ECE1004 Signals and Systems
Fall Semester 16-17
Module 3: Fourier Analysis of CT Signals
Faculty: Dr. R. Neelakandan
Key Points
Note
• The material is NOT SUFFICIENT for examination.
• This contains only the key points and summary of the lectures.
• This material is COMPLEMENTARY WITH the class notes.
• This material does not contain the problems solved in the class.
• Practice Set problems at the end of each material is STRONGLY RECOMMENDED TO GET
A (JUST) PASS IN THE COURSE.
1
Topics
1. Quadrature Fourier Series
2. Exponential Fourier Series
3. Relation between the Coefficients
4. Important Identities
5. Parseval’s Theorem
6. Examples
2
Quadrature Fourier Series
Any periodic signal x(t) can be represented by Fourier series expansion. There are several forms of
Fourier series. In this section, the Quadrature Fourier series expansion is given.
x(t) = a0 +
∞
X
ak cos(2πkf0 t) +
k=1
∞
X
bk sin(2πkf0 t)
(1)
k=1
Z
1
x(t)dt
T0 T0
Z
2
ak =
x(t)cos(2πkf0 t)dt
T0 T0
Z
2
bk =
x(t)sin(2πkf0 t)dt
T0 T0
a0 =
(2)
(3)
(4)
where f0 = T10 is the fundamental frequency. The term a0 is referred as DC component of the signal.
1
3
Exponential Fourier Series
Another form of Fourier representation, Exponential Fourier series is given below.
x(t) =
∞
X
ck ej2πkf0 t
(5)
x(t)e−j2πkf0 t dt
(6)
k=−∞
ck =
4
1
T0
Z
T0
c−k = c∗k
(7)
|ck | = |c−k |
(8)
Relation between the Coefficients
c0 = a 0
ak − jbk
ck =
2
ak + jbk
c−k =
2
ak = 2Re{ck }
(9)
(10)
(11)
(12)
bk = −2Im{ck } = 2Im{c−k }
(13)
Note: If both the quadrature and exponential Fourier series coefficients have to be computed, find the
one which is easy to compute and for another, use the relation.
5
Important Identities
Z
cos(2πmf0 t)dt = 0
(14)
sin(2πmf0 t)dt = 0
(15)
T
Z 0
T0
Z
cos(2πmf0 t)cos(2πnf0 t)dt = 0,
m 6= n
(16)
sin(2πmf0 t)sin(2πnf0 t)dt = 0,
m 6= n
(17)
Z T0
ZT0
T0
6
cos(2πmf0 t)sin(2πnf0 t)dt = 0, ∀m, n
Z
T0
cos2 (2πmf0 t)dt =
2
Z T0
T0
sin2 (2πmf0 t)dt =
2
T0
Z
T0 m = n
ej2π(m−n)f0 t =
0
m 6= n
T0
(18)
(19)
(20)
(21)
Parseval’s Theorem
The Parseval’s theorem relates the power of a periodic signal to its individual frequency components.
The theorem is given below.
2
Theorem 1. For any periodic signal x(t), the power of a signal is equal to the sum of powers of its
individual frequency components, i.e.
P =
1
T0
Z
∞
X
|x(t)|2 dt =
T0
|ck |2
(22)
k=−∞
Proof :
∞
X
x(t) =
ck ej2πkf0 t
(23)
k=−∞
"
∗
x (t) =
#∗
∞
X
ck e
j2πkf0 t
∞
X
=
k=−∞
c∗k e−j2πkf0 t
(24)
k=−∞
Z
1
|x(t)|2 dt
T0 T0
Z
1
x(t)x∗ (t)dt |z|2 = zz ∗ = z ∗ z
=
T0 T0
( ∞
)
Z
X
1
∗ −j2πkf0 t
=
ck e
dt from (24)
x(t)
T0 T0
P =
(25)
(26)
(27)
k=−∞
Interchanging the order of summation and integration
Z
∞
X
1
=
c∗k
x(t)e−j2πkf0 t dt
T0 T0
=
k=−∞
∞
X
∞
X
c∗k ck =
k=−∞
|ck |2
(28)
(29)
(30)
k=−∞
Therefore
P =
7
1
T0
Z
∞
X
|x(t)|2 dt =
T0
|ck |2
(31)
k=−∞
Calculation of Fourier Series Coefficients
7.1
Find the quadrature Fourier series coefficients for the signal shown in Fig. 1.
x(t) = a0 +
∞
X
ak cos(2πkf0 t) +
k=1
∞
X
bk sin(2πkf0 t)
(32)
k=1
Z
1
x(t)dt T0 = 0.5
T0 T0
Z 0.5
0.5
=2
e−t dt = −e−t 0 = 1 − e−0.5 = 0.79,
0
Z
2
ak =
x(t)cos(2πkf0 t)dt
T0 T0
Z 0.5
= 4.
e−t cos(4πkt)dt
a0 =
(33)
(34)
(35)
(36)
0
The integral of the above form can be simplified using
Z
eat eat cos(bt)dt = 2
acos(bt)
+
bsin(bt)
a + b2
3
(37)
1
0.9
0.8
exp(-t)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1.5
-1
-0.5
0
0.5
1
1.5
Figure 1: Periodic exponential signal
By noting that a = −1, b = 4πk,
#
0.5
e−t
− cos(4πkt) + (4πk)sin(4πkt)
ak = 4.
1 + (4πk)2
0
"
#
4
−0.5
0
=
e
[−cos(2πk) + (4πk)sin(2πk)] − e [−cos(0) + 4πksin(0)]
1 + (4πk)2
h
i
4
−0.5
0
e
[−1
+
0]
−
e
[−1
+
0]
=
1 + (4πk)2
4
(1 − e(−0.5) )
=
1 + (4πk)2
1.576
=
1 + (4πk)2
"
Similarly, the coefficients bk can be computed as,
Z
2
bk =
x(t)sin(2πkf0 t)dt
T0 T0
Z 0.5
= 4.
e−t sin(4πkt)dt
(38)
(39)
(40)
(41)
(42)
(43)
(44)
0
The integral of the above form can be simplified using
Z
eat eat sin(bt)dt = 2
asin(bt)
−
bcos(bt)
a + b2
4
(45)
Again by noting that a = −1, b = (4πk),
#
0.5
e−t
bk = 4.
− sin(4πkt) − (4πk)cos(4πkt)
1 + (4πk)2
0
#
"
−4
−0.5
0
e
[sin(2πk) + (4πk)cos(2πk)] − e [sin(0) + 4πkcos(0)]
=
1 + (4πk)2
h
i
−4
−0.5
=
e
(4πk)
−
1(4πk)
1 + (4πk)2
4
(4πk)(1 − e−0.5 )
=
1 + (4πk)2
6.32πk
=
1 + (4πk)2
"
(46)
(47)
(48)
(49)
(50)
Therefore the periodic signal x(t) can be written as
x(t) = 0.79 +
∞
X
k=1
∞
X 6.32πk
1.576
cos(4πkt) +
sin(4πkt)
2
1 + (4πk)
1 + (4πk)2
(51)
k=1
7.2
Find the Quadrature and Exponential Fourier series coefficients for the signal given below. Verify the
Parseval’s theorem for the given signal. x(t) = 4 + 3cos(4πt) − 5sin(20πt).
Solution:
Given Signal: x(t) = 4 + 3cos(4πt) − 5sin(20πt).
1. Addition of DC constant will not affect the period of the signal.
2. If the sum of periodic signal is periodic, 1 the overall fundamental period T0 = LCM (T1 , T2 ), f0 =
GCD(f1 , f2 ).
1
3. For the given signal, T1 = 0.5, T2 = 0.1. Therefore T0 = LCM ( 21 , 10
) = 12 , f0 = GCD(2, 10) = 2.
4. Write the signal in terms of the integer multiple of its fundamental frequency components. Here
f0 = 2
x(t) = 4 + 3cos(4πt) − 5sin(20πt)
(52)
x(t) = 4 + ak cos(2πkf0 t) + bk sin(2πkf0 t)
(53)
x(t) = 4 + 3cos(2π.1.2.t)−5sin(2π.5.2.t)
(54)
Compare with the general form of Quadrature Fourier series and it can be seen that
a0 = 4
(55)
a1 = 3
(56)
b5 = −5
(57)
5. Now from the relation between the coefficients, find out the exponential Fourier series coefficients.
ak − jbk
2
ak + jbk
c−k =
2
3
3
j5
−j5
c0 = 4, c1 = , c−1 = , c5 = , c−5 =
2
2
2
2
ck =
To verify the Parseval’s theorem:
1 Sum of CT periodic signals may not be periodic sometimes. Check the material for Module-1 for more details.
5
(58)
(59)
(60)
1. First compute the power of the signal x(t) using the general formula:
1
T0
Z T0
1
T0
Z T0
1
=
T0
(Z
P =
=
|x(t)|2 dt
(61)
0
2
[4 + 3cos(4πt) − 5sin(20πt)] dt
(62)
0
T0
Z T0
16dt + 9
0
Z T0
2
cos (4πt)dt + 25.
0
sin2 (20πt)dt
[Note: 4πt = 2π.1.f0 t, 20πt = 2π.5.f0 t]
0
Z T0
Z T0
Z T0
)
cos(4πt)dt −2.3.5
cos(4πt)sin(20πt)dt −2.5.4
sin(20πt)dt
0
0
|
{z
}
{z
}
{z
}
|
|
=0
=0
=0
(
)
T0
1
T0
16T0 + 9. + 25.
( See the Identities for the values of the integrals)
=
T0
2
2
+ 2.4.3
(63)
0
P = 16 +
9 25
+
= 33Watts
2
2
(64)
(65)
2. Now compute the power using the Exponential Fourier series coefficients:
P =
∞
X
|ck |2
(66)
k=−∞
= |c0 |2 + |c1 |2 + |c−1 |2 + |c5 |2 + |c−5 |2
P = 16 +
9 9 25 25
+ +
+
= 33Watts
4 4
4
4
(67)
(68)
3. From (65) and (68), Parseval’s Theorem is verified.
8
Convergence Rate of Fourier Series Coefficients
1. The Fourier series coefficients of a periodic signal usually decay with the harmonic index k (k th
frequency component) as k1n .
2. Decay rate of the FS coefficients can be directly determined without computing the coefficients.
3. FS coefficients decay at a rate of k1n where n is the required number of differentiations of the
signal until impulses to first appear.
4. Larger n means smoother signal. n = 1 means, the signal has jumps or discontinuities.
5. For periodic signals with jumps, the decay rate is the k1 as the first differentiation itself results in
impulses and n = 1.
Examples
Find the convergence rate of the follwing signals.
1. x(t) = rect(t − 0.5).
2. x(t) = tri(t − 0.5)
Solution:
The plot of the signals and their successive differentiation is shown in the Fig. 2. (Refer Material for
Module 1 to see how to draw these pulses). Therefore for
x(t) = rect(t − 0.5), n = 1 and the decay rate is k1 and
for x(t) = tri(t − 0.5) n = 2 and the decay rate is k12 .
6
Figure 2: Signals and their successive differentiation
9
Gibbs Phenomenon
1. Fourier series coefficients are given for infinite length, i.e. k = 0 to ∞,
x(t) = a0 +
∞
X
ak cos(2πkf0 t) +
∞
X
bk sin(2πkf0 t)
k=1
k=1
2. In practice, only finite number of coefficients can be stored, say N coefficients.
3. Now the question is, can the signal x(t) be reconstructed perfectly with only a finite number of
coefficients?
4. The answer is YES BUT EXCEPT THE SIGNALS WITH JUMPS (DISCONTINUITIES). All
the signals which do not have jumps can be perfectly reconstructed with a finite number of
coefficients.
5. PERFECT RECONSTRUCTION IS NOT POSSIBLE FOR SIGNALS WITH JUMPS.
6. In previous section it is mentioned that decay rate of a signal equals to k1n where n is the number
of derivatives to produce the first impulse. Therefore, it can be said that for signals with FS
coefficients with decay rate k1 , perfect reconstruction is not possible.
Now let us see what happens in the reconstructed signal when the signal has discontinuities. Let us
write the reconstructed signal with N coefficients as xN (t). The following Fig. 3 shows the effect of a
rectangular pulse reconstruction. The reconstructed signal xN (t) is given by
xN (t) = a0 +
N
X
ak cos(2πkf0 t) +
k=1
N
X
bk sin(2πkf0 t)
k=1
i.e. instead of infinity coefficients, only a finite number N coefficients are used to reconstruct the signal.
The reconstructed signal (blue color) for various values of N is given in the Fig. 3 and Fig. 4.
The following are the inferences made, which are referred to as Gibbs effect.
1. The reconstructed signal has ripples in the flat portion of the signal.
7
N= 1
N= 3
2
x N(t)
x N(t)
2
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
-0.5
Time
N= 7
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
2
x N(t)
x N(t)
2
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
-0.5
Time
N= 49
0
Time
N= 100
2
x N(t)
2
x N(t)
0
Time
N= 19
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
Time
-0.5
0
Time
Figure 3: Signals with Jumps and Gibbs effect
2. The ripples settle down to the original signal value as N increases. See in the Fig. 4 and for higher
values of N the red and blue color lines merge with each other on the flat portion of the pulse.
3. Near the discontinuities, there are ’spikes’ upward and downward in the reconstructed signal. The
height of the spike is 9% of the discontinuity.
4. In the example shown in the figures, the height of the discontinuity is 2. Hence there are four
spikes, two at t = −1 and two at t = 1 where the discontinuity occurs. The height of the spike is
2(0.09) = 0.18, therefore the downward spikes go upto −0.18 and upward spikes go upto 2.18.
5. The reconstructed signal converges to the midpoint of the discontinuity at the points of discontinuity. See Fig. 3, in each plot the blue line crosses the original signal in red line exactly at the
mid point. For higher values of N it is not directly visible.
6. For any larger value of N the SPIKES DO NOT DISAPPEAR. See Fig. 4.
Example
For the signal x(t) = t.rect(t − 0.5), reconstructed with finite number of Fourier series coefficients,
answer the following questions.
1. How many spikes are there in the reconstructed signal? Mention the locations and their heights.
2. What is the value of the reconstructed signal at t = 1?
3. Draw the reconstructed signal with labels.
Solution:
The given signal is shown in Fig. 5.
1. There is a discontinuity at t = 1. Hence there are two spikes upward and downward at t=1. The
height of the spike is 9% of the discontinuity height. Here the height of the discontinuity is 1 and
therefore the spikes go upward to 1.09 and downward to -0.09.
8
N= 100
N= 200
2
x N(t)
x N(t)
2
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
-0.5
Time
N= 300
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
2
x N(t)
x N(t)
2
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
-0.5
Time
N= 500
0
Time
N= 1000
2
x N(t)
2
x N(t)
0
Time
N= 400
1
0
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
Time
-0.5
0
Time
Figure 4: Signals with jumps and Gibbs effect
2. Reconstructed signal converges to the original signal value at all points other than discontinuity.
At discontinuity, the reconstructed signal converges to midpoint of the discontinuity. Therefore
the reconstructed signal converges xN (t) = 0.5 at t = 1.
3. The reconstructed signal is given in Fig. 6.
10
Practice Set
10.1
Fourier Series I
Find the Quadture and Exponential Fourier series coefficients for the following signals.
1. x(t) = 2 − 4cos(8πt) + 10sin(12πt)
2
2. x(t) = [cos(t) + 2cos(2t)]
3. x(t) = 4 + 2sin(4πt) + 3sin(16πt) + 4cos(16πt)
10.2
Fourier Series II
the Sawtooth
signal
shown in Fig. 7, compute the exponential Fourier series coefficients. Hint: Use
RFor at
te dt = eat at − a12
10.3
Parseval’s Relation
For each of the Signal, compute their Exponential Fourier series coefficients. Verify the Parsevals
relation.
9
Figure 5: Signal for Gibbs example
Figure 6: Solution: Reconstructed signal for Gibbs example
1. x(t) = 2 − 4cos(8πt) + 10sin(12πt)
2
2. x(t) = [cos(t) + 2cos(2t)]
3. x(t) = 4 + 2sin(4πt) + 3sin(16πt) + 4cos(16πt)
10.4
Convergence Rate and Gibbs Effect
Each of the following signals has fundamental period of T0 = 3 and the description of the signal over
one period is given below.
−t
e rect(t − 0.5) 0 ≤ t ≤ 1
1. x(t) =
0
1≤t≤3
tri(t) 0 ≤ t ≤ 1
2. x(t) =
0
1≤t≤3
u(t) − 2u(t − 1) 0 ≤ t ≤ 2
3. x(t) =
0
2≤t≤3
For each of the signal:
• draw the periodic signal.
• compute the convergence rate and state whether Gibbs effect is present or not.
• If Gibbs effect is present, mention the location of the spikes, height of the spike and convergence
value at the point of discontinuity.
10
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.5
1
1.5
2
2.5
3
3.5
Figure 7: Sawtooth signal
• draw the reconstructed signal with necessary labels.
.
11
4
4.5
5