1
2
3
4
5
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
0.4pt0pt
6.003,Signals and Systems Fall 2003
Quiz 1
Tuesday,October 14,2003
Directions,The exam consists of 5 problems on pages 2 to 19 and work space on pages 20 and
21,Please make sure you have all the pages,Tables of Fourier series properties
are supplied to you at the end of this booklet,Enter all your work and your
answers directly in the spaces provided on the printed pages of this booklet,
Please make sure your name is on all sheets,DO IT NOW!,All sketches must
be adequately labeled,Unless indicated otherwise,answers must be derived or
explained,not just simply written down,This examination is closed book,but
students may use one 8 1/2 × 11 sheet of paper for reference,Calculators may not
be used,
NAME,SOLUTIONS
Check your section Section Time Rec,Instr,
1 10-11 Prof,Zue
2 11-12 Prof,Zue
3 1- 2 Prof,Gray
4 11-12 Dr,Rohrs
5 12- 1 Prof,Voldman
6 12- 1 Prof,Gray
7 10-11 Dr,Rohrs
8 11-12 Prof,Voldman
Please leave the rest of this page blank for use by the graders:
Grader
18
20
20
21
21
100
Problem No,of points Score
Total
PROBLEM 1 (18%)
For the questions in this problem,no explanation is necessary,
Consider the following three systems,
SYSTEM A,y(t) = x(t + 2) sin(?t + 2),where? = 0 →
1
n
SYSTEM B,y[n] = (x[n] + 1)?
2
n
SYSTEM C,y[n] =
x
2
[k + 1]? x[k]
k=1
where x and y are the input and output of each system.
Circle YES or NO for each of the following questions for each of these three systems.
SYSTEM A SYSTEM B SYSTEM C
Is the system linear?
Is the system time invariant?
Is the system causal?
Is the system stable?
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
2
Fall 2003,Quiz 1 NAME:
Work Page for Problem 1
Linearity
System A,YES
x
1
(t)? y
1
(t) = x
1
(t + 2)sin(wt + 2)
x
2
(t)? y
2
(t) = x
2
(t + 2)sin(wt + 2)
x
3
(t) = ax
1
(t) + bx
2
(t)? y
3
(t) = x
3
(t + 2)sin(wt + 2)
= ax
1
(t + 2)sin(wt + 2) + bx
2
(t + 2)sin(wt + 2)
= ay
1
(t) + by
2
(t)
System B,NO if x[n] = 0,then y[n] = 0 →
2
System C,NO not linear due to the x [k + 1] term,
Time Invarience
System A,NO the output has time varying gain
System B,NO the output has time varying gain
System C,NO the number of summed terms depends on n
Causality
System A,NO y(t) depends on x(t + 2)
System B,YES y[n] depends only on the current value of x[n]
System C,NO y[n] depends on the future value of x[n]
Stability
System A,YES bounded x(t) bounded y(t)
1 n
System B,NO (?
2
) √ as n√
System C,NO if x[n] =? 1 y[n] =
n
k=1
2,y √ as n √
3
1?2
PROBLEM 2 (20%)
Consider a DT LTI system,H
2
with a unit sample response h
2
[n] = h[n] h[n + 1],as shown?
below,where h[n] =?[n][n? 1],You may remember from one of the lectures that h[n]
can be viewed as the unit sample response of a DT LTI system that acts as an edge detector,
The purpose of this problem is to develop an edge detector that is robust against additive
noise,
h
2
[n]
1 1
System H
2
x[n]
+
0
h
2
[n] np[n] y[n]
1 2?2?1
d[n]?2
Part a,Assume that the input to the system,p[n] is as shown below,and there is no noise,
i.e.,d[n] = 0 and p[n] = x[n],Provide a labeled sketch of y[n],the output of the system,
2 2 2 2
p[n]
2?1 0 1 2 3?3?4
n
2 3 4 5 6 7?1?3?4?5?6?7
y[n]
n
2
2?2
2
4
2?1 0 1
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
x[n] = p[n],y[n] = x[n] h
2
[n]?
Using DT convolution sum,y[n] is the sum of 3 copies of p[n] shifted and scaled according
to h
2
[n]
2?1 0 1 2 3?3?4
p[n
n
2 2 22 + 1]
+
2 3?3?4
2p[n]
n
4?4?4?4
+
2?1 0 1 2 3?3?4
p[n? 1]
n
22 2 2
The resulting y[n] is shown on page 4,
5 Problem 2 continues on the following page,
1?2
Part b,For the same input signal as Part a.,now assume that the noise signal is
d[n] =[n + 1],
Provide a labeled sketch of the output y[n],i.e.,the response to x[n] = p[n] + d[n],
2 3 4 5 6 7?1?3?4?5?6?7
y[n]
n
2 2
1
2
2
3
6
1
2 0
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
y[n] = x[n] h
2
[n]?
= (p[n] + d[n])? h
2
[n]
= (p[n] h
2
[n]) + (d[n] h
2
[n])
We have already found (p[n] h
2
[n]) in Part a,Now we need d[n] h
2
[n].
d[n]
2 0 1
n
1
1 1
n
d[n]? h
s
[n]
1
2
1
Adding the above (d[n] h
2
[n]) to the answer in Part a,we get y[n] for this part which is
shown on page 6,
7 Problem 2 continues on the following page.
1?2
Part c,In order to use system H
2
as a part of an edge detector,we would like to add an
LTI system H
s
whose unit sample response,h
s
[n] is shown below,System H
s
smoothes out
effect of noise on x[n],The overall system can be represented as below,
2
h
s
[n]
2?1 0 1
n
1 1
System H
s
System H
2
p[n] +
x[n]
h
s
[n] h
2
[n] y
s
[n]
d[n]
Provide a labeled sketch of the overall output y
s
[n],when p[n] and d[n] are exactly the same
as in Part b,
y
s
[n]
2 3 4 5 6 7?1?3?4?5?6?7
n
2
1
2?2
2 2
3
8
1?2
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
y
s
[n] = x[n] h
s
[n] h
2
[n]
= (x[n] h
2
[n])? h
s
[n]?
x[n] = p[n] + d[n] is as de?ned in Part b,Therefore,we already know (x[n] h
2
[n]) from?
Part b,
2 3 4 5 6 7?1?3?4?5?6?7
x[n]? h
2
[n]
n
2 2
1
2
2
3
Now we just need to convolve the above signal with h
s
[n],We can do this convolution with
ipping and sliding h
s
[n],
h
s
[n? k]
2
k
1 1
n? 1
n
n + 1
The result is the desired y
s
[n] and is shown on page 8,
9
PROBLEM 3 (20%)
Consider the CT LTI system whose impulse response is given as,
h(t)
x(t) h(t) y(t)
t
1
0 1?1
The following two parts can be done independently,
Part a,The input x(t),an impulse train starting at t = 2,is depicted below,
a0a1a0a1a0a2a0a1a0
x(t)
(1) (1) (1) (1)
t
0 1 2 3 4 5?1
Provide a labeled sketch of the corresponding output y(t),
y(t)
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
2
1
10
Fall 2003,Quiz 1 NAME:
Work Space for Problem 3
Using CT convolution sum (summation of shifted and scaled impulse response),we can?nd
y(t),
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
,
,
Adding the shifted and scaled pulses,we get y(t) as shown on page 10,
11 Problem 3 continues on the following page,
Part b,For this part,the output y(t) is periodic and is depicted below:
3?3?6
6 9
2?2
4 8
y(t)
t
······
2
2
Provide a labeled sketch of the input x(t) that produces this y(t),
x(t)
1 2 3 4 5 6 7 8 9 10?1?2?3?4?5?6?7?8?9?10
t
1
1
12
9
Fall 2003,Quiz 1 NAME:
Work Page for Problem 3
There are different ways to solve this problem,and here is one of them:
Let’s take the derivative of the output,
dy(t)
,and relate it to the impulse response of the
dt
system,
y˙(t)
6
6 9
2?2
4 8
t
······
2
2
It can be seen that y˙(t) is the sum of a number of scaled and shifted versions of h(t),Alter-
natively,,y˙(t) = p(t)?h(t),where p(t) is shown below,
p(t)
6
6
2?2
4 8
(2)
(?2)
(2)
t
······
2
2
We know that y(t) = x(t)?h(t)? x(t) = u
1
(t)?p(t)
t
x(t) =
p(v)dv = x(t) + c,where the constant c restores the DC level of the periodic
signal,
x(t)
1 2 3 4 5 6 7 8 9 10?1?2?3?4?5?6?7?8?9?10
t
c + 1
c? 1
To?nd the value of the constant c,we compare the value of y(0) with x(t)?h(t)
t=0
,i.e,the|
area under the curve of x(t)h(-t),
y(0) = 2 = x(t)?h(t)
t=0
= (1 + c)(2) = 2? c = 0.|
13
PROBLEM 4 (21%)
Consider the following periodic triangular wave shown below,
x(t)
1?1
1
1
t
1/4
3/4 1/4
3/4
··· ···
Part a,Determine the Fourier series coef?cients,a
k
for x(t),
0,k = 0
a
k
=
4j
sin(k
2
),k = 0
k
2
2
→
14
Fall 2003,Quiz 1 NAME:
Work Page for Problem 4
x(t)
1?1
1
1
t
1/4
3/4 1/4
3/4
··· ···
Period= T = 1?
0
= 2?/T = 2?,
1
1
a
0
= x(t)dt =
T
(Area under the curve for one period)= 0 = a
0
T T
Finding a
k=0?
using the analysis equation will be tedious,Instead,we will use the integration
property of the Fourier series,
¨x(t)
8
8
t
3/4
1/4
1/4
3/4
··· ···
Let ¨x(t)? b
k
1
1/2
¨
T
b
k
= x(t)e
jk?
0
t
dt = [8?(t + 1/4)? 8?(t? 1/4)] e
jk?
0
t
dt
T?1/2
= 8e
jk?
0
(?1/4)
8e
jk?
0
(1/4)
= 16j sin(k?
0
/4)
= 16j sin(k ),
2
1
a
k
= b
k
(integration property)
(jk?
0
)
2
16j
=
k
2
(2?)
2
sin(k ) =
k
2
4
j
2
sin(k ),
2 2
As a double check,note that a
k
= a because x(t) is real,
k
15 Problem 4 continues on the following page,
Part b,Consider a causal LTI system,S,whose input-output relation is characterized by the
following stable linear constant coef?cient differential equation,
d
2
y dy
+ 4? + 4?
2
y(t) = 4?
2
x(t),
dt
2
dt
where x(t) is the input and y(t) is the output of the system,Suppose x(t) shown on the
previous page is applied to the system S as an input,Let b
k
be the Fourier coef?cients of the
corresponding output y(t),Find b
3
and b
3
,
b
4j 4j
3
=
9?
2
(?8+j6)
b
3
=
9?
2
(8+j6)
,
16
|
Fall 2003,Quiz 1 NAME:
Work Page for Problem 4
First,we?nd H(j?) by inspection from the coef?cients of the differential equation,
d
2
y dy
+ 4? + 4?
2
y(t) = 4?
2
x(t),
dt
2
dt
4?
2
1
H(j?) = =,
(j?)
2
+ 4?j? + 4?
2
1 + j?/
2
/4?
2
b
k
= a
k
H(jk?
0
),where?
0
= 2?,therefore,
b
3
= a
k
H(jk?
0
) = a
3
H(j6?)
k=3
4j
1 4j 1
= sin 3 =
(3)
2
2
2 1 + j6?/ (6?)
2
/4?
2
9?
2
(1 + j6? 9)
4j
b
3
=,
9?
2
(?8 + 6j)
1 4j
b
3
= b
=
4j
=,
3
9?
2
8? 6j 9?
2
(8 + 6j)
17
PROBLEM 5 (21%)
You are given the following facts about a discrete time sequence x[n],
(a) x[n] is real and odd,
(b) x[n] is periodic with period N = 6,
1
(c) |x[n]
2
= 10,
N
|
n=<N >
(d) (?1)
n/3
x[n] = 6j,
n=<N >
(e) x[1] > 0,
Find an expression of x[n] in the form of sines and cosines,
There are two possible answers
(depending on whether condition (d) was used to?nd a
1
or a
1
),
x[n] = 4 sin(
2
3
n) + 2 sin(
n)
3
or
x[n] = 4 sin(
2?
n)? 2 sin(
n)
3 3
18
| | | | | | | | | | | |
| | |
| | | | |
Fall 2003,Quiz 1 NAME:
Work Page for Problem 5
To Find an expression of x[n] in the form of sines and cosines,we?rst need to?nd a
k
,the
Fourier series coef?cients of x[n],
Here we will show the information we can extract from each given fact,
(a) x[n] is real and odd,? a
k
is pure imaginary and odd,and a
k
= a
k
a
k
is odd a
0
= 0?
(b) x[n] is periodic with period N = 6?
0
= 2?/6 =?/3.?
N = 6 a
3
= a
3
(5.1)
a
k
= a
k+N
a
k
is odd? a
k
=?a
k
a
3
=?a
3
(5.2)
From (5.1) and (5.2) we can conclude that a
3
= a
3
= 0,
Another result that we can conclude in this part is that a
k
has only 6 distinct values,So
the only values we still need to?nd are a
±2
and a
±1
1
(c) |x[n]
2
= 10,
N
|
n=<N>
By Perseval’s therom,
1
x[n]
2
= |a
k
|
2
N
n=<N>
| |
k=<N>
2 2 2 2 2 2
a
2
+ a
1
+ a
0
+ a
1
+ a
2
+ a
3
= 10
2 2 2 2
a
2
+ a
1
+ a
1
+ a
2
= 10
2
a
k
= a
k
2|a
2
2
+ 2 a
1
= 10 (?)| | |
(d) (?1)
n/3
x[n] = 6j,
n=<N>
±j?
Remembering that e =?1
±j?
)
n/3
(?1)
n/3
x[n] = (e x[n]
n=<N> n=<N>
±j?n/3
x[n]= e
n=<N>
1
= N x[n] e
jk?
0
n
N
n=<N>
k=?1
= Na
1
= 6a
1
= 6j
a
1
= j and a
±1
= a
1
= ±j and a
1
=?jj
From (*),2 a
2
2
+ 2 a
1
2
= 2 a
2
2
+ 2(1)
2
= 10? a
2
= 2 | | | | | | | |
19
Fall 2003,Quiz 1 NAME:
(e) x[1] > 0,
a
k
is odd? a
k
= a
k
.?
a
2
= 2? either a
2
= 2j? a
2
=?2j or a
2
=?2j? a
2
= 2j| |
To?nd which choice is the right one,let’s designate p to be the sign of a
2
,
a
2
= 2jp,where p = 1 if a
2
= 2j and p =?1 if a
2
=?2j,
x[n] = a
k
e
jk?
0
n
k=<N>
= a
2
e
j(?2)?
0
n
+ a
1
e
j(?1)?
0
n
+ a
1
e
j(1)?
0
n
+ a
2
e
j(2)?
0
n
j2?
0
n
a
1
e
j?
0
n
+ a
1
e
j?
0
n
= a
2
e + a
2
e
j2?
0
n
j2?
0
n
a
2
e
j2?
0
n
+ a
1
e
j?
0
n
a
1
e
j?
0
n
= a
2
e
j2?
0
n)
j?
0
n
= a
2
e
j2?
0
n
e + a
1
e
j?
0
n
e
= a
2
(2j) sin(2?
0
n) + a
1
(2j) sin(?
0
n)
= (2jp)(2j) sin(2?
0
n) + (±j)(2j) sin(?
0
n)
=?4p sin(2?
0
n)? 2 sin(?
0
n)
x[1] =?4p sin(2?
0
)? 2 sin(?
0
) =?4p sin(2?/3)? 2 sin(?/3)
3
= (?4p? 2)
2
x[1] > 04p?2 > 0? p =?1,regardless of the sign used for the second term,
x[n] = 4 sin
2?
n
2 sin n
3 3
20
Fall 2003,Quiz 1 NAME:
Work Page
21
Fall 2003,Quiz 1 NAME:
Work Page
22
2
3
4
5
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
0.4pt0pt
6.003,Signals and Systems Fall 2003
Quiz 1
Tuesday,October 14,2003
Directions,The exam consists of 5 problems on pages 2 to 19 and work space on pages 20 and
21,Please make sure you have all the pages,Tables of Fourier series properties
are supplied to you at the end of this booklet,Enter all your work and your
answers directly in the spaces provided on the printed pages of this booklet,
Please make sure your name is on all sheets,DO IT NOW!,All sketches must
be adequately labeled,Unless indicated otherwise,answers must be derived or
explained,not just simply written down,This examination is closed book,but
students may use one 8 1/2 × 11 sheet of paper for reference,Calculators may not
be used,
NAME,SOLUTIONS
Check your section Section Time Rec,Instr,
1 10-11 Prof,Zue
2 11-12 Prof,Zue
3 1- 2 Prof,Gray
4 11-12 Dr,Rohrs
5 12- 1 Prof,Voldman
6 12- 1 Prof,Gray
7 10-11 Dr,Rohrs
8 11-12 Prof,Voldman
Please leave the rest of this page blank for use by the graders:
Grader
18
20
20
21
21
100
Problem No,of points Score
Total
PROBLEM 1 (18%)
For the questions in this problem,no explanation is necessary,
Consider the following three systems,
SYSTEM A,y(t) = x(t + 2) sin(?t + 2),where? = 0 →
1
n
SYSTEM B,y[n] = (x[n] + 1)?
2
n
SYSTEM C,y[n] =
x
2
[k + 1]? x[k]
k=1
where x and y are the input and output of each system.
Circle YES or NO for each of the following questions for each of these three systems.
SYSTEM A SYSTEM B SYSTEM C
Is the system linear?
Is the system time invariant?
Is the system causal?
Is the system stable?
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
YES NO
2
Fall 2003,Quiz 1 NAME:
Work Page for Problem 1
Linearity
System A,YES
x
1
(t)? y
1
(t) = x
1
(t + 2)sin(wt + 2)
x
2
(t)? y
2
(t) = x
2
(t + 2)sin(wt + 2)
x
3
(t) = ax
1
(t) + bx
2
(t)? y
3
(t) = x
3
(t + 2)sin(wt + 2)
= ax
1
(t + 2)sin(wt + 2) + bx
2
(t + 2)sin(wt + 2)
= ay
1
(t) + by
2
(t)
System B,NO if x[n] = 0,then y[n] = 0 →
2
System C,NO not linear due to the x [k + 1] term,
Time Invarience
System A,NO the output has time varying gain
System B,NO the output has time varying gain
System C,NO the number of summed terms depends on n
Causality
System A,NO y(t) depends on x(t + 2)
System B,YES y[n] depends only on the current value of x[n]
System C,NO y[n] depends on the future value of x[n]
Stability
System A,YES bounded x(t) bounded y(t)
1 n
System B,NO (?
2
) √ as n√
System C,NO if x[n] =? 1 y[n] =
n
k=1
2,y √ as n √
3
1?2
PROBLEM 2 (20%)
Consider a DT LTI system,H
2
with a unit sample response h
2
[n] = h[n] h[n + 1],as shown?
below,where h[n] =?[n][n? 1],You may remember from one of the lectures that h[n]
can be viewed as the unit sample response of a DT LTI system that acts as an edge detector,
The purpose of this problem is to develop an edge detector that is robust against additive
noise,
h
2
[n]
1 1
System H
2
x[n]
+
0
h
2
[n] np[n] y[n]
1 2?2?1
d[n]?2
Part a,Assume that the input to the system,p[n] is as shown below,and there is no noise,
i.e.,d[n] = 0 and p[n] = x[n],Provide a labeled sketch of y[n],the output of the system,
2 2 2 2
p[n]
2?1 0 1 2 3?3?4
n
2 3 4 5 6 7?1?3?4?5?6?7
y[n]
n
2
2?2
2
4
2?1 0 1
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
x[n] = p[n],y[n] = x[n] h
2
[n]?
Using DT convolution sum,y[n] is the sum of 3 copies of p[n] shifted and scaled according
to h
2
[n]
2?1 0 1 2 3?3?4
p[n
n
2 2 22 + 1]
+
2 3?3?4
2p[n]
n
4?4?4?4
+
2?1 0 1 2 3?3?4
p[n? 1]
n
22 2 2
The resulting y[n] is shown on page 4,
5 Problem 2 continues on the following page,
1?2
Part b,For the same input signal as Part a.,now assume that the noise signal is
d[n] =[n + 1],
Provide a labeled sketch of the output y[n],i.e.,the response to x[n] = p[n] + d[n],
2 3 4 5 6 7?1?3?4?5?6?7
y[n]
n
2 2
1
2
2
3
6
1
2 0
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
y[n] = x[n] h
2
[n]?
= (p[n] + d[n])? h
2
[n]
= (p[n] h
2
[n]) + (d[n] h
2
[n])
We have already found (p[n] h
2
[n]) in Part a,Now we need d[n] h
2
[n].
d[n]
2 0 1
n
1
1 1
n
d[n]? h
s
[n]
1
2
1
Adding the above (d[n] h
2
[n]) to the answer in Part a,we get y[n] for this part which is
shown on page 6,
7 Problem 2 continues on the following page.
1?2
Part c,In order to use system H
2
as a part of an edge detector,we would like to add an
LTI system H
s
whose unit sample response,h
s
[n] is shown below,System H
s
smoothes out
effect of noise on x[n],The overall system can be represented as below,
2
h
s
[n]
2?1 0 1
n
1 1
System H
s
System H
2
p[n] +
x[n]
h
s
[n] h
2
[n] y
s
[n]
d[n]
Provide a labeled sketch of the overall output y
s
[n],when p[n] and d[n] are exactly the same
as in Part b,
y
s
[n]
2 3 4 5 6 7?1?3?4?5?6?7
n
2
1
2?2
2 2
3
8
1?2
Fall 2003,Quiz 1 NAME:
Work Page for Problem 2
y
s
[n] = x[n] h
s
[n] h
2
[n]
= (x[n] h
2
[n])? h
s
[n]?
x[n] = p[n] + d[n] is as de?ned in Part b,Therefore,we already know (x[n] h
2
[n]) from?
Part b,
2 3 4 5 6 7?1?3?4?5?6?7
x[n]? h
2
[n]
n
2 2
1
2
2
3
Now we just need to convolve the above signal with h
s
[n],We can do this convolution with
ipping and sliding h
s
[n],
h
s
[n? k]
2
k
1 1
n? 1
n
n + 1
The result is the desired y
s
[n] and is shown on page 8,
9
PROBLEM 3 (20%)
Consider the CT LTI system whose impulse response is given as,
h(t)
x(t) h(t) y(t)
t
1
0 1?1
The following two parts can be done independently,
Part a,The input x(t),an impulse train starting at t = 2,is depicted below,
a0a1a0a1a0a2a0a1a0
x(t)
(1) (1) (1) (1)
t
0 1 2 3 4 5?1
Provide a labeled sketch of the corresponding output y(t),
y(t)
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
2
1
10
Fall 2003,Quiz 1 NAME:
Work Space for Problem 3
Using CT convolution sum (summation of shifted and scaled impulse response),we can?nd
y(t),
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
1 2 3 4 5 6 7?1?2?3?4?5?6?7
t
1
+
,
,
Adding the shifted and scaled pulses,we get y(t) as shown on page 10,
11 Problem 3 continues on the following page,
Part b,For this part,the output y(t) is periodic and is depicted below:
3?3?6
6 9
2?2
4 8
y(t)
t
······
2
2
Provide a labeled sketch of the input x(t) that produces this y(t),
x(t)
1 2 3 4 5 6 7 8 9 10?1?2?3?4?5?6?7?8?9?10
t
1
1
12
9
Fall 2003,Quiz 1 NAME:
Work Page for Problem 3
There are different ways to solve this problem,and here is one of them:
Let’s take the derivative of the output,
dy(t)
,and relate it to the impulse response of the
dt
system,
y˙(t)
6
6 9
2?2
4 8
t
······
2
2
It can be seen that y˙(t) is the sum of a number of scaled and shifted versions of h(t),Alter-
natively,,y˙(t) = p(t)?h(t),where p(t) is shown below,
p(t)
6
6
2?2
4 8
(2)
(?2)
(2)
t
······
2
2
We know that y(t) = x(t)?h(t)? x(t) = u
1
(t)?p(t)
t
x(t) =
p(v)dv = x(t) + c,where the constant c restores the DC level of the periodic
signal,
x(t)
1 2 3 4 5 6 7 8 9 10?1?2?3?4?5?6?7?8?9?10
t
c + 1
c? 1
To?nd the value of the constant c,we compare the value of y(0) with x(t)?h(t)
t=0
,i.e,the|
area under the curve of x(t)h(-t),
y(0) = 2 = x(t)?h(t)
t=0
= (1 + c)(2) = 2? c = 0.|
13
PROBLEM 4 (21%)
Consider the following periodic triangular wave shown below,
x(t)
1?1
1
1
t
1/4
3/4 1/4
3/4
··· ···
Part a,Determine the Fourier series coef?cients,a
k
for x(t),
0,k = 0
a
k
=
4j
sin(k
2
),k = 0
k
2
2
→
14
Fall 2003,Quiz 1 NAME:
Work Page for Problem 4
x(t)
1?1
1
1
t
1/4
3/4 1/4
3/4
··· ···
Period= T = 1?
0
= 2?/T = 2?,
1
1
a
0
= x(t)dt =
T
(Area under the curve for one period)= 0 = a
0
T T
Finding a
k=0?
using the analysis equation will be tedious,Instead,we will use the integration
property of the Fourier series,
¨x(t)
8
8
t
3/4
1/4
1/4
3/4
··· ···
Let ¨x(t)? b
k
1
1/2
¨
T
b
k
= x(t)e
jk?
0
t
dt = [8?(t + 1/4)? 8?(t? 1/4)] e
jk?
0
t
dt
T?1/2
= 8e
jk?
0
(?1/4)
8e
jk?
0
(1/4)
= 16j sin(k?
0
/4)
= 16j sin(k ),
2
1
a
k
= b
k
(integration property)
(jk?
0
)
2
16j
=
k
2
(2?)
2
sin(k ) =
k
2
4
j
2
sin(k ),
2 2
As a double check,note that a
k
= a because x(t) is real,
k
15 Problem 4 continues on the following page,
Part b,Consider a causal LTI system,S,whose input-output relation is characterized by the
following stable linear constant coef?cient differential equation,
d
2
y dy
+ 4? + 4?
2
y(t) = 4?
2
x(t),
dt
2
dt
where x(t) is the input and y(t) is the output of the system,Suppose x(t) shown on the
previous page is applied to the system S as an input,Let b
k
be the Fourier coef?cients of the
corresponding output y(t),Find b
3
and b
3
,
b
4j 4j
3
=
9?
2
(?8+j6)
b
3
=
9?
2
(8+j6)
,
16
|
Fall 2003,Quiz 1 NAME:
Work Page for Problem 4
First,we?nd H(j?) by inspection from the coef?cients of the differential equation,
d
2
y dy
+ 4? + 4?
2
y(t) = 4?
2
x(t),
dt
2
dt
4?
2
1
H(j?) = =,
(j?)
2
+ 4?j? + 4?
2
1 + j?/
2
/4?
2
b
k
= a
k
H(jk?
0
),where?
0
= 2?,therefore,
b
3
= a
k
H(jk?
0
) = a
3
H(j6?)
k=3
4j
1 4j 1
= sin 3 =
(3)
2
2
2 1 + j6?/ (6?)
2
/4?
2
9?
2
(1 + j6? 9)
4j
b
3
=,
9?
2
(?8 + 6j)
1 4j
b
3
= b
=
4j
=,
3
9?
2
8? 6j 9?
2
(8 + 6j)
17
PROBLEM 5 (21%)
You are given the following facts about a discrete time sequence x[n],
(a) x[n] is real and odd,
(b) x[n] is periodic with period N = 6,
1
(c) |x[n]
2
= 10,
N
|
n=<N >
(d) (?1)
n/3
x[n] = 6j,
n=<N >
(e) x[1] > 0,
Find an expression of x[n] in the form of sines and cosines,
There are two possible answers
(depending on whether condition (d) was used to?nd a
1
or a
1
),
x[n] = 4 sin(
2
3
n) + 2 sin(
n)
3
or
x[n] = 4 sin(
2?
n)? 2 sin(
n)
3 3
18
| | | | | | | | | | | |
| | |
| | | | |
Fall 2003,Quiz 1 NAME:
Work Page for Problem 5
To Find an expression of x[n] in the form of sines and cosines,we?rst need to?nd a
k
,the
Fourier series coef?cients of x[n],
Here we will show the information we can extract from each given fact,
(a) x[n] is real and odd,? a
k
is pure imaginary and odd,and a
k
= a
k
a
k
is odd a
0
= 0?
(b) x[n] is periodic with period N = 6?
0
= 2?/6 =?/3.?
N = 6 a
3
= a
3
(5.1)
a
k
= a
k+N
a
k
is odd? a
k
=?a
k
a
3
=?a
3
(5.2)
From (5.1) and (5.2) we can conclude that a
3
= a
3
= 0,
Another result that we can conclude in this part is that a
k
has only 6 distinct values,So
the only values we still need to?nd are a
±2
and a
±1
1
(c) |x[n]
2
= 10,
N
|
n=<N>
By Perseval’s therom,
1
x[n]
2
= |a
k
|
2
N
n=<N>
| |
k=<N>
2 2 2 2 2 2
a
2
+ a
1
+ a
0
+ a
1
+ a
2
+ a
3
= 10
2 2 2 2
a
2
+ a
1
+ a
1
+ a
2
= 10
2
a
k
= a
k
2|a
2
2
+ 2 a
1
= 10 (?)| | |
(d) (?1)
n/3
x[n] = 6j,
n=<N>
±j?
Remembering that e =?1
±j?
)
n/3
(?1)
n/3
x[n] = (e x[n]
n=<N> n=<N>
±j?n/3
x[n]= e
n=<N>
1
= N x[n] e
jk?
0
n
N
n=<N>
k=?1
= Na
1
= 6a
1
= 6j
a
1
= j and a
±1
= a
1
= ±j and a
1
=?jj
From (*),2 a
2
2
+ 2 a
1
2
= 2 a
2
2
+ 2(1)
2
= 10? a
2
= 2 | | | | | | | |
19
Fall 2003,Quiz 1 NAME:
(e) x[1] > 0,
a
k
is odd? a
k
= a
k
.?
a
2
= 2? either a
2
= 2j? a
2
=?2j or a
2
=?2j? a
2
= 2j| |
To?nd which choice is the right one,let’s designate p to be the sign of a
2
,
a
2
= 2jp,where p = 1 if a
2
= 2j and p =?1 if a
2
=?2j,
x[n] = a
k
e
jk?
0
n
k=<N>
= a
2
e
j(?2)?
0
n
+ a
1
e
j(?1)?
0
n
+ a
1
e
j(1)?
0
n
+ a
2
e
j(2)?
0
n
j2?
0
n
a
1
e
j?
0
n
+ a
1
e
j?
0
n
= a
2
e + a
2
e
j2?
0
n
j2?
0
n
a
2
e
j2?
0
n
+ a
1
e
j?
0
n
a
1
e
j?
0
n
= a
2
e
j2?
0
n)
j?
0
n
= a
2
e
j2?
0
n
e + a
1
e
j?
0
n
e
= a
2
(2j) sin(2?
0
n) + a
1
(2j) sin(?
0
n)
= (2jp)(2j) sin(2?
0
n) + (±j)(2j) sin(?
0
n)
=?4p sin(2?
0
n)? 2 sin(?
0
n)
x[1] =?4p sin(2?
0
)? 2 sin(?
0
) =?4p sin(2?/3)? 2 sin(?/3)
3
= (?4p? 2)
2
x[1] > 04p?2 > 0? p =?1,regardless of the sign used for the second term,
x[n] = 4 sin
2?
n
2 sin n
3 3
20
Fall 2003,Quiz 1 NAME:
Work Page
21
Fall 2003,Quiz 1 NAME:
Work Page
22
