EE − 210. Signals and Systems
Solutions of homework 3∗
Spring 2010
Exercise Due Date
Week of 8th Feb.
Problems
Q1 (a) Find the fundamental period T , the fundamental frequency ω0 , and the Fourier series
coefficients ak of the following periodic signal x(t). Express x(t) as a Fourier series.
x(t)
1
...
...
-1
-0.5
0
0.5
1
t
-1
ANSWER The fundamental period is T=1, hence ω0 = 2π. First of all, the average over one
∗ LUMS
School of Science & Engineering, Lahore, Pakistan.
1
period is 0, so a0 = 0. For k 6= 0,
Z
1 1
ak =
x(t)e−jk2πt dt
T 0
Z 1
=
(1 − 2t)e−jk2πt dt
0
Z 1
−1
1
−jk2πt 1
[(1 − 2t)e
]0 −
e−jk2πt dt
=
jk2π
jkπ 0
1
1
=
+
jk2π jk2π
1
=
jkπ
−j
=
kπ
Note that the coefficients are purely imaginary and form an odd sequence, which is
consistent with our real, odd signal. The Fourier series representation of x(t) is
x(t) =
+∞
X
jk2π
ak e
k=−∞
+∞
X
−j jk2πt
=
e
kπ
k=−∞
k6=0
(b) Find the coefficients of the real form of the Fourier series of x(t):
x(t) = a0 + 2
+∞
X
[Bk cos(kω0 t) − Ck sin(kω0 t)]
k=1
ANSWER Recall that the Ck coefficients are the imaginary parts of the ak ’s. Hence
Ck = −
1
,
kπ
Bk = 0,
a0 = 0
(c) Find the total average power of the second the third harmonic components (taken
together) of the signal x(t).
ANSWER
2
2
−j −j 1
2
Pavg = 2|a2 |2 + 2|a3 |2 = 2 + 2 =
+ 2
2
2π
3π
2π
9π
13
=
18π 2
Q2 Fourier series of the output voltage of an ideal full-wave Diode Rectifier
The following circuit is an ideal full-wave rectifier. It is often used as a first stage to
generate a constant voltage from 60Hz sinusoidal line voltage for all kinds of electronic
devices
The input and output voltages are related through the memoryless, time-invariant, nonlinear system v(t) = |vin (t)|. The input voltage is as given in Q1: vin (t) = x(t)
2
(a) Sketch the rectified voltage signal v(t). Find the fundamental period T and its fundamental frequency ω0 .
ANSWER Fundamental period: T = 1, fundamental frequency ω0 = 2π
v(t)
1
...
-1.5
...
1
-0.5
0
0.5
1
1.5 t
(b) Compute and sketch the Fourier series of coefficients of v(t)
1
ANSWER First of all, the average over one period is 0.5, so a0 = . For k 6= 0,
2
Z T2
1
ak =
v(t)e−jkω0 t dt
T − T2
Z 0
Z 12
−jk2πt
=
(2t + 1)e
dt +
(1 − 2t)e−jk2πt dt
− 12
0
k
= ... =
1 − (−1)
(kπ)2
Thus the Fourier series expension of the full-wave rectified voltage is:
+∞
X
1 − (−1)k −jk2πt
e
(kπ)2
v(t) = 0.5 +
k=−∞
k6=0
(c) What is the total average power of the output voltage v(t)
3
ANSWER The total average power of the output voltage is
Z 0
Z 12
Z T2
1
2
2
|v(t)| dt =
(2t + 1) dt +
(1 − 2t)2 dt
Pout =
T − T2
− 12
0
Z 12
=2
(1 − 2t)2 dt (by symmetry)
0
1
2
1
=
[(1 − 2t)3 ]02 =
3
−6
(d) Express v(t) as its real Fourier series of the form
v(t) = a0 + 2
+∞
X
[Bk cos(kω0 t) − Ck sin(kω0 t)]
k=1
ANSWER Note that the Fourier series coefficients of v(t) are real and even. Hecne
v(t) = a0 +
= a0 +
+∞
X
k=1
+∞
X
(ak ejkω0 t + a−k e−jkω0 t )
ak (ejkω0 t + e−jkω0 t )
k=1
+∞
X
= a0 + 2
ak cos(kω0 t)
k=1
and we can identify the coefficients Bk = ak , Ck = 0. Note that v(t) is an even
function, hence it must be an infinite sum of even functions only (cosines and a
constant). This is why the Ck coefficients are 0 (the sine function is odd!!!).
Q3 Fourier Series of the output voltage of an ideal half-wave diode rectifier
The following circuit is an ideal half-wave rectifier.
The output voltage is given by the memoryless, nonlinear, time-invariant system:
vin (t),
vin > 0
v(t) =
0,
vin (t) ≤ 0
Suppose that the periodic input voltage is vin (t) = A sin(ω1 (t) volts, ω1 = 2π/T1 .
4
(a) Sketch the half-wave rectified voltage signal v(t). Find its fundamental period T and
its fundamental frequency ω0
ANSWER The fundamental period (T ) is T1 and the fundamental frequency (ω0 ) is ω1
(b) Compute the Fourier series coefficients of v(t) and write the voltage as Fourier series
ANSWER
1
ak =
T
=
Z
T
−jkω0 t
v(t)e
0
A
2jT
Z
A
2jT
Z
T
2
A
dt =
T
Z
T
2
sin(ω0 t)e−jkω0 t dt
0
(ejω0 t − e−jω0 t )e−jkω0 t dt
0
π
ω0
(ejω0 (1−k)t − e−jω0 (1+k)t dt
0
A (−1)k + 1
= ... =
2π
1 − k2
=
Note that we have to make sure that this expression is finite for k = ±1 (L’Hopital’s
rule)


d −jkπ
+ 1)
A −jπe−jkπ
jA
A  dk (e

=
=−
a1 =

 d
2π
−2k
4
2π
k=1
(1 − k 2 )
dk
k=1
jA
A −jπe−jkπ
=
a−1 =
2π
−2k
4
k=−1
Notice that these are imaginary whereas all the other coefficients are real !! Thus the
Fourier series expension of the half-wave rectified sinusoid is
+∞
X
A (−1)k + 1 jkω0 t
e
2π
1 − k2
v(t) =
k=−∞
Q4 Fourier series of a rectangular impulse train
Consider the rectangular impulse train
x[n]
1
...
...
-10
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
n
(a) Compute the Fourier series coefficients of the signal. Plot its magnitude and phase.
5
ANSWER Fourier series coefficients are
4
X
X[k] =
e−jk(2π/10)n
n=0
1 − e−jk(2π/10)5
1 − e−jk(2π/10)
sin πk/2
= e−j(4πk/10)
sin πk/10
=
(b) Plot the Fourier Series of the signal for different number of terms considered in the
summation. What is the minimum number of terms for which the Fourier series
resembles the original signal?
z[n]
1
0
1
2
3
4
5
6
7
8
9
n
(c) What will be the Fourier series of the signal: x[n]z[n]?
ANSWER The Fourier coefficients of the signal y[n]=x[n]z[n] can be computed by
Yk =
X
Xl Zk−l
l=<5>
The Fourier series coefficients of x[n] and z[n] are equal. Using the results from part
(a) of the problem, Fourier series of y[n] can be calculated as:
X
Yk =
Xl Xk−l
l=<5>
"
=
=
X
4
X
l=<5>
n=0
!
−jl(2π/10)n
e
4
X
!#
−j(k−l)(2π/10)n
e
n=0
X sin (πl/2)
sin (π[k − l]/2)
e−j(4π/10)l
e−j(4π/10)[k−l]
sin (πl/10)
sin (π[k − l]/10)
l=<5>
= e−j(4πk/10)
X
l=<5>
sin (πl/2) sin (π[k − l]/2)
sin (πl/10) sin (π[k − l]/10)
6