1. No category

Sample Final Exam Covering Chapters 10-17 (finals01)
Sample Final Exam (finals01)
Covering Chapters 10-17 of Fundamentals of Signals & Systems
Problem 1 (15 marks)
Consider the system depicted below used for discrete-time processing of continuous-time signals. The
−6
sampling period is 1 microsecond ( T = 10 s).
x c (t )
xd [ n ]
CT/DT
y c (t )
yd [ n ]
H d ( z)
DT/CT
Τ
Τ
The discrete-time filter H d ( z ) is given by the following block diagram:
xd [ n]
2
z −1
0
+
yd [ n ]
+
+
+
z −1
2
(a) [5 marks] Write the transfer function H d ( z ) of the filter, specify its ROC, and sketch its pole-zero
plot. Compute the DC gain of H d ( z ) .
1
Sample Final Exam Covering Chapters 10-17 (finals01)
Answer:
H ( z ) = 2 + 2 z −2 =
2z2 + 2
,
z2
z > 0.
Im{z}
1
ROC: all z
except 0
1
Re{z}
-1
DC gain is
H (1) = 4 .
(b) [5 marks] Compute the frequency response H (e
jΩ
) of the filter and sketch its magnitude and its
phase over Ω ∈ [ −π , π ] .
Answer:
(
H (e jΩ ) = 2 + 2e − j 2Ω = 4e − jΩ 0.5e jΩ + 0.5e − jΩ
= 4 cos Ωe − jΩ
H (e jΩ ) = 4 cos Ω
π

Ω≤
 −Ω,
2

π

∠H (e jΩ ) = −Ω + π ,
< Ω≤π
2

π

 −Ω − π , −π < Ω < − 2

2
)
Sample Final Exam Covering Chapters 10-17 (finals01)
H (e j Ω )
Ω
∠H (e jΩ )
Ω
(c) [5 marks] Suppose that the continuous-time signal to be filtered is given by:
xc (t ) = cos(500000π t ) + 250000 cos(500000π t )sinc(250000t ) . Sketch the spectra of the
continuous-time and discrete-time versions of the input signal X c ( jω ) and X d (e
spectra Yd (e
jΩ
jΩ
) . Sketch the
) of the discrete-time output signal and Yc ( jω ) of the continuous-time output signal
yc (t ) .
Answer:
sin π x
and the perfect unit-magnitude lowpass filter of bandwidth W has
πx
W
W
sin Wt
impulse response:
.
sinc( t ) =
π
π
πt
Recall that sinc( x ) =
3
Sample Final Exam Covering Chapters 10-17 (finals01)
X c ( jω )
1/2
−1000000π −750000π −500000π −250000π
250000π 500000π 750000π 1000000π
ω
X d (e jΩ )
500000
−π
−.75π
.25π
−.5π
.25π
.5π
.75π
π
.75π
π
Ω
Yd (e jΩ )
1000000 2
−π
−.75π
.25π
−.5π
.25π
.5π
Ω
Yc (e jω )
2
−1000000π −750000π −500000π
−250000π
250000π 500000π 750000π
1000000π
ω
Problem 2 (15 marks)
Design an envelope detector to demodulate the AM signal:
y (t ) = [1 + 0.3x(t )]cos(2π 106 t )
where x(t ) is the periodic modulating signal shown below. That is, draw a circuit diagram of the
envelope detector and compute the values of the circuit components. Justify all of your approximations
and assumptions. Provide rough sketches of the carrier signal, the modulated signal and the signal at
the output of the detector. What is the modulation index m of the AM signal?
4
Sample Final Exam Covering Chapters 10-17 (finals01)
x (t ) = sin(1000π t )
1
−0.001
0.002
0.001
t (s)
-1
Answer:
An envelope detector can be implemented with the following simple RC circuit with a diode.
+
+
y (t )
R
-
C
w( t )
-
The output voltage of the detector, when it goes from one peak at voltage v1 to the next when it
intersects the modulated carrier at voltage v 2 after approximately one period T = 1µs of the carrier,
is given by:
v 2 ≅ v1e − T / RC
Since the time constant τ = RC of the detector should be large with respect to T = 1µs , we can use
a first-order approximation of the exponential such that
v 2 ≅ v1 (1 − T RC ) .
This is a line of negative slope −
v1
between the initial voltage v1 and the final voltage v 2 so that
RC
v 2 − v1
≅ − v1 RC .
T
This slope must be more negative than the maximum negative slope of the envelope of y (t ) , which is
maximized as follows:
d (0.3x(t ))
d (0.3sin(1000π t ))
= min
t
dt
dt
= min 300π cos(1000π t ) = −300π
min
t
t
5
Sample Final Exam Covering Chapters 10-17 (finals01)
Taking the worst-case v1 = 0.7 that will make − v1 RC the largest, we must have
0.7
< −300π
RC
⇔
0.7
= 0.00074
RC <
300π
−
We could take R = 7 k Ω, C = 0.1µ F to get RC = 0.0007 .
Let K be the maximum amplitude of 0.3 x(t ) , i.e., 0.3 x(t ) < 0.3 = K and here A = 1 . The
modulation index m is the ratio m = K A = 0.3 .
Problem 3 (15 marks)
The system shown below is a lower single-sideband, suppressed-carrier AM modulator implemented
in discrete time.
The message signal x (t ) is corrupted by an additive noise n(t ) :
xn (t ) = x(t ) + n(t ) before sampling. We want the modulator to operate at a carrier frequency
ω c = 2π × 106 rd/s .
The antialiasing filter H a ( jω ) is a perfect unity-gain lowpass filter with cutoff frequency
ωa .
The
antialiased signal w(t ) is first converted to a discrete-time signal wd [ n] via the CT/DT operator.
Signal wd [ n] is modulated with frequency Ω c and bandpass filtered by H bp (e
jΩ
) to create the lower
sidebands. Finally, the DT/CT operator produces the continuous-time lower SSB/SC AM signal y (t ) .
x (t )
+
x n (t ) H ( j ω ) w ( t )
a
+
n(t )
CT /DT
wd [ n ]
×
vd [ n ]
cos(Ωc n)
T
jΩ
H bp (e )
y (t )
yd [ n ]
CT /DT
T
The modulating (or message) signal x (t ) has spectrum X ( jω ) as shown below. The spectrum
N ( jω ) of the noise signal is also shown.
6
Sample Final Exam Covering Chapters 10-17 (finals01)
2 X ( jω )
−2000π
2000π
N ( jω )
ω
1
−2000π
2000π
(a) [5 marks] The antialiasing filter's cutoff frequency is given as
ω
ω a = 3000π . Sketch the spectra
X n ( jω ) and W ( jω ) of signals xn (t ) and w(t ) . Indicate the important frequencies and
magnitudes on your sketch.
Answer:
X n ( jω )
2
1
−2000π
2000π
ω
W ( jω )
2
1
−3000π −2000π
2000π 3000π
ω
(b) [10 marks] Find the minimum sampling frequency
ωs
and corresponding sampling period T that
will produce the required modulated signal. Find the discrete-time modulation frequency Ω c that
will result in a continuous-time carrier frequency
ω c = 2π × 106 rd/s . Give the cutoff frequencies
Ω1 < Ω 2 of the bandpass filter to obtain a lower SSB/SC AM signal. Using these frequencies,
jΩ
jΩ
jΩ
sketch the spectra Wd (e ), Vd (e ), Yd (e ) and Y ( jω ) .
7
Sample Final Exam Covering Chapters 10-17 (finals01)
Answer:
In order to find the minimum sampling rate, we need to look at the desired lower SSB signal at the
output. Its highest frequency component has to be at
ω c = 2π × 106 rd/s , and therefore the sampling
frequency has to be greater than 2ω c = 4π × 10 rd/s , which is the minimum.
6
ω s = 2(ω c ) = 4π 106 rd/s , T =
The carrier frequency: Ω c =
2π
ωs
2π
ωs
=
2π
10−6
=
= 5 × 10−7
2
4π 106
ωc = π
Bandpass filter's frequencies:
 2π
Ω1 = π − 
 ωs
Ω2 = π

4π 2 103
3
2
π
π
=
−
= π − π 10−3 = 0.999π
 10
6
4π 10

(assume unity gain)
Wd (e jΩ )
4x106
−2π
1.999π
3π/2000
π/1000
Vd ( e j Ω )
4x106
−π
−0.999π
0.999π
π
Yd (e jΩ )
4x106
π
−π
8
2π
Sample Final Exam Covering Chapters 10-17 (finals01)
Y ( jω )
2
2000000π
−2000000π
Problem 4 (25 marks)
Consider the maximally decimated multirate filter bank shown below, used for voice data compression.
This system transmits a signal x[ n] over two low-bit-rate channels. The sampled voice signal is
bandlimited to
ωM =
jω
x[n]
H lp (e )
jω
H hp (e )
2π
.
5
x1[n]
x2 [n]
channel 1
↓ N1
v1[n]
v2 [n]
↓ N2
↑ N2
channel 2
Numerical values:
Input lowpass filters' cutoff frequency
↑ N1
ω clp =
Input highpass filter's cutoff frequency
π
3
ω chp =
w1[n]
w2 [n]
,
π
3
,
5
2π

1−
ω, ω ≤

 2π
5
jω
signal's spectrum over [ −π , π ] : X (e ) = 
.
2π
 0,
< ω <π

5
9
H1 (e jω )
jω
H 2 (e )
y1[n]
+
y2 [n] +
y[n]
Sample Final Exam Covering Chapters 10-17 (finals01)
X (e jω ) , X 1 (e jω ) , X 2 (e jω ) , indicating the important frequencies
(a) [5 marks] Sketch the spectra
and magnitudes.
X (e jω )
1
−π
−0.4π
0.4π
1
π
ω
π
ω
X 1 (e jω )
1/6
−π
−π/3
π/3
X 2 (e jω )
1
1/6
−π
−2π/5 −π/3
π/3 2π/5
π
ω
(b) [10 marks] Find the maximum decimation factors N1 and N 2 avoiding aliasing and sketch the
V1 (e jω ) , V2 (e jω ) , W1 (e jω ) , W2 (e jω ) , indicating the important
jω
jω
frequencies and magnitudes. Specify what the ideal output filters H1 (e ) and H 2 (e ) should
corresponding
spectra
be for perfect signal reconstruction. Sketch their frequency responses, indicating the important
frequencies and magnitudes.
Maximum decimation factors: N1 = 3 and N 2 = 12
1/3
−π
V1 (e jω )
−0.5π
0.5π
π
ω
X 2 p ( e jω )
1/72
−π
−2π/3
−π/3
π/3
10
π/15
2π/5
2π/3
π
ω
Sample Final Exam Covering Chapters 10-17 (finals01)
V2 (e jω )
1/72
−π −4π/5
4π/5 π
ω
π
ω
π
ω
W1 (e jω )
1/3
1/18
−π
−π/3
π/3
W2 (e jω )
1/72
−π
−2π/3
−π/3
π/3
π/15
Filter
2π/3
2π/5
H1 (e jω ) should be lowpass and filter H 2 (e jω ) should be bandpass:
3
−π
−π/3
H1 (e jω )
π/3
π
ω
H 2 (e jω )
12
−π
−2π/5 −π/3
π/3 2π/5
11
π
ω
Sample Final Exam Covering Chapters 10-17 (finals01)
(c) [6 marks] Compute the percentage ratio of energies at the output between the high-frequency
y[n] , i.e., compute r := 100
subband y2 [ n] and the output signal
given by E { y} :=
+∞
∑
E { y2 }
E { y}
, where the energy is
2
y[n] .
n =−∞
Answer:
Using Parseval's relationship:
+∞
∑
E { y} :=
n =−∞
1
=
2
y[n] =
2π / 5
∫
π
1
2π
π
∫π Y (e
0
2
) dω
−
2
1−
jω
5
1
ω dω =
2π
π
2π / 5
∫
0
25 2 
 5
 1 − ω + 2 ω  dω
4π
 π

2π / 5
1
5 2
25 3 
= ω −
ω +
ω 
2π
12π 2
π
0
2π / 5
1  2π 2π
25(8π 3 ) 
= 
−
+

5 12π 2 (125) 0
π 5
1  2π  2
= = 0.1333
π  15  15
=
1
E { y2 } =
2π
=
1
π
π
∫π Y (e
2
jω
2
) dω
−
2π / 5
∫
π /3
2
5
ω dω
1−
2π
2π / 5
r = 100
=
1
5 2
25 3 
ω +
ω 
ω −
π
2π
12π 2
π / 3
=
1  2π 2π
25(8π 3 ) π 5π
25(π 3 ) 
−
+
−
+
−


π 5
5 12π 2 (125) 3 18 12π 2 (27) 
=
1  2π π
25π  2 1 25
− −
= − −
= 0.000617

π  15 18 12(27)  15 18 324
E { y2 }
E { y}
= 100
0.000617
= 0.46%
0.1333
(d) [4 marks] Now suppose each of the two signal subbands v1[ n] and v2 [ n] are quantized with a
different number of bits to achieve good data compression, without losing intelligibility of the voice
message. According to your result in (c), it would seem to make sense to use less bits to quantize
the high-frequency subband. Suppose that the filter bank operates at an input/output sample rate
of 10kHz, and you decide to use 12 bits to quantize v1[ n] and 4 bits to quantize v2 [ n] . Compute
the bit rates of channel 1 and channel 2, and the overall bit rate of the system.
12
Sample Final Exam Covering Chapters 10-17 (finals01)
Answer:
After decimation by 3, channel 1 operates at 3.333 kilosamples/s which results in a
bit rate of 3333.33333*12 = 40 000 bits/s . And after decimation by 2,
channel 2 operates at 5 kilosamples/s which results in a bit rate of 5000*12 = 60 000 bits/s .
Therefore, the overall bit rate of the filter bank is 100 000 bits/s.
Problem 5 (10 marks)
(a) [5 marks] A causal recursive DLTI system described by its zero-state impulse response given
below has an initial condition y[ −1] = 5 . Calculate the zero-input response y zi [ n] of the system
for n ≥ 0 using the unilateral z-transform.
h[n] = (0.2) n u[n] + δ [n]
Answer:
Taking the unilateral z-transform, we get:
H ( z) =
1
2 − 0.2 z −1
+
1
=
, z > 0.2
1 − 0.2 z −1
1 − 0.2 z −1
which yields the difference equation:
y[n] − 0.2 y[n − 1] = 2 x[n] − 0.2 x[n − 1] .
Taking the unilateral z-transform, we can include the initial condition:
Y ( z ) − 0.2 z −1Y ( z ) − 0.2 y[−1] = 2X ( z ) − 0.2 z −1X ( z )
2 − 0.2 z −1
0.2
y[−1],
X ( z) +
−1
1 − 0.2 z
1 − 0.2 z −1
1


y zi [n] = Z −1 
= (0.2) n u[n]
−1 
1 − 0.2 z 
Y ( z) =
z > 0.2
(b) [2 marks] Determine if the following signal is periodic. If it is, find its fundamental period N and
fundamental frequency Ω 0 . If it is not periodic, explain why.
x[n] = e − j 2π n sin(
Answer:
0.13π
n + 0.13π ) .
3
0.13π
(n + 3))
3
0.13π 13π 13(2π )
=
=
ω0 =
3
300
600
x[n] = sin(
Yes the signal is periodic.
The fundamental period is N = 600 , fundamental frequency is Ω 0 =
13
2π
π
=
600 300
Sample Final Exam Covering Chapters 10-17 (finals01)
(c) [3 marks] Compute the zero-state response of the system in (a) for the input signal in (b).
Answer:
Using the TF found in (a), we have the frequency response:
H (e jω ) =
So
2e jω − 0.2
e jω − 0.2
 0.13π

(n + 3) + ∠H (e jω0 ) 
y[n] = H (e jω0 ) sin 
 3

0.13π
ω0 =
3
H (e jω0 ) =
2e
e
j
0.13π
3
0.13π
j
3
⇒
− 0.2
= 2.2457 − 0.0422 j = 2.246e − j 0.01877
− 0.2
 0.13π

(n + 3) − 0.01877 
y[n] = 2.246sin 
 3

Problem 6 (20 marks)
The causal continuous-time LTI system given by its transfer function G ( s ) =
s+2
is discretized with
s +1
sampling period T = 0.1 s for simulation purposes. The "c2d" transformation as shown below is used
first, and then it is compared to the bilinear transformation.
c 2d
ud [n]
u (t )
DT /ZOH
s+2
s +1
G (s)
yd [ n ]
y (t )
CT /DT
Recall that the DT/ZOH operator is defined as a mapping from a discrete-time signal to an impulse
train (consisting of forming a train of impulses occurring every T seconds, with the impulse at time
nT having an area ud [n] ), followed by a "hold" function that holds each value for a period T .
ud [ n]
conversion of
discrete-time
sequence to
impulse train
u p (t )
1
h (t )
T t
DT/ ZOH
14
u (t )
Sample Final Exam Covering Chapters 10-17 (finals01)
(a) [5 marks] Find a state-space realization of G ( s ) =
s+2
, Re{s} > −1 .
s +1
Answer:
The transfer function corresponding to the system is
s+2
, Re{s} > −1 .
s +1
Partial fraction expansion:
s+2
1
, Re{s} > −1 .
= 1+
s +1
s +1
Corresponding state-space system:
x = − x + u
y = x+u
Thus, A = −1, B = 1, C = 1, D = 1 .
Alternate solution using the controllable canonical state-space realization:
A = −1, B = 1, C = 1, D = 1 .
b1
sW ( s )
U (s)
+
-
1
s
+
X1 (s)
b0
+
Y ( s)
a0
Where a0 = 1, b0 = 2, b1 = 1 . State-space system:
x = −a0 x + u = −
N1 x + 1N u
A
B
y = (b0 − a0b1 ) x + b1u = 1N x + 1N u
C
D
where A = −1, B = 1, C = 1, D = 1 .
(b) [8 marks] Write the mathematical relationships mapping the continuous-time state-space matrices
( A, B, C , D) into the discrete-time state-space matrices ( Ad , Bd , Cd , Dd ) in the "c2d"
discretization technique. Compute the discrete-time state-space system ( Ad , Bd , Cd , Dd ) for
G ( s ) and its associated transfer function Gc 2 d ( z ) , specifying its ROC.
15
Sample Final Exam Covering Chapters 10-17 (finals01)
Answer:
Discretized state-space system using "c2d":
Ad = e AT = e −0.1 = 0.9048
Bd := A−1 e AT − I n  B = −(e −0.1 − 1)(1) = 0.0952
Cd := C = 1
Dd := D = 1
transfer function:
Gc 2 d ( z ) = Cd ( zI n − Ad ) −1 Bd + Dd
= 1( z − 0.9048) −1 (0.0952) + 1 =
=
z − 0.8096
, z > 0.9048
z − 0.9048
0.0952
+1
z − 0.9048
(c) [7 marks] Find the bilinear transformation Gbilin ( z ) of G ( s ) , specifying its ROC. Compute the
difference between the two frequency responses obtained with c2d and the bilinear
transformation, i.e., compute E (e
jω
) := Gbilin (e jω ) − Gc 2 d (e jω ) . Evaluate this difference at dc
and at the highest discrete-time frequency.
Answer:
The bilinear transformation is s =
2 1 − z −1
1 − z −1
. Thus
=
20
T 1 + z −1
1 + z −1
1 − z −1
20(1 − z −1 ) + 2 + 2 z −1
+
2
−1
s+2
1 + z −1
Gbilin ( z ) =
= 1 + z −1
=
−1
s + 1 s = 201− z
1− z
20(1 − z −1 ) + 1 + z −1
−1
20
+
1
1+ z
1 + z −1
1 + z −1
20(1 − z −1 ) + 2 + 2 z −1 −18 z −1 + 22
=
=
20(1 − z −1 ) + 1 + z −1
−19 z −1 + 21
22 z − 18
19
=
, z >
21z − 19
21
z − 0.8182
19
= 1.048
, z >
z − 0.9048
21
20
Alternate solution:
Discretized state-space system using "bilinear transf.":
16
Sample Final Exam Covering Chapters 10-17 (finals01)
−1
−1
 T   T   0.1   0.1  0.95
= 0.9048
Ad :=  1 − A   1 + A  =  1 +
 1 −
=
2  
2  
2  
2  1.05

−2
 T 
 0.1 
Bd := T  1 − A  B = 0.1 1 +

2 
2 


Cd := C = 1
−2
= 0.0907
−1
−1
T  T 
 0.1 
Dd := D + C  1 − A  B = 1 + 0.05(1) 1 +
 = 1.0476
2 
2 
2 

transfer function:
Gbilin ( z ) = Cd ( zI n − Ad ) −1 Bd + Dd
0.0907
+ 1.0476
z − 0.9048
1.0476 z − 0.8572 1.0476( z − 0.8182)
=
=
, z > 0.9048
z − 0.9048
z − 0.9048
= 1( z − 0.9048) −1 (0.0907) + 1.0476 =
z − 0.8182 z − 0.8096
−
z − 0.9048 z − 0.9048
0.048 z − 0.8575 + 0.8096 0.048 z − 0.048
0.048(e jω − 1)
=
=
E (e jω ) = jω
,
z − 0.9048
z − 0.9048
e − 0.9048
0.048( z − 1)
=
,
z − 0.9048
E ( z ) := Gbilin ( z ) − Gc 2 d ( z ) = 1.048
E (e j 0 ) = 0
E (e jπ ) =
0.048(−2)
= 0.05
−1.9048
END OF EXAMINATION
17