Sessions Nos. 13-22 Module - II
10/02/2025
ECT204 – Signals and Systems
Module-II
Learning Outcomes
By attending this session, students are expected to
 Understand the content of the second module
 Understand the importance of spectral transformation
 Define spectral analysis
 Understand the role of periodicity property in spectral analysis
 Understand the need for transforms
 List the transforms used for signal analyses
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
2
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Module II (11 Hours)
Fourier Analyses
Laplace Transform
 Frequency domain representation of
 Laplace Transform
continuous time signals
 Region of Convergence
 Continuous Time Fourier Series (CTFS)
 Inverse Laplace transform
 Properties
 Properties
 Convergence
 Unilateral Laplace transform
 Continuous Time Fourier Transform (CTFT)  Relation between Fourier and
Laplace transforms
 Properties
12 February 2025
ECT204 - Signals and Systems
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Signals and Spectra
 Signals are usually described in time domain where the independent
variable is time.
 However, for the analysis purpose it becomes necessary to deal with the
signal location in frequency domain.
 For this, a signal x(t) or a sequence x(n) is to be mapped to the frequency
domain using a suitable transformation.
 The description of waveforms in the frequency domain and the study of
the correspondence between the frequency domain description and the
time domain description is known as Spectral or Fourier analysis.
 The analysis stems on the two classes of signals – periodic and aperiodic.
 Signals that are periodic with a finite power within each period can be
represented by the Fourier Series.
 The frequency domain representation of aperiodic energy signals is using
the Fourier Transform.
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
4
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Need for Transforms
 Better representation of signals (domain change)
Laplace Transform
t
CT Fourier Transform
Sampling
Z Transform
n
DT Fourier Transform
Discrete Fourier Transform
s
Ω
z
ω
k
 Analysis of signals (components, bandwidth, characteristics etc.)
 Synthesis of signals (based on spectral components)
 Simplification of processing (based on the properties of transforms)
 To study the correlation between signals (comparison)
 Compression (Audio, Image & Video)
12 February 2025
ECT204 - Signals and Systems
5
Fourier Analyses
Nature of Signal
 Continuous time, periodic
 Continuous time aperiodic
 Discrete time, periodic
 Discrete time, aperiodic
Making periodic
Periodic sequence
Fourier Representation
 Continuous time Fourier Series
(Discrete spectra)
 Continuous time Fourier Transform
(Continuous spectra)
 Discrete time Fourier Series
(Discrete spectra)
 Discrete time Fourier Transform
(Continuous spectra)
Sampling
Discrete Fourier Transform
Efficient Implementation
Fast Fourier Transform
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
6
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Continuous Time Fourier Series
A periodic signal in the interval (–∞, +∞) with fundamental period T0 can be
represented as an infinite sum of sinusoidal waveforms
∞
∞
n=1
n=1
x(t) = A0 + ∑ An cos 2πnf0t + ∑ Bn sin 2πnf0t — 3
where f0 = 1/T0 is the fundamental frequency,
T0/2
A0 = Average or d.c. value of x(t) = (1/T0) ʃ x(t) dt
T0/2
An = (2/T0) ʃ x(t) cos 2πnf0t dt ;
–T0/2
T0/2
and Bn = (2/T0) ʃ x(t) sin 2πnf0t dt ;
–T0/2
n = 1, 2, … ∞
n = 1, 2, … ∞
–T0/2
Equation 3 can be conveniently written as
∞
x(t) = A0 + ∑ (An cos 2πnf0t + Bn sin 2πnf0t)
n=1
The above equation is called continuous time Fourier series representation.
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ECT204 - Signals and Systems
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CTFS (Contd…)
Contd…)
• Equation 3 is called the trigonometric Fourier series. The term A0 in the
equation is the d.c. component of the series and is the average value of x(t)
over a period.
• The term An cos 2πnf0t + Bn sin 2πnf0t is called the nth harmonic. The first
harmonic is obtained when n = 1. This is also called fundamental
harmonic with the fundamental frequency of Ω0 (= 2πf0).
• When n = 2, we have the second harmonic and so on.
• An alternate representation of trigonometric Fourier series is
∞
x(t) = C0 + ∑ Cn cos (2πnf0t – ϕn) —
4
n=1
where C0 = A0 = the d.c. value,
Cn = √ An2 + Bn2 : Amplitude spectrum
ϕn = tan–1(Bn/An) : Phase spectrum
Amplitude spectrum, lacking phase information does not specify the signal.
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
8
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Contd…)
CTFS (Contd…)
One-sided amplitude spectrum :
second harmonic
Cn
d.c. value
first harmonic
0
f0
2f0
3f0
4f0
5f0
6f0
7f0
8f0
f
• The CTFS can also be expressed in exponential form as :
+∞
n= –∞
x(t) e –j2πnf0t dt ;
T0/2
where Xn = (1/T0) ʃ
x(t) = ∑ Xn e j2πnf0t —
5
n = –∞, ... –2, –1, 0, 1, 2, … ∞
–T0/2
• Exponential Fourier coefficients are related to trigonometric ones by
X0 = C0 = A0 and
Xn = (Cn/2) e –jϕn
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ECT204 - Signals and Systems
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CTFS (Contd…)
Contd…)
Two-sided exponential Fourier series coefficients :
d.c. value
│Xn│
–4f0 –3f0 –2f0 –f0
f0
0
2f0
3f0
4f0
f
Properties of CTFS :
1) If x(t) is real, then Xn = X–n*
2) If x(t) is real & even, then Img(Xn) = 0
3) If x(t) is real & odd, then Rea(Xn) = 0
4) For x(t) real or complex,
T0/2
2
+∞
2
(1/T0) ʃ │x(t)│ dt = ∑ │Xn│
–T0/2
12 February 2025
ECT204 – Signals & Systems
n= –∞
— Parseval’s Power Theorem
ECT204 - Signals and Systems
10
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFS – Problems
1) Find the Fourier Series representation of an impulse train,
+∞
x(t) = A ∑ δ(t – kT0)
k= –∞
x(t)
–4T0 –3T0 –2T0 –T0
Solution :
A
0
T0 2T0 3T0 4T0
T0/2
Xn = (1/T0) ʃ A δ(t) e–j2πnf0t dt = A/T0
t
+∞
x(t) = (A/T0) ∑ e j2πnf0t
n= –∞
+∞
–T0/2
Hence the Fourier Representation is, Xn = (A/T0) ∑ δ(f – nf0)
Xn
n= –∞
A/T0
Preserves shape !
–4f0 –3f0–2f0 –f0
12 February 2025
0 f0 2f0 3f0 4f0
f
ECT204 - Signals and Systems
11
CTFS – Problems (Contd...)
2) Find the Fourier series representation of the periodic pulse train :
x(t)
–2T0
τ /2
Xn = (1/T0) ʃ
–T0
A
–τ/2 0 τ/2
T0
2T0
t
A e–j2πnf0t dt
Xn = (A/T0) [sin(πnf0τ) / πnf0]
Defining sinc function as sinc(x) = sin(πx)/πx,
Xn = (Aτ/T0) sinc(nf0τ)
• τ/T0 is called the duty cycle of the pulse train
Taking τ/T0 = ¼, we have 1/τ = 4f0
nf0 = 2/τ
• Spectral Separation = f0 Xn A/4
nf0 = 1/τ
–τ/2
–4f0
12 February 2025
ECT204 – Signals & Systems
0
4f0
ECT204 - Signals and Systems
8f0
nf0 = 3/τ
12f0
f
12
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFS – Problems (Contd...)
 Hence the Fourier series representation of the periodic pulse train is found
to be a series of spectral amplitudes which form an envelope of a sinc
function
 Now, for the spectral analysis, we need to change the parameters of pulse
width and fundamental time period and study the correspondence
x(t)
–2T0
–T0
A
–τ/2 0 τ/2
nf0 = 2/τ
Xn A/4
–4f0
12 February 2025
2T0
T0
nf0 = 1/τ
4f0
0
t
nf0 = 3/τ
12f0
f
8f0
ECT204 - Signals and Systems
13
CTFS – Problems (Contd...)
• T0 same, τ increased :
Xn
A/2
–2f0
2f0
0
Xn
• When τ = T0
width of main lobe decreases
amplitude of side lobes increases
6f0
4f0
f
A
only d.c. Component
Xn = A δ(f)
–f0
12 February 2025
ECT204 – Signals & Systems
0
f0
2f0
3f0
ECT204 - Signals and Systems
f
14
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFS – Problems (Contd...)
Xn
• T0 same, τ decreased :
Aτ/T0
width of main lobe increases
amplitude of side lobes decreases
1/τ
0
f
Xn
• T0 same, τ → 0│Aτ = 1
1/T0
–f0
0
f0
impulse train
f
2f0 3f0
+∞
Xn = (1/T0) ∑ δ(f – nf0)
n= –∞
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ECT204 - Signals and Systems
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CTFS – Problems (Contd...)
• T0 increased, τ same :
Xn
spectral lines come closer
Aτ/T0
nf0 = 1/τ
nf0 = 2/τ
0
nf0 = 3/τ
f
∆f
• T0 → ∞, τ same :
(Fourier Transform)
continuous spectra
∆f → 0
X(f)
Aτ
0
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
f
16
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFS – Problems (Contd...)
3) Find the Fourier series of x(t) = A sin Ω0t, where Ω0 = 2πf0
x(t)
–T0
–T0/2
A
0
T0/2
T0
t
x(t) = A (e j2πf0t – e –j2πf0t)/2j
= (–Aj/2) e j2π(f0)t + (Aj/2) e j2π(–f0)t
The exponential Fourier series representation of x(t) is,
Xn
x(t) = .... + X–1e j2π(–f0)t + X0 + X1e j2π(f0)t + ....
Comparing, we have X–1 = Aj/2
+Aj/2
and X1 = –Aj/2
f0
–2f0
–f0
0
Xn for other values n are zeros.
–Aj/2
Xn = – (Aj/2) δ(f–f0) + (Aj/2) δ(f+f0).
12 February 2025
2f0
ECT204 - Signals and Systems
f
17
CTFS – Problems (Contd...)
4) Find the Fourier series of x(t) = A cos Ω0t, where Ω0 = 2πf0
x(t)
–T0
–T0/2
A
0
T0/2
T0
x(t) = A (e j2πf0t + e –j2πf0t)/2
= (A/2) e j2π(f0)t + (A/2) e j2π(–f0)t
The exponential Fourier series representation of x(t) is,
Xn
x(t) = .... + X–1e j2π(–f0)t + X0 + X1e j2π(f0)t + ....
Comparing, we have X–1 = A/2
A/2
and X1 = A/2
–2f0
–f0
0
Xn for other values n are zeros.
Xn = (A/2) δ(f–f0) + (A/2) δ(f+f0).
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
t
f0
2f0
f
18
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Continuous Time Fourier Transform
• As discussed, a periodic signal can be expressed as a sum of spectral
components. These spectral components have finite amplitudes and are
separated by finite frequency intervals, f0 = 1/T0.
• As T0 → ∞, the spacing between spectral components becomes
infinitesimal.
• Then the frequency of the spectral components nf0 becomes a continuous
variable f.
• It can be seen that the normalized energy of the aperiodic signal remains
finite, but its normalized power becomes infinitesimal. The spectral
amplitudes similarly become infinitesimal.
+∞
• Hence the the Fourier series for the periodic signal x(t) = ∑ Xn e j2πnf0t
n= –∞
+∞
becomes the Fourier Transform, given by x(t) = ʃ X(f) e j2πft df
–∞
where X(f) is the continuous time Fourier transform of x(t).
12 February 2025
ECT204 - Signals and Systems
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CTFT (Contd
Contd…)
…)
T0/2
• We have Xn = (1/T0) ʃ x(t) e –j2πnf0t dt
T0/2
–T0/2
XnT0 = ʃ x(t) e –j2πnf0t dt
∞
–T0/2
As T0 → ∞, f0 → 0 and nf0 → f. Therefore Lim (Xn/f0) = ʃ x(t) e –j2πft dt
f0 → 0
Lim
The Fourier transform X(f) is defined as
(Xn/f0).
f0 → 0
• Hence the Fourier transform pair is
∞
X(f) = ʃ x(t) e –j2πft dt —
–∞
∞
5
–∞
&
x(t) = ʃ X(f) e j2πft df — 6
–∞
• Equation 5 is called the analysis transform and equation 6 is called
the synthesis transform.
• Note that both x(t) and X(f) are continuous and aperiodic. The CTFT of
the impulse response h(t) of a system is known as Frequency Response.
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
20
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Contd…)
…)
CTFT (Contd
For an aperiodic signal to be Fourier transformable, the following criteria,
called Dirichlet’s conditions are to be satisfied.
∞
1) x(t) should be absolutely integrable, i.e., ʃ │x(t)│dt < ∞
–∞
2) x(t) should have a finite number of local maxima, minima and
discontinuities in any finite interval
3) The size of each discontinuity should be finite.
Properties of CTFT :
1) If x(t) is real, then X(f) = X*(–f)
∞
2) If x(t) is having even symmetry, X(f) = 2 ʃ x(t) cos2πft dt
0
∞
3) If x(t) is having odd symmetry, X(f) = –2j ʃ x(t) sin2πft dt
0
4) The energy of the aperiodic
signal is given
by (Parseval’s or Rayleigh’s
∞
∞
energy theorem)
ʃ │x(t)│2dt = ʃ │X(f)│2df
–∞
12 February 2025
–∞
ECT204 - Signals and Systems
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Properties of CTFT (Contd
Contd…)
…)
5) Duality theorem : Let CTFT{x(t)} = X(f).
If there exist a signal z(t) = X(t), then CTFT{z(t)} = Z(f) = x(–f).
6) Time delay : Let CTFT{x(t)} = X(f).
Then CTFT{x(t–td) = e–j2πf td X(f)
7) Convolution : Let CTFT{x1(t)} = X1(f) and CTFT{x2(t)} = X2(f).
Then CTFT{x1(t) * x2(t)} = X1(f) X2(f).
(Multiplicative Property)
8) Differentiation :
Let CTFT{x(t)} = X(f).
d
Then CTFT{ dt [x(t)]} = j2πf X(f)
9) Integration :
Let CTFT{x(t)} = X(f).
∞
Then CTFT{ ʃ x(λ) dλ} = (1/j2πf) X(f)
–∞
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
22
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
x(t)
CTFT (Problems)
A
–τ/2 0
t
τ/2
1) Compute and plot the CTFT of x(t) = A rect(t|τ).
+τ/2
+τ/2
–τ/2
–j2πf –τ/2
∞
–j2πft
 We have, X(f) = ʃ x(t) e –j2πft dt = ʃ A e –j2πft dt = A e
–∞
= A sin πf τ
πf
main lobe
= Aτ sinc fτ
X(f)
f = 2/τ
Aτ
side lobes
f = 3/τ
f = 1/τ
0
f
 When τ increases, height of the main lobe increases while zero crossings (ZC)
come closer to the origin and vice versa.
12 February 2025
ECT204 - Signals and Systems
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CTFT (Problems – Contd
Contd…)
…)
x(t)
–τ/2
x(t)
A
0
X(f)
0
τ/2
t
A
t
0
Aτ
X(f)
A
f
f
0
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
24
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
contd…)
…)
CTFT (Problems – contd
δ(t)
1
2) Compute and plot the CTFT of δ(t).
t
0
∞
 We have, ʃ δ(t) dt = Area of unit impulse = 1
–∞
∞
 Therefore, the CTFT of δ(t) = ∆(f) = ʃ δ(t) e –j2πft dt = 1
–∞
∆(f)
–f
1
0
 Thus the spectral components of δ(t) extend with uniform amplitude and
phase over the entre frequency domain
 This substantiates the importance and the role of unit impulse δ(t) in the
analysis of physical systems.
12 February 2025
ECT204 - Signals and Systems
25
CTFT (Problems – contd…)
3) Compute and plot the CTFT of the signum function
sgn(t)
+1
0
–1
Solution :
∞
FT{sgn(t)} = ʃ sgn(t) e –j2πft dt
–∞
∞
=
FT{sgn(t)} =
e –j2πft
–j2πf 0
+
1
jπf
0
f
t
∞
= ʃ (–1) e –j2πft dt + ʃ (+1) e –j2πft dt
–∞
e –j2πft
∞
–j2πf 0
0
=
0–1
–j2πf
+
0–1
–j2πf
X(f)
(no dc component)
–f
12 February 2025
ECT204 – Signals & Systems
0
ECT204 - Signals and Systems
f
26
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
u(t)
+1
contd…)
…)
CTFT (Problems – contd
0
t
4) Compute and plot the CTFT of the unit step signal, u(t).
Solution :
u(t) can be expressed in terms of signum function as u(t) =
1 + sgn(t)
2
Therefore the spectra of u(t) is, U(f) = CTFT{sgn(t)/2} + CTFT{1/2}
Hnece U(f) =
1
1
+ 2 δ(f)
j2πf
X(f)
0.5
–f
12 February 2025
0
f
ECT204 - Signals and Systems
27
CTFT (Problems)
5) If x(t) = A cos Ωct, where Ωc = 2πfc, compute and plot X(f).
Solution :
∞
∞
We have, X(f) = ʃ x(t) e –j2πft dt = ʃ A cos 2πfct e –j2πft dt
–∞
–∞
∞
= (A/2) ʃ [e j2πfct + e –j2πfct ] e –j2πft dt
–∞
∞
∞
= (A/2) ʃ e –j2π(f – fc)t dt + (A/2) ʃ e –j2π(f + fc)t dt
–∞
–∞
∞
We know that the FT of a dc signal is, ʃ 1. e –j2πft dt = δ(f)
X(f)
–∞
Therefore, X(f) = (A/2) δ(f – fc) + (A/2) δ(f + fc)
–fc
Therefore, A cos Ωct has both CTFS and CTFT representations
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
A/2
0
fc
f
28
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
contd…)
…)
CTFT (Problems – contd
6) If x(t) = m(t) cos 2πfct, find X(f) in terms of the Fourier transform of m(t),
i.e., M(f).
∞
∞
Solution : X(f) = ʃ m(t) cos 2πfct e –j2πft dt = (1/2) ʃ m(t) [e j2πfct + e –j2πfct ] e –j2πft dt
–∞
–∞
∞
∞
–∞
–∞
= (1/2) ʃ m(t) e –j2π(f – fc)t dt + (1/2) ʃ m(t) e –j2π(f + fc)t dt
∞
We know that, ʃ m(t) e –j2πft dt = M(f)
M(f)
–∞
M0
Therefore, X(f) = (1/2) M(f – fc) + (1/2) M(f + fc)
X(f)
M0/2
–fc – fm
–fc
0
–fc + fm
12 February 2025
-fm
0
fm
fc
fc – fm
fc + fm
ECT204 - Signals and Systems
f
f
29
CTFT (Problems – contd
contd…)
…)
 The above spectra is that of a variant of Aamplitude Modulation, namely
Double Sidebad Suppressed Carrier AM (DSBSC-AM)
 As a special case, if we have m(t) = A cos 2πfmt, then we have the single-tone
modulation. Then we have, M(f) = (A/2) δ(f – fm) + (A/2) δ(f + fm)
 Then, the spectra is, X(f) = (A/4) δ(f – fc – fm) + (A/4) δ(f – fc + fm)
Spectra of single-tone
DSB-SC AM
+ (A/4) δ(f + fc – fm) + (A/4) δ(f + fc + fm)
X(f)
A/4
–fc – fm –fc
12 February 2025
ECT204 – Signals & Systems
–fc + fm
0
ECT204 - Signals and Systems
fc – fm
fc
fc + fm
f
30
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFT (Problems – contd
contd…)
…)
7) Compute and plot the spectra of a standard AM wave.
Solution :
We have, x(t) = Vc [1+ka m(t)] cos 2πfct = Vc cos 2πfct + Vc ka m(t) cos 2πfct
Therefore, X(f) = X1(f) + X2(f), where
X1(f) = (Vc/2) δ(f – fc) + (Vc/2) δ(f + fc)
Carrier Component
X2(f) = (Vcka /2) M(f – fc) + (Vcka /2) M(f + fc)
Lower Side Band
X(f)
Vc/2
Vc/2
Upper Side Band
kaVc/2
–fc – fm
–fc
0
–fc + fm
12 February 2025
fc
fc – fm
fc + fm
f
ECT204 - Signals and Systems
31
CTFT (Problems – contd
contd…)
…)
8) Compute and plot the Fourier series of a pulsed RF waveform and hence
deduce the Fourier transform of a single RF pulse.
x(t)
A
A cos Ωct
0
t
τ
Solution : Method 1
τ /2
T0
2A τ/2
We have, Xn = (1/T0) ʃ A cos 2πfct e–j2πnf0t dt = T ʃ cos 2πfct cos 2πnf0t dt
0 0
τ/2
–τ/2
= 2A ʃ [cos 2π(nf0 – fc)t + cos 2π(nf0 + fc)t] dt
T0 0
= 2A
T0
12 February 2025
ECT204 – Signals & Systems
τ/2
τ/2
sin 2π(nf 0– fc)t
2A sin 2π(nf0 + fc)t
+
T0 2π(nf0 + fc)
2π(nf0 – fc) 0
0
ECT204 - Signals and Systems
32
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFT (Problems – contd
contd…)
…)
sin π(nf0 – fc) τ
2A sin π(nf0 + fc) τ
= 2A
+
T0
2π(nf0 – fc)
T0
2π(nf0 + fc)
Xn = Aτ [sinc(nf0 – fc)τ + sinc(nf0 + fc)τ]
2T0
Xn
–fc
Aτ
2T0
0
fc
f
 Hence the spectra of a pulsed RF waveform consist of two discrete sinc
functions located at +fc and –fc
 The spectral separation is given by f0
12 February 2025
ECT204 - Signals and Systems
33
CTFT (Problems – contd
contd…)
…)
Xn
f → 0 f0
To deduce the Fourier transform a single RF pulse, we have X(f) = Lim
Such that nf0 is a continuous variable, f
Therefore, X(f) =
0
Aτ [sinc(f – f )τ + sinc(f + f )τ]
c
c
2
Solution : Method 2
We have, x(t) = A cos Ωct . rect(t|τ, T0)
Hence, by the multiplicative property of DTFS, the spectra is a convolved one.
We have, the spectra of rect(t|τ , T0) is (Aτ/ T0) sinc nf0τ and that of cos 2πfct is
(1/2) [δ(f – fc) + δ(f + fc)]
Hence the CTFS Xn is, (Aτ/ T0) sinc nf0τ, halved and translated to –fc and +fc
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
34
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
CTFT (Problems – contd
contd…)
…)
9) Compute and plot the CTFT of a single RF pulse.
x(t)
A cos Ωct
A
0
Solution : Method 1
t
τ
∞
τ/2
We have, the CTFT, X(f) = ʃ x(t) e –j2πft dt = ʃ A cos Ωct e –j2πft dt
–τ/2
–∞
τ/2
τ/2
–τ/2
τ/2
0
= A ʃ cos 2πfct e –j2πft dt = 2A ʃ cos 2πfct cos 2πft dt
= A ʃ [cos 2π(f – fc)t + cos 2π(f + fc)t] dt
0
= A sin 2π(f – fc)t
2π(f – fc)
12 February 2025
τ/2
0
sin 2π(f + fc)t
+ A
2π(f + fc)
τ/2
0
ECT204 - Signals and Systems
35
CTFT (Problems – contd
contd…)
…)
A sin π(f – fc) τ + A sin π(f + fc) τ
2π(f – f )
2π(f + f )
=
c
X(f) =
c
Aτ [sinc(f – f )τ + sinc(f + f )τ]
c
c
2
Solution : Method 2
We have, x(t) = A cos Ωct . rect(t|τ)
Therefore, by the multiplicative property of DTFT, the spectra is a convolved one.
We have, the spectra of rect(t|τ) is Aτ sinc fτ and that of cos 2πfct is (1/2) [δ(f – fc)
+ δ(f + fc)]
Hence the resultant spectra is, Aτ sinc fτ, halved and translated to –fc and +fc
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
36
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
contd…)
…)
CTFT (Problems – contd
X(f)
fc + 1/τ
Aτ/2
–fc
+fc
f
 Whenever a function is multiplied by an infinite cosine waveform, the resultant
spectra will be the original spectra, halved and translated to –fc and fc
 The waveform given in Problem No. 9 is associated with amplitude shift
keying (ASK), which is a digital modulation scheme
 The minimum bandwidth required to pass the components of such a signal
would be of 2/τ centered at fc
12 February 2025
ECT204 - Signals and Systems
37
CTFT (Problems – contd
contd…)
…)
10) Find the CTFT of a sinc pulse, z(t) = A sinc(2Wt).
z(t) A
0
Solution :
–1/2W
t
1/2W
We have, the CTFT of x(t) = B rect(t|τ) is X(f) = Bτ sinc fτ. Then by duality
theorem, we have, the spectra of X(t) = x(–f)
A
Now, z(t) = A (2W) sinc(2Wt). Therefore, z(t) = X(f), with B =
and τ = 2W
2W
2W
Z(f)
Hence, by duality theorem, Z(f) = x(–f) = B rect(–f|τ)
Z(f) = A rect(f|2W)
2W
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
A/2W
–W
0
W
f
38
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
contd…)
…)
CTFT (Problems – contd
 Note that the spectra a sinc pulse is a rectangular function
 If this sinc pulse happens to be the impulse response h(t) of a system, the
corresponding frequency response is, H(f) = a rectangular function
 Since the impulse response is a non-causal signal, the system is not a physically
realizable one.
 It could also be noted from the curves, that the frequency response is that of an
ideal filter, which is never practical.
 To get an implementable systems, we should go for causal impulse responses
 However, such signals would have a spectra, which are smoothened functions,
differing from the above rectangular function
12 February 2025
ECT204 - Signals and Systems
39
x(t) A
CTFT (Problems – contd
contd…)
…)
t
0
2τ
11) Compute and plot the CTFT of a triangular pulse, x(t) = A tri(t|τ).
A tri(t|τ) =
1–|t|/τ
Solution :Method 1
0;|t|>τ
for | t | < τ
Starting from a single square wave z(t),
z(t)
A
τ
–τ
z(t) = A rect[(t + τ/2)|τ) – A rect[(t – τ/2)|τ)
t
–A
Therefore, Z(f) = e –j2πf(–τ/2 ) R(f) – e –j2πf(τ/2 ) R(f) ; where R(f) is the CTFT of A rect(t|τ)
Hence, Z(f) = R(f) [e j2πf(τ/2 ) - e –j2πf(τ/2 )] = R(f) 2j sin πfτ, where R(f) = Aτ sinc fτ
= (Aτ sinc fτ) (2j sin πfτ) = (Aτ sinc fτ) (2j πfτ) sinc fτ
= (j2πfτ) Aτ sinc2fτ
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
40
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
contd…)
…)
CTFT (Problems – contd
Now, we have, x(t) = A tri(t|τ) = ∫ z(t) dt
τ
1
1
Therefore, X(f) = τ
Z(f)
j2πf
1
= τ τ Aτ sinc2 fτ
FT{A tri(t|τ)} = Aτ sinc2 fτ
X(f)
Aτ
f = 1/τ
f
0
12 February 2025
ECT204 - Signals and Systems
41
CTFT (Problems – contd
contd…)
…)
Solution :Method 2 – Simpler ?
 Starting from a two rectangular pulses of same duration τ and hight A,
z1(t) = A rect[(t|τ) & z2(t) = A rect[(t|τ)
z(t)
A
–τ/2 0
τ/2
t
 We have, their convolution yields a triangular waveform of τ duration and A2τ
height.
 Hence according to the convolution property of CTFT, the required spectra is
(1/Aτ) {Product of the individual spectra}
 Therefore, X(f) = (1/Aτ) (Aτ sinc fτ) (Aτ sinc fτ) = Aτ sinc2 fτ
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
42
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Signal & Spectra – Philosophy Behind
 Newton - White Light – Prism - VIBGYOR
 Behaviour & Character
 Observing for a period of time
 Frequent happenings
Behaviour
Character
Mapping
(Time)
(Frequency)
 Behaviour is to be analyzed to yield the character (constituents)
 Character (constituents) decides behaviour
 Time domain aspects decides the frequency domain aspects and vice versa
12 February 2025
ECT204 - Signals and Systems
43
Signal & Spectra – An analogical treatment
 Food and ingredients analogy
 What are the ingredients used in our kitchen ?
 Tea : Water, Tea, Sugar, Milk & Cardamom in different proportions
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
Item
Ingredients
44
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Spectra – Analogy
Item
12 February 2025
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
See the relative amplitudes of the basis
ECT204 - Signals and Systems
Ingredients
45
Spectra – Analogy (contd
(contd…)
…)
Item
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
See the relative amplitudes of the basis
Ingredients
46
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Spectra – Analogy (contd
(contd…)
…)
Item
12 February 2025
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
See the relative amplitudes of the basis
ECT204 - Signals and Systems
Ingredients
47
Spectra – Analogy (contd
(contd…)
…)
Item
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
See the relative amplitudes of the basis
Ingredients
48
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Spectra – Analogy (contd
(contd…)
…)
Item
12 February 2025
Rice
Cardamom
Mustard
Chilli
Coconut
Coffee
Tea
Milk
Sugar
Water
See the relative amplitudes of the basis
Ingredients
ECT204 - Signals and Systems
49
The Laplace Transform & ss-domain
 The CTFT provides us with a representation for signals as linear combinations
of complex exponentials of the form est with s = jΩ
 Observing that s can take any arbitrary value, not only purely imaginary, there
leads to a generalization of the CTFT, known as the Laplace transform (LT)
 LT can be applied in some very important contexts in which Fourier transform
cannot !
 LT can be applied to the analysis of many unstable systems and consequently,
play an important role in the investigation of the stability of systems
 The algebraic properties of LT leads to a very important set of tools for system
analysis and in particular, for the analysis of feedback systems.
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
50
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Laplace Transform (LT) : L
The Laplace transform of a signal x(t) is expressed as :
+∞
X(s) = ʃ x(t) e–st dt — 1
–∞
where s is a complex quantity, s = σ + j Ω
σ+j∞
The inverse Laplace Transform is defined by : x(t) = (1/2πj) ʃ X(s) est ds — 2
σ–j∞
Equations 1 & 2 form a Laplace Transform pair. It is also expressed by
X(s) = L [x(t)]
x(t) = L-1[X(s)]
The transform defined by eq. (1) is often called the bilateral Laplace transform, to
distinguish it from the unilateral Laplace transform which will be defined later.
12 February 2025
ECT204 - Signals and Systems
51
Region of Convergence
 In general, the range of values of s for which the integral converges is referred
to as the region of convergence (RoC) of the Laplace transform.
 That is, the RoC consists of those values of s = σ + j Ω for which the Fourier
transform of x(t)e-σt converges
 The RoC of X(s) consists of strips parallel to the jΩ axis in s-plane
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
52
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Region of Convergence (RoC
(RoC))
jΩ
Region of Convergence (RoC)
0
σ1
s - plane
–σ
σ2
σ
–jΩ
12 February 2025
ECT204 - Signals and Systems
53
Some Standard Results
• L [δ(t)] = 1
• L [U(t)] = 1/s
• L [eat] = 1/(s – a)
•
•
•
•
L [t] = 1/s2
L [sin Ωt] = Ω / (s2 + Ω2)
L [cos Ωt] = s / (s2 + Ω2)
L [d(x(t)/dt] = s X(s) – x(0)
t
• L [ ʃ x(t) dt] = X(s)/s
0
• L [x(t – T] = e–sT X(s)
• L [t x(t)] = – d X(s)/ds
∞
• L [(1/t) x(t)] = ʃ X(s) ds
s
• Initial Value Theorem :
• Final Value Theorem :
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
Lim
x(t) =
t→ 0
Lim
x(t) =
t→∞
Lim
s X(s)
s→∞
Lim
s→0 s X(s)
54
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Properties of LT – Consolidated
12 February 2025
ECT204 - Signals and Systems
55
Standard Results of LT
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
56
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Standard Results of LT
12 February 2025
ECT204 - Signals and Systems
57
Unilateral Laplace Transform
 The unilateral Laplace transform of a signal x(t) is expressed as :
∞
X(s) = ʃ x(t) e–st dt — 3
0
 We see that the difference in the definitions of the unilateral and bilateral
Laplace transform lies in the lower limit on the integral
 The bilateral transform depends on the entire signal from t = -∞ to t = ∞,
whereas the unilateral transform depends only on the signal from t = 0- to ∞.
 Consequently, two signals that differ for t < 0, but that are identical for t ≥ 0,
will have different bilateral LTs, but identical unilateral LTs.
 Similarly, any signal that is identically zero for t < 0 has identical bilateral and
unilateral transforms
 Since the unilateral LT of x(t) is identical to the bilateral LT of the signal
obtained from x(t) by setting its value to 0 for all t < 0, many of the concepts
and results pertaining to bilateral LTs can be readily adapted to unilateral also
12 February 2025
ECT204 – Signals & Systems
ECT204 - Signals and Systems
58
Dr. A. Ranjith Ram – arr@gcek.ac.in
Sessions Nos. 13-22 Module - II
10/02/2025
Relation between FT & LT
We have the Laplace transform as :
+∞
X(s) = ʃ x(t) e–st dt
–∞
where the Laplacian variable s is a complex quantity, s = σ + j Ω
+∞
+∞
–∞
–∞
Now, substituting for s, we have X(s) = ʃ x(t) e– (σ + jΩ)t dt = ʃ x(t) e– σ t e– jΩt dt
Now, assuming σ = 0 (i.e., s is having only the imaginary part – along y-axis),
+∞
X(s) = ʃ x(t) e– jΩt dt
–∞
Now remembering that Ω = 2πf where f is in Hertz,
+∞
X(s) = ʃ x(t) e– j2πft dt … (Fourier Transform)
–∞
That is, evaluating Laplace transform along jΩ axis yields the Fourier transform.
Or, X(Ω) = X(s) s = jΩ
12 February 2025
ECT204 - Signals and Systems
59
End of ModuleModule-II !
The endend-product of
education should be a
free creative man,
who can battle
against historical
circumstances and
adversities of nature
ECT204 – Signals & Systems
Dr. A. Ranjith Ram – arr@gcek.ac.in