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    Comm 602 L2

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    Comm 602 L2

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    COMM 602 L2

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    Comm 602 L2

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    of 24

    COMM 602

    Digital Signal Processing

    Lecture 2

    Dr. Engy Aly Maher


    Spring 2020
    Remember
    Properties of Linear Time-Invariant (LTIS) Systems

    Convolution

    • For a Linear Time Invariant (LTI) system, the output can be


    represented by its impulse response and input as follows:

    Input : xn    xk  n  k 
    k  
     

    Output : yn   T xn   T   xk  n  k 

    k   
    • T and Σ are linear operators, therefore y(n) can be written as:
     
    y n    xk T  n  k    xk hn  k 
    k   k  

     x ( n) * h( n)
    • h(n) is called the impulse response of the LTI system
    Impulse Response of a LTIS

    T{.}
    Convolution Properties:
    Interconnection between Systems
    Causal LTI System

    • In LTIS, causality can be expressed in terms of Impulse response


    as follows: consider the LTI system having an output at time
    n=no:

    y (n o )   h( k ) x ( n
    k  
    o  k)
     1
      h(k ) x(no  k )   h( k ) x ( n
    o  k)
    k 0 k  

     [h(0) x(no )  h(1) x(no  1)  h(2) x(no  2)]  ....


    [......  h(2) x(no  2)  h(1) x(no  1)]
    Must be zero
    The terms in the first sum involve x(no ), x(no  1),...which are the
    present and past values of the input signal. The second sum
    involves the future signal components.
    • Clearly, for a LTI system to be causal, the impulse
    response of the system must satisfy :

    h(n)  0, n  0

    • Then for a causal LTIS, the summation of the


    convolution formula is modified to be:


    y ( n )   h( k ) x ( n  k )
    k 0
    Convolution Formula for IIR and FIR LTI Systems

    (Causal FIR)


    y ( n )   h( k ) x ( n  k ) (Causal IIR)
    k 0
    Stable LTIS

    • In General: A system is said to be stable if every bounded


    input produces a bounded output
    xn    yn  
    For the LTIS to be stable:

    yn    hk  xn  k 


    k 0 Causal System
    If x(n) is bounded for all n

    xn   M  yn   M  hk 


    k 0

    For y(n) to be bounded


     hk   
    k 0
    Unstable System Stable System

    hn  e hn  e n ,   0
    h(n) n h(n)
    ,  0

    . . . . . . . . . .
    n n

     

     hn  
    n 0
     hn  
    n 0
    Characterization of Digital System

    • Any digital system can be characterized by one of the


    following:

    1. Difference Equation

    2. Impulse Response

    3. Transfer Function

    4. Frequency Response
    1. Difference Equation
    • A LTI system can be described by a linear constant coefficient difference equation
    of the form:

    a0  1
    N M

     a yn  i    b xn  i 
    i 0
    i
    i 0
    i

    OR
    M N
    yn    bi xn  i    ai yn  i 
    i 0 i 1

    •This equation describes a recursive approach for computing the current output
    given the input values and previously computed output values.

    ai and bi are constants independent on the time


    1. Difference Equation (Contd.)
    Non Recursive LTI Systems
    • This is also called Finite Impulse Response (FIR).

    • For time invariant (FIR), the difference equation has the


    form:
    M
    yn    bi xn  i  Always
    Stable
    i 0

    M is called the order of the system and


    bi 0M are the constant coefficients of the system Verify
    this?
    1. Difference Equation (Contd.)
    Recursive LTI Systems
    • This is also called Infinite Impulse Response (IIR) system.
    Can be
    • It has the following difference equation: Stable or
    Unstable
    M N
    yn    bi xn  i    ai yn  i  Verify
    i 0 i 1
    this?

    M is called the order of the all zero part The meaning will be
    Clear after you study
    N is called the order of the all pole part Z-transform
    2. The Impulse Response
    If the input to the filter is the unit impulse δ(n)

    The output y(n) is said to be the impulse response


    i.e., the response to the impulse excitation.
    1. For Non Recursive (FIR)
    M
    yn    bi n  i   n  Digital Filter hn 
    i 0

    2. For Recursive (IIR)


    M N
    yn    bi n  i    ai yn  i 
    i 0 i 1
    Example:

    • Find the impulse response of the causal system:


    y(n)  x(n)  e y(n  1)
    solution
    The input is: x(n)   (n) .

    y (n)   (n)  e y (n  1) 0

    n0 y (0)  1  e y (1)  1


    n 1 y (1)  0  e y (0)  e
    n2 y (2)  0  e y (1)  e .e  e 2
    n3 y (3)  0  e y (2)  e 2 .e  e3
    Then y (n)  h(n)  e n
    Is this System Stable??? Plot h(n) for   0.5
    3. Transfer Function of Digital System
    The transfer function or the system function H(z)

    Y z 
    H z  
    X z 

    X(z) Y(z) =X(z) H(z)


    H(z)

    X(z) and Y(z) are the Z-Transform of x(n) and y(n)

    Z yn  Y z  Z xn  X z 
    4. The Frequency Response
    •The discrete-time Fourier transform of an impulse
    response is called the frequency response
    or the system function.

    Remember:

    j
    X (e )   x[
    n  
    n ] e  j n

     

    H (e j )   h[
    n  
    n ] e  j n Assume:  xn  
    n  

     
    X e jw
    H(ejw)
     
    Y e jw
    LTIS
    Example:

    Given h(n)  a nu (n) where a  1 find


    the frequency responseof the system
    Solution

    H ( e j )   h[
    n  
    n ] e  j n
    Remember

    1
     

    a n
    u ( n ) e  j n 
    n 0
    n
    a 
    1 a
    n  
     
      a n e  j n   ( a e  j ) n
    n 0 n 0
    | e  j | 1

    1
    j
    H (e ) 
    1  a e  j
    Example:

    j 1
    H (e ) 
    1  a e  j
    1

    1  (a cos( )  ja sin( )
    Magnitude response:
    j 1
    H (e ) 
    [1  a cos( )]2  [a sin( )]2
    1

    [1  a 2 cos 2 ( )  a 2 sin 2 ( )  2a cos( )
    1

    [1  a 2  2a cos( )
    Phase response:
    j  a sin( ) 
    Angle [ H (e )]   tan 1
    
     1  a cos( ) 
    Example:

    • Consider the following difference equation for a causal LTI System


    yn  yn  1  0.9yn  2  xn
    a- Calculate and plot the impulse response h(n) for n=0, 1, 2,..….5
    b- Calculate and plot the unit step response for n=0,1, 2,…...5

    Solution
    (a) The system is causal, therefore h(n) =0 for all n< 0

    hn  hn  1  0.9hn  2   n


    n  0  h0  h 1  0.9h 2   0  h0   0  1

    n  1  h1  h0  0.9h1   1  h1  h0  1


    n  2  h2  h1  0.9h0   2  h2  h1  0.9 h0  0.1

    n  3  h3  h2  0.9h1   3  h3  h2  0.9 h1  0.1  0.9  0.8
    yn  yn  1  0.9  yn  2  un
    (b)

    n  0  y0  y 1  0.9 y 2  u0  y0  u0  1

    n  1  y1  y0  0.9 y 1  u1  y1  y0  u1  1  1  2

    n  2  y2  y1  0.9 y0  u1  y2  y1  0.9  y0  u1  1  0.9  1  0.9

    Plot the output???


    Example:

    • Using the recursive method, check the stability for the following system:
    y(n)  y 2 (n  1)  x(n) Where x(n)  C  (n)
    Assume that y(-1)=0 and C is constant and C>1.
    Solution
    The output sequence is:
    n  0, y (0)  y 2 (1)  C (0)  0  C  C
    n  1, y (1)  y 2 (0)  C (1)
    C2  0  C2
    n  2, y (2)  y 2 (1)  C (2)
     (C 2 ) 2  0  (C 2 ) 2  C 4
    n  3, y (3)  y (2)  0  (C )  C
    2 4 2 8

    2n
    Then y (n)  C
    The output is unbounded , so the system is unstable.

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