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Net 222: Communications and networks fundamentals ( Practical Part)

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Net 222: Communications and networks fundamentals ( Practical Part). Tutorial 5 : Matlab – Aljabric equations. – convolution . Lecture Contents. Solving Algebraic equations: Simple equations. Quadratic equations. Plotting Symbolic Equations. Computing derivatives. Integration. - PowerPoint PPT Presentation
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NET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS (PRACTICAL PART) Tutorial 5 : Matlab Aljabric equations. – convolution Networks and Communication Department 1
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Page 1: Net 222: Communications and networks fundamentals  ( Practical Part)

NET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS (PRACTICAL PART)Tutorial 5 : Matlab – Aljabric equations. – convolution

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Page 2: Net 222: Communications and networks fundamentals  ( Practical Part)

Lecture Contents

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Solving Algebraic equations: Simple equations. Quadratic equations. Plotting Symbolic Equations. Computing derivatives. Integration.

Convolution.

Page 3: Net 222: Communications and networks fundamentals  ( Practical Part)

Simple equations3

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Example

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Solve : x+5=0

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Quadratic equations5

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Example

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Solve :

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Plotting Symbolic Equations7

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Example 1:

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Example 2:

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Computing derivatives10

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Example

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Find the derivative for :

f = sin(5*x)

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Integration12

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Example 1:

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Find the integration for x^2 :

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Example 2:

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Convolution15

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Convolution SumThe Convolution sum:

The equation below defines the convolution of two sequences and denoted by:

(The convolution sum or superposition sum)And the operation on the right hand side (equation in bold ) is known as the convolution of the sequence and h.

It is commonly called the convolution sum. Thus, again, we have the fundamental result that the output of any discrete-time LTI system is the convolution of the input with the impulse response of the system.

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Convolution Sum (Cont.) The Figure below illustrates the definition of the impulse response h[n] and the

relationship of

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Example

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Consider an LTI system with impulse response h[n] and input x[n].

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x = [ 0 . 5 2 0 0 0 ] ;h = [ 1 1 1 0 ] ;y = c o n v ( x , h ) ;

n 1 = 0 : ( l e n g t h ( x ) - 1 ) ;s u b p l o t ( 2 , 2 , 1 ) ;s t e m ( n 1 , x , ’ L i n e W i d t h ’ , 3 ) ;x l a b e l ( ‘ n ’ ) ;y l a b e l ( ‘ x [ n ] ’ ) ;

n 2 = 0 : ( l e n g t h ( h ) - 1 ) ;s u b p l o t ( 2 , 2 , 2 ) ;s t e m ( n 2 , h , ’ L i n e W i d t h ’ , 3 ) ;x l a b e l ( ‘ n ’ ) ;y l a b e l ( ‘ h [ n ] ’ ) ;

n 3 = 0 : ( l e n g t h ( y ) - 1 ) ;s u b p l o t ( 2 , 2 , [ 3 , 4 ] ) ;s t e m ( n 3 , y, ’ L i n e W i d t h ’ , 3 ) ;x l a b e l ( ‘ n ’ ) ;y l a b e l ( ‘ y [ n ] ’ ) ;

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Any Questions ?The End


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