MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003: Signals and Systems|Fall 2000
Final Exam Review Packet
Quiz Date:
Time:
Location:
Coverage:
Notes:
Monday, December 18, 2000
9:00 am { noon
Rockwell
All material covered in the course.
The quiz is closed book except for three 8:500 1100 two-sided sheet of notes. No
calculators are allowed. We will provide copies of the Tables of CT and DT Fourier
Series properties on pages 206 and 221 of O&W, the Tables of CT Fourier transform
pairs and properties on pages 328 and 329, the Tables of DT Fourier transform
pairs and properties on pages 391 and 392, the Tables of Laplace transform pairs
and properties on pages 691 and 692, and the Tables of z -transform pairs and
properties on pages 775 and 776.
Marathon Oce Hours: The TAs will jointly hold oce hours Thursday-Friday, December 1415. A schedule will be posted on the web by Tuesday morning, December 14.
Final Exam Review Session: There will be a two-hour nal exam review session given by the
TAs. In this session, a set of problems from previous exams will be solved. The review session is
completely optional. Details for the nal exam review are:
Date:
Thursday, December 14
Time:
7:30{9:30 pm
Location: 34{101
Practice Problems: The attached set of problems should provide you with ample opportunity to
exercise your understanding of the material covered on the nal. This review packet consists of two
sets of problems. The rst part contains the nal exam review problems, which will be covered in
the review session. The solutions to these problems will be posted on the 6.003 web page after the
review session on December 14, and copies will also be available at the review. The second part of
this packet is the 6.003 Final Exam (and solutions) from Spring 1999.
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2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003: Signals and Systems|Fall 2000
Final Exam Review Problems
Problem #1 The purpose of this problem is to test your understanding of continuous-time convolution.
A CT LTI system has input x(t), impulse response h(t), and output y(t) as shown below. Note
that the scales for x(t) are not necessarily the same as for h(t) and y(t).
x(t)
h(t)
A
y(t)
12
1
9
6
3
-T
T
t
-6
-4
-2
2
4
6
t
-6
-4
-2
2
4
6
t
Determine the values of the parameters of x(t): A and T.
Problem #2 The purpose of this problem is to test your understanding of discrete-time sam-
pling.
Suppose that we have two discrete-time signals, x1 [n] and x2 [n], that we wish to transmit simultaneously using frequency-division multiplexing. The problem is that each of the signals lls the
entire frequency band. In particular, suppose that X1 (ej! ) and X2 (ej! ), the DTFTs of x1 [n] and
x2 [n], are as shown below.
X1(ej ω)
−2π
−π
0
π
2π
π
2π
ω
X2(ej ω)
−2π
−π
0
3
ω
To perform the frequency-division multiplexing, a system with the following structure is proposed,
x1[n]
z1[n]
Insert
Zeros
Lowpass
Filter
y[n]
x2[n]
z2[n]
Insert
Zeros
Highpass
Filter
where the lowpass and highpass lters, H1 (ej! ) and H2 (ej! ) respectively, are as shown below.
H1 (ej ω)
1
−π −π/2 0
−2π
π/2
π
2π
π
2π
ω
H2(ej ω)
1
−π −π/2 0
−2π
π/2
ω
The signals z1 [n] and z2 [n] are obtained by inserting zeros between successive values of x1 [n] and
x2 [n], respectively. This can be mathematically expressed as follows,
z1 [n] =
z2 [n] =
8
<
x1
:
8
<
x2
:
n
0
2
n
0
2
n even
n odd
n even
n odd
(a) Sketch the DTFTs of z1 [n], z2 [n], and y[n].
(b) Suppose that y[n] is passed through another lowpass lter whose frequency response is H1 (ej! )
given above. This is illustrated in the gure below. It is claimed that x1 [n] can be recovered
from the lter output w[n]. Show that the claim is valid, and describe how x1 [n] can be
recovered.
y[n]
H1 (ej ω)
4
w[n]
Problem #3 The purpose of this problem is to test your understanding of continuous-time modulation.
xm(t)
y(t)
Nonlinear
No Memory
Filter
H(j ω )
z(t)
w(t)
cos(ωc t)
In the modulator shown above, the modulating signal xm (t) and a sinusoid at the intended carrier
frequency are added to produce y(t) = xm (t) + cos(!ct), which is then passed through a non-linear
device to yield
z (t) = 5y(t) + y2(t):
(a) Assume xm (t) is a real, even function having the spectrum shown below, where W !c.
Xm(j ω )
1
-W
W
ω
Make a carefully labeled sketch of Z (j!) over the range ,3!c < ! < 3!c.
(b) Describe the frequency response H (j!) of the lter such that w(t) has the form of xm (t)
double-sideband amplitude-modulated (with carrier) on a carrier at !c.
Problem #4 The purpose of this problem is to test your understanding of continuous-time mod-
ulation.
We would like to transmit the signal x(t) with the Fourier transform depicted on the left side of the
gure below. Unfortunately, the only available communications channels have limited bandwidth.
Speci cally, each such channel can be viewed as an LTI system with frequency response H (j!)
depicted on the right side of the gure below.
-2W
-W
X(j ω )
H(j ω )
1
1
W
2W
ω
-W
W
ω
Fortunately, we have two such channels at our disposal, and thus, it is possible to design systems
S1 and S2 , depicted below, so that z (t) = x(t).
H(j ω )
x(t)
S1
S2
H(j ω )
5
z(t)
Both S1 and S2 can be constructed using:
(1) signal generators that can produce signals of the form cos(!0 t) at any xed frequency !0 ;
(2) multipliers and adders;
(3) ideal lters.
Specify the designs of S1 and S2 .
Problem #5 The purpose of this problem is to test your understanding of the Laplace transform.
An LTI system with system function H (s) has input x(t) and output y(t). It is known that:
When x(t) = e,t u(t), then
,
y(t) = K e,3t u(t) + etu(,t) ;
where K is a constant that you will need to determine to solve the problem.
When x(t) = 1 for all t, then y(t) = 83 for all t.
Find H (s) including its region of convergence (ROC).
Problem #6 The purpose of this problem is to test your understanding of z-transforms.
Determine the DT signal x[n] given that the z -transform is
,1
X (z) = 1 + 13z+,13z+ 2z ,2 ;
for 1 < jz j < 2.
Problem #7 The purpose of this problem is to test your understanding of discrete-time system
functions.
The system function of a DT LTI system is
4
H (z) = z4 z, a4 ;
where a is real and positive. It is known that
1
X
n=,1
jh[n]j < 1
and that the unit-sample response of the system h[n] = 0 for all n < N for some value of N .
(a) Sketch the pole/zero diagram of H (z ).
(b) Determine the region of convergence of H (z ) consistent with the information given.
(c) Determine the range of a that is consistent with the information given.
6
Problem #8 The purpose of this problem is to test your understanding of pole-zero diagrams
and Bode plots for continuous-time systems.
For each pole-zero plot shown in Figure 1 below, nd the frequency response, among those given
in Figure 2, that could result from the pole-zero plot. The correct answers are among those given,
and the same answer does not apply to more than one pole-zero plot.
jω
jω
jω
A
B
σ
C
σ
jω
σ
jω
D
E
σ
σ
Figure 1: Pole-zero diagrams for H (s).
40
15
1
20
10
2
10
3
5
−20
Magnitude (dB)
Magnitude (dB)
Magnitude (dB)
0
5
0
0
−5
−40
−5
−60
−80 −2
10
−1
10
0
10
ω (rad/sec)
10
1
2
−10 −2
10
3
10
10
40
20
30
10
4
20
−10
−1
10
0
10
ω (rad/sec)
10
1
2
−15 −2
10
3
10
10
−1
10
10
0
1
ω (rad/sec)
10
2
3
10
10
50
40
5
0
6
30
0
−10
Magnitude (dB)
−10
Magnitude (dB)
Magnitude (dB)
10
−20
−30
−20
−40
−30
−50
−40
−60
−50 −2
10
−70 −2
10
20
10
0
−1
10
0
10
ω (rad/sec)
10
1
2
10
3
10
−10
−1
10
0
10
ω (rad/sec)
10
1
2
10
3
10
−20 −2
10
−1
10
10
0
1
ω (rad/sec)
Figure 2: Bode plot magnitudes (i.e. 20 log 10 jH (j!)j versus ! on a log scale).
7
10
2
10
3
10
Problem #9 The purpose of this problem is to test your ability to determine the response of an
LTI system to an input.
(a) The input of a CT LTI system is x(t) = e,t for all t, and the impulse response of the system
is h(t) = (t) , 2e,2t u(t). Determine the output y(t).
(b) The input of a CT LTI system is
t) ;
x(t) = sin(4
4t
and the impulse response is
2
sin(2
(
t
,
2))
:
h(t) =
2(t , 2)
Determine the output y(t).
(c) The input to a DT LTI system is x[n] = 2n u[,n], and the unit sample response is h[n] = 0:5n u[n].
Determine the output y[n].
(d) The input to the feedback system shown below, is a step, i.e. x(t) = u(t). Determine the
steady-state value of the output, i.e. tlim
y(t).
!1
x(t)
3
s
y(t)
100
(s+2)(s+10)
(e) The input x(t) and the unit impulse response h(t) = et+1 u(,t , 1) of a CT LTI system are
shown below. Determine the output y(t) for t = 1.
x(t)
h(t)
1
1
1
3
t
-1
t
Problem #10 The purpose of this problem is to test your understanding of discrete-time systems.
A causal DT LTI system has input sequence x[n] = (1=2)n u[n], output sequence y[n], and system
function H (z ), where
1 , z ,1
H (z ) =
:
1 , 2z ,1 , 41 z ,2 + 12 z ,3
Determine the value of y[1]. [HINT: the solution to this problem using partial fraction expansion
is more time-consuming than using other methods.]
8
Problem #11 The purpose of this problem is to test your understanding of continuous-time
feedback.
A causal CT LTI feedback system is shown below along with a root-locus diagram as K varies from
0 to +1. The arrowheads on the locus indicate the direction of increasing values of K . The \2"
near the pole indicates that this is a second-order pole.
2
1.5
X(s)
1
Y(s)
Imaginary Axis
K
(s-1) 2
s+1
0.5
2
0
−0.5
−1
−1.5
−2
−3.5
−3
−2.5
−2
−1.5
−1 −0.5
Real Axis
0
0.5
1
1.5
(a) Find the closed-loop system function H (s) = Y (s)=X (s).
(b) Find the range of values of K > 0 for which the system is BIBO stable.
(c) For some value of K , the impulse response of the system has the form h(t) = Ate, t u(t).
Find the corresponding values of K , , and A.
Problem #12 The purpose of this problem is to test your understanding of discrete-time feedback.
Consider the causal discrete-time feedback system shown below.
x[n]
z
1+ 21
y[n]
1
z
1
z
K
1
1
2
1
z
1
(a) Determine the overall system function H (z ).
(b) Sketch the locus of the poles of H (z ) for positive values of K and determine the range of
positive values of K for which the overall system is stable.
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10
NAME:
Spring 1999 Final Examination
11
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering
and Computer Science
6.003 Signals and Systems | Spring 1999
Final Examination | May 18, 1999
Please make sure your name is on all00 sheets. DO IT NOW!
This is a closed book exam, but three 8 12 1100 sheets of notes (both sides) are allowed.
Calculators may not be used.
Enter all your work and your answers directly in the spaces provided in this exam
booklet.
All sketches must be adequately labeled. Answers must be derived or explained, not just
simply written down. A correct answer does not guarantee full credit, and a wrong answer
does not guarantee loss of credit. You should clearly but concisely indicate your reasoning and
show all relevant work. Your grade on each problem will be based on our best assessment
of your level of understanding as re ected by what you have written in this exam booklet.
Transform tables are included at the end of this quiz.
Check your section Section
1
2
3
4
5
6
7
8
Time
10-11
11-12
12-1
1-2
12-1
10-11
11-12
1-2
Room
36-839
36-839
34-304
36-153
36-839
36-144
34-302
36-144
Rec. Instr.
Weiss
Weiss
Massaquoi
Massaquoi
Freeman
Smith
Smith
Freeman
TA
Bharti
Bharti
Poyneer
Poyneer
Carr
Fan
Fan
Carr
Please leave the rest of the page blank for use of the graders:
Problem 1
Problem 6
Problem 2
Problem 7
Problem 3
Problem 8
Problem 4
Problem 9
Problem 5
GRADE
12
Spring 1999 Final Examination
Problem 1 (12 points).
Circle either True or False and brie y justify your answer. If false, showing a counterexample is
sucient (but not necessary) justi cation.
1a. True or False : Every continuous-time signal that has a Laplace transform also has a Fourier
transform.
1b. True or False : Every continuous-time, right-sided signal has a Laplace transform.
1c. True or False : If the Fourier transform of a signal contains just impulses, then the signal
must be periodic.
Spring 1999 Final Examination
NAME:
13
1d. True or False : The system function H (s) and its associated ROC determine the output of a
continuous time, linear, time-invariant system for all possible input signals.
1e. True or False : A discrete-time system whose output y[n] is equal to the even part of its
input x[n] is linear and time-invariant.
1f. True or False : If a signal is purely imaginary, then the real part of its Fourier transform is
odd.
14
Spring 1999 Final Examination
Problem 2 (11 points).
Two discrete-time, linear, time-invariant systems have unit sample responses g[n] and g[,n] and
are connected in cascade as follows:
1
x[n]
g[n]
g[−n]
g[n]
y[n]
1
−4 −3
−2
−1
−½
The signal g[n] is equal to zero except at n = 0 and n = 1.
Part 2a. For the cascaded system, determine the system function
Y (z )
H (z ) = X
(z )
including its region of convergence, and plot a pole-zero diagram.
H (z ) =
ROC:
Pole-zero diagram:
2
3
4
5
n
Spring 1999 Final Examination
NAME:
Part 2b. Find an input x[n] where x[n] = 0 for n < 0, such that the output y[n] = [n].
15
16
Spring 1999 Final Examination
Problem 3 (10 points).
The Fourier transforms of three signals shown below are related by
X3 (j!) =
Z
1
,1
X1 (j )X2 (j (! , ))d :
X1(jω)
X2(jω)
2
−2
X3(jω)
a
2
ω
2
−2
2
b
ω
Part 3a. Determine the values of a, b, and c.
a=
b=
c=
−c
− ½c
½c
c
ω
Spring 1999 Final Examination
NAME:
Part 3b. Determine an expression for x3(t).
17
18
Spring 1999 Final Examination
Problem 4 (10 points).
A sampling system consists of three major components:
C/D: a continuous to discrete converter that samples its input with a periodic impulse train
with period T and maps the resulting impulses to samples,
D/C: a discrete to continuous converter that implements ideal band-limited reconstruction
for a signal sampled with sampling period T , and
h[n]: a linear, time-invariant, discrete-time lter that is an ideal lowpass lter, whose parameters you are free to choose.
x(t)
y(t)
C/D
h[n]
D/C
T
T
The input x(t) has a real-valued Fourier transform X (j!) shown below.
X(jω)
a
½a
−9c
−8c
½a
−7c
−c
7c
c
8c
9c
ω
Is it possible to adjust the sampling period T so that the Fourier transform Y (j!) of the output
y(t) has the following shape?
Y(jω)
a
½a
−3c
−2c
yes:
−c
½a
c
no:
2c
3c
ω
NAME:
Spring 1999 Final Examination
If yes, determine the sampling period T as a function of the parameters of X (j!).
T=
If no, explain why not.
19
20
Spring 1999 Final Examination
Problem 5 (15 points).
Consider the following continuous-time LTI system:
LTI
system
H(jω)
x(t)
y(t) = cos(2π.100.t)
For exactly one of the following input waveforms, there is a stable frequency response H (j!) such
that the output y(t) = cos(2100t). Identify that waveform. For each waveform that could not
give rise to y(t), brie y explain why it could not. Waveforms 1-4 are periodic; waveform 5 is not.
x1(t)
−0.01
Could give rise to y(t)?
YES
NO
0.01
0.02
t(s)
x2(t)
cosine
−0.01
0.01
0.02
t(s)
If NO, briefly
explain why not.
NAME:
Spring 1999 Final Examination
x3(t)
−0.01
Could give rise to y(t)?
YES
NO
0.01
0.02
t(s)
x4(t)
semi-circle
− 0.005
0.005
t(s)
x5(t)
0.03
t(s)
21
If NO, briefly
explain why not.
22
Spring 1999 Final Examination
Problem 6 (10 points).
The rst 26 values of the unit sample response h[n] and the z -transform H (z ) of h[n] for a discretetime, causal, linear, time-invariant system are given below.
2
h[n]
1
H(z) =
z−1
1 − 1.6 z−1 + 0.89 z−2
n
0
0
25
Part 6a. Determine a di erence equation that relates the input x[n] and output y[n] of the system
described above.
Spring 1999 Final Examination
NAME:
23
Part 6b. Is the system described above stable in the bounded-input bounded-output sense?
Explain.
24
Spring 1999 Final Examination
Problem 7 (10 points).
The Laplace transform of the impulse response g(t) of a linear, time-invariant, continuous time
system is given by
G(s) = s(s 1, 2) :
(1)
Determine which of the following time waveforms could represent g(t). The thin horizontal line in
each panel represents the time axis and marks where the ordinate is zero. The thin vertical line
indicates the time t = 0. All signals are piecewise combinations of constants and exponentials of
the form 12 e2t and 21 .
Is Equation 1 the
Laplace transform
of this time function?
YES NO
1
−
2
(i)
t
1
−−
2
(ii)
t
Indicate the ROC
for Laplace transform
of this time function.
NAME:
Spring 1999 Final Examination
Is Equation 1 the
Laplace transform
of this time function?
YES NO
(iii)
t
1
−−
2
1
−
2
(iv)
t
(v)
t
1
−−
2
25
Indicate the ROC
for Laplace transform
of this time function.
26
Spring 1999 Final Examination
Problem 8 (10 points).
Two linear, time-invariant, causal systems are represented by boxes labeled with their system
functions and are connected in a feedback loop, as follows.
X(s) +
+
−
1
s(s−2)
Y(s)
Ks+1
Part 8a. For what range of values of K is the feedback system stable? Explain.
Range of K:
NAME:
Spring 1999 Final Examination
27
Part 8b. Provide a labeled sketch of the magnitude and angle of the frequency response
Y (j!)
H (j!) = X
(j!)
on the axes below for the case K = 2:1. Clearly label the axes and indicate the scales. The scale
for the vertical axis should be in dB for the magnitude plot and radians for the angle plot.
0.1
1
10
ω
0.1
1
10
ω
28
Spring 1999 Final Examination
Problem 9 (12 points).
The following transmission system is intended to allow a signal x(t) to be transmitted through a
\channel" that also carries other signals represented by z (t).
x(t)
HLP(jω)
Channel
p(t)
q(t)
+
r(t)
y(t)
HLP(jω)
C
z(t)
−ωf
ωf
0
cos ωct
Both x(t) and z (t) are bandlimited, and their Fourier transforms X (j!) and Z (j!) are real, as
sketched below. Notice that the bandwidth !z of Z (j!) is much greater than the bandwidth !x of
X (j!).
X(jω)
Z(jω)
Α
−ωx
0
ωx
Β
ω
−ωz
0
ωz
ω
Part 9a. We wish to determine parameters for the transmission system so that the output y(t) is
equal to the input x(t). For this part of the problem, determine the range of values of !c for which
y(t) can be made equal to x(t). Explain.
Range of !c :
ω
Spring 1999 Final Examination
NAME:
29
Part 9b. Given a value of !c in the range speci ed in Part 9a, determine the range of values of
!f and the value of C for which y(t) = x(t). Your expressions may contain !c and/or parameters
of the Fourier transforms X (j!) and Z (j!). Brie y explain your reasoning.
Range of !f :
C=
30
Spring 1999 Final Examination
Part 9c. Consider next what would happen if the \channel" also had appreciable delay, as in the
following diagram.
x(t)
p(t)
Channel
+
Delay
by T
q(t)
r(t)
HLP(jω)
y(t)
HLP(jω)
C
z(t)
−ωf
cos ωct
Find an expression for the frequency response
Y (j!)
H (j!) = X
(j!)
assuming that the parameters are chosen as in Parts 9a and 9b.
0
ωf
ω
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003: Signals and Systems|Fall 2000
Solutions to Spring 1999 Final Exam
Problem 1
1a. False
Counter example: x(t) = e2t u(t).
x(t) has a Laplace Transform X (s) = s,1 2 for > 2; but it does not have a Fourier transform.
A common and wrong answer for this problem was: \If the j!-axis is not contained in the
ROC of X (s), X (j!) does not exist." This is a false statement! What about u(t)?
1b. False
CounterZexample:
x(t) = et2 u(t). Plugging x(t) into the Laplace transform equation gives
1 2
X (s) =
et e,t dt: This cannot converge for any because the et2 grows too fast for any
,1
e,t to compensate for it.
1c. False
Counter example 1: x(t) = cos t + cos 2t
x(t) is not periodic because there is no common frequency between them. The Fourier transform of x(t), however, is X (j!) = [(! + 1) + (! , 1) + (! + 2) + (! , 2)]. Thus, we
have found a signal that is not periodic but whose Fourier transform contains only impulses.
Counter example 2: x[n] = cos n
x[n] is not periodic since n is restricted to be an integrer. The Fourier transform of this signal
1
X
j!
is an impulse train X (e ) = (! , 1 + 2l) + (! + 1 + 2l). Once again, we have
l=,1
found a signal that is not periodic but whose Fourier transform contains only impulses.
A common and wrong answer for this problem was: \x(t) = 1 in not periodic but X (j!) =
2(!) is just an impulse." The fallacy in this logic is that x(t) = 1 is periodic.
1d. True
Given H (s) and its ROC, h(t) is uniquely determined. Furthermore, Given an LTI system
and its impulse response h(t), the response of the system can be determined for any input
using convolution.
1e. False
Counter example: If the input into the system is x[n] = [n], then the output is y[n] = [n].
The output of the system when x[n] = [n , 1] is y[n] = 12 [n , 1] + 21 [n + 1]. Thus, since
the output of the system to a delayed input is not the output delayed, the system is not time
invariant.
31
1f. True
Given that x(t) is real, then X (j!) is conjegate symmetric, so the RefX (j!)g is even and
the I mfX (j!)g is odd. Therefore, if y(t) = jx(t), it is purely imaginary and RefY (j!)g =
I mfX (j!)g, so the real part is odd.
Problem 2
a. By inspection of g[n] we can nd its z -transform, which is G(z ) = 1 , 21 ; for jz j > 0:
Furthermore, from the z -transform tables we know that g[,n] Z! G(z ,1 ). Therefore, if we
say call the impulse response of the complete system h[n], we know that h[n] = g[n] g[,n].
Therefore, we can nd H (z ) as follows:
H (z ) = G
(z )G(z ,1) = 1 , 12 z ,1 1 , 12 z
= 54 , 21 z ,1 , 12 z
"
#
z 2 , 52 z + 1
1
= ,2
z
"
#
(z , 2)(z , 21 )
1
= ,2
z
"
(z , 2)(z , 12 )
H (z ) = , 21
z
#
ROC: 0 < jz j < 1
6I mfzg
1
Pole-Zero Diagram:
h
@,
-1
-1
h
1
2
ROC: 0 < jz j < 1
b. Using the fact that y[n] = [n] Z! Y (z ) = 1, we can nd X (z ) as
Y (z )
X (z ) = H
(z )
,2z
=
(z , 2)(z , 12 )
,2z,1
=
(1 , 2z ,1 )(1 , 21 z ,1 )
= 1 ,A2z ,1 + B1 ,1
1 , 2z
32
-
Refzg
where
,2z,1
= , 43
1
,
1
1 , 2 z z,1 = 1
2
,1
,
2
z
B = 1 , 2z,1
= 34
z,1 =2
4
4
X (z ) = 1 ,,23z,1 + 31 ,1
1 , 2z
A =
Since we want x[n] = 0 for n < 0, we must choose the ROC of X(z) to be jz j > 2. Therefore,
n
4
4
n
x[n] = , 3 2 u[n] + 3 21 u[n]
Problem 3
a. X3 (j!) will be the area under the product of X1 (j ) and X2 (j (! , )).
c: The value of \c" can be found by determining the value for ! where the two functions
X1 (j ) and X2 (j (! , )) begin to overlap. This rst occurs when ! + 2 = ,2, which
means c = 4.
a: The value of \a" is X3 (j!)j!=0 , which corresponds to the area under X1 (j ) X2 (,j ).
This product looks like the triangle of X1 (j!) scaled by a factor of 2. Therefore, the
area will be a = 8.
b: The value of \b" is X3 (j!)j!=2 . This is just the area under X1 (j ) X2 (j (2 , )), which
looks like:
4
6
SS
,! b = 4.
S2
b. First we realize that X1 (jw) can be decomposed into a convolution of X0 (j!) with itself where
X1 (j!) = X0 (j!) * X0 (j!)
1
;
j
!
j
<
1
2
X0 = 0; otherwise
,@
1*
, @ =
-2
2 !
Using the Tables, we can nd the following relations:
x3 (t) = 2x1 (t)x2 (t)
Z!
-1 1 ! -1 1 !
X3 (j!) = X1 (j!) X2 (j!)
x1 (t) = 2x20 (t) Z! X1 (j!) = X0 (j!) X0 (j!)
t
x0 (t) = sin
t
t)
x2 (t) = 2 sin(2
t
2 t) sin(2t)
Therefore, we can nd x3 (t) to be x3 (t) = 8 sin (t
.
3
33
Problem 4 The C/D converter performs the following operations:
C/D Converter
x(t)
- l xp (t) 6
s(t)
=
X1
k=,1
Impulse to
Sequence
Converter
(t
-
xd [n]
, kT )
The multiplacation by the impulse train that yields s(t) causes the X (j!) (scaled by T1 ) to replicate
every 2T frequency units. The conversion from an impulse train to a sequence scales the frequency
axis by T . The lter h[n] will then reject high frequencies. The conversion back to a continuous
time lter will rescale the frequency by T1 and the amplitude by T and remove all the replicas.
Therefore, if it is possible to generate Y (j!), it must occur in the C/D conversion since Y (j!)
is going to have the same shape as Xp (j!). If we pick the period to be T = 3c (or !s = 6c), then
Xp (j!) will be:
Xp (j!) 6Ta
BB
BBB
BBB
B
B
BB2aT
TT B TT TT B TT TT B B T T B BB TT TT
B T T
B TT
!
? 2c
-6c
-4c
-2c
4c
6c
If h[n] is picked so that its cut-o frequency is c = , then it will pass all frequencies. Then when
the signal is converted back to continuous time, the replicas will be removed and scaled correctly
so that the desired Y (j!) is obtained.
It is also possible to pick T = 5c (or !s = 10c). Then if you pick the cut-o frequency of the
low-pass lter h[n] to be c > 35 , then you can also obtain Y (j!) on the output.
So the answer is YES with T = 3c or 5c .
Problem 5
1. NO ,! The period of x1 (t) is T1 = 0:02, so the fundamental frequency is !0 = 2T = 2 50.
According to Table 5.2, a square wave has a spectrum with impulses at the the odd harmonics
of !0 . Therefore, it has no energy at 2!0 = 2 100. Since the input into the LTI system has
no energy at 2 100, the output cannot have energy at 2 100, so x1 (t) cannot give rise to
y(t).
2. NO ,! The period of x2 (t) is T2 = 0:02, so the frequency is !0 = 2 50. Once again,
because the input does not contain any energy at 2 100, the output cannot have any energey
at 2 100, so x2 (t) cannot give rise to y(t).
3. YES
4. NO ,! The period of x4 (t) is T4 = 0:005, so the fundamental frequency is !0 = 2 200.
Since it is periodic, the signal can only have energy at integer multiples of the fundamental,
so x4 (t) does not have any energy at 2 100.
34
5. NO ,! X5 (j!) = A 2sin(0!:015!) e,j 0:015! . There are no impulses in X5 (j!), so the lter
H (j!) would need to have impulses to produce the cosine wave y(t). However, if H (j!) has
impulses, then h(t) would oscillate and be unstable.
Problem 6
a. To determine the di erence equation, we use H (z ) as follows:
z,1
Y (z ) =
H (z ) = X
(z )
1 , 1:6z ,1 + 0:89z ,2
,
1
,
2
Y (z ) 1 , 1:6z + 0:89z
= z ,1 X (z )
Taking inverse z -transform of both sides yeilds the di erence equation:
y[n] , 1:6y[n , 1] + 0:89y[n , 2] = x[n , 1]
.
b. To check stability we can check the pole locations of H (z ).
,1
H (z) = 1 , 1:6z ,z1 + 0:89z,2
= z 2 , 1:6zz + 0:89
Therefore, using the quadratic formula, the pole locations are going to be
p
2:56 , 3:56
1
:
6
z =
2
= 0p:8 0:5j
jzj = p (0:8)2 + (0:5)2
= 0:89
Since both poles have a magnitude less than 1, both poles are inside the unit circle. Furthermore, the system is causal, so the ROC is outside the radius of hte outermost pole. Therefore,
the ROC contains the unit circle, so the system is STABLE .
Problem 7 We rst nd the partial fraction expansion of G(s) to be
1
1
G(s) = s(s 1, 2) = ,s2 + s ,2 2
i. NO . This system has a Laplace transform of
1 u(,t) , 1 e,2t u(t) L! , 12 , 12
2
2
s s+2
with an ROC of ,2 < Refg < 0. The ROC is
plotted to the right, and one can see that the Laplace
transform of this signal does not match G(s).
Im{}
11111
00000
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
0
1
0
1
00000
11111
0
1
00
1
-2
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
ROC
35
Re{}
ii. YES . This system has a Laplace transform of
Im{}
1
1
, 12 u(t) + 12 e2t u(t) L! ,s2 + s ,2 2
with an ROC of Refg > 2. This matches G(s), and
the ROC is plotted to the right.
Re{}
11111
00000
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
0 11111
1
0
1
00000
00 11111
1
0
1
00000
11111
2
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
ROC
iii. NO . This system has a Laplace transform of
1
2
1
2
, 12 u(,t) + 21 e2t u(,t) L! s , s , 2
with an ROC of Refg < 0. This does not match
G(s). The ROC is plotted to the right.
11111 Im{}
00000
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
0
1
0
1
00000
11111
00
1
0
1
2
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
00000
11111
ROC
iv. NO . This system has a Laplace transform of
Im{}
1 u(t) + 1 e2t u(,t) L! 12 , 12
2
2
s s,2
with an ROC of 0 < Refg < 2. This does not match
G(s). The ROC is plotted to the right.
Re{}
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0
1
0
1
0000
1111
00
1
0
1
2
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
ROC
36
Re{}
v. NO . This system has a Laplace transform of
Im{}
1
1
, 12 e,2t u(t) , 12 e2t u(,t) L! s,+22 + s ,2 2
with an ROC of ,2 < Refg < 2. This does not
match G(s). The ROC is plotted to the right.
Re{}
111111
000000
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
0
1
0
1
000000
111111
0
0
1
-21
2
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
ROC
Problem 8
a. We can nd the transfer function from input to output by calculating it from the block
diagram.
Y (s) = s(s 1, 2) (X (s) , (Ks + 1)Y (s))
Ks
+
1
Y (s) 1 + s(s , 2) = s(Xs (,s)2)
Y (s) = s(s1,2)
+1
X (s)
1 + sKs
(s,2)
= s2 + (K 1, 2)s + 1
Therefore, if K > 2 then all coecients of the denominator will be greater than zero and the
system will be stable. So the answer is K > 2 .
b. When K = 2:1, then H (s) = s2 +01:1s+1 . This is a second order system with !n = 1 and
= 0:05. Since < 1, the poles are complex. Furthermore, for ! 1, this magnitude of the
system is 1 and phase is 0. For ! 1, the magnitude roles o as !12 (-40 dB/dec) and the
phase approaches ,. This has been plotted below.
37
20log10|H(jω)|
Magnitude (dB)
40
20
0
−20
−40
−1
10
0
1
10
∠ H(jω)
10
Phase (degrees)
0
−45
−90
−135
−180
−1
10
0
1
10
Frequency ω (rad/sec)
10
Problem 9
a. The modulation of the x(t) by cos !c t will
shift X (jw) to be centered around !c (as
shown in the gure to the right), so as long as
!z < !c , !x then there will be no overlap between Z (j!) and P (j!). Therefore, the range
is !c > !z + !x .
Q(j!)
BE
E
%%
%@B@
0
@
E
A
2
!z
!c
!c , !x
R(j!)
b. The low-pass lter should only pass the replic
cate of X (j!) centered at ! = 0. Therefore,
A
%
@
2
!x < !f < !c , !z . Furthermore, from the
% @@
BEE !f
%
graph to the right you can see that if c = 2
0
then y(t) = x(t).
c. The delay gives us:
P (j!) = 12 [X (j (! , !c)) + X (j (! + !c))]
Q(j!) = e,j!T (P (j!) + Z (j!))
R(j!) = 21 Q(j!) [(! , !c) + (! + !c)]
= 12 Q(j (! , !c)) + 21 Q(j (! + !c ))
= 12 e,j (!,!c )T [P (j (! , !c)) + Z (j (! , !c))] +
38
BE
!
B
%@@
%
@
%
2
!c
!
+ 12 e,j (!+!c )T [P (j (! + !c)) + Z (j (! + !c))]
The lowpass lter HLP (j!) then passes only the image X (j!) centered at ! = 0, so
Y (j!) = 2c 12 X (j!)e,j(!,!c )T + 12 X (j!)e,j(!+!c )T
j!c T
,j!cT c
e
+
e
,
j!T
= 2 X (j!)e
2
c
= 2 X (j!)e,j!T cos !cT
Therefore, H (j!) = 2c e,j!T cos !cT
39