

i. Discrete Time Periodic Signals: DTFS
ii. Continuous-Time Periodic Signals: CTFS
iii. Discrete-Time Non Periodic Signals: DTFT
iv. Continuous-Time Non Periodic Signals:CTFT

Properties of Fourier Representation.

Linearity and Symmetry Properties.

Convolution Properties.

2
3.1 Introduction.

Superposition of complex sinusoidal

3
3.2 Complex Sinusoidal and
Frequency Response of LTI System.
 Euler’s Identity:

jy
1. e = cos(y) + jsin(y)
-jy
2. e = cos(y) - jsin(y)
1 jy -jy
3. cos(y) = [e e ]
+
2
1 jy
4. sin(y) = [e - e-jy]
2j
4
3.2 Complex Sinusoidal and
Frequency Response of LTI System.
 The response of an LTI system to a sinusoidal input that lead to
the characterization of the system behavior that is called
frequency response of the system.

5
Cont’d…
Discrete-Time (DT):
IF impulse response h[n] and the complex sinusoidal input x[n]=ejWn,


 The output derivation: yn   hk xn  k 

k  

yn   hk 
e jW  n  k 

k  

factor e jWn out


yn  e jWn  hk 
e  jWk

k  

 
 H e jW e jWn

   hk e

 Frequency Response: H e jW   jWk

k   6
Cont’d…
Continuous Time (CT):
IF impulse response h(t) and the complex sinusoidal input x(t)=ejωt,

 The output derivation: y t    h e j t  d


 e jt  h e  j d


 H  j e jt


 Frequency Response: H  j    h e  j d


7
Cont’d…

System Response
y[n]

 The output of a complex sinusoidal input to an LTI system is a

complex sinusoid of the same frequency as the input, multiplied
by the frequency response of the system.

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3.3 Fourier Representation of
Four Class of Signals.

 There are 4 distinct Fourier representations, class of signals;

i. Discrete Time Fourier Series: DTFS
ii. Continuous-Time Fourier Series: CTFS
iii. Discrete-Time Fourier Transform: DTFT
iv. Continuous-Time Fourier Transform:CTFT

9
Cont’d…

Time
Property Periodic Nonperiodic

Discrete Time Fourier Discrete Time Fourier

Discrete (n) Series (DTFS) Transform (DTFT)
[Chapter 3.6.1] [Chapter 3.6.3]
Continuous Time Continuous Time Fourier
Continuous (t) Fourier Series (CTFS) Transform (CTFT)
[Chapter 3.6.2] [Chapter 3.6.4]

Table 3.1: Relation between Time Properties of a Signal an the

Appropriate Fourier Representation.
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3.4 Periodic Signals:
Fourier Series Representation

DISCRETE-TIME SIGNAL
 If x[n] is a discrete-time signal with fundamental period, N

Thus, x[n] of DTFS is represents as,

FS of x[n] xˆn   Ak e jkWo n

 where Wo= 2p/N is the fundamental frequency of x[n].

 The frequency of the kth sinusoid in the superposition is kWo
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Cont’d…

CONTINUOUS-TIME SIGNAL
 If x(t) is a continuous-time signal with fundamental period T,
x(t) of FS is represents as

FS of x(t) xˆ t    Ak e jko t

 where o= 2p/T is the fundamental frequency of x(t). The

frequency of the kth sinusoid is ko and each sinusoid has a
common period, T.

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Cont’d…

 A sinusoid whose frequency is an integer multiple of a fundamental

frequency is said to be a harmonic of a sinusoid at the
fundamental frequency.

For example, ejk0t is the kth harmonic of ej0t.

where k = 0,1,2,….

 The variable k indexes the frequency of the sinusoids, A[k] is the

function of frequency.
 Same goes for DTFS.
13
Cont’d…

 The relation of N-periodic in k index in complex sinusoids can

also be shown as,

For example, where k = 0,1,2,…. harmonic of ejkW0n.

 jkWo n  jk 2p jkWo n
e e e
jkW o n
e

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3.5 Nonperiodic Signals:
Fourier-Transform Representation.
 The Fourier Transform representations employ complex sinusoids
having a continuum of frequencies.
 The signal is represented as a weighted integral of complex
sinusoids where the variable of integration is the sinusoid’s
frequency.
 Continuous-time sinusoids are used to represent continuous
signal in FT.

xˆ t    
1

jt
X j  e d
2p 

where X(j)/(2p) is the “weight” or coefficient applied to a

sinusoid of frequency  in the FT representation
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Cont’d…

 Discrete-time sinusoids are used to represent discrete time signal

in DTFT.
 It is unique only a 2p interval of frequency. The sinusoidal
frequencies are within 2p interval.

p
xˆn  p X e e
1 jW jWn
dW
2p 

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3.6.1 Discrete Time Fourier Series
 The DTFS representation;
N 1
xn   X k e jkWo n
k 0

 where x[n] is a periodic signal with period N and fundamental

frequency Wo=2p/N. X[k] is the DTFS coefficient of signal x[n].
N 1
X k    
1
N
 x
n 0
n e  jkW o n

 The relationship of the above equation,

xn X k 
DTFS ;Wo
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Cont’d…

 Given N value of x[n] we can find X[k].

 Given N value of X[k] we can find x[n] vise-versa.

 The X[k] is the frequency-domain representation of x[n].

 DTFS is the only Fourier representation that can be numerically

evaluated using computer, e.g. used in numerical signal analysis.

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Example 3.1: Determining DTFS Coefficients.
Find the frequency-domain representation of the signal in Figure 3.2
below.

Figure 3.2: Time Domain Signal.

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Example 3.1: Determining DTFS Coefficients.
Find the frequency-domain representation of the signal in Figure 3.2
below.

Figure 3.2: Time Domain Signal.

Solution:
Step 1: Determine N and W0.
The signal has period N=5, so W0=2p/5.
Also the signal has odd symmetry, so we sum over n = -2 to n = 2
from equation

20
Cont’d…

Step 2: Solve for the frequency-domain, X[k].

From step 1, we found the fundamental frequency, N =5, and we sum
over n = -2 to n = 2 .

N 1
X k    xne
1  jkW o n

N n 0

1 2
X k    xne  jk 2pn / 5
5 n  2


 x 2e jk 4p / 5  x 1e jk 2p / 5  x0e j 0  x1e  jk 2p / 5  x2e  jk 4p / 5
1
5


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Cont’d…
From the value of x{n} we get,
1  1 jk 2p / 5 1  jk 2p / 5 
X k   1  e  e 
5 2 2 
 1  j sin k 2p / 5
1
5

TRY a different range of n,

WHAT WILL HAPPEN??

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Cont’d…
TRY to find X[k] using 0≤n ≤4 for the limit of the sum.

X k  
1
5

x 0e j 0  1e  j 2p / 5  2e  jk 4p / 5  3e  jk 6p / 5  4e  jk 8p / 5 
1  1  jk 2p / 5 1  jk 8p / 5 
X k   1  e  e 
5 2 2 

# Note that,
e  jk 8p / 5  e  jk 2p e jk 2p / 5
 e jk 2p / 5

Thus, -2≤n ≤2 and 0≤n ≤4, yield equivalent expression for the DTFS
coefficient.

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