Lectures in signals and systems

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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