Final Sample Exam Covering Chapters 10-17 (finals02)
Sample Final Exam (finals02)
Covering Chapters 10-17 of Fundamentals of Signals & Systems
Problem 1 (30 marks)
Consider the system depicted below used for discrete-time processing of continuous-time signals. The
sampling period is 100 milliseconds ( T = 0.1 s).
x c (t )
xd [ n ]
CT/DT
y c (t )
yd [ n ]
H d ( z)
DT/CT
The discrete-time filter H d ( z ) is given by the following causal difference equation initially at rest:
y[n ] + a1 y[n 1] + a2 y[n 2] = b0 x[n ] + b1 x[n 1] + b2 x[n 2]
(a) [8 marks] Find the controllable canonical state-space realization of the filter H d ( z ) , i.e., sketch the
block diagram and give the state-space equations.
Answer:
u[n ]
+
b0
+
y[ n ]
+
+
z 1
-
-
b1
a1
z 1
b2
a2
The state equation:
1
+
+
Sample Final Exam Covering Chapters 10-17 (finals02)
x1 [n + 1] 0
x [n + 1] = a
2
2
1 x1 [n ] 0
+
u[n ] .
a1 x2 [n ] 1
B
A
The output equation:
y[n ] = [b2 a2 b0
b1 a1b0 ] x[n ] + b0 u[n ]
D
C
(b) [10 marks] The filter H d ( z ) is designed to approximate the unity-gain, second-order, continuous-
2
, Re{s} > 1 . Find the values of the parameters
s + 3s + 2
{a1 , a2 , b0 , b1 , b2 } of H d ( z ) using the "c2d" transformation. Specify the ROC of H d ( z ) .
time, causal LTI filter H ( s ) =
2
Answer:
The state equation for H ( s ) :
x1 (t ) 0 1 x1 (t ) 0
x (t ) = 2 3 x (t ) + 1 u (t ) .
2
2
A
B
The output equation for H ( s ) :
y ( t ) = [ 2 0] x ( t )
C
Diagonalize to compute matrix exponential:
1 0
1 1
A1 = V 1 AV =
, V = 1 2
0 2
0 2 1
1 1 e T
Ad = e AT = Ve A1T V 1 =
0 e 2T 1 1
1 2
T
T
T
2 T
e 2e e
e T e 2T 0.9909 0.0861
1 1 2e
=
2T e 2T = 2e T + 2e 2T e T + 2e 2T = 0.1722 0.7326
1 2 e
e T e 2T 0
3 / 2 1/ 2 2e T e 2T 1
e AT I 2 B =
Bd = A
0 2e T + 2e 2T e T + 2e 2T 1 1
1
1
1
3 / 2 1/ 2 e T e 2T e T + e 2T + 0.0045
=
2
2 =
=
0.0861
0 e T + 2e 2T 1
T
2 T
1
e e
Cd = C
1
Transfer function:
2
Sample Final Exam Covering Chapters 10-17 (finals02)
H ( z ) = Cd ( zI n Ad ) 1 Bd + Dd
1
z 0.9909 0.0861 0.0045
= [ 2 0]
z 0.7326 0.0861
0.1722
0.0045( z 0.7326) + (0.0861)(0.0861)
1
[ 2 0]
( z 0.9909)( z 0.7326) + 0.0148
( 0.1722)(0.0045) + (0.0861)( z 0.9909)
0.009( z 0.7326) + 2(0.0861)(0.0861)
=
( z 0.9909)( z 0.7326) + 0.0148
0.009 z + 0.0082
=2
z 1.7235z + 0.7407
0.009 z 1 + 0.0082 z 2
=
1 1.7235z 1 + 0.7407 z 2
=
Hence, b0 = 0, b1 = 0.009, b2 = 0.0082, a1 = 1.7235, a2 = 0.7407
(c) [5 marks] Sketch the pole-zero plot of H d ( z ) . Compute the gain of H d ( z ) at the highest
frequency.
Answer:
0.009 z + 0.0082
z 1.7235z + 0.7407
0.009( z + 0.915)
,
=
( z 0.8184)( z 0.9051)
H d ( z) =
2
.
z > 0.9051
Im{z}
1
0.905
0.915
0.818
-1
3
1
Re{z}
Sample Final Exam Covering Chapters 10-17 (finals02)
Gain at highest frequency: H d ( 1) =
0.009 + 0.0082
= 0.00023
1 + 1.7235 + 0.7407
(d) [7 marks] Suppose that the continuous-time signal to be filtered is given by:
xc (t ) = cos(6 t ) cos(4 t ) . Sketch the spectra of the continuous-time and discrete-time
versions of the input signal X c ( j ) and X d (e
j
) . Sketch the spectra Yd (e j ) of the discretetime output signal and Yc ( j ) of the continuous-time output signal yc (t ) . Finally, give an
expression for the output signal yc (t ) .
Answer:
We need to compute
0.009e j 0.4 + 0.0082
H d (e
) = j 0.8
= 0.0106 + j 0.0044 = 0.0115e j 2.7521
j 0.4
1.7235e
+ 0.7407
e
.
j 0.6
+ 0.0082
0.009e
j 0.6
j 2.3717
= 0.0032 + j 0.0031 = 0.0045e
H d (e
) = j1.2
1.7235e j 0.6 + 0.7407
e
j 0.4
X c (e j )
1/2
10
8
6
4
2
4
2
6
8
1 0
0.6
0.8
1 2
-1/2
X d (e j )
5
0.8
0.6
0.4
0.2
0.4
0.2
-5
Yd (e j )
0.0225e j 2.37
0.8
0.6
0.0575e j 0.39
0.4
0.0575e j 2.752 3.142
= 0.0575e j 0.39
0.2
0.2
4
0.4
0.0225e j 2.37
0.6
0.8
Sample Final Exam Covering Chapters 10-17 (finals02)
Yc ( j )
10
8
6
0.0575e j 0.39
0.0575e j 0.39
0.0225e j 2.37
2
4
2
4
0.0225e j 2.37
6
8
1 0
yc (t ) = 0.0045cos(6 t + 2.3717) 0.0115cos(4 t + 2.752)
Problem 2 (15 marks)
Suppose we want to design a causal, stable, first-order high-pass filter of the type
H ( z) =
with -3dB cutoff frequency
of the filter is
c =
B
,z>a
1 az 1
2
(i.e., frequency where the magnitude of the frequency response
3
1
) and a real.
2
(a) [2 marks] Express the real constant B in terms of the pole a to obtain unity gain at the highest
frequency = .
Ans:
The gain at = is
H (e j ) = H ( 1) =
and unity gain is obtained for B = 1 + a .
B
,
1+ a
(b) [10 marks] "Design" the filter, i.e., find the numerical values of the pole a and the constant B .
Answer:
2
1
(1 + a ) 2
= H (e j c ) =
2
(1 a cos c ) 2 + a 2 sin 2 c
(1 + a ) 2
=
(1 + 0.5a ) 2 + 0.75a 2
This yields the quadratic equation
5
Sample Final Exam Covering Chapters 10-17 (finals02)
2(1 + a ) 2 = (1 + 0.5a ) 2 + 0.75a 2
a 2 + 3a + 1 = 0
whose solutions are a1,2 =
circle: a =
3 5
. But for the filter to be stable, we select the pole inside the unit
2
3 + 5
= 0.382 . Finally B = 0.618 , and the high-pass filter is
2
1+ 5
0.618
2
H ( z) =
=
, z > 0.382
3 5 1
1 + 0.382 z 1
1+ 2 z
(c) [3 marks] Sketch the magnitude of the filter's frequency response.
Answer:
1
H (e j )
0.9
0.8
0.7
0.6
0.5
0.4
0
0.5
1
1.5
2
2.5
6
3
3.5
Sample Final Exam Covering Chapters 10-17 (finals02)
Problem 3 (20 marks)
Consider the following sampling system where the sampling frequencies are
x p (t )
H lp1 ( j )
x
x (t )
p (t ) =
wp (t )
w(t )
x
KH lp 2 ( j )
p (t ) =
(t kT )
+
(t kT )
1
k =
y (t )
+
CT/DT
2
2
, s2 =
.
T1
T2
yd [n ]
T1
2
k =
s1 =
The spectrum X ( j ) of the input signal
x(t ) , and the frequency responses of the two ideal lowpass
filters, are shown below. The gain of the second lowpass filter is K > 0 .
X ( j )
1
W
1
W
c1
c1
(a) [10 marks] For what range of sampling frequencies
first sampler (from
H lp 2 ( j )
H lp1 ( j )
1
c 2
s1
c 2
is the sampling theorem satisfied for the
x(t ) to x p (t ) )? Suppose that the cutoff frequencies of the lowpass filters are
c1 = 3W , c 2 = W . For what range of sampling frequencies s 2 is the sampling
theorem satisfied for the second sampler (from w(t ) to wp (t ) )? Choosing the lowest sampling
given by
frequencies in the ranges that you found for the two samplers, sketch the spectra X p ( j ) ,
W ( j ) , W p ( j ) , and Y ( j ) . Find the gain K of the second filter that leads y to (t ) = x(t ) .
Answer:
The sampling theorem is satisfied for
spectra, we set
s1 = 2W
s1 > 2W
so that T1 =
W
and
for the first sampler, and for
s 2 = 6W
so that T1 =
3W
s 2 > 6W
:
X p ( j )
W
2s1
-3W
s1
-W
W
7
s1
3W
2 s1
. For the
Sample Final Exam Covering Chapters 10-17 (finals02)
W ( j )
W
-3W
-2W
3W
4W
3W
2W
W
-W
2 s1
W p ( j )
3W 2
2
2s1
-3W
s1
s1
W
-W
Y ( j )
3W 2
2
-W
Finally K =
2
3W 2
K
W
.
s1 = 2W and s 2 = 3W and that
the cutoff frequencies of the lowpass filters are given by c1 = 3W , c 2 = W . Find the signals
(b) [10 marks] Assume that the sampling frequencies are set to
wp (t ) and y (t ) and sketch them. Use the value for the gain K that you found in (a). Find and
sketch the discrete-time signal yd [n ] .
Answer:
The spectrum of w p (t ) is constant:
3W
W p ( j )
2
2
-4W
-3W
-2W
-W
W
8
2W
3W
4W
Sample Final Exam Covering Chapters 10-17 (finals02)
Therefore w p (t ) =
3W 2
2
(t ) .
w p (t )
3W 2
2
t
After filtering with the second lowpass with gain K =
2
3W 2
, we get
Y ( j )
1
-W
Which has the inverse FT:
y (t ) =
W
W
sinc
W
t.
y (t )
W
After the CT/DT operation, we obtain:
W
W
t
yd [n ] = y ( nT1 ) = y ( n
W
)=
W
sinc ( n ) =
W
...
...
-7 -6 -5 -4 -3 -2 -1
1
2
9
345
67
n
W
[n ]
Sample Final Exam Covering Chapters 10-17 (finals02)
Problem 4 (20 marks)
Space rendez-vous
Consider the spacecraft shown below which has to maneuver in order to dock on a space station.
space
station
spacecraft
For simplicity, we consider the one-dimensional case where the state of each vehicle consists of its
position and velocity along axis z.
Assume that the space station moves autonomously according to the state-space system
xs (t ) = As xs (t )
zs
0 1
, and As = 0 0 ,
zs
where xs =
and the spacecraft's equation of motion is:
xc (t ) = Ac xc (t ) + Bc uc (t )
zc
0 1
0
, uc (t ) is the thrust, Ac = 0 0 and Bc = 0.1 .
zc
where xc =
(a) [5 marks] Write down the state-space system of the state error e := xc xs which describes the
evolution of the difference in position and velocity between the spacecraft and the space station.
The output is the difference in position.
Answer:
zc z s
, so that
zc z s
Let e := xc xs =
10
Sample Final Exam Covering Chapters 10-17 (finals02)
e(t ) = Ac ( xc (t ) xs (t )) + Bc uc (t )
= Ac e(t ) + Bc uc (t )
0 1
0
=
e(t ) + 0.1 uc (t )
0 0
y (t ) = [1 0] e(t )
(b) [5 marks] A controller is implemented in a unity feedback control system to drive the position
difference to zero for docking. The controller is given by:
K ( s) =
100( s + 1)
, Re{s} > 100
0.01s + 1
ydes ( t ) = 0
+
-
uc ( t )
K ( s)
y(t )
G ( s)
Find G ( s ) and assess the stability of this feedback control system (hint: one of the closed-loop poles
is at 10 .)
Answer:
G ( s ) = Ce ( sI Ae ) 1 Be + De
1
s 1 0
= [1 0]
0 s 0.1
0.1
= 2 , Re{s} > 0
s
Closed-loop characteristic polynomial with coprime numerators and denominators:
p ( s ) = nG nK + d G d K = 0.1(100s + 100) + s 2 (0.01s + 1)
= 0.01s 3 + s 2 + 10s + 10 = 0.01( s 3 + 100s 2 + 1000s + 1000)
Thus,
s 3 + 100s 2 + 1000s + 1000 = ( s + a )( s + b)( s + 10)
= s 3 + (10 + a + b) s 2 + (10a + 10b + ab) s + 10ab
By identifying coefficients, we find three equations in two unknowns (one is redundant):
10 + a + b = 100
10a + 10b + ab = 1000
10ab = 1000
We combine the first and third equations to find a quadratic equation:
11
Sample Final Exam Covering Chapters 10-17 (finals02)
a 2 90a + 100 = 0
a1 = 45 + 10 4.52 1 = 88.8748
a2 = 45 10 4.52 1 = 1.1252
It turns out that for the choice a = 88.8748 b = 1.1252 and for the choice
a = 1.1252 b = 88.8748 . Therefore the three closed-loop poles are at
10, 1.1252, 88.8748 , and the closed-loop system is stable.
(c) [7 marks] Find the loop gain, sketch its Bode plot, and compute the phase margin of the closedloop system. Assuming for the moment that the controller would be implemented on earth, what
would be the longest communication delay that would not destabilize the automatic docking
system?
Answer:
Loop gain:
L( s ) =
10( s + 1)
s (0.01s + 1)
2
150
100
50
0
-50
-100
-2
10
-1
10
0
1
10
10
2
10
3
10
-100
-110
-120
-130
-140
-150
-160
-170
-180
-2
10
-1
10
0
1
10
10
12
2
10
3
10
Sample Final Exam Covering Chapters 10-17 (finals02)
Phase margin:
The crossover frequency is approximately
co = 10rd/s . The phase margin is given by:
L( j co ) 102 m = 78 . From the broken line approximation, we find a phase margin of
L( j co ) 90 m = 90 . Using the latter, we compute the maximum time-delay that can be
tolerated:
co =
=
180
m
90
= 0.1571s
m =
180 co
180 10
(d) [3 marks] Compute the sensitivity function of the system and give the steady-state error to a unit
step disturbance on the output.
Answer:
S ( s) =
1
=
1 + L( s )
s 2 (0.01s + 1)
1
=
10( s + 1)
0.01s 3 + s 2 + 10s + 10
1+ 2
s (0.01s + 1)
The Laplace transform of the error signal is
e( s ) =
1
0
S ( s) , and from the final value theorem, we have lim e(t ) = S (0) =
=0
t +
s
10
Problem 5 (15 marks)
Consider a fourth-order ( M = 4 ) causal moving average filter H ( z ) .
(a) [5 marks] Compute H ( z ) and give its pole-zero plot including the ROC.
Answer:
The z-transform of this filter is given by
1 1 1 1 2 1 3 1 4 1 z 4 + z 3 + z 2 + z + 1
+z+z+z+z=
55
5
5
5
5
z4
with ROC {z , z 0} , i.e., the whole complex plane excluding 0.
H ( z) =
Four poles at 0, and zeros are on the unit circle at
2 4 6 8
,
,,
.
5555
Im{z}
1
Re{z}
4 poles
-1
1
-1
13
Sample Final Exam Covering Chapters 10-17 (finals02)
(b) [5 marks] Compute the filter's frequency response
Answer:
H ( e j ) and give its magnitude and phase.
1 1 j 1 j 2 1 j 3 1 j 4
+e +e
+e
+e
55
5
5
5
1
= e j 2 ( e j 2 + e j + 1 + e j + e j 2 )
5
2
1 2
= e j 2 + cos( ) + cos(2 )
5
5 5
12
2
H ( e j ) = + cos( ) + cos(2 )
55
5
H ( e j ) =
Phase in the passband is sufficient here: H ( e
j
2 2
) = 2 ,
,
, but the complete
5 5
answer is:
2 2
2 , 5 , 5
2 4
2 , 5 , 5
4
,
H ( e j ) = 2 ,
5
4
2 , ,
5
4 2
,
2 ,
5
5
(c) [5 marks] Compute and sketch the unit step response of the filter.
Answer:
The step response is the running sum of the impulse response:
1
1
1
1
1
h[n ] = [n ] + [n 1] + [n 2] + [n 3] + [n 4]
5
5
5
5
5
1
1
y[n ] = ( n + 1)u[n ] ( n 5 + 1)u[n 5]
5
5
1
4/5
3/5
2/5
1/ 5
...
...
-7 -6 -5 -4 -3 -2 -1
1
2
345
END OF EXAMINATION
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67
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