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Introduction to signals and systems
Aliraqia university
Network and software engineers /third class Lecturer : Marwa Moutaz
Lectures in signal and systems
Lecture: One
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This series of lectures was prepared for the third class of both
software engineering and network engineering / Aliraqia University/
Baghdad/ Iraq.
In prepared these lectures, I depend on the book “signals and
systems” / Simon Haykin, and some of my lectures (4, 6, 8, 10), I
depend on some lectures of my professor Dr. Emad Shehab
/University of Technology/ Baghdad/ Iraq /
Forgive me for anything wrong by mistake, I wish you can profit
from these lectures
My regard
Marwa Moutaz/ M.Sc. studies of Communication Engineering /
University of Technology/ Bagdad / Iraq.
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1-1what is a system?
Why it is important to study signals and systems?
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1-2 classification of signals
Here we will study one dimensional signals only
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Figure 1:2
………………………………(1.1)
…………………………………...(1.2)
Example:- which one of the 2 signals shown below is odd or even?
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………………………….(1.3)
(1.3)
………………………..(1.4)
……………………......(1.5)
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(1.3) is called an aperiodic or non-periodic signal
Figure 1.3
……………...(1.6)
The smallest value of N which satisfies eq (1.6) is called the fundamental
period of x(n), the fundamental angular
…….………………(1.7)
Example: - what is the fundamental frequency of the discrete time
signal below?
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The square wave and the rectangular pulse shown in figure (1.3) is example of
deterministic pulse.
Before its actual occurrence, the noise in the amplifier and television is an
example of random signal, its amplitude is fluctuate between + and – in a random
value, the signal in the figure below is an example of random signal
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……………………….(1.8)
For a resistance of 1 ohm , the instantaneous power of the signal is given by :
……………………...(1.9)
………… (1.10)
…….(1.11)
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Problems:-
1- What is the total energy of the rectangular pulse shown in figure (1.3)b
2- What is the average power of the square wave shown in figure (1.3)a
3- What is the average power of the square wave shown in the figure below
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2.1 Elementary signals
………….(2.1)
In radian, figure below
Figure (2.1)
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……..(2.2)
……….(2.3)
(2.2)
(2.3)
Example
Solving eq (2.3), we get
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……….(2.4)
(2.2).
Figure (2.2)
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…………. (2.5)
Figure below illustrate the decaying and growing form of discrete time
Figure (2.3)
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………(2.6)
Damped sinusoidal function of equation (2.6) is defined by :
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……………(2.7)
(2.7)
………….(2.8)
Rectangular function
……………………..(2.9)
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Triangular function
…………………………(2.10)
Or, equivalently, as the convolution of two identical unit rectangular functions:
……...........(2.11)
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Example:- find the mathematical expression of xn if :-
……….(2.12)
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2.2: Basic operations on signals
………..(2.13)
……………..(2.14)
………(2.15)
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……………………(2.16)
………………….(2.17)
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Problems:-
*
*
*
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In the previous lecture we studied the operations performed on depended
variables, here we will study the operations on independent variables:
-rated in figure below
Figure (3.1)
Figure (3.2)
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……………………………………
Example: a
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b
Figure (3.3)
Example:-
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Example: the figure below shows a rectangular pulse of unit amplitude
and unit duration.
Figure (3.4)
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Example:-
…………….(3.1)
….....................(3.2)
…………………(3.3)
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Figure (3.5):
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(3.5),
(3.5)b,
(3.5)c
(3.5)c,
(3.2)&(3.3)
………………
Example:-
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Figure (3.6):
…………………………….
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Problems:-
…………………………
Find the time shifted signal y[n] =x [n - 5]
…………………………
Find y[n] = x [3n-7]
……………………………………
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Convolution integral
Example: - Find the output of the system if the input and the response of
the system is given as shown in the figure below:-
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Circular convolution
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Digital convolution method 3
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Problems:-
Find the convolution between x(t) and y(t)
*Find the output of the system below by convolution methods
X(n)= 3,-6,-9,4,3 for -2≤ 𝑛 ≤ 2
H(n) =6,-8,-1,5,6 for -2≤ 𝑛 ≤ 2
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…………………(5.1)
And for discreet time signal
…………….(5.2)
Figure (5.1)
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Example:-
Show that the following system is BIBO
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below
……..
= ( )..(5.3)
In this parallel system the response of the system has additive property
…..(5.4)
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…….(5.5) where
Substitute in equation (5.5) yields
In this cascade system, the system response has commutative property
where
, This yields
……..(5.6
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Time domain analysis
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2- The step response
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Problems
………………………………………..
Find the four impulse response & step response for the following
system
h(n)=δ(n)+δ(n-1)+δ(n-2)
……………………………………..
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DIFFERENCE EQUATIONS
LTI system can be expressed using linear constant coefficient difference equation, or
LCCDE. The general form of a LCCDE is
………………(7.1)
Where the coefficients a(k) and h(k) are constants that define the system. If the
difference equation has one or more terms a(k) that are nonzero, the difference
equation is said to be recursive. On the other hand, if all of the coefficients a(k) are
equal to zero, the difference equation is said to be non-recursive.
Difference equations provide a method for computing the response of a system, y(n),
to an arbitrary input x(n). Before these equations may be solved, however, it is
necessary to specify a set of initial conditions. For
example, with an input x(n) that begins at time n = 0 , the solution to Eq. (7.1) at time
n = 0 depends on the values of y ( - l ) , . . . , y ( - p ) . Therefore, these initial conditions
must be specified before the solution for n ≥ 0 may be found. When these initial
conditions are zero, the system is said to be in initial rest.
There are several different methods that one may use to solve LCCDEs for a general
input x(n). The first is to simply set up a table of input and output values and evaluate
the difference equation for each value of n. This approach would be appropriate if
only a few output values needed to be determined. Another approach is to use z-
transforms. The third is the classical approach of finding the homogeneous and
particular solutions, which we now describe.
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Given an LCCDE, the general solution is a sum of two parts,
……………………(7.2)
Where 𝑦ℎ (𝑛) is known as the homogeneous solution and 𝑦𝑝 (𝑛) is the particular
solution. The homogeneous solution is the response of the system to the initial
conditions, assuming that the input x(n) = 0. The particular solution is the response
of the system to the input x(n), assuming zero initial conditions.
The homogeneous solution is found by solving the homogeneous difference equation
……………………(7.3)
The solution to Eq. (7.3) may be found by assuming a solution of the form
Substituting this solution into Eq. (7.3) we obtain the polynomial equation
The polynomial in braces is called the characteristic polynomial. Because it is of
degree p, it will have p roots, which may be either real or complex. If the coefficients
a(k) are real-valued, these roots will occur in complex conjugate pairs (i.e., for each
complex root z, there will be another that is equal to z*). If the p roots zi are
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Distinct, 𝑧𝑖 ≠ 𝑧𝑘 for k ≠ i , the general solution to the homogeneous difference
equation is
……………………..(7.4)
Where the constants Ak are chosen to satisfy the initial conditions. For repeated roots,
the solution must be modified as follows If 𝑧𝑖 is a root of multiplicity m with the
remaining p - m roots distinct, the homogeneous solution becomes
……….(7.5)
For the particular solution, it is necessary to find the sequence 𝑦𝑝 (𝑛) that satisfies
the difference equation for the given x(n). In general, this requires some creativity
and insight. However, for many of the typical inputs that we are interested in, the
solution will have the same form as the input. Table7-1 lists the particular solution
for some commonly encountered inputs. For example, if x(n) = 𝑎𝑛u(n), the particular
solution will be of the form
Provided a is not a root of the characteristic equation. The constant C is found by
substituting the solution into the difference equation. Note that for x(n) = C δ(n) the
particular solution is zero. Because x(n) = 0 for n > 0, the unit sample only affects the
initial condition of y(n).
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Example:-
……..(7.6)
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….(7.7)
The constants A, and A2 must now be found so that the total solution satisfies the
given initial conditions, y(-1) = 1 and y(-2) = 0. Because the solution given in Eq.
(7.7) only applies for n 0, we must derive an equivalent set of initial conditions for
y(0) and y(1). Evaluating Eq. (7.6) at n = 0 and n = 1. we have
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Example:-
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In this lecture
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Table (9.1)
In the previous lecture, we studied the (DTFS) & (DTFT) for discrete
signals, in this lecture, we will study (FS) & (FT) for continuous signal.
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………..(9.1)
……….(9.2)
X[K] is the coefficient of the signal in the frequency domain .
(9.1)
(9.2),
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Fig (9.1)
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Fig (8.5)
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…………..(9.3)
………..(9.4)
(9.4)
(9.3),
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(9.3)
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Fig (9.3)
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(9.3)
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Table (9.4)
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Table (9.5)
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Introduction to signals and systems
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Introduction to signals and systems
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Introduction to signals and systems
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