304-303B Solutions to Problem Assignment 4
Problem 1: Find the Fourier transforms of the following two signals.
(a)
(b)
x1 ( t )
x2 ( t )
1
-2
1
2
-1
t
t
-1
-1
(a) The signal x1 ( t ) is a combination of two time-shifted pulses.
Let
z (t ) 
1, t  0.5
R
, a unit pulse centered at t  0 . We know that its Fourier transform is
S
T0, otherwise
a sinc function Z ( j )  sinc(

) . The signal x1 ( t )
2
can be written as
x1 (t )  z(t  0.5)  z(t  15
. ).
Using the time-shifting property of the Fourier transform, we obtain
X 1 ( j )  e  j 0.5 Z ( j )  e  j 1.5 Z ( j )


)  e  j1.5sinc( )
2
2

 e  j [ e j 0.5  e  j 0.5 ]sinc( )
2

 je  j sinc 2 ( )
2
 e  j 0.5sinc(
(b) The signal x2 (t ) is the integral of x1 (t ) after a time-shift to the left.
z
t
x2 (t ) 
x1 (  2)d

Thus,
1
Prof. Benoit Boulet
February 20, 1999
1
X 1 ( j )e j 2   X 1 ( j )e j 2
 0
j
1

X 1 ( j )e j 2
j
X 2 ( j )  

e j 2

je  j sinc 2 ( )
j
2
  e j sinc 2 (
 sinc 2 (

)
2
 j (   )
)e
2
Problem 2: Find the inverse Fourier transform
shown below.
x (t ) of X ( j ) whose magnitude and phase are
X ( j )
X ( j )

W
-W
-W



W
W
Y ( j ) be the Fourier transform of a rectangular window of unit magnitude and zero phase
(i.e., it is real) from -W to W. Then Fourier transform of x ( t ) is
X ( j )  jY ( j ) .
Using the differentiation property, the signal x ( t ) is given by
dy (t )
x (t ) 
dt
W
d sinc( W t )
 
dt
W d sin(Wt )

 dt Wt
W W 2 t cos(Wt )  Wsin(Wt )


W 2t 2
Let
L
M
N

O
P
Q
W cos(Wt ) sin(Wt )

t
t 2
2
Prof. Benoit Boulet
February 20, 1999
Problem 3: Consider the followin system.
H ( j )
x(t )  u(t )
1
10
y (t )
j
1
j
10
t
(a) Calculate and sketch the magnitude of
H ( j ) . If this system is a filter, what type is it?
H ( j ) 

  100
2
This is a highpass filter. The gain at high frequencies is
lim H ( j )  lim
 
 

  100
2
 1,
and the dc gain is 0:
H ( j 0) 
0
0  100
 0.
SKETCH of magnitude:
1.2
H ( j )
1
0.8
0.6
0.4
0.2
0
-300
-200
-100
0
3
100
200

300
Prof. Benoit Boulet
February 20, 1999
(b) Find the output signal
y (t ) using the Fourier transform, and sketch it.
Y ( j ) 
F
G
H
j
1
  ( )
j  10 j
IJ
K

1
j

 ( )
j  10 j  10   0

1
j  10
Using the table of Fourier transform pairs, we obtain
y(t )  e 10t u(t ) .
y (t )
1
0.1
t
Problem 4: Consider a mechanical system consisting of a mass attached to a spring of stiffness
k and a viscous damper (dashpot) of damping factor b , both rigidly connected to ground. This
basic system model is quite useful to study a number of systems, including a car's suspension, or
a flexible robot link.
x(t ) (force input)
force applied on mass by spring and gravity:
y(t ) (mass position wrt rest)
m
force applied on mass by damper:
Fb (t )  by(t )
0
k
Fk (t )  ky (t )
b
Assume that the mass-spring-damper system is initially at rest, which means that the spring
generates a force equal to the force of gravity to support the mass. The balance of forces on the
mass causing motion is
x (t )  Fk (t )  Fb (t )  my(t )
(a) Write the differential equation governing the motion of the mass.
4
Prof. Benoit Boulet
February 20, 1999
my(t )  by(t )  ky (t )  x (t )
(b) Find the frequency response of the system. Express it in the form
H ( j ) 
A 2n
.
( j ) 2  2 n ( j )   2n
1
m( j )  bj  k
(1 k )( k m)

b
k
( j ) 2  j 
m
m
H ( j ) 
2
Note that  (zeta) is called the damping ratio of the system, and  n is called the undamped
natural frequency of the system. This is the frequency that the mass would oscillate at would
there be no viscous damper attached to it. What is the damping ratio  for this mechanical
n ?
system? What is its undamped natural frequency
The natural frequency is given by
 2n 
k
k
 n 
m
m
Note: the larger the mass, the lower the undamped natural frequency; the stiffer the spring, the
higher the undamped natural frequency. The damping ratio of the system is then

b
m
2 n

b
m
2
k
m

b
.
2 mk
For a given dashpot, the larger the mass and/or spring, the less damped the system will be.
(c) Let the physical constants have numerical values m  1 kg , k  10 N m , and b  3 N
m
s
.
Suppose that the applied force is sinusoidal and has the form x (t )  sin(10t ) N and the system
is in steady-state. What is the mass position at time t ?
With the numerical values given, the damping ratio and undamped natural frequencies are:
kgm
2
10 ms
k
n 

 10 rds
m
1 kg
3 mN/ s
b


2 mk 2 10 kg
N
m

/s
3 kgm
m/ s
2
2
2 10 kg mm/ s
2
 0.4743
and the frequency response is
1
.
( j )  3 j  10
1 j10t
Write the sinusoidal input force as x (t )  sin(10t ) 
(e  e  j10t ) . Recall that complex
2j
H ( j ) 
2
exponentials are eigenfunctions of LTI systems and thus
5
Prof. Benoit Boulet
February 20, 1999
y (t ) 

1 j10t
[e H ( j10)  e  j10t H (  j10)]
2j
H ( j10) j10t jH ( j10)
[e e
 e  j10t e  jH ( j10) ]
2j
 H ( j10) sin(10t  H ( j10))
where we have used the properties that the magnitude of the frequency response of a real
system (real coefficients in the differential equation) is even and its phase is odd.
The magnitude and phase at 10 rd/s is
1
( j10)  30 j  10
1 30

3  j
H ( j10) 

2
1
j arctan
e
1 I
F
G
H3 JK
30 10
1

e  j 2.8198
30 10
Hence,
1
sin(10t  2.8198) m
30 10
 0.01054 sin(10t  2.8198) m
y (t ) 
We can think of the sinusoidal input force as a vibration force that produces a sinusoidal
displacement of the mass of amplitude 1cm.
(d) With the numerical values given in (c), what is the position response of the mass
Newton step input? Sketch it.
y (t ) to a 1-
The Fourier transform of the step response is given by
Y ( j ) 
F
I
1
  ( )J
G
Hj
K( j )
2
1
 3 j  10
1
1

  ( )
2
j ( j )  3 j  10 10
A
B
C 



  ( )
31
31
j 10
j  15
. j 2
j  15
. j 2

the coefficients are given by:
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Prof. Benoit Boulet
February 20, 1999
A


1
1
s s  15
. j
31
2 s 1.5 j
31
2
1
( 15
. j
(j
31
2
)(  j 31)
1
31  312 )
3
2
 0.05  j 0.026941
B


1
1
s s  15
. j
31
2 s 1.5 j
31
2
1
( 15
. j
(  j 23
31
2
)( j 31)
1
31  312 )
 0.05  j 0.026941
C
1
1

s  3s  10 s0 10
2
Thus,
Y ( j ) 
0.05  j 0.026941 0.05  j 0.026941
1



  ( ) .
31
31
10 j 10
j  15
. j 2
j  15
. j 2
Taking the inverse Laplace transform of the partial fraction, we get
( 1.5 j
31
)t
2
 ( 0.05  j 0.026941)e
31
t  2 .6474 )
2
e
j(
y (t )  ( 0.05  j 0.026941)e
 0.056796e 1.5t e
 j(
 011359
.
e 1.5t cos(
31
2
31
t  2 .6474 )
2
t  2.6474)u(t ) 
u( t ) 
( 1.5 j
31
)t
2
u( t ) 
1
u( t )
10
1
u( t )
10
1
u( t ) m
10
This step response is plotted below.
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
5
10
7
15
20
25
Prof. Benoit Boulet
February 20, 1999
Problem 5: Consider the sampling/modulation system whose block diagram is given below. This
system is used to transmit over a radio link a signal x ( t ) whose spectrum is shown below.
X ( j ) 1
 M
x (t )
M
Hbp ( j )
x
y (t )
receiver

receiving
antenna
emitting
antenna

p( t ) 
y (t )
  (t  kT )
k 
Let the sampling frequency be
s 
given by:
Hbp ( j ) 
2
. The ideal bandpass filter has a frequency response
T
1, 10      10
R
S
T0, otherwise
s
1
s
 1
(a) According to the sampling theorem, for what range of sampling frequencies
 s , and
 1 can this system transmit a signal y (t ) containing all the
information needed to recover x ( t ) undistorted, while minimizing the bandwidth of the filter?
bandpass filter frequency
The sampling operation produces an infinite number of replicas at frequencies k s of the
original spectrum
X ( j ) , scaled by 1 T .
X P ( j ) 
1 X ( j )
 M
M
1
1/2
*
 s

2
P( j ) 
T
1 
 X ( j (  k s ))
T k 
s  M

  (  k
s
)

s
s  M
2
k 
2 s  s
 s 2 s 
No information will be lost if the sampling frequency
select the bandpass filter frequency as
1   M
s
is greater than 2 M . In this case, if we
we will have minimized the bandwidth
(passband) of the filter while keeping two complete replicas of
illustrated in the figure below for positive frequencies.
8
X ( j ) at   10 s . This is
Prof. Benoit Boulet
February 20, 1999
Hbp ( j )
1
s
10 s
10 s   M
(b) Suggest a general design (block diagram) for the receiver to recover

10 s   M
x (t ) .
Because other radio signals at other frequencies will also be received by the antenna, we need a
bandpass filter at the receiver too. Then we need to demodulate the signal, i.e., bring it down to
frequencies centered around zero. We can do this by sampling at the same rate  s or by
multiplication by a complex exponential (or sinusoid) signal of frequency 10 s . There are of
course many different ways to demodulate the signal. In the sampling case, two replicas will be
superimposed around every integer multiple of the sampling frequency. Then we lowpass filter
the resulting signal to get rid of spectral replicas at higher frequencies.
x (t )
Hbp ( j )
receiving
antenna
x
T2
2
X ( j )
Hlp ( j )
 M
p( t ) 
1
M


  (t  kT )
k 
9
Prof. Benoit Boulet
February 20, 1999