Fourier Series and Parseval’s Relation
Çağatay Candan
Dec. 22, 2013
We study the exam problem (EE 301 MT2, Fall2013-14) in some detail to illustrate some
connections between Fourier series, Parseval’s relation and RMS values.
Q1. (20 pts)
a) The signal x (t ) = sin(2π t ) is the
input to a half-wave rectifier circuit with
the following input-output relationship:
 x(t ), x(t ) ≥ 0
xhalf (t ) = 
.
other
0 ,
si n(2: t)
1
0
-1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
1.4
1.6
1.8
2
1.4
1.6
1.8
2
Half-wave rectified
1
0.5
Find the Fourier series coefficients of
xhalf (t ). (Hint: You may use Euler’s
0
0
0.2
0.4
0.6
0.8
1
1.2
Full-wave rectified
1
relation to express sin( 2πt ) .)
0.5
0
0
0.2
0.4
0.6
0.8
1
t (sec.)
1.2
b) The signal x(t ) = sin(2π t ) is the input to a full-wave rectifier circuit with the following
 x(t ), x(t ) ≥ 0
input-output relationship: x full (t ) = x(t ) = 
.
− x(t ), other
Express the Fourier series coefficients of x full (t ) in terms of the Fourier series coefficients found
in part (a). (Note: You can call coefficients in part (a), ak , and solve part (b) using ak ’s .)
Solution: (The solution is more detailed than it needs to be.)
a) The signal x(t ) is periodic with T = 1 . The Fourier series coefficients can be
1
2π
x(t )e jkωot dt where ω o =
= 2π rad/sec.
∫
T 0
T
T
found through the relation a k =
1
1/ 2
a k = ∫ x(t )e − j 2πkt dt = ∫ sin(2πt )e − j 2πkt dt
0
=
=
0
∫ (e
1
2j
1/ 2
1
2j
1/ 2
j 2πt
)
− e − j 2πt e − j 2πkt dt
0
∫ (e
− j 2π ( k −1) t
)
− e − j 2π ( k +1) t dt
(∗)
0
↓t =1 / 2
1  e − j 2π ( k −1) t
e − j 2π ( k +1)t 


=
−
2 j  − j 2π (k − 1) − j 2π (k + 1)  ↓t =0
(∗∗)
 e − jπ ( k −1) − 1 e − jπ ( k +1) − 1 


−
k + 1 
 k −1
At this point, it should be noted that k index of the sequence a k is an integer and
=
1
4π
therefore, e − jπk = (−1) k . Substituting e − jπk = (−1) k into the last relation, we can get the
following:
1  e − jπ ( k −1) − 1 e − jπ ( k +1) − 1 


ak =
−
4π  k − 1
k + 1 
=
1
4π
 1 + (− 1)k 1 + (− 1)k
−
+

k −1
k +1





1 + (− 1)  1
1 
=
−


4π  k + 1 k − 1 
k
1 + (− 1)
1
=
2π 1 − k 2
1

k : even

= π 1 − k 2

0
k : odd
k
(
)
It appears from the last relation that for k = ±1 , we have a k = 0 . This is not true, since
the equation (**) given above shows that the derived a k values are only valid when
k ≠ ±1 . We need to examine the cases of k = ±1 separately.
From the equation (*), given above, we can express the coefficient a1 as
a1 =
1
2j
∫ (e
1/ 2
− j 2π ( k −1) t
− e − j 2π ( k +1)t
)
↓ k =1 dt =
0
show that a −1 =
1
2j
∫ (1 − e
1/ 2
0
− j 4πt
)dt = 4j1 = −4 j . Similarly, we can
j
. Given all, the FS coefficients can be written as:
4
1

k : even
π 1 − k 2

k =1 .
ak =  − j / 4
 j/4
k = −1

0
k : other

(
)
Now, we can write the Fourier series expansion of the half-wave rectified sinusoidal
signal:
x Half (t ) =
=
∞
∑a e ω
j
k = −∞
o kt
k
(
)
∞
1 jω o t
e
− e − jωot + ∑ a k e jωo kt
4j
k = −∞ ,
k :even
(
=
∞
1
(2 j sin(ω o t ) ) + a0 + ∑ a k e jωokt + e − jωokt
4j
k = 2,
=
sin(ω o t ) 1 2
cos(ω o kt )
+ + ∑
2
π π k = 2, 1 − k 2
)
k :even
∞
k :even
It is always useful (and fun) to verify the expansion with a few lines of Matlab code:
t=linspace(-2.5,2.5,1024);
FSterms=10;
T=1;
w0=2*pi/T;
out = 1/pi + 1/2*sin(w0*t); %First few terms
for k=2:2:FSterms, %Remaining terms in the series
ak = 1/pi/(1-k^2);
out = out + ak*2*cos(w0*k*t);
end;
plot(t,sin(w0*t),'-.'); hold all
plot(t,out); hold off;
legend('Sine wave','Half wave rectified');
1.5
Sine wave
Half wave rectified
1
0.5
0
-0.5
-1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Figure: The plot generated by the given Matlab code
b) If x Half (t ) ↔ a k , the full wave rectified sinusoid signal can be expressed as follows.
x Full (t ) = x Half (t ) + x Half (t − 1 / 2) =
∞
∑a e
k = −∞
jk 2πt
k
+
∞
∑ae
k = −∞
k
jk 2π ( t −1 / 2 )
=
∞
∑a
k = −∞
k
(1 + e − jk 2π / 2 )e jk 2πt
So,
x Full (t ) =
∞
∑ a k (1 + e − jkπ )e jk 2πt =
k = −∞
∞
∑ 2a e
k = −∞
k even
jk 2πt
k
We should be a little careful in this calculation, since the period of x Half (t ) is 1 second,
while the period of full wave rectified sine signal is 1/2. For T0 = 1 / 2 , the corresponding
fundamental frequency is ω 0 =
2π
= 4π .
T0
When writing
x Full (t ) =
∞
∑ 2a e
k = −∞
k even
k
jk 2πt
2a k , for k even
,
0 , else
, with coefficients c k = 
(1)
it is implicitly assumed that x Full (t ) is also periodic with 1 second. It is indeed true that
xFull (t ) is periodic with 1 second, but this is not the fundamental period ( T0 = 1 / 2 ) and 2π
is not the fundamental frequency ( ω 0 = 4π ). 2π is equal to half of the fundamental
frequency ; therefore, ck ’s are not the FS coefficients of x Full (t ) . By substituting k = 2k ′
in (1), one can easily find the FS coefficients bk of xFull (t ) :
x Full (t) =
∞
∑ 2a e
k
k = −∞
k even
jk 2πt
∞
= ∑ 2a2ke
jk 4πt
=
k =−∞
∞
∑b e
jk 4πt
k
bk = 2 a2 k , ∀k
,
(2)
k = −∞
[Note that for any periodic signal x p (t ) with fundamental frequency ω0 , one can express
x p (t ) as the sum of complex sinusoidals at all frequencies that are multiples of ω 0 / L ,
for L=2, 3, …; however, corresponding coefficients, ck , are all zero whenever k ≠ mL
(m is an integer). Above example corresponds to L = 2.
In order to preserve the uniqueness of the FS representation x p (t ) ↔ a k , and to safely
2
2
talk about the k’th harmonic power of x p (t ) looking at ak + a−k ; only the FS coefficients in the expansion with respect to “complex sinusoidals at all multiples of frequencies ω 0 , where ω 0 is the fundamental period” are called the FS coefficients of x p (t ) .]
In order to obtain the FS expansion in detail, ak ’s found in part (a) can be substituted
into (2).
x Full (t) =
∞
∑b e
jk 4πt
k
k =−∞
=
∞
∑ 2a
2k
e j 4πkt
k =−∞
=
=
2
∞
1
∑ 1 − 4k
π k =−∞
2
e j 4πkt
∞

2
1
cos(4
π
kt)
1+ 2∑

π
1 − 4k 2

k =1
It should be noted that the last relation is the conventional expansion of x Full (t ) and
2 /π
corresponds to the coefficient of the k’th harmonic ( e j 4πkt ). We can use
1 − 4k 2
Matlab to verify our findings.
t=linspace(-2.5,2.5,1024);
FSterms=10;
out = 2/pi; %DC term
for k=1:FSterms, %Remaining terms in the series
ak = 4/pi/(1-4*k^2);
out = out + ak*cos(4*pi*k*t);
end;
plot(t,sin(w0*t),'-.'); hold all
plot(t,out); hold off;
legend('Sine wave','Full wave rectified');
1
Sine wave
Full wave rectified
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
Figure: The plot generated by the given Matlab code
2
2.5
Parseval’s Relation, RMS Values and Harmonic Series:
Let’s remember the definition for the RMS value of a periodic waveform:
x RMS =
1
T
T
∫ [x(t )] dt .
2
0
In the terminology of electrical engineering, x (t ) is considered to be the periodic
waveform representing either current or voltage of an R Ω resistor. Then, the average
[v ]2
2
power dissipated over the resistor becomes, PAVG = RMS or PAVG = R[i RMS ] .
R
Some of the typical periodic waveforms utilized in circuit applications are given in
figure below. (This figure is provided to refresh our memory on RMS calculations.)
Square Waveform
Sinousoidal Waveform
Sawtooth Waveform
1
1
1
0.8
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
0
0
-0.2
-0.2
-0.2
-0.4
-0.4
-0.4
-0.6
-0.6
-0.6
-0.8
-0.8
-0.8
-1
-1
-1
0
5
0
5
0
5
Figure: Some periodic waveforms typically utilized in circuit applications
We know that the square wave with the amplitude A has the RMS value of A;
A
sinusoidal waveform has the RMS value of
and the sawtooth waveform has the
2
 1 1 
A
value of
. The coefficients 1,
,
 scaling the amplitude in this description
3
 2 3
are called the RMS scaling factors.
In the rest of this section, we calculate the RMS scaling factor for the half wave
rectified sinusoidal signal and establish a connection with the Parseval’s relation.
We start with the calculation of the RMS value. It should be clear that x Half (t )
provides half the average power of a sinusoidal signal; hence its scaling factor should
1
1
1
be
x
= . We can easily verify this guess with the following
2
2 2
1
2
1
2
1
[sin(2πt )]2 dt = 1 ∫ 1 − cos(4πt ) dt = 1 . (Very similarly,
∫
10
10
2
2
calculation: x Half , RMS =
the full wave rectified signal has the RMS scaling factor of
1
2
.)
∞
1
2
2
Parseval’s relation states that ∫ x(t ) dt = ∑ a k . Given our refreshed knowledge
T 0
k = −∞
on the RMS values, we can also write the Parseval’s relation as
T
∞
1
2
2
2
x
(
t
)
dt
=
a k = ( x RMS ) .
∑
∫
T 0
k = −∞
T
Half-wave rectified signal: We have previously found the RMS value of the unit
1
amplitude half-wave rectified signal as . In addition, we have also found the FS
2
coefficients of this signal as
1

k : even
π 1 − k 2

k =1 .
ak =  − j / 4
 j/4
k = −1

0
k : other

(
1
T
Then Parseval’s relation
)
∞
T
∫
∑a
x(t ) dt =
2
k = −∞
0
= ( x RMS )
2
k
2
gives us the following
identity:
∞
1
= .
4
k = −∞
The summation given above can be rewritten as follows:
∞
1
2
2
2
a 0 + 2 a1 + 2 ∑ a k = .
4
k =2,
∑a
2
k
k :even
Substituting a 0 =
1
π
also be written as
, a1 =
2
π
2
1
, we reach
4
∞
1
∑ (1 − k
k = 2,
k :even
2 2
)
=
∞
∑a
k = 2,
k :even
2
k
=
1 1
. The same relation can
−
8 π2
1 1
and simplified to the following,
−
8 π2
∞
1
∑ (1 − k
k = 2,
k :even
2 2
)
=
π 2 −8
16
and the summation on the left hand side can be compactly
expressed as:
∞
1
∑ (1 − 4k
k =1
2 2
)
=
π 2 −8
16
.
The end result of this calculation is the summation identity given above. These
identities involving reciprocal of integer powers are difficult to prove with elementary
means. Therefore, we can not provide any other arguments for the validity of this
identity; but we can always use Matlab to numerically examine the correctness of the
this identity:
>> kvec=1:10;
>> partial_sum = sum(1./(1-4*kvec.^2).^2)
partial_sum =
0.1168
>> (pi^2-8)/16
ans =
0.1169
It seems that everything is in order.