University Carleton Department of Systems and Computer Engineering SYSC-4405: Introduction to Digital Signal processing Winter Semester 2009 Midterm 2 March 17, 2009 Duration: 60 minutes. ______________________________________________________________________________ Answer five of the following questions
Question 1 Determine the signal x(n) if its z-transform is given by: 1.5z 1 1 X ( z ) = ------------------------------------------------- with a region of convergence -- z 2 2 1 1 z 1 ( 1 2z 1 ) - 2 solution 1 1 1 1 1.5z 1 X ( z ) = ------------------------------------------------- = ------------------ ------------------- with a ROC -- z 2 1 1 2 1 2z 1 1 z 1 ( 1 2z 1 ) 1 -- z 1 -2 2 1 The first factor above has a ROC z 2 while the second factor has a ROC -- z . Therefore the 2 inverse trasform must be: 1 x ( n ) = 2 n u ( n 1 ) -- u ( n ) 2 Question 2 Use the z-transform to determine the convolution (as a function of the time index n) of the following two sequences: x 1 ( n ) = { 1, 1, 1, 1 } and x 2 ( n ) = { 1, 1, 1 } Solution 2 Y 1 ( z ) = X 1 ( z )X 2 ( z ) = ( 1 + z 1 + z 2 + z 3 ) ( 1 + z 1 + z 2 ) = ( 1 + 2z 1 + 3z 2 + 3z 3 + 2z 4 + z 5 ) Therefore y ( n ) = ( n ) + 2 ( n 1 ) + 3 ( n 2 ) + 3 ( n 3 ) + 2 ( n 4 ) + ( n 5 )
n
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Question 3 An LTI system has the property that when the input is x ( n ) = u ( n ) , and all initial conditions are 3 n zeros, the output is given by y ( n ) = 1 -- u ( n ) . 4 a) Determine the transfer function of the system. b) Indicate whether this system is causal and stable. Give a brief justification for your answer. Solution 3 a) The transfer function is 1 z 1 Y(z) 1 1 3 H ( z ) = ---------- = ------------------- --------------- = ------------------- with a ROC z -4 X( z) 3 1 1 z 1 3 1 1 -- z 1 -- z 4 4 3 b) The system is causal because the ROC is given by z -- . It is stable because the pole is inside 4 the unit circle in the z-plane. Question 4 Cosider the system shown below: x(n) z 2 system 1 system 2 z 2 z 2 means two sample delay
Show that the impulse response, h(n), of this system is given by the convolution of the following sequences:
y(n) 1 -4
x1 ( n ) = ( n ) + ( n 2 ) 1 n 1 n x 2 ( n ) = C 1 -- + C 2 -- u ( n ) where C 1, C 2 are constants which you are not required to deter 2 2
mine. solution 4 The system can be thought of as being made of two systems in cascade. system 1
h1 ( n ) = Z 1 { H1 ( z ) } = Z 1 { 1 + z 2 } = ( n ) + ( n 2 )
system2
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h2 ( n ) =
Z 1 { H
2(z)}
=
Z 1
C2 C1 1 1 ------------------- + ------------------- = ----------------------- = Z 1 1 z 1 1 + 1 z 1 1 1z 2 ---2 2 4
1 n 1 n C 1 -- + C 2 -- u ( n ) 2 2
Therefore, the impulse response of the original system is given by
1 n 1 n h ( n ) = conv [ ( n ) + ( n 2 ) ], C 1 -- + C 2 -- u ( n ) 2 2
Question 5
A sequence x(n) is obtained by sampling a continuous-time signal x ( t ) = 2 cos 2000t + -- + sin ( 4000t ) 4 The sampling rate was f s = 8000 samples per sec ond . x(n) is applied to an LTI system with a frequency response H ( e j ) = cos ( e ) j . Write an expression for the system's output y(n) in the steady state. Solution 5 First wright an expression for the sampled sequence 4000n n n 2000n - - x ( n ) = 2 cos ------------------ + -- + sin ------------------ = 2 cos ----- + -- + sin ----- 8000 8000 4 4 2 4 - Now compute the frequency response at = -- , -4 2
j - j -1- j -H e 4 = cos e 4 = ------ e 4 - 4 2 j - j -H e 2 = cos e 2 = 0 - 2
The steady state output os given by 1n n - - - y ( n ) = ------ 2 cos ----- + -- -- + 0 sin ----- -- = 4 4 4 2 2 2 n 2 cos ----4
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Question 6
Consider the causal system defined by the pole-zero pattern shown below, and a gain of 1 at = 0.
Im(z)
Re(z)
-1.25
0.8
a) Write an expression for the frequency response of this system. Show the mgnitude and phase reponses using separate equations. b) Give an approximate sketch for the phase response as a function of for the range 0 2 . solution 6 a) The transfer function is of the form z + 1.25 H ( z ) = g ------------------ where g is a constant chosen to give a gain of 1 when = 0 . z + 0.8 1.25 + e j 1 + 0.8e jH ( e j ) = g ----------------------- = 1.25ge j -------------------------0.8 + e j 1 + 0.8e j H ( e j0 ) = 1.25g = 1 g = 0.8 Therefore, the frequency response can be written as 1 + 0.8e jH ( e j ) = e j -------------------------1 + 0.8e j The magnitude and phase responses are 0.8 sin H ( e j ) = + 2 atan ----------------------------1 + 0.8 cos ( 1 + 0.8 cos ) 2 + ( 0.8 sin ) 2 H ( e j ) = ------------------------------------------------------------------------------ = 1 ( 1 + 0.8 cos ) 2 + ( 0.8 sin ) 2
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phase angle versus omega 1 0 -1 -2 -3 -4 -5 -6 -7
0
1
2
3
4
5
6
7
input form
C Cn Can Ccos(n)
particular solution
C1 C1n+C2 C1an C3cos(n) + C4sin(n) or C1cos(n+C2)
Csin(n)
C3cos(n) + C4sin(n) or C1cos(n+C2)
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input form
Cancos(n)
particular solution
C3ancos(n) +C4ansin(n) or C1ancos(n+C2)
Cansin(n)
C3ancos(n) +C4ansin(n) or C1ancos(n+C2)
AnM AnnM C(n)
K0nM+K1nM-1+...+KM An(K0nM+K1nM-1+...+KM) ---------
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time sequence x(n) (n) u(n) anu( n ) na n u ( n ) an u ( n 1 ) na n u ( n 1 )
z-Transform X(z) 1 1 --------------1 z 1 1 -----------------1 az 1 az 1 -------------------------( 1 az 1 ) 2 1 -----------------1 az 1 az 1 -------------------------( 1 az 1 ) 2 1 z 1 cos 0 -----------------------------------------------1 2z 1 cos 0 + z 2 z 1 sin 0 -----------------------------------------------1 2z 1 cos 0 + z 2 1 az 1 cos 0 --------------------------------------------------------1 2az 1 cos 0 + a 2 z 2 az 1 sin 0 --------------------------------------------------------1 2az 1 cos 0 + a 2 z 2 A A ------------------ + --------------------1 pz 1 1 p z 1
ROC allz z >1 z > a z > a z < a z < a
[ cos 0 n ]u ( n ) [ sin 0 n ]u ( n ) [ a n cos 0 n ]u ( n ) [ a n sin
0 n ]u ( n )
z >1 z >1
z > a
z > a z > a
2 A p n cos ( n p + A ) u ( n )
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