+? +?
+? +? +? +?
+? +?
+? +?
+? +?
+? +?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 3 Solution
Home Study Exercise
(E1) O&W 3.46 (a) and (c)
x(t) and y(t) are continuous-time periodic signals with a period = T
0
and Fourier series
representations given by
b
k
e
jk?
0
t
x(t) = a
k
e
jk?
0
t
y(t) =
k= k=
(a) Show that the Fourier series coe?cients of the signal
c
k
e
jk?
0
t
z(t) = x(t)y(t) =
+?
are given by the discrete convolution
k=
c
k
=
+?
a
n
b
k?n
(multiplication property),
n=
b
m
e
jm?
0
t
= a
n
b
m
e
jn?
0
t jm?
0
t
z(t) = x(t)y(t) = a
n
e
jn?
0
t
e
n= m= n= m=
= a
n
b
m
e
j(n+m)?
0
t
n= m=
= a
n
b
k?n
e
jk?
0
t
Let k = n + m? m = k? n? z(t)
n= k=
a
n
b
k?n
jk?
0
t
Interchange the summations order? z(t) = e
k= n=
z(t) = x(t)y(t) = c
k
e
jk?
0
t
,where c
k
= a
n
b
k?n
,
k= n=
(c) Suppose that y(t) = x
(t),Express b
k
in terms of a
k
,and use the result of part(a)
to prove Parseval’s relation for periodic signals,
1
+?
+?
+?
+? +?
+? +? +? +?
+?
+? +?
jk?
0
t
k
e
jk?
0
t
y(t) = x
(t) = a
k
e = a
k= k=
jk?
0
t
Let k =?k? y(t) = a
k
e
k=
b
k
= a
k
Now to prove Parseval’s relation for period signals we can use the results above
as follows:
|x(t)|
2
= x(t)x
(t) = x(t)y(t)
= c
k
e
jk?
0
t
,where c
k
= a
n
b
k?n
,from part(a)
k= n=
= a
n
b
k?n
e
jk?
0
t
= a
n
a
jk?
0
t
,as proven earlier,
n?k
e
k= n= k= n=
A common mistake is to using the substitution b
k?n
= a
(i.e,just negating
k+n
the index n) which would be correct if the relationship was shifting in frequency
and then taking the conjugate,while the case at hand is taking the conjugate?rst
and then shifting in frequency to perform the convolution,
1
T
0
1
T
0
+?
T
0 0
|
T
0 0
k=
n?k
e
jk?
0
t
dt
P
ave
= |x(t)
2
dt = a
n
a
n=
T
0
1
a
n
a
=
T
0
n?k
e
jk?
0
t
dt
0
n= k=
Notice that the integration has a value of T
0
only when k = 0 and zero for all the
other values of k,as seen below,This rule,usually referred to as orthogonality,
holds for any complex exponential signal,integrated over one period,
T
0
1
T
0
1
jk?
0
t
For k √= 0,
0
e
jk?
0
t
dt =
jk?
0
e
=
jk?
0
(e
jk?
0
T
0
e
j0
)
0
1
= (e
jk(2?)
e
j0
) = 0
jk?
0
T
0
T
0
T
0
For k = 0, e
jk?
0
t
dt = e
j(0)?
0
t
dt = (1)dt = T
0
,
0 0 0
1
T
0?
+?
1
+?
Therefore we have,P
ave
= a
n
a
T
0
=
2
= x(t)|
2
dt,
T
0
n
|a
n
|
T
0 0
|
n= n=
2
0
Problem 1 (O&W 3.22 (a) - only the signal in Figure p3.22 (c)) Determine the
Fourier series representation for the signal x(t),
x(t)
2
··· ···
x(t) =
2 + t,for?2 t← ←
2?2t,for 0 t 1.← ←
5?4?3?2?1 0 1 2 3 4
t
T
2?
x(t) periodic with period T = 3?
0
=
2
T
=
3
A goal of this problem solution is to show di?erent ways to reaching the same answer,
Finding the Fourier series coe?cients of a signal using the analysis equation usually requires
the most e?ort,but can be reverted to if everything else fails,Oftentimes,a signal can be
dissected into simpler signals that are easier to analyze or can be derived from a simpler
signal by integration,di?erentiation,time shifting,or any combination of the properties of
the Fourier series (see Table 3.1,O&W,p.206),
We will start with?nding a
0
,which is usually straight-forward and doesn’t require much
e?ort,and then explore the di?erent methods for?nding a
k=0
:
√
1 1 1
a
0
= x(t)dt = (the total area under the curve for one period) = (2 + 1) = 1,
T
T
3 3
The following are four possible methods to calculate a
k=0
,the Fourier series coe?cients of
√
x(t) for k = 0,√
Method (a),Using the integration property,
Let g(t) =
dx(t)
x(t) = g(t)dt + p,where p is the value of x(t) at the beginning of
dt
the period,and it equals to zero for the period we selected that starts at t =?2,Note
that,since we are trying to?nd a
k=0
,the value of p is not important because it only
√
a?ects the DC level of x(t) and we have already calculated it by?nding a
0
,
g(t)
5?4?3
2 3 4
1
t
2
1
··· ···
3
?
Note that g(t) must have a zero DC level,otherwise a ramping signal will be included
in x(t) making it non-periodic,and unbounded,By de?nition,g(t) should have a
zero DC level because the derivative operation eliminates it,so this can be used as a
double-check,
After?nding b
k
,the Fourier series coe?cients for g(t),we can use the Fourier series
properties to?nd a
k
,the Fourier series coe?cients for x(t)
1
?
1
g(t)e
jk?
0
t
dt =
1
0
(1)e
jk?
0
t
dt + (?2)e
jk?
0
t
dtb
k
=
T
T
3
2 0
0
1
1 1
1
= e
jk?
0
t
2
jk?
0
e
jk?
0
t
1
jk?
0
2
2e
jk?
0
= + 2
3?jk?
0
2
3jk?
0
1? e
0
1
jk?
0
2
+ 2e
jk?
0
3
= e,
3jk?
0
1
a
k
= b
k
(from the Integration property,Table 3.1,O &W,p.206 )
jk?
0
1 1
jk?
0
2
+ 2e
jk?
0
3
1
3? 2e
jk?
0
jk?
0
2
= e =
3k
2
2
e
jk?
0
3jk?
0
0
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
1
1? e
jk
4
3
1
1? e
jk
2?
= =
3
,
k
2
2
k
2
2
0 0
Method (b),Using the integration property twice,
Let’s de?ne v(t) as the following,
d
2
x(t) dg(t)
v(t) =
dt
2
=? x(t) = v(t) dt dt + p = g(t) dt + p
dt
Similar to the discussion in Method(a) of the DC level of g(t),v(t) must have a zero DC
level,In addition,its limited integration over one period must also have a zero DC level,
We can?nd v(t) by di?erentiating g(t),However,in our case,but not always,we
can?nd v(t) directly from x(t) in one step,by placing an impulse at each point of time
where the slope of x(t) changes abruptly,The value of that impulse (i.e its area) is the
change in slope of x(t) at that point,
4
5?4?3?2
1 0
1 2 3 4
t
v(t)
3
··· ···
3 3
To?nd c
k
,the Fourier series coe?cients of v(t),let’s take the period between -1 and
2,which contains two impulses,
Note that we can also take the period between -2 and 1,but we have to be careful not
include the impulses at both -2 and 1,In other words,we can take the period between
2 +? and 1 +? or the period between?2 and 1,
c
k
=
1
v(t)e
jk?
0
t
dt =
1
2
[?3?(t) + 3?(t?1)] e
jk?
0
t
dt
T
T
3
1
2
= [(t) +?(t?1)] e
jk?
0
t
dt
1
=?e
jk?
0
(0)
+ e
jk?
0
(1)
= e
jk?
0
1,
Now to?nd a
k
,we just need to use the integration property two times,
1 1
a
k
= c
k
(from the Integration property,Table 3.1,O &W,p.206 )
jk?
0
jk?
0
=
1
e
jk?
0
(jk?
0
)
2
1
=
1
1?e
jk?
0
k
2
2
0
1
3
= 1?e
jk
2?
,which is the same answer found in Method(a),
k
2
2
0
5
Before exploring the other methods,let’s?rst?nd the Fourier series for y(t),shown
below,which is a periodic triangular function with a period of T,y(t) will be useful
for the Method(c),
y(t)
t
1
T
1
T
1
T T
··· ···
2 2
z(t) =
dy(t)
dt
1
T
1
···
T
1
···
T?T
1
T t
2 2
1
T
1
F
Let z(t) =
dy(t)
,z(t) e
k
,and y(t)
F
=
jk?
0
e
k
dt
d
k
1
We will?nd the Fourier series for z(t) and from it,we will?nd the Fourier series for
y(t),as follows,
T
1
1
z(t) e
jk?
0
t
dt =
1
0
1
)e
jk?
0
t
dt + (
1
)e
jk?
0
t
dte
k
= (
T
T
T
T
1
T
1 0
T
1
1 1
1
TT
1
(?jk?
0
)
e
jk?
0
t 0
e
jk?
0
t T1
1
jk?
0
T
1
+ 1?e
jk?
0
T
1
= =
0
|
T
1
|
TT
1
(?jk?
0
)
1?e
1
=
1
2?(e
jk?
0
T
1
+ e
jk?
0
T
1
)
=
TT
1
jk?
0
[2?2 cos (jk?
0
T
1
)],
TT
1
jk?
0
d
Thus,
k
= e
k
(
1
) =
2?2 cos (k?
0
T
1
)
,
jk?
0
TT
1
k
2
2
0
6
Method (c),By dissecting the signal into simpler components,
Here,we will dissect x(t) into x
1
(t) and x
2
(t) which we know their Fourier Series (using
the result of d
k
above and the time-shifting property),
x
1
(t)
2
4? 3? 2? 1 0 1 2 3 4
t
x
2
(t)
4? 3? 2? 1 0 1 2 3 4
t
1
x(t) = x
1
(t) + x
2
(t),and let x
1
(t)
F
and x
2
(t)
F
b
k
c
k
a
k
= b
k
+ c
k
= (2)
2? 2 cos (k?
0
(1))
+ (1)
2? 2 cos (k?
0
(1))
e
jk?
0
(?1)
(3)(1)k
2
2
(3)(1)k
2
2
0 0
=
2? 2 cos k?
0
(2 + e
jk?
0
).
3k
2
0
2
Although this result looks di?erent from those found in the previous methods,further
simpli?cation will show that they are identical,
1
a
k
=
2? 2 cos k?
0
(2 + e
jk?
0
) =
3k
2
2
(2? 2 cos k?
0
)(2 + e
jk?
0
)
3k
2
2
0 0
1
=
3k
2
0
2
(2? e
jk?
0
e
jk?
0
)(2 + e
jk?
0
)
1
4 + 2e
jk?
0
2e
jk?
0
e
jk?
0
2
2e
jk?
0
0
=
3k
2
2
e
0
1
1
=
jk?
0
2
2e
jk?
0
=
jk?
0
2
2e
jk?
0
1
3k
2
0
2
4? e
3k
2
0
2
3? e
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
=
1
1? e
jk
4
3
,which is the same answer found in previous methods,
k
2
2
0
7
?
Method (d),using the analysis equation,
In the process of evaluating the analysis equation,the following integral will save us a
lot of derivation steps,
t 1
at
te
at
dt =
a
a
e,for any a = 0
2
√
1
1
1
x(t)e
jk?
0
t
dt = x(t)e
jk?
0
t
dt
a
k
=
T 3
2
1
1
0
(2? 2t)e
jk?
0
t
dt= (2 + t)e
jk?
0
t
dt +
3
0?2
1
1
1
0
e
jk?
0
t
dt + te
jk?
0
t
dt? 2 te
jk?
0
t
dr= 2
3
0?2?2
0
e
jk?
0
t
1
1
1
1 t
1 t
= 2
+
e
jk?
0
t
3?jk?
0
k
2
2
jk?
0
e
jk?
0
t
2
k
2
2
jk?
0
0 ?2 0 0?2
1?2
e
jk?
0
(1)
e
jk?
0
(?2)
1 1 (?2)
e
jk?
0
(?2)
= +
3 jk?
0
k
2
0
2
k
2
2
jk?
0
0
11 (1)
e
jk?
0
(1)
2
k
2
2
jk?
0
k
2
0
2
0
1?2
e
jk?
0
2
k
2
1
0
2
k
2
1
0
2
jk?
0
2
2
jk?
0
jk?
0
2
= + e
jk?
0
2
+ e e
3 jk?
0
jk?
0
2 2 2
k
2
0
2
e
jk?
0
+ e
jk?
0
+
jk?
0
k
2
2
0
1 21?2
e
jk?
0
2
e
jk?
0
2
e
jk?
0
= + + +
3 jk?
0
k
2
0
2
jk?
0
k
2
2
k
2
2
0 0
2 1 2
+ e
jk?
0
2 jk?
0
2 jk?
0
2
e e
jk?
0
k
2
0
2
jk?
0
1 2 3 1
+ e
jk?
0
2
=
3
k
2
0
2
e
jk?
0
k
2
0
2
k
2
0
2
1
= 3? 2e
jk?
0
jk?
0
2
3k
2
0
2
e
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
=
1
1? e
jk
4
3
,which is the same answer found in previous methods,
k
2
2
0
=
9
1? e
jk
4
3
,
4k
2
2
8
Problem 2 O & W 3.23 (a)
Given a
k
,the Fourier series coe?cients of a periodic continuous time signal with period 4,
determine the signal x(t).
The Fourier series coe?cients a
k
are given as follows:
a
0,k = 0
k
=
(j)
k
sin k?/4
,otherwise,
k?
Here are some of the facts we know about x(t),
a
0
= 0? no DC component in x(t)
T = 4?
0
= 2?/4 =?/2
k
(j)
k
sin(? k?/4) 1? sin(k?/4)
=a
k
=
k? j? k?
(? j)
k
sin(k?/4)
= = a
k
,
k?
Thus x(t) is a real signal (O&W,Section 3.5.6,p.204),
e
jk?
0
(?1)
Noting that j = e
j?/2
(j)
k
= e
j?/2
k
= e
jk?/2
= e
jk?
0
=,we can consider x(t)
to be a time-shifted version of another signal y(t) such that,
F
x(t) = y(t + 1),where y(t) b
0
= 0,b
k=0
=
sin k?/4
and a
k
= b
k
e
jk?
0
(1)
√
k?
By backtracking the derivation equation of b
k
,we can?nd the signal?
b
y(t) which has the same
k
but can have a di?erent DC level (i.e,b
0
),
b
k=0√
= b
k=0
=
sin k?/4
=
1 e
jk?/4
e
jk?/4
√
k? k? 2j
1 1
=
(4) jk(
e
jk?/4
e
jk?/4
2
)
1
2
)
1
2
1 1
jk?
0
(
1
e
jk?
0
(?
2
)
1
(1)e
jk?
0
t
dt.= e =
1
2
T jk?
0
T
The integration above suggests that
1
y?(t) =
1,?
2
< t <
1
2
0,elsewhere in the same period T=4,
9
y?(t)
3?2?1 0 1 2 3
t
1
T
2
T
2
··· ···
Note that the same conclusion can be reached by noticing that?y(t) is the same signal in
Example 3.5 (O& W,p.193) with T
1
1
=
2
and T = 4.
To?nd y(t),which has b
0
= 0,we?rst calculate
b
0
and then subtract it from?
y(t),
1 1
1/2
1
b
0
= y?(t)dt = (1)dt =
T
T
4
1/2
4
1
2 2
3 1 1
< t <
y(t) = y?(t)?
4
y(t) =
4
,?
1 1
4
,<|t| < 2.
2
3
1.5 < t <?0.5
x(t) = y(t + 1) =
4
,
1
4
,?0.5 < t < 2.5?
Sketches of y(t) and x(t) are shown below,
y(t)
3?2?1
1 2 3
t
3/4
1/4
··· ···
x(t)
3?2
1 2 3
t
3/4
1/4
1
··· ···
10
Problem 3 Determine the Fourier series coe?cients for the periodic signal x[n] depicted
below,Plot the magnitude and phase of these coe?cients,
x[n]
2
12?6 0 6 12
n
1
··· ···
1
Fundamental period N = 6?
0
=
2
6
=,
3
5 2
1
1 1
a
k
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
N 6 6
n=<N> n=0 n=?3
Notice that the last two expressions will give the same result,but the latter would take
advantage of the symmetry of some of the samples to combine them into sinusoids,
1
a
k
= (0)e
jk?
0
(?3)
+ (1)e
jk?
0
(?2)
+ (2)e
jk?
0
(?1)
+ (1)e
jk?
0
(0)
+
6
+(2)e
jk?
0
(1)
+ (?1)e
jk?
0
(2)
=
1
e
jk?
0
(?2)
e
jk?
0
(2)
+ 2e
jk?
0
(?1)
+ 2e
jk?
0
(1)
+ 1
6
1
= [(2j) sin k?
0
2 + 2(2) cos k?
0
+ 1]
6
1 2 j
= + cos k?
0
+ sin k?
0
2
6 3 3
1 2 1 2?
a
k
= + cos k + j sin k
6 3 3 3 3
1 2?
= 1 + 4 cos k + j2 sin k
6 3 3
11
Here are the values of a
k
for one period of six consecutive points (from k=-2 to k=3),
1
2?
a
2
=
6
1 + 4 cos (?2)
3
+ 2j sin (?2)
3
=
1
1 + 4 cos
2? 4?
=
1
1 + 4
1?
3
6 3
2j sin
3 6 2
2j
2
1
= (?1 + j
3) =?0.1667 + j0.2887
6
(?1)
2
+ (
3)
2
1 2?
j atan(
3,?1)
1
j
2?
3
= e = e a
2
=
3
,
a
2
=,
6 3
| |
3
Similarly,for the magnitude and phase of a
k
for k =?1 3 which are summarized in the?
table below,
k a
k
-2 -0.1667 + j 0.2887
-1 0.5000 - j 0.2887
0 0.8333
1 0.5000 + j 0.2887
2 -0.1667 -j 0.2887
3 -0.5000
a
k
||
1/3
1/
3
5/6
1/
3
1/3
3/2
a
k
2?/3
/6
0
/6
2?/3
12
The magnitude and phase of the Fourier series coe?cients were plotted below,using MAT-
LAB,
Magnitude of a
k
0
0.2
0.4
0.6
0.8
1
|a
k
|
15?10?5 0 5 10 15
k
Phase of a
k
—
a
k
(radians)
1
0.5
0
0.5
1
15?10?5 0 5 10 15
k
For your reference,the MATLAB code used to compute and plot the magnitude and phase
of the Fourier series coe?cients is shown below,
MATLAB Code:
A=inline(’1/6 +2/3*cos(k*pi/3)+j/3*sin(k*2*pi/3)’);
k=-12:12;a=A(k);am=abs(a);ap=angle(a); subplot(2,1,1);stem(k,am);grid
on;xlabel(’k’);ylabel(’|a_k|’);title(’Magnitude of a_k’);
subplot(2,1,2);stem(k,ap/pi);grid on; xlabel(’k’);ylabel(’\angle a_k
(radians)’);title(’Phase of a_k’);
MATLAB tip,you can use TEX expressions in the text of?gures,
13
Problem 4 O & W 3.29 (a)
Given a
k
,the Fourier series coe?cients of a periodic discrete time signal with period 8,
determine the signal x[n].
The Fourier coe?cients are given as follows:
k? 3k?
a
k
= cos + sin,
4 4
N=8? there are only 8 samples to compute in x[n],some of which can have a zero value,
0
= 2?/8 =?/4,
k? 3k?
a
k
= cos
4
+ sin
4
= cos k?
0
+ sin 3k?
0
=
1
2
e
jk?
0
+
1
2
e
jk?
0
+
1
2j
e
j3k?
0
1
2j
e
j3k
=
1
8
4 e
jk?
0
(?1)
+ 4 e
jk?
0
(1)
+
4
j
e
jk(?3)?
0
4
j
e
jk(3)
1
(4) e
jk?
0
(?1)
+ (4) e
jk?
0
(1)
+ (?4j) e
jk(?3)?
0
+ (4j) e
jk(3)
=
8
4
1
1
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
N 8
n=<N> n=?3
4j,n =?3
4j,n = 3
By matching the expressions of a
k
x[n] =?
4,n = ±1
0,n = 0,±2,4
4j
4 x[n] 4
4j
n
6?5?4
3?2
1 0 1 2 3 4
5 6
··· ···
4j?4j
14
Problem 5 Consider the following CT periodic signals,x(t),y(t),and z(t),
(a) Determine the fundamental frequency,period,and Fourier series coe?cients, a
k
,for
x(t),
x(t)
6?5?4?3?2?1 0 1 2 3 4 5 6
1
t
Fundamental period of x(t) = T = 4?
0
= 2?/4 =?/2.?
1 1
1
2 1
a
0
= x(t) dt = x(t) dt = =,
T
T
4
1
4 2
1
1
1
x(t) e
jk?
0
t
dt =
1
1
(1) e
jk?
0
t
dt = e
jk?
0
t
a
k
=
T
T
4
1
jk?
0
4
1
1
(e
jk?
0
e
jk?
0
) =
sin (k?
0
) sin (k
2
)
= =,
jk2? k? k?
The same result can also be found directly using Example 3.5 ( O & W,P.193),
(b) Determine the fundamental frequency,period,and Fourier series coe?cients,b
k
,for
y(t).
y(t)
6?4?2 2 4 6
1
1
t
Fundamental period of y(t) = T
= 2?
0
= 2?/2 =?.
1
b
0
= y(t)dt = 0 (? y(t) has no DC component),
T
T
15
+?
0
t j(?1)?
b
k
e
jk?
e
j
0
t
0
t
y(t) = sin?
j(1)?
0
t =
0
t
e
j?
=
j
e?
j
e
0
t
=
2j 2 2
k=
2
j
,k = 1
j
b
k
=
2
,k =?1
0,otherwise
(c) Determine the fundamental frequency and period for z(t),Also,using the results of
parts (a) and (b),determine the Fourier series coe?cients,c
k
for z(t),
z(t)
6?5
4?3
2?1
1 2 3 4 5 6
1
1
t
Fundamental period of z(t)=Fundamental period of x(t) = T = 4?
0
= 2?/4 =
.?
1
c
0
= z(t)dt = 0 (? z(t) has no DC component),
T
T
Noticing that z(t) = x(t)y(t),we can?nd c
k
using the multiplication property,How-
ever,the fundamental frequencies of x(t) and y(t) must be identical in order for the
Fourier coe?cients to match (i.e,to represent the same frequencies),The fundamen-
tal period of y(t) is 2,but if we de?ne it to be 4,then we only need to scale the
frequency components accordingly to keep the value of k?
0
constant,In our case,for
y(t),?
0
=
=?
0
/2
2
2
j
,k = 2
j
b
=
2
,k =?2
k
0,otherwise
16
2
+?
Using the multiplication property:
c
k
= a
n
b
k?n
,which looks like the discrete-time convolution,in frequency,
n=
k?n
= 0 only when k? n =Note that a
n
b
±2√
c
k
= a
k?2
b
j j
2
+ a
k+2
b
= a
k?2
j
+ a
k+2
2
2 2
=
2
(a
k+2
a
k?2
)
sin (k+2)
2
)
2
a
k+2
=
1
(k+2)?
,k =?2
sin (k
k =?2
=
(k+2)?
,√ √
1
2
,k =?2
2
,k =?2
a
k?2
=
sin (k?2)
,√
=
sin (k
√
2
)
(k?2)?
2
k = 2
(k?2)?
,k = 2
1 1
2
,k = 2
2
,k = 2
17
Problem 6 Let x(t) be a periodic signal with fundamental period T and Fourier series
coe?cients a
k
,Derive the Fourier series coe?cients of each of the following signals in terms
of a
k
,
(a) O d{ x(t? T/2)}
2
= a
k
e
jk?
0
T
(Time Shifting Property) x(t? T/2) b
k
= a
k
e
jk?
= a
k
(e
j?
)
k
= a
k
(? 1)
k
If we assume that x(t) is real,then,
= j → m{ b
k
} (Even-Odd Decomposition of Real Signals O d{ x(t? T/2)} c
k
Propriety,Table 3.1,O & W,p,206)
= j → m{ a
k
(? 1)
k
} = (? 1)
k
j→ m{ a
k
},
However,the question didn’t specify x(t) to be real,so assuming that x(t) is complex,
we will just use the general formula for?nding the Odd part of a signal,
=
1
[x(t)? x(? t)] (O &W,Sec,1.2.3,and speci?cally eq.(1.19),p.14) O d{ x(t)}
2
1 1
O d{ x(t? T/2)} = [x(t? T/2)? x(? t? T/2)] d
k
= [a
k
(? 1)
k
a
k
(? 1)
k
]
2 2
1
= (? 1)
k
(a
k
a
k
),
2
Note that for real x(t),a
k
= a
k
a
k
a
k
= 2j → = d
k
.m{ a
k
}? c
k
(b) x(T/4? t)
= a
k
(Time-Reversal Property) x(? t) c
k
= c
k
e
jk?
0
T/4
(Time-Shift in the positive time direction,i.e,delay) x(T/4? t) d
k
2
= c
k
e
jk
= c
k
(? j)
k
= a
k
(? j)
k
,
18
Problem 7 O & W 3.31 (also determine a
0
)
1,0 n 7
Let x[n] =
← ←
,x[n], periodic,N = 10,Fourier series coe?cients,a
k
,
0,8 n 9← ←
Also,let g[n] = x[n]?x[n?1],
x[n]
n
1
··· ···
0 1 2 3 4 5 6 7 8 9 10?3?2?1
g[n] 1
n
3?2?1
0 1 2 3 4 5 6
7 8
9 10
1
··· ···
1?1
x[n]?1
n
3?2?1
0 1 2 3 4 5 6 7 8
9 10
··· ···
1?1?1?1
Fundamental Period = N = 10?
0
=
.
5
9
1
1 1
a
0
= x[n] = x[n] = [1(8) + 0(2)] = 8/10 = 4/5,
N 10 10
n=<N > n=0
(a) Show that g[n] has a fundamental period of 10,
g[n + N] = x[n + N]?x[n + N?1]
x[n + N] = x[n]? g[n + N] = x[n] [n?1] = g[n]
x
[n] has a fundamental period of N = 10.? g
19
(b) Determine the Fourier series coe?cients of g[n].
1,n = 0
g[n] = x[n]? x[n? 1] = 0,1 n 7
← ←
n = 8.? 1,
1
b
k
= g[n]e
jk?
0
n
N
n=<N>
7
1
= g[n]e
jk?
0
n
,the limits were chosen to use the non-zeros near the origin
10
n=?2
1
= (? 1)e
jk?
0
(?2)
+ (1)e
jk?
0
(0)
=
1
1? e
jk?
0
2
=
1
e
jk?
0
e
jk?
0
jk?
0
)
e
10 10 10
e
jk?
0
e
jk?
0
j2
jk?
0
5
= e =
j
e
jk?
0
sin k?
0
=
j
e
jk
sin k
10 2j 5 5 5
(c) Using the Fourier series coe?cients of g[n] and the First-Di?erence property in Table 3.2,
determine a
k
for k = 0.√
From Table 3.2 (O & W,p,221),x[n]? x[n? 1] (1? e
jk(2?/N)
) a
k
= b
k
1
(1? e
jk?
0
2
b
k
= ) = (1? e
jk?
0
) a
k
10
1 1? e
jk?
0
2
1 (1? e
jk?
0
)(1 + e
jk?
0
) (1 + e
jk?
0
)? 1
a
k
= = =
1? e
jk?
0
e
jk?
0
(e
jk?
0
1) e
jk?
0
10 10 10
e
jk?
0
1
2
3 1? 1
(e
jk?
0
1
jk?
0
1
=
2
+ e
2
) =
1
e
jk?
0
2
cos k?
0
e
jk?
0
10 5 2
1
jk
3?
= e
10
cos k,
5 10
Let’s double check the result,and at the same time use another route to?nd a
k
,
Note that x[n]? 1 would have the same a
k
(only a
0
changes with a change in the DC
level of a signal),
0,0 n 7 0,0 n 7
x[n]? 1 =
← ←
=
← ←
1,8 n 9? 1,? 2← ← ← n ←? 1
1
1)e
jk?
0
n
1
(? 1)e
jk?
0
(?2)
+ (? 1)e
jk?
0
(?1)
=
1
e
jk?
0
2
+ e
jk?
0
a
k
= (x[n] =
N 10 10
n=<N>
2
)
+ e
jk?
0
(
1
jk
3?
jk?
0
(3/2)
jk?
0
(
1
1
=
1
e e
2
)
=
1
e
jk?
0
(3/2)
cos (k?
0
) =
1
e
10
cos k,
10 5 2 5 10
20
Problem 8 O & W 3.51
x[n], periodic signal with period N = 8 and Fourier series coe?cients a
k
=? a
k?4
.
y[n], periodic signal with period N = 8 and Fourier series coe?cients b
k
,
y[n] =
1 + (? 1)
n
x[n? 1]
2
Find a function f[k] such that b
k
= f[k]a
k
,
1 1
y[n] =
1 + (? 1)
n
x[n? 1] = x[n? 1] + (? 1)
n
x[n? 1],
2 2 2
n? 1] a
k
e
jk?
0
(1)
x[ (Time Shifting Property) (1)
= e
j4(
2
8
Note that (? 1)
n
= (e
j?
)
n )n
= e
j4?
0
n
,?
0
= 2?/8 =
4
(? 1)
n
x[n? 1] = e
j4?
0
n
x[n? 1] a
k?4
e
j(k?4)?
0
(Frequency Shifting Property) (2)
From (1) and (2),y[n] =
2
x[n? 1] +
11
2
(? 1)
n
x[n? 1] b
k
=
1
a
k
e
jk?
0
+
1
a
k?4
e
j(k?4)?
0
.
2 2
Substituting a
k
=? a
k?4
,
1 1
a
k
e
jk?
0
(? a
k
) e
jk?
0
b
k
= + e
j4?
0
2 2
1
= a
k
e
jk?
0
(1? e
j4?
0
) =
(1? e
j4?
0
)
e
jk?
0
a
k
2 2
(1? e
j4
4
) 1? (? 1)
e
jk
4
a
k
e
jk
=
4
a
k
=
2 2
4
a
k
e
jk
=
f[k] = e
jk
4
.?
21
+? +? +? +?
+? +?
+? +?
+? +?
+? +?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 3 Solution
Home Study Exercise
(E1) O&W 3.46 (a) and (c)
x(t) and y(t) are continuous-time periodic signals with a period = T
0
and Fourier series
representations given by
b
k
e
jk?
0
t
x(t) = a
k
e
jk?
0
t
y(t) =
k= k=
(a) Show that the Fourier series coe?cients of the signal
c
k
e
jk?
0
t
z(t) = x(t)y(t) =
+?
are given by the discrete convolution
k=
c
k
=
+?
a
n
b
k?n
(multiplication property),
n=
b
m
e
jm?
0
t
= a
n
b
m
e
jn?
0
t jm?
0
t
z(t) = x(t)y(t) = a
n
e
jn?
0
t
e
n= m= n= m=
= a
n
b
m
e
j(n+m)?
0
t
n= m=
= a
n
b
k?n
e
jk?
0
t
Let k = n + m? m = k? n? z(t)
n= k=
a
n
b
k?n
jk?
0
t
Interchange the summations order? z(t) = e
k= n=
z(t) = x(t)y(t) = c
k
e
jk?
0
t
,where c
k
= a
n
b
k?n
,
k= n=
(c) Suppose that y(t) = x
(t),Express b
k
in terms of a
k
,and use the result of part(a)
to prove Parseval’s relation for periodic signals,
1
+?
+?
+?
+? +?
+? +? +? +?
+?
+? +?
jk?
0
t
k
e
jk?
0
t
y(t) = x
(t) = a
k
e = a
k= k=
jk?
0
t
Let k =?k? y(t) = a
k
e
k=
b
k
= a
k
Now to prove Parseval’s relation for period signals we can use the results above
as follows:
|x(t)|
2
= x(t)x
(t) = x(t)y(t)
= c
k
e
jk?
0
t
,where c
k
= a
n
b
k?n
,from part(a)
k= n=
= a
n
b
k?n
e
jk?
0
t
= a
n
a
jk?
0
t
,as proven earlier,
n?k
e
k= n= k= n=
A common mistake is to using the substitution b
k?n
= a
(i.e,just negating
k+n
the index n) which would be correct if the relationship was shifting in frequency
and then taking the conjugate,while the case at hand is taking the conjugate?rst
and then shifting in frequency to perform the convolution,
1
T
0
1
T
0
+?
T
0 0
|
T
0 0
k=
n?k
e
jk?
0
t
dt
P
ave
= |x(t)
2
dt = a
n
a
n=
T
0
1
a
n
a
=
T
0
n?k
e
jk?
0
t
dt
0
n= k=
Notice that the integration has a value of T
0
only when k = 0 and zero for all the
other values of k,as seen below,This rule,usually referred to as orthogonality,
holds for any complex exponential signal,integrated over one period,
T
0
1
T
0
1
jk?
0
t
For k √= 0,
0
e
jk?
0
t
dt =
jk?
0
e
=
jk?
0
(e
jk?
0
T
0
e
j0
)
0
1
= (e
jk(2?)
e
j0
) = 0
jk?
0
T
0
T
0
T
0
For k = 0, e
jk?
0
t
dt = e
j(0)?
0
t
dt = (1)dt = T
0
,
0 0 0
1
T
0?
+?
1
+?
Therefore we have,P
ave
= a
n
a
T
0
=
2
= x(t)|
2
dt,
T
0
n
|a
n
|
T
0 0
|
n= n=
2
0
Problem 1 (O&W 3.22 (a) - only the signal in Figure p3.22 (c)) Determine the
Fourier series representation for the signal x(t),
x(t)
2
··· ···
x(t) =
2 + t,for?2 t← ←
2?2t,for 0 t 1.← ←
5?4?3?2?1 0 1 2 3 4
t
T
2?
x(t) periodic with period T = 3?
0
=
2
T
=
3
A goal of this problem solution is to show di?erent ways to reaching the same answer,
Finding the Fourier series coe?cients of a signal using the analysis equation usually requires
the most e?ort,but can be reverted to if everything else fails,Oftentimes,a signal can be
dissected into simpler signals that are easier to analyze or can be derived from a simpler
signal by integration,di?erentiation,time shifting,or any combination of the properties of
the Fourier series (see Table 3.1,O&W,p.206),
We will start with?nding a
0
,which is usually straight-forward and doesn’t require much
e?ort,and then explore the di?erent methods for?nding a
k=0
:
√
1 1 1
a
0
= x(t)dt = (the total area under the curve for one period) = (2 + 1) = 1,
T
T
3 3
The following are four possible methods to calculate a
k=0
,the Fourier series coe?cients of
√
x(t) for k = 0,√
Method (a),Using the integration property,
Let g(t) =
dx(t)
x(t) = g(t)dt + p,where p is the value of x(t) at the beginning of
dt
the period,and it equals to zero for the period we selected that starts at t =?2,Note
that,since we are trying to?nd a
k=0
,the value of p is not important because it only
√
a?ects the DC level of x(t) and we have already calculated it by?nding a
0
,
g(t)
5?4?3
2 3 4
1
t
2
1
··· ···
3
?
Note that g(t) must have a zero DC level,otherwise a ramping signal will be included
in x(t) making it non-periodic,and unbounded,By de?nition,g(t) should have a
zero DC level because the derivative operation eliminates it,so this can be used as a
double-check,
After?nding b
k
,the Fourier series coe?cients for g(t),we can use the Fourier series
properties to?nd a
k
,the Fourier series coe?cients for x(t)
1
?
1
g(t)e
jk?
0
t
dt =
1
0
(1)e
jk?
0
t
dt + (?2)e
jk?
0
t
dtb
k
=
T
T
3
2 0
0
1
1 1
1
= e
jk?
0
t
2
jk?
0
e
jk?
0
t
1
jk?
0
2
2e
jk?
0
= + 2
3?jk?
0
2
3jk?
0
1? e
0
1
jk?
0
2
+ 2e
jk?
0
3
= e,
3jk?
0
1
a
k
= b
k
(from the Integration property,Table 3.1,O &W,p.206 )
jk?
0
1 1
jk?
0
2
+ 2e
jk?
0
3
1
3? 2e
jk?
0
jk?
0
2
= e =
3k
2
2
e
jk?
0
3jk?
0
0
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
1
1? e
jk
4
3
1
1? e
jk
2?
= =
3
,
k
2
2
k
2
2
0 0
Method (b),Using the integration property twice,
Let’s de?ne v(t) as the following,
d
2
x(t) dg(t)
v(t) =
dt
2
=? x(t) = v(t) dt dt + p = g(t) dt + p
dt
Similar to the discussion in Method(a) of the DC level of g(t),v(t) must have a zero DC
level,In addition,its limited integration over one period must also have a zero DC level,
We can?nd v(t) by di?erentiating g(t),However,in our case,but not always,we
can?nd v(t) directly from x(t) in one step,by placing an impulse at each point of time
where the slope of x(t) changes abruptly,The value of that impulse (i.e its area) is the
change in slope of x(t) at that point,
4
5?4?3?2
1 0
1 2 3 4
t
v(t)
3
··· ···
3 3
To?nd c
k
,the Fourier series coe?cients of v(t),let’s take the period between -1 and
2,which contains two impulses,
Note that we can also take the period between -2 and 1,but we have to be careful not
include the impulses at both -2 and 1,In other words,we can take the period between
2 +? and 1 +? or the period between?2 and 1,
c
k
=
1
v(t)e
jk?
0
t
dt =
1
2
[?3?(t) + 3?(t?1)] e
jk?
0
t
dt
T
T
3
1
2
= [(t) +?(t?1)] e
jk?
0
t
dt
1
=?e
jk?
0
(0)
+ e
jk?
0
(1)
= e
jk?
0
1,
Now to?nd a
k
,we just need to use the integration property two times,
1 1
a
k
= c
k
(from the Integration property,Table 3.1,O &W,p.206 )
jk?
0
jk?
0
=
1
e
jk?
0
(jk?
0
)
2
1
=
1
1?e
jk?
0
k
2
2
0
1
3
= 1?e
jk
2?
,which is the same answer found in Method(a),
k
2
2
0
5
Before exploring the other methods,let’s?rst?nd the Fourier series for y(t),shown
below,which is a periodic triangular function with a period of T,y(t) will be useful
for the Method(c),
y(t)
t
1
T
1
T
1
T T
··· ···
2 2
z(t) =
dy(t)
dt
1
T
1
···
T
1
···
T?T
1
T t
2 2
1
T
1
F
Let z(t) =
dy(t)
,z(t) e
k
,and y(t)
F
=
jk?
0
e
k
dt
d
k
1
We will?nd the Fourier series for z(t) and from it,we will?nd the Fourier series for
y(t),as follows,
T
1
1
z(t) e
jk?
0
t
dt =
1
0
1
)e
jk?
0
t
dt + (
1
)e
jk?
0
t
dte
k
= (
T
T
T
T
1
T
1 0
T
1
1 1
1
TT
1
(?jk?
0
)
e
jk?
0
t 0
e
jk?
0
t T1
1
jk?
0
T
1
+ 1?e
jk?
0
T
1
= =
0
|
T
1
|
TT
1
(?jk?
0
)
1?e
1
=
1
2?(e
jk?
0
T
1
+ e
jk?
0
T
1
)
=
TT
1
jk?
0
[2?2 cos (jk?
0
T
1
)],
TT
1
jk?
0
d
Thus,
k
= e
k
(
1
) =
2?2 cos (k?
0
T
1
)
,
jk?
0
TT
1
k
2
2
0
6
Method (c),By dissecting the signal into simpler components,
Here,we will dissect x(t) into x
1
(t) and x
2
(t) which we know their Fourier Series (using
the result of d
k
above and the time-shifting property),
x
1
(t)
2
4? 3? 2? 1 0 1 2 3 4
t
x
2
(t)
4? 3? 2? 1 0 1 2 3 4
t
1
x(t) = x
1
(t) + x
2
(t),and let x
1
(t)
F
and x
2
(t)
F
b
k
c
k
a
k
= b
k
+ c
k
= (2)
2? 2 cos (k?
0
(1))
+ (1)
2? 2 cos (k?
0
(1))
e
jk?
0
(?1)
(3)(1)k
2
2
(3)(1)k
2
2
0 0
=
2? 2 cos k?
0
(2 + e
jk?
0
).
3k
2
0
2
Although this result looks di?erent from those found in the previous methods,further
simpli?cation will show that they are identical,
1
a
k
=
2? 2 cos k?
0
(2 + e
jk?
0
) =
3k
2
2
(2? 2 cos k?
0
)(2 + e
jk?
0
)
3k
2
2
0 0
1
=
3k
2
0
2
(2? e
jk?
0
e
jk?
0
)(2 + e
jk?
0
)
1
4 + 2e
jk?
0
2e
jk?
0
e
jk?
0
2
2e
jk?
0
0
=
3k
2
2
e
0
1
1
=
jk?
0
2
2e
jk?
0
=
jk?
0
2
2e
jk?
0
1
3k
2
0
2
4? e
3k
2
0
2
3? e
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
=
1
1? e
jk
4
3
,which is the same answer found in previous methods,
k
2
2
0
7
?
Method (d),using the analysis equation,
In the process of evaluating the analysis equation,the following integral will save us a
lot of derivation steps,
t 1
at
te
at
dt =
a
a
e,for any a = 0
2
√
1
1
1
x(t)e
jk?
0
t
dt = x(t)e
jk?
0
t
dt
a
k
=
T 3
2
1
1
0
(2? 2t)e
jk?
0
t
dt= (2 + t)e
jk?
0
t
dt +
3
0?2
1
1
1
0
e
jk?
0
t
dt + te
jk?
0
t
dt? 2 te
jk?
0
t
dr= 2
3
0?2?2
0
e
jk?
0
t
1
1
1
1 t
1 t
= 2
+
e
jk?
0
t
3?jk?
0
k
2
2
jk?
0
e
jk?
0
t
2
k
2
2
jk?
0
0 ?2 0 0?2
1?2
e
jk?
0
(1)
e
jk?
0
(?2)
1 1 (?2)
e
jk?
0
(?2)
= +
3 jk?
0
k
2
0
2
k
2
2
jk?
0
0
11 (1)
e
jk?
0
(1)
2
k
2
2
jk?
0
k
2
0
2
0
1?2
e
jk?
0
2
k
2
1
0
2
k
2
1
0
2
jk?
0
2
2
jk?
0
jk?
0
2
= + e
jk?
0
2
+ e e
3 jk?
0
jk?
0
2 2 2
k
2
0
2
e
jk?
0
+ e
jk?
0
+
jk?
0
k
2
2
0
1 21?2
e
jk?
0
2
e
jk?
0
2
e
jk?
0
= + + +
3 jk?
0
k
2
0
2
jk?
0
k
2
2
k
2
2
0 0
2 1 2
+ e
jk?
0
2 jk?
0
2 jk?
0
2
e e
jk?
0
k
2
0
2
jk?
0
1 2 3 1
+ e
jk?
0
2
=
3
k
2
0
2
e
jk?
0
k
2
0
2
k
2
0
2
1
= 3? 2e
jk?
0
jk?
0
2
3k
2
0
2
e
=
1
1? e
jk?
0
2
(remember that e
jk?
0
= e
jk?
0
2
for T = 3)
k
2
2
0
=
1
1? e
jk
4
3
,which is the same answer found in previous methods,
k
2
2
0
=
9
1? e
jk
4
3
,
4k
2
2
8
Problem 2 O & W 3.23 (a)
Given a
k
,the Fourier series coe?cients of a periodic continuous time signal with period 4,
determine the signal x(t).
The Fourier series coe?cients a
k
are given as follows:
a
0,k = 0
k
=
(j)
k
sin k?/4
,otherwise,
k?
Here are some of the facts we know about x(t),
a
0
= 0? no DC component in x(t)
T = 4?
0
= 2?/4 =?/2
k
(j)
k
sin(? k?/4) 1? sin(k?/4)
=a
k
=
k? j? k?
(? j)
k
sin(k?/4)
= = a
k
,
k?
Thus x(t) is a real signal (O&W,Section 3.5.6,p.204),
e
jk?
0
(?1)
Noting that j = e
j?/2
(j)
k
= e
j?/2
k
= e
jk?/2
= e
jk?
0
=,we can consider x(t)
to be a time-shifted version of another signal y(t) such that,
F
x(t) = y(t + 1),where y(t) b
0
= 0,b
k=0
=
sin k?/4
and a
k
= b
k
e
jk?
0
(1)
√
k?
By backtracking the derivation equation of b
k
,we can?nd the signal?
b
y(t) which has the same
k
but can have a di?erent DC level (i.e,b
0
),
b
k=0√
= b
k=0
=
sin k?/4
=
1 e
jk?/4
e
jk?/4
√
k? k? 2j
1 1
=
(4) jk(
e
jk?/4
e
jk?/4
2
)
1
2
)
1
2
1 1
jk?
0
(
1
e
jk?
0
(?
2
)
1
(1)e
jk?
0
t
dt.= e =
1
2
T jk?
0
T
The integration above suggests that
1
y?(t) =
1,?
2
< t <
1
2
0,elsewhere in the same period T=4,
9
y?(t)
3?2?1 0 1 2 3
t
1
T
2
T
2
··· ···
Note that the same conclusion can be reached by noticing that?y(t) is the same signal in
Example 3.5 (O& W,p.193) with T
1
1
=
2
and T = 4.
To?nd y(t),which has b
0
= 0,we?rst calculate
b
0
and then subtract it from?
y(t),
1 1
1/2
1
b
0
= y?(t)dt = (1)dt =
T
T
4
1/2
4
1
2 2
3 1 1
< t <
y(t) = y?(t)?
4
y(t) =
4
,?
1 1
4
,<|t| < 2.
2
3
1.5 < t <?0.5
x(t) = y(t + 1) =
4
,
1
4
,?0.5 < t < 2.5?
Sketches of y(t) and x(t) are shown below,
y(t)
3?2?1
1 2 3
t
3/4
1/4
··· ···
x(t)
3?2
1 2 3
t
3/4
1/4
1
··· ···
10
Problem 3 Determine the Fourier series coe?cients for the periodic signal x[n] depicted
below,Plot the magnitude and phase of these coe?cients,
x[n]
2
12?6 0 6 12
n
1
··· ···
1
Fundamental period N = 6?
0
=
2
6
=,
3
5 2
1
1 1
a
k
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
N 6 6
n=<N> n=0 n=?3
Notice that the last two expressions will give the same result,but the latter would take
advantage of the symmetry of some of the samples to combine them into sinusoids,
1
a
k
= (0)e
jk?
0
(?3)
+ (1)e
jk?
0
(?2)
+ (2)e
jk?
0
(?1)
+ (1)e
jk?
0
(0)
+
6
+(2)e
jk?
0
(1)
+ (?1)e
jk?
0
(2)
=
1
e
jk?
0
(?2)
e
jk?
0
(2)
+ 2e
jk?
0
(?1)
+ 2e
jk?
0
(1)
+ 1
6
1
= [(2j) sin k?
0
2 + 2(2) cos k?
0
+ 1]
6
1 2 j
= + cos k?
0
+ sin k?
0
2
6 3 3
1 2 1 2?
a
k
= + cos k + j sin k
6 3 3 3 3
1 2?
= 1 + 4 cos k + j2 sin k
6 3 3
11
Here are the values of a
k
for one period of six consecutive points (from k=-2 to k=3),
1
2?
a
2
=
6
1 + 4 cos (?2)
3
+ 2j sin (?2)
3
=
1
1 + 4 cos
2? 4?
=
1
1 + 4
1?
3
6 3
2j sin
3 6 2
2j
2
1
= (?1 + j
3) =?0.1667 + j0.2887
6
(?1)
2
+ (
3)
2
1 2?
j atan(
3,?1)
1
j
2?
3
= e = e a
2
=
3
,
a
2
=,
6 3
| |
3
Similarly,for the magnitude and phase of a
k
for k =?1 3 which are summarized in the?
table below,
k a
k
-2 -0.1667 + j 0.2887
-1 0.5000 - j 0.2887
0 0.8333
1 0.5000 + j 0.2887
2 -0.1667 -j 0.2887
3 -0.5000
a
k
||
1/3
1/
3
5/6
1/
3
1/3
3/2
a
k
2?/3
/6
0
/6
2?/3
12
The magnitude and phase of the Fourier series coe?cients were plotted below,using MAT-
LAB,
Magnitude of a
k
0
0.2
0.4
0.6
0.8
1
|a
k
|
15?10?5 0 5 10 15
k
Phase of a
k
—
a
k
(radians)
1
0.5
0
0.5
1
15?10?5 0 5 10 15
k
For your reference,the MATLAB code used to compute and plot the magnitude and phase
of the Fourier series coe?cients is shown below,
MATLAB Code:
A=inline(’1/6 +2/3*cos(k*pi/3)+j/3*sin(k*2*pi/3)’);
k=-12:12;a=A(k);am=abs(a);ap=angle(a); subplot(2,1,1);stem(k,am);grid
on;xlabel(’k’);ylabel(’|a_k|’);title(’Magnitude of a_k’);
subplot(2,1,2);stem(k,ap/pi);grid on; xlabel(’k’);ylabel(’\angle a_k
(radians)’);title(’Phase of a_k’);
MATLAB tip,you can use TEX expressions in the text of?gures,
13
Problem 4 O & W 3.29 (a)
Given a
k
,the Fourier series coe?cients of a periodic discrete time signal with period 8,
determine the signal x[n].
The Fourier coe?cients are given as follows:
k? 3k?
a
k
= cos + sin,
4 4
N=8? there are only 8 samples to compute in x[n],some of which can have a zero value,
0
= 2?/8 =?/4,
k? 3k?
a
k
= cos
4
+ sin
4
= cos k?
0
+ sin 3k?
0
=
1
2
e
jk?
0
+
1
2
e
jk?
0
+
1
2j
e
j3k?
0
1
2j
e
j3k
=
1
8
4 e
jk?
0
(?1)
+ 4 e
jk?
0
(1)
+
4
j
e
jk(?3)?
0
4
j
e
jk(3)
1
(4) e
jk?
0
(?1)
+ (4) e
jk?
0
(1)
+ (?4j) e
jk(?3)?
0
+ (4j) e
jk(3)
=
8
4
1
1
= x[n]e
jk?
0
n
= x[n]e
jk?
0
n
N 8
n=<N> n=?3
4j,n =?3
4j,n = 3
By matching the expressions of a
k
x[n] =?
4,n = ±1
0,n = 0,±2,4
4j
4 x[n] 4
4j
n
6?5?4
3?2
1 0 1 2 3 4
5 6
··· ···
4j?4j
14
Problem 5 Consider the following CT periodic signals,x(t),y(t),and z(t),
(a) Determine the fundamental frequency,period,and Fourier series coe?cients, a
k
,for
x(t),
x(t)
6?5?4?3?2?1 0 1 2 3 4 5 6
1
t
Fundamental period of x(t) = T = 4?
0
= 2?/4 =?/2.?
1 1
1
2 1
a
0
= x(t) dt = x(t) dt = =,
T
T
4
1
4 2
1
1
1
x(t) e
jk?
0
t
dt =
1
1
(1) e
jk?
0
t
dt = e
jk?
0
t
a
k
=
T
T
4
1
jk?
0
4
1
1
(e
jk?
0
e
jk?
0
) =
sin (k?
0
) sin (k
2
)
= =,
jk2? k? k?
The same result can also be found directly using Example 3.5 ( O & W,P.193),
(b) Determine the fundamental frequency,period,and Fourier series coe?cients,b
k
,for
y(t).
y(t)
6?4?2 2 4 6
1
1
t
Fundamental period of y(t) = T
= 2?
0
= 2?/2 =?.
1
b
0
= y(t)dt = 0 (? y(t) has no DC component),
T
T
15
+?
0
t j(?1)?
b
k
e
jk?
e
j
0
t
0
t
y(t) = sin?
j(1)?
0
t =
0
t
e
j?
=
j
e?
j
e
0
t
=
2j 2 2
k=
2
j
,k = 1
j
b
k
=
2
,k =?1
0,otherwise
(c) Determine the fundamental frequency and period for z(t),Also,using the results of
parts (a) and (b),determine the Fourier series coe?cients,c
k
for z(t),
z(t)
6?5
4?3
2?1
1 2 3 4 5 6
1
1
t
Fundamental period of z(t)=Fundamental period of x(t) = T = 4?
0
= 2?/4 =
.?
1
c
0
= z(t)dt = 0 (? z(t) has no DC component),
T
T
Noticing that z(t) = x(t)y(t),we can?nd c
k
using the multiplication property,How-
ever,the fundamental frequencies of x(t) and y(t) must be identical in order for the
Fourier coe?cients to match (i.e,to represent the same frequencies),The fundamen-
tal period of y(t) is 2,but if we de?ne it to be 4,then we only need to scale the
frequency components accordingly to keep the value of k?
0
constant,In our case,for
y(t),?
0
=
=?
0
/2
2
2
j
,k = 2
j
b
=
2
,k =?2
k
0,otherwise
16
2
+?
Using the multiplication property:
c
k
= a
n
b
k?n
,which looks like the discrete-time convolution,in frequency,
n=
k?n
= 0 only when k? n =Note that a
n
b
±2√
c
k
= a
k?2
b
j j
2
+ a
k+2
b
= a
k?2
j
+ a
k+2
2
2 2
=
2
(a
k+2
a
k?2
)
sin (k+2)
2
)
2
a
k+2
=
1
(k+2)?
,k =?2
sin (k
k =?2
=
(k+2)?
,√ √
1
2
,k =?2
2
,k =?2
a
k?2
=
sin (k?2)
,√
=
sin (k
√
2
)
(k?2)?
2
k = 2
(k?2)?
,k = 2
1 1
2
,k = 2
2
,k = 2
17
Problem 6 Let x(t) be a periodic signal with fundamental period T and Fourier series
coe?cients a
k
,Derive the Fourier series coe?cients of each of the following signals in terms
of a
k
,
(a) O d{ x(t? T/2)}
2
= a
k
e
jk?
0
T
(Time Shifting Property) x(t? T/2) b
k
= a
k
e
jk?
= a
k
(e
j?
)
k
= a
k
(? 1)
k
If we assume that x(t) is real,then,
= j → m{ b
k
} (Even-Odd Decomposition of Real Signals O d{ x(t? T/2)} c
k
Propriety,Table 3.1,O & W,p,206)
= j → m{ a
k
(? 1)
k
} = (? 1)
k
j→ m{ a
k
},
However,the question didn’t specify x(t) to be real,so assuming that x(t) is complex,
we will just use the general formula for?nding the Odd part of a signal,
=
1
[x(t)? x(? t)] (O &W,Sec,1.2.3,and speci?cally eq.(1.19),p.14) O d{ x(t)}
2
1 1
O d{ x(t? T/2)} = [x(t? T/2)? x(? t? T/2)] d
k
= [a
k
(? 1)
k
a
k
(? 1)
k
]
2 2
1
= (? 1)
k
(a
k
a
k
),
2
Note that for real x(t),a
k
= a
k
a
k
a
k
= 2j → = d
k
.m{ a
k
}? c
k
(b) x(T/4? t)
= a
k
(Time-Reversal Property) x(? t) c
k
= c
k
e
jk?
0
T/4
(Time-Shift in the positive time direction,i.e,delay) x(T/4? t) d
k
2
= c
k
e
jk
= c
k
(? j)
k
= a
k
(? j)
k
,
18
Problem 7 O & W 3.31 (also determine a
0
)
1,0 n 7
Let x[n] =
← ←
,x[n], periodic,N = 10,Fourier series coe?cients,a
k
,
0,8 n 9← ←
Also,let g[n] = x[n]?x[n?1],
x[n]
n
1
··· ···
0 1 2 3 4 5 6 7 8 9 10?3?2?1
g[n] 1
n
3?2?1
0 1 2 3 4 5 6
7 8
9 10
1
··· ···
1?1
x[n]?1
n
3?2?1
0 1 2 3 4 5 6 7 8
9 10
··· ···
1?1?1?1
Fundamental Period = N = 10?
0
=
.
5
9
1
1 1
a
0
= x[n] = x[n] = [1(8) + 0(2)] = 8/10 = 4/5,
N 10 10
n=<N > n=0
(a) Show that g[n] has a fundamental period of 10,
g[n + N] = x[n + N]?x[n + N?1]
x[n + N] = x[n]? g[n + N] = x[n] [n?1] = g[n]
x
[n] has a fundamental period of N = 10.? g
19
(b) Determine the Fourier series coe?cients of g[n].
1,n = 0
g[n] = x[n]? x[n? 1] = 0,1 n 7
← ←
n = 8.? 1,
1
b
k
= g[n]e
jk?
0
n
N
n=<N>
7
1
= g[n]e
jk?
0
n
,the limits were chosen to use the non-zeros near the origin
10
n=?2
1
= (? 1)e
jk?
0
(?2)
+ (1)e
jk?
0
(0)
=
1
1? e
jk?
0
2
=
1
e
jk?
0
e
jk?
0
jk?
0
)
e
10 10 10
e
jk?
0
e
jk?
0
j2
jk?
0
5
= e =
j
e
jk?
0
sin k?
0
=
j
e
jk
sin k
10 2j 5 5 5
(c) Using the Fourier series coe?cients of g[n] and the First-Di?erence property in Table 3.2,
determine a
k
for k = 0.√
From Table 3.2 (O & W,p,221),x[n]? x[n? 1] (1? e
jk(2?/N)
) a
k
= b
k
1
(1? e
jk?
0
2
b
k
= ) = (1? e
jk?
0
) a
k
10
1 1? e
jk?
0
2
1 (1? e
jk?
0
)(1 + e
jk?
0
) (1 + e
jk?
0
)? 1
a
k
= = =
1? e
jk?
0
e
jk?
0
(e
jk?
0
1) e
jk?
0
10 10 10
e
jk?
0
1
2
3 1? 1
(e
jk?
0
1
jk?
0
1
=
2
+ e
2
) =
1
e
jk?
0
2
cos k?
0
e
jk?
0
10 5 2
1
jk
3?
= e
10
cos k,
5 10
Let’s double check the result,and at the same time use another route to?nd a
k
,
Note that x[n]? 1 would have the same a
k
(only a
0
changes with a change in the DC
level of a signal),
0,0 n 7 0,0 n 7
x[n]? 1 =
← ←
=
← ←
1,8 n 9? 1,? 2← ← ← n ←? 1
1
1)e
jk?
0
n
1
(? 1)e
jk?
0
(?2)
+ (? 1)e
jk?
0
(?1)
=
1
e
jk?
0
2
+ e
jk?
0
a
k
= (x[n] =
N 10 10
n=<N>
2
)
+ e
jk?
0
(
1
jk
3?
jk?
0
(3/2)
jk?
0
(
1
1
=
1
e e
2
)
=
1
e
jk?
0
(3/2)
cos (k?
0
) =
1
e
10
cos k,
10 5 2 5 10
20
Problem 8 O & W 3.51
x[n], periodic signal with period N = 8 and Fourier series coe?cients a
k
=? a
k?4
.
y[n], periodic signal with period N = 8 and Fourier series coe?cients b
k
,
y[n] =
1 + (? 1)
n
x[n? 1]
2
Find a function f[k] such that b
k
= f[k]a
k
,
1 1
y[n] =
1 + (? 1)
n
x[n? 1] = x[n? 1] + (? 1)
n
x[n? 1],
2 2 2
n? 1] a
k
e
jk?
0
(1)
x[ (Time Shifting Property) (1)
= e
j4(
2
8
Note that (? 1)
n
= (e
j?
)
n )n
= e
j4?
0
n
,?
0
= 2?/8 =
4
(? 1)
n
x[n? 1] = e
j4?
0
n
x[n? 1] a
k?4
e
j(k?4)?
0
(Frequency Shifting Property) (2)
From (1) and (2),y[n] =
2
x[n? 1] +
11
2
(? 1)
n
x[n? 1] b
k
=
1
a
k
e
jk?
0
+
1
a
k?4
e
j(k?4)?
0
.
2 2
Substituting a
k
=? a
k?4
,
1 1
a
k
e
jk?
0
(? a
k
) e
jk?
0
b
k
= + e
j4?
0
2 2
1
= a
k
e
jk?
0
(1? e
j4?
0
) =
(1? e
j4?
0
)
e
jk?
0
a
k
2 2
(1? e
j4
4
) 1? (? 1)
e
jk
4
a
k
e
jk
=
4
a
k
=
2 2
4
a
k
e
jk
=
f[k] = e
jk
4
.?
21
