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    Exam With Solutions

    Uploaded by

    Kayan Sami

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Exam With Solutions

    Uploaded by

    Kayan Sami
    11 views4 pages

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    © © All Rights Reserved

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    Copyright:

    © All Rights Reserved
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    11 views4 pages

    Exam With Solutions

    Uploaded by

    Kayan Sami

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
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    Question 1

    a) The difference equation for a linear time-invariant system is

    y (n) − 1.4142 y (n − 1) + y (n − 2) = x(n) − 0.7071x(n − 1)

    Calculate the transfer function for the system. (5 marks)


    y (n) − 1.4142 y (n − 1) + y (n − 2) = x(n) − 0.7071x(n − 1)

    Y ( z ) − 1.4142 z −1Y ( z ) + z −2Y ( z ) = X ( z ) − 0.7071z −1 X ( z )

    [ ] [
    Y ( z ) 1 − 1.4142 z −1 + z −2 = X ( z ) 1 − 0.7071z −1 ]
    H ( z) =
    Y ( z)
    =
    [
    1 − 0.7071z −1 ]
    =
    z 2 − 0.7071z[ ]
    [
    X ( z ) 1 − 1.4142 z −1 + z −2 ] [
    z 2 − 1.4142 z + 1 ]

    b) A system is defined with an impulse response

    −n
    1
    h( n ) =   −∞≤n≤0
    2
    =0 n>0

    Is the system both stable and noncausal? Explain your reason. (5 marks)
    The system is noncausal because it begins before n=0.
    0
    The system is stable because ∑ h ( n) < ∞ .
    n = −∞

    −n
    0
    1
    ∑  
    n = −∞  2 
    <∞
    c) A signal has a frequency representation that is given as

    1 1
    X ( f ) = δ ( f + 1000) + δ ( f ) + δ ( f − 1000)
    2 2

    i) Is the time representation continuous or discrete-time? Explain your


    reasons. (5 marks)
    The time representation is continuous.

    ii) Calculate the time representation of the signal. (5 marks)


    ∞ j 2πft
    x(t ) = ∫ X ( f )e df
    −∞

    j 2πft j 2πft j 2πft


    ∞ 1 ∞ 1∞
    =∫ δ ( f − 1000)e df + ∫ δ ( f )e df + ∫ δ ( f + 1000 )e df
    −∞ 2 −∞ −∞ 2

    1 j 2π 1000t 1
    = e + 1 + e − j 2π 1000t
    2 2
    = 1 + cos 2π 1000t −∞ ≤t ≤ ∞
    Question 2

    The transfer function of a linear time-invariant system is given as

    z2 − z
    H ( z) =
    3 1
    z2 − z +
    4 8

    a) From the transfer function, is the system continuous or discrete-time? Explain


    your reasons. (5 marks)
    The system is discrete-time because the transfer function is in z domain. For
    continuous time, the transfer function is represented in s domain (Laplace
    domain).

    b) Draw the pole-zero plot for the system. (5 marks)


    z ( z − 1)
    H ( z) =
    (z − 1 2)(z − 1 4)
    Poles at z=0.5, z=0.25
    Zeros at z=0, z=1

    c) Is the system stable? Explain your reasons. (5 marks)


    The system is stable because the poles are inside the unit circle.

    d) Calculate the output y(n) if the input is

    x(n) = u (n − 5) (10 marks)


    Let x(n) = u (n)
    z
    X ( z) =
    z −1
    Y ( z) = X ( z) • H ( z)
    z z ( z − 1)
    = •
    z − 1 (z − 1 2 )( z − 1 4 )
    z2
    =
    (z − 1 2)(z − 1 4)
    Y ( z) z A B
    = ≡ +
    z (z − 1 2)(z − 1 4) (z − 1 2) (z − 1 4 )
    Y ( z)
    A= (z − 1 2) z =1 2 = 2
    ( z)
    Y ( z)
    B= (z − 1 4) z =1 4 = −1
    ( z)
    Y ( z) 2 −1
    = +
    z ( z − 1 2 ) (z − 1 4 )
    2z −z
    Y ( z) = +
    (z − 1 2) (z − 1 4 )
      1 n  1 n 
    y (n) = 2  −   u (n)
      2   4  

    When x(n) = u (n − 5) ,

      1  n − 5  1  n −5 
    y (n) = 2  −   u (n − 5)
      2   4  

    d) Calculate the frequency representation of the system. (5 marks)


    z2 − z
    H ( z) =
    3 1
    z2 − z +
    4 8
    H (e j 2πf ) = H ( z ) z = exp( j 2πf )

    =
    (e ) − e j 2πf 2 j 2πf

    (e ) − 34 (e ) + 18
    j 2πf 2 j 2πf

    =
    (cos 4πf
    − cos 2πf ) + j (sin 4πf − sin 2πf )
     3 1  3 
     cos 4πf − cos 2πf +  + j  sin 4πf − sin 2πf 
     4 8  4 

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