SOLUTIONS MANUAL FOR
SIGNALS AND SYSTEMS
A MATLAB® Integrated Approach
by
Oktay Alkin
CRC Press
Taylor & Francis Group
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Version Date: 20140611
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2
Contents
1 Signal Representation and Modeling
5
2 Analyzing Continuous-Time Systems in the Time Domain
41
3 Analyzing Discrete-Time Systems in the Time-Domain
85
4 Fourier Analysis for Continuous-Time Signals and Systems
121
5 Fourier Analysis for Discrete-Time Signals and Systems
149
6 Sampling and Reconstruction
189
7 Laplace Transform for Continuous-Time Signals and Systems
207
8 z-Transform for Discrete-Time Signals and Systems
269
9 State-Space Analysis of Systems
331
10 Analysis and Design of Filters
367
11 Amplitude Modulation
401
3
4
CONTENTS
Chapter 1
Signal Representation and Modeling
replacemen
1.1.
The signal xa (t)
3
a.
Amplitude
2
1
0
−1
−1
1
0
2
3
4
5
t (sec)
The signal xb (t)
3
b.
Amplitude
2
1
0
−1
−2
−1
0
1
3
2
4
5
t (sec)
1.2.
a.
b.

0,





 2t + 2 ,
−t + 2 ,
x a (t ) =



1,



−t + 3 ,
t < −1 or t > 3
−1 < t < 0
0<t <1
1<t <2
2<t <3





0,
1.5t + 1.5 ,
x a (t ) =

−1.5t
+ 1.5 ,



1.5t − 4.5 ,
5
t < −1 or t > 3
−1 < t < 0
0<t <2
2<t <3
6
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
6
1.3.
The signal g1 (t)
4
a.
t < −1 or t > 3
−1 < t < 0
0<t <1
1<t <2
2<t <3
Amplitude
3

0,





3.5t
+ 3.5 ,

−2.5t + 3.5 ,
g 1 (t ) =



−1.5t
+ 2.5 ,



0.5t − 1.5 ,
2
1
0
−1
−2
−1
0
1
2
3
4
2
3
4
2
3
4
t (sec)
The signal g2 (t)
4
b.
0,
2
3t + 6t + 3 ,
1.5t 2 − 4.5t + 3 ,
g 2 (t ) =



−1.5t + 1.5 ,



−1.5t 2 + 9t − 13.5 ,
t < −1 or t > 3
−1 < t < 0
0<t <1
1<t <2
2<t <3
Amplitude
3







2
1
0
−1
−2
−2
−1
0
1
t (sec)
The signal g3 (t)
7
c.
3,
2.5t + 5.5 ,
−0.5t + 5.5 ,
g 3 (t ) =



1.5t + 3.5 ,



−3.5t + 13.5 ,
t < −1 or t > 3
−1 < t < 0
0<t <1
1<t <2
2<t <3
Amplitude
6







5
4
3
2
−2
−1
0
1
t (sec)
7
1.4.
The signal g1 (t)
2
Time reversal
g 1 (t ) =x (−t )
Amplitude
1
a.
0
−1
−2
−10
−5
0
5
10
5
10
5
10
5
10
t (sec)
The signal g2 (t)
2
Time scaling
Amplitude
1
b.
0
−1
g 2 (t ) =x (2t )
−2
−10
−5
0
t (sec)
The signal g3 (t)
2
c.
Time scaling
µ ¶
t
g 3 (t ) =x
2
Amplitude
1
0
−1
−2
−10
−5
0
t (sec)
The signal g4 (t)
2
d.
g 4a (t ) =x (−t )
Step 2: Time shifting
g 4 (t ) =g 4a (t − 3) = x (−t + 3)
Amplitude
1
Step 1: Time reversal
0
−1
−2
−10
−5
0
t (sec)
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
8
The signal g5 (t)
e.
2
Step 1: Time scaling
1
Amplitude
µ ¶
t
g 5a (t ) =x
3
−1
Step 2: Time shifting
g 5 (t ) =g 5a (t − 1) = x
0
µ
(t − 1)
3
¶
−2
−10
−5
0
5
10
5
10
5
10
t (sec)
The signal g6 (t)
2
f.
g 6a (t ) =x (4t )
Amplitude
1
Step 1: Time scaling
0
−1
Step 2: Time shifting
g 6 (t ) =g 6a (t − 3/4) = x (4t − 3)
−2
−10
−5
0
t (sec)
g.
The signal g7 (t)
Step 1: Time scaling
Step 2: Time reversal
µ
¶
t
g 7b (t ) =g 7a (−t ) = x =
3
Step 3: Time shifting
¶
µ
t
g y (t ) =g 7b (t − 3) = x 1 −
3
1
Amplitude
µ ¶
t
g 7a (t ) =x
3
2
0
−1
−2
−10
−5
0
t (sec)
9
replacemen
1.5.
The signal x (t)
0.4
Amplitude
0.3
0.2
0.1
0
−0.1
−0.2
−10
−5replacemen 0
5
10
t (sec)
The signal g1 (t)
a.
0.4
Amplitude
g 1 (t ) =x (2t − 1)
Step 1: Time scaling
g 1a (t ) =x (2t )
0.3
0.2
0.1
0
−0.1
Step 2: Time shifting
−0.2
−10
1)
g 1 (t ) =g 1a (t − 0.5) = x (2t −replacemen
−5
0
5
10
5
10
5
10
t (sec)
The signal g2 (t)
b.
0.4
Amplitude
g 2 (t ) =x (−t + 2)
Step 1: Time reversal
g 2a (t ) =x (−t )
0.3
0.2
0.1
0
−0.1
Step 2: Time shifting
−0.2
−10
g 2 (t ) =g 2a (t − 2) = x (−t + 2)
−5
0
t (sec)
replacemen
c.
The signal g3 (t)
g 3 (t ) =x (−3t + 5)
g 3a (t ) =x (3t )
Step 2: Time reversal
g 3b (t ) =g 3a (−t ) = x (−3t )
Step 3: Time shifting
g 3 (t ) =g 3b (t − 5/3) = x (−3t + 5)
0.4
Amplitude
Step 1: Time scaling
0.3
0.2
0.1
0
−0.1
−0.2
−10
−5
0
t (sec)
replacemen
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
10
d.
The signal g4 (t)
t −1
3
¶
0.4
0.3
Amplitude
g 4 (t ) =x
µ
Step 1: Time scaling
µ ¶
t
g 4a (t ) =x
3
0.2
0.1
0
−0.1
Step 2: Time shifting
µ
t −1
g 4 (t ) =g 4a (t − 1) = x
3
−0.2
−10
¶
−5
1.6.
Let q (t ) be a rectangular pulse with height 1/a and width a.
µ ¶
t
1
q (t ) = Π
a
a
and the unit-impulse function can be obtained through
£
¤
δ (t ) = lim q (t ) = lim
a→∞
It follows that
Let ã = a/b so that
a→∞
£
·
¤
δ (bt ) = lim q (bt ) = lim
a→∞
a→∞
µ ¶¸
t
1
Π
a
a
·
µ ¶¸
bt
1
Π
a
a
µ ¶
µ ¶
t
1
1
bt
Π
=
Π
ã b
ã
a
a
Therefore
·
µ ¶¸
1
t
Π
a→∞ ã b
ã
·
µ ¶¸
1
1
t
= lim
Π
b a→∞ ã
ã
δ (bt ) = lim
1
= δ (t )
b
1.7.
Given that
0
t (sec)

µ ¶  1
,
1
t
a
q (t ) = Π
=

a
a
0,
a
a
<t <
2
2
otherwise
−
the time shifted pulse q (t − t 1 ) is

¶  1
µ
,
1
t − t1
a
q (t − t 1 ) = Π
=

a
a
0,
a
a
< t < t1 +
2
2
otherwise
t1 −
5
10
11
and the integral can be written as
ˆ ∞
1
f (t ) q (t − t 1 ) d t =
a
−∞
ˆ t1 +a/2
f (t ) d t
t 1 −a/2
If f (t ) is continuous in the vicinity of t = t 1 then, for small values of a, the value of the integral above is
approximately equal to the area of a rectangle with height equal to f (t 1 ) and width equal to a, that is,
1
a
ˆ t1 +a/2
t 1 −a/2
f (t ) d t ≈
¤
1£
f (t 1 ) a = f (t 1 )
a
1.8.
a.
x (t)
1
−1
−0.5
0
1
1
0.5
1
1.5
2
2.5
3
1.5
2
2.5
3
1.5
2
2.5
3
t (sec)
b.
x (t)
1
1
−1
−0.5
0
0.5
1
t (sec)
c.
x (t)
e−1
−1
−0.5
0
0.5
1
t (sec)
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
12
d.
x (t)
1.5
1
0.5
0.3679
0
−0.5
−1
−0.5
0
0.5
1
1.5
2
2.5
3
t (sec)
e.
x (t)
...
−1
0
−0.5
0.5
1
1.5
2
t (sec)
1.9.
a.
The signal x1 (t)
3
Amplitude
2
1
0
−1
−2
−1
0
1
2
t (sec)
3
4
5
13
b.
The signal x2 (t)
3
Amplitude
2
1
0
−1
−2
−1
0
1
2
3
4
5
4
5
4
5
t (sec)
c.
The signal x3 (t)
3
Amplitude
2
1
0
−1
−2
−1
0
1
2
3
t (sec)
d.
The signal x4 (t)
3
Amplitude
2
1
0
−1
−2
−1
0
1
2
t (sec)
3
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
14
e.
The signal x5 (t)
3
Amplitude
2
1
0
−1
−2
−1
0
2
1
3
4
5
t (sec)
1.10.
a.
x a (t ) = u (t + 2) + 0.5 u (t + 1) − 2 u (t − 1.5) + 0.5 u (t )
b.
x b (t ) = 1.5 u (t + 1) − 2.5 u (t ) + u (t − 1)
1.11.
a.
b.
¡
¢
¡
¢
x a (t ) = Π (t + 1.5) + 1.5 Π t −0.25
− 0.5 Π t −2.25
2.5
1.5
x b (t ) = 1.5 Π (t + 0.5) − Π (t − 0.5)
1.12.
a.
x (t ) = −r (t + 1) + 3.5 r (t ) − 3 r (t − 1) − 0.5 r (t − 2) + r (t − 3)
b.
x (t ) = −Λ(t ) + 1.5 Λ(t − 1) + Λ(t − 2)
1.13.
For each data point (t i , x i ) use a triangle
x i Λs (t , a i , b i ) with parameters a i = t i − t i −1
and b i = t i +1 − t i . The signal x (t ) can be expressed as
x (t ) =
=
X
x (t)
(t6 , x6 )
(t2 , x2 )
(t4 , x4 )
(t1 , x1 )
x i Λs (t , a i , b i )
i
x i Λs (t , t i − t i −1, t i +1 − t i )
(t7 , x7 )
(t8 , x8 )
i
X
(t5 , x5 )
(t3 , x3 )
t
15
1.14.
Integration by parts:
ˆ b
a
′
¯b ˆ b
¯
v (t ) d u (t )
u (t ) d v (t ) = u (t ) v (t )¯ −
a
a
Let u (t ) = f (t ) and d v (t ) = δ (t ) d t :
ˆ ∞
−∞
ˆ ∞
¯∞
¯
f (t ) δ (t ) d t = f (t ) δ (t )¯
−
δ (t ) f ′ (t ) d t
′
′
−∞
−∞
Using the sifting property of the unit impulse function yields
ˆ ∞
−∞
f (t ) δ′ (t ) d t = − f ′ (0)
1.15.
Using
u (t ) =
½
1,
0,
t >0
t <0
and
u (−t ) =
½
1,
0,
t <1
t >0
the signum function can be written as
sgn (t ) = −u (−t ) + u (t )
or as
sgn (t ) = −1 + 2 u (t )
1.16.
a.
e j a = cos (a) + j sin (a)
e − j a = cos (a) − j sin (a)
Therefore
1 ja 1 −ja
e + e
= cos (a)
2
2
b.
e j a = cos (a) + j sin (a)
e − j a = cos (a) − j sin (a)
Therefore
1 ja 1 −ja
e − e
= j sin (a)
2
2
⇒
sin (a) =
1 ja 1 −ja
e −
e
2j
2j
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
16
c.
·
¸
j
j
1
j −ja
d 1 ja 1 −ja
= e j a − e−j a = − e j a +
e + e
e
= − sin (a)
da 2
2
2
2
2j
2j
d.
1
1
cos (a + b) = e j (a+b) + e − j (a+b)
2
2
i
1h
= e j a e jb + e−j a e−jb
2
¤£
¤ £
¤£
¤i
1 h£
=
cos (a) + j sin (a) cos (b) + j sin (b) + cos (a) − j sin (a) cos (b) − j sin (b)
2
= cos (a) cos (b) − sin (a) sin (b)
e.
·
¸
d 1 ja 1 −ja
j
j
1
j −ja
= e j a − e−j a = − e j a +
e + e
e
= − sin (a)
da 2
2
2
2
2j
2j
d.
1 j (a+b) 1 − j (a+b)
e
e
−
2j
2j
i
1 h ja jb
=
e e − e−j a e−jb
2j
¤£
¤ £
¤£
¤i
1 h£
=
cos (a) + j sin (a) cos (b) + j sin (b) − cos (a) − j sin (a) cos (b) − j sin (b)
2j
sin (a + b) =
= sin (a) cos (b) + cos (a) sin (b)
f.
cos2 (a) =
·
¸
1 ja 1 −ja 2
e + e
2
2
1 1
1
= e j 2a + + e − j 2a
4
2 4
1 1
= + cos (2a)
2 2
1.17.
a. Periodic.
2π f 0 = 2
b. Periodic.
p
2π f 0 = 20
⇒
⇒
f0 =
1
Hz ,
π
T0 =
p
20
5
=
Hz ,
f0 =
2π
π
p
1
= π sec
f0
T0 =
1
π
= p sec
f0
5
17
c. Not periodic due to the factor u (t ).
d. Periodic.
2π f 0 = 3
3
Hz ,
2π
f0 =
⇒
1
2π
=
sec
f0
3
T0 =
e. Not periodic due to the factor e −|t | .
f. Not periodic.
g. Periodic.
x (t ) = cos (2t + π/10) + j si n (2t + π/10)
2π f 0 = 2
1
Hz ,
π
f0 =
⇒
1
= π sec
f0
T0 =
h. Not periodic.
1.18.
a.
f1 =
5
Hz,
2π
f2 =
5
Hz
2π
⇒
f0 =
5
Hz,
2π
T0 =
2π
sec
5
b.
f 1 = 5 Hz,
f 2 = 15 Hz
c.
f 0 = 5 Hz,
⇒
T0 =
1
= 0.2 sec
5
p
2
2
Hz, f 2 =
Hz
2π
2π
For periodicity we require two integers m 1 and m 2 to be found such that
f1 =
m1 m2
=
f1
f2
⇒
m1 m2
p =
2
2
No two integers can be found; therefore the signal is not periodic.
d.
f 1 = 22.5 Hz,
f 2 = 27.5 Hz
⇒
f 0 = 2.5 Hz,
T0 =
1
= 0.4 sec
2.5
1.19.
a. The energy of the signal x (t ) is
Ex =
ˆ ∞
−∞
| x (t )|2 d t < ∞
The energy of g (t ) is found as
Eg =
ˆ ∞
−∞
¯
¯2
= ¯ g (t ) ¯ d t =
ˆ ∞
−∞
2
2
| A x (t ) | d t = |A|
ˆ ∞
−∞
| x (t )|2 d t
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
18
Thus we have
E g = |A|2 E x
If x (t ) is an energy signal, x (t ) is also an energy signal.
b. The power in the signal x (t ) is
D¯
¯2 E
P x = ¯x (t )¯
The power in g (t ) is found as
Thus we have
D¯
D¯
¯2 E
¯2 E
¯2 E D¯
P g = ¯g (t )¯ = ¯ A x (t )¯ = |A|2 ¯x (t )¯
P g = |A|2 P x
If x (t ) is a power signal, x (t ) is also a power signal.
1.20.
a.
We will assume that parameters A and B are real-valued. Given that E x =
attempting to compute the normalized energy of the signal g (t ) results in
Eg =
ˆ ∞
−∞
¯
¯
¯ A x (t ) + B ¯2 d t
which does not converge for B 6= 0. Therefore g (t ) is a power signal.
b. The normalized average power in g (t ) is
D¯
¯2 E
P g = ¯g (t )¯
¢¡
¢∗ ®
A x (t ) + B A x (t ) + B
­
®
­
®
= A 2 |x (t )|2 + B 2 + 2AB Re {x (t )}
=
­¡
Since x (t ) is an energy signal, we have
Pg = B 2
1.21.
a. The signal x (t ) can be written as
x (t ) = e
−2|t |
½
e 2t ,
e −2t ,
t <0
t >0
ˆ 0
e 4t d t +
ˆ ∞
=
The energy of the signal is
Ex =
ˆ ∞
−∞
|x (t )|2 d t =
−∞
0
e −4t d t =
1
2
´∞
−∞ |x (t )|
2
dt,
19
b.
Ex =
ˆ ∞
−∞
c.
¯ −2t ¯2
¯e ¯ u (t ) d t =
ˆ ∞
0
e −4t d t =
1
4
¯
¯
¯
¯ ¯
¯
¯x (t )¯2 = ¯e −2t ¯2 ¯ cos (5t )¯2 u (t )
= e −4t cos2 (5t ) u (t )
Remembering that
cos2 (5t ) =
1 1
+ cos (10t )
2 2
we have
¯
¯
¯x (t )¯2 = 1 e −4t u (t ) + 1 e −4t cos (10t ) u (t )
2
2
1
1
1
= e −4t u (t ) + e −4t e j 10t u (t ) + e −4t e − j 10t u (t )
2
4
4
1
1
1
= e −4t u (t ) + e (−4+ j 10) t u (t ) + e (−4− j 10) t u (t )
2
4
4
The normalized energy is
Ex =
ˆ ∞
=
1
2
=
1 1
+
= 0.1422
8 58
−∞
¯
¯
¯x (t )¯2 d t
ˆ ∞
0
e −4t d t +
1
4
ˆ ∞
0
e (−4+ j 10) t d t +
1
4
ˆ ∞
e (−4− j 10) t d t
0
1.22.
a.
Ex =
b.
ˆ 0
−1


2t + 2 ,






−t + 2 ,


1,
x a (t ) =




−t + 3 ,




 0; ,
(2t + 2)2 d t +
ˆ 1
0
x b (t ) =
(−t + 2)2 d t +


1.5t + 1.5 ,




 −1.5t + 1.5 ,

1.5t − 4.5 ,





0,
−1 < t < 0
0<t <1
1<t <2
2<t <3
otherwise
ˆ 2
1
(1)2 d t +
ˆ 3
−1 < t < 0
0<t <2
2<t <3
otherwise
2
(−t + 3)2 d t = 5
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
20
Ex =
ˆ 0
−1
2
(1.5t + 1.5) d t +
ˆ 2
0
2
(−1.5t + 1.5) d t +
ˆ 3
2
(1.5t − 4.5)2 d t = 3
1.23.
a.
ˆ 0.5
D¯
¯2 E
1
P x = ¯x (t )¯ =
(1)2 d t =
3
0
b.
ˆ 1
¯1
D¯
¯2 E
t 3 ¯¯
2
¯
¯
=
P x = x (t )
t d t = ¯ = 0.5
3 0
0
c.
Px =
ˆ 1
0
2
sin (πt ) d t =
ˆ 1µ
0
¶
1
1 1
− cos (2πt ) d t =
2 2
2
1.24.
The two terms with the same frequency need to be combined. Consider that
¡
¢
¡
¢
¡
¢
A cos 2π f 1 t + θ = A cos (θ) cos 2π f 1 t − A sin (θ) sin 2π f 1 t
Let A cos (θ) = 2 and −A sin (θ) = 3. Solving the two equations we get
A = 3.6056
The signal is
and its RMS value is
a.
Even
b.
Odd
c.
Neither even nor odd
d.
Even
e.
Odd
f.
Neither even nor odd
θ = −0.9828 radians
¡
¢
¡
¢
x (t ) = 3.6056 cos 2π f 1 t − 0.9858 + 6 cos 2π f 2 t
x R MS =
1.25.
and
s
3.60562 62
+
= 4.9497
2
2
21
1.26.
a.
ˆ λ
−λ
x (t ) d t =
ˆ 0
x (t ) d t
ˆ λ
x (t ) d t
0
−λ
For the first integral, apply the variable change t = −α to obtain
ˆ λ
−λ
x (t ) d t =
ˆ 0
λ
=−
=
x (−α) (−d α) +
ˆ 0
x (−α) d α +
λ
ˆ λ
x (−α) d α +
0
ˆ λ
x (t ) d t
0
ˆ λ
x (t ) d t
0
ˆ λ
x (t ) d t
0
Since x (t ) is even, x (−α) = x (α) and
ˆ λ
−λ
x (t ) d t =
ˆ λ
0
=2
x (α) d α +
ˆ λ
ˆ λ
x (t ) d t
0
x (t ) d t
0
b. From part (a) we have
ˆ λ
−λ
x (t ) d t =
ˆ λ
0
x (−α) d α +
ˆ λ
x (t ) d t
ˆ λ
x (t ) d t
0
Since x (t ) is odd, x (−α) = −x (α) and
ˆ λ
−λ
x (t ) d t = −
ˆ λ
0
x (α) d α +
0
=0
1.27.
a.
x (−t ) = x 1 (−t ) x 2 (−t ) = x 1 (t ) x 2 (t ) = x (t )
b.
x (−t ) = x 1 (−t ) x 2 (−t ) = [−x 1 (t )] [−x 2 (t )] = x 1 (t ) x 2 (t ) = x (t )
c.
x (−t ) = x 1 (−t ) x 2 (−t ) = x 1 (t ) [−x 2 (t )] = −x 1 (t ) x 2 (t ) = −x (t )
1.28.
x E (t ) =
=
1 (−2+ j 10π)t 1 (−2− j 10π)(−t )
+ e
e
2
2
¢
1 ¡ −2t
e
+ e 2t e j 10πt
2
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
22
2
π
2
1.5
1
0
−
0.5
0
−0.6
−0.4
0
−0.2
0.2
0.4
π
2
−π
0.6
−0.6
−0.4
−0.2
t (sec)
(a)
(b)
=
2
π
2
Phase (rad)
Magnitude
6
π
1.5
1
0
0.2
0.4
0.6
−π
−0.6
−0.4
−0.2
0
(c)
(d)
1.29.
a.
x o (t ) =
=
0.4
0.6
π
2
t (sec)
=
0.2
xO (t)
t (sec)
x e (t ) =
0.6
0
−
0.5
−0.2
0.4
¢
1 ¡ −2t
e
− e 2t e j 10πt
2
|xO (t)|
−0.4
0.2
1 (−2+ j 10π)t 1 (−2− j 10π)(−t )
− e
e
2
2
2.5
−0.6
0
t (sec)
xO (t ) =
0
xE (t)
6
π
Phase (rad)
Magnitude
|xE (t)|
2.5
1 −5t
1
e
sin (t ) u (t ) + e 5t sin (−t ) u (−t )
2
2
£
¤
1
sin (t ) e −5t u (t ) − e 5t u (−t )
2
1 −5t
1
e
sin (t ) u (t ) − e 5t sin (−t ) u (−t )
2
2
£
¤
1
sin (t ) e −5t u (t ) + e 5t u (−t )
2
23
b.
x e (t ) =
1
1 −3|t |
e
cos (t ) + e −3|−t | cos (−t )
2
2
= e −3|t | cos (t )
x o (t ) =
1 −3|t |
1
e
cos (t ) − e −3|−t | cos (−t )
2
2
=0
c.
x e (t ) =
1 −3|t |
1
e
sin (t ) + e −3|−t | sin (−t )
2
2
=0
x o (t ) =
1
1 −3|t |
e
sin (t ) − e −3|−t | sin (−t )
2
2
= e −3|t | sin (t )
d.
x e (t ) =
¢
¢
1¡
1 ¡ −3t
te
+ 2 u (t ) +
−t e 3t + 2 u (−t )
2
2
1
1
= t e −3t u (t ) − t e 3t u (−t ) + 1
2
2
¡
¢
¢
1¡
1
t e −3t + 2 u (t ) −
−t e 3t + 2 u (−t )
x o (t ) =
2
2
e.
1 −2|t −1| 1 −2|−t −1|
+ e
e
2
2
1 −2|t −1| 1 −2|−t −1|
x o (t ) = e
− e
2
2
x e (t ) =
1.30.
The signal x(t)
2
−1 < t < 0
0<t <1
1<t <2
2<t <3
otherwise
1
Amplitude

−t − 1 ,





2.5
t −1 ,

−0.5 t + 2 ,
x (t ) =


 −t + 3 ,



0,
0
−1
−2
−5
0
t (sec)
5
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
24
The signal x(−t)
2
1
−3 < t < −2
−2 < t < −1
−1 < t < 0
0<t <1
otherwise
Amplitude

t +3 ,





 0.5 t + 2 ,
−2.5 t − 1 ,
x (−t ) =



t −1 ,



0,
0
−1
−2
−5
0
5
t (sec)
The signal xe (t)
2
x e (t ) =
Amplitude
1
x (t ) + x (−t )
2
0
−1
−2
−5
0
5
t (sec)
The signal xo (t)
2
x o (t ) =
Amplitude
1
x (t ) − x (−t )
2
0
−1
−2
−5
0
t (sec)
1.31.
◦
a.
X = 3e j 0 ,
b.
x (t ) = 7 cos (100πt − π/2)
ω0 = 200π rad/s
⇒
X = 7 e − j π/2 ,
ω0 = 100π rad/s
c.
x (t ) = 2 cos (10πt − π/2) + 5 cos (10πt + π)
X =2 e − j π/2 + 5 e j π
=5.3852 e − j 2.7611 ,
ω0 = 10π rad/s
5
25
1.32.
f 0 = 10 Hz
⇒
ω0 = 20π rad/s
a.
n
o
¡
¢
x (t ) = Re X e j ω0 t = 5 cos 20πt + 14◦
b.
n
o
n
o
n
o
◦
◦
x (t ) = Re X e j ω0 t = Re 2 e j 28 e j 20πt + Re 3 e j 18 e j 20πt
Alternatively
¡
¢
¡
¢
=2 cos 20πt + 28◦ + 3 cos 20πt + 18◦
◦
X 1 = 2 e j 28 ,
c.
X 2 = 3 e j 18
◦
X = X 1 + X 2 = 4.9817 e j 22
and
◦
o
o
n
n
¢
¡
◦
x (t ) = Re X e j ω0 t = Re 4.9817 e j 22 e j 20πt = 4.9817 cos 20πt + 22◦
o
o
n
o
n
n
◦
◦
x (t ) = Re X e j ω0 t = Re 2 e j 28 e j 20πt − Re 3 e j 18 e j 20πt
Alternatively
¡
¢
¡
¢
=2 cos 20πt + 28◦ − 3 cos 20πt + 18◦
◦
X 1 = 2 e j 28 ,
X 2 = −3 e j 18
◦
X = X 1 + X 2 = 1.0873 e j 179.4
and
◦
n
o
n
o
¡
¢
◦
x (t ) = Re X e j ω0 t = Re 1.0873 e j 179.4 e j 20πt = 1.0873 cos 20πt + 179.4◦
1.33.
a.
The signal g1 [n]
Time shifting
g 1 [n] =x[n − 3]
Amplitude
5
0
−5
−10
−5
0
Sample index n
5
10
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
26
b.
The signal g2 [n]
g 2 [n] =x[2n − 3]
Amplitude
5
0
−5
−10
−5
0
5
10
5
10
5
10
5
10
Sample index n
c.
The signal g3 [n]
Time reversal
g 3 [n] =x[−n]
Amplitude
5
0
−5
−10
−5
0
Sample index n
d.
The signal g4 [n]
Step 1: Time reversal
g 4a [n] =x[−n]
Step 2: Time shifting
Amplitude
5
g 4 [n] =g 4a [n − 2] = x[2 − n]
0
−5
−10
−5
0
Sample index n
e.
The signal g5 [n]
g 5 [n] =
½
x[n/2] ,
0,
if n/2 is integer
otherwise
Amplitude
5
0
−5
−10
−5
0
Sample index n
27
f.
The signal g6 [n]
Amplitude
5
g 7 (t ) =x[n] δ[n − 3]
½
x[n] ,
n=0
=
0,
otherwise
0
−5
−10
−5
0
5
10
5
10
5
10
Sample index n
g.
The signal g7 [n]
g 7 (t ) =x[n] δ[n]
½
x[n] ,
=
0,
Amplitude
5
n=3
otherwise
0
−5
−10
−5
0
Sample index n
h.
The signal g8 [n]
5
Amplitude
©
ª
g 8 [n] =x[n] u[n + 2] − u[n − 2]
½
x[n] ,
n = −2, . . . , 1
=
0,
otherwise
0
−5
−10
−5
0
Sample index n
1.34. We have 2πF0 = 2π/23, therefore the normalized frequency is
F0 =
3
46
The period of the signal is
N=
46k
k
=
F0
3
if it can be made integer. For k = 3 we get N = 46. The signal is periodic with a period of 46 samples.
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
28
replacemen
1.35.
a.
The signal x[n]
6
Amplitude
4
x[n] = { 1, −1, 3, −2, 5 }
↑
2
0
−2
−4
−6
−5
5
0
10
15
Sample index n
replacemen
b.
The signal x[n]
Amplitude
1
x[n] = (0.8)n u[n]
0.8
0.6
0.4
0.2
0
−0.2
−5
0
5
10
15
10
15
Sample index n
replacemen
c.
The signal x[n]
x[n] =u[n] − u[n − 10]
½
1,
n = 0, . . . , 9
=
0,
otherwise
Amplitude
1
0.8
0.6
0.4
0.2
0
−0.2
−5
0
5
Sample index n
29
d.
The signal x[n]
6
5
Amplitude
x[n] =r [n] − 2r [n − 5] + r [n − 10]

n,
n = 0, . . . , 5

n = 6, . . . , 10
= 10 − n ,

0,
otherwise
4
3
2
1
0
−1
−5
0
5
Sample index n
1.36.
a.
E x = (1)2 + (−1)2 + (3)2 + (−2)2 + (5)2 = 40
b.
Ex =
c.
∞
X
(0.8)2n
∞
X
n=0
n=0
Ex =
d.
Ex =
5
X
n=0
(64)n =
9
X
n=0
1
= 2.7708
1 − 0.64
(1)2 = 10
n2 +
10
X
n=6
(10 − n)2
Using the variable change m = 10 − n in the second summation we obtain
Ex =
5
X
n=0
n2 +
=x[5] + 2
0
X
m=4
4
X
n=0
m2 =
5
X
n=0
n2 +
4
X
n2
n=0
£
¤
n 2 = 2 (5)2 + (0)2 + (1)2 + (2)2 + (3)2 + (4)2 = 85
1.37.
a.
M
X
n=−M
x[n] =
−1
X
n=−M
x[n] + x[0] +
In the first summation, apply the variable change n = −k:
−1
X
n=−M
x[n] =
1
X
k=M
x[−k]
M
X
n=1
x[n]
10
15
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
30
Since x[n] is even, x[−k] = x[k], and therefore
−1
X
n=−M
x[n] =
M
X
x[k]
k=1
It follows that
M
X
n=−M
x[n] =
M
X
n=1
x[n] + x[0] +
M
X
=x[0] + 2
b.
M
X
n=−M
x[n] =
−1
X
n=−M
M
X
x[n]
n=1
x[n]
n=1
x[n] + x[0] +
M
X
x[n]
n=1
If x[n] is odd, then x[−0] = −x[0], and consequently we must have x[0] = 0. Thus
M
X
n=−M
x[n] =
−1
X
n=−M
x[n] +
M
X
x[n]
n=1
In the first summation, apply the variable change n = −k:
−1
X
n=−M
x[n] =
1
X
x[−k]
k=M
Since x[n] is even, x[−k] = −x[k], and therefore
−1
X
n=−M
It follows that
M
X
n=−M
x[n] = −
x[n] = −
M
X
n=1
M
X
x[k]
k=1
x[n] +
M
X
n=1
x[n] = 0
1.38.
a. Since x1 [n] and x2 [n] are both even, we have
x 1 [−n] = x 1 [n]
and
x 2 [−n] = x 2 [n]
and
x[−n] = x 1 [−n] x 2[−n] = x 1 [n] x 2[n] = x[n]
Therefore, x[n] is even.
b. Since x1 [n] and x2 [n] are both odd, we have
x 1 [−n] = −x 1 [n]
and
x 2 [−n] = −x 2 [n]
31
and
Therefore, x[n] is even.
¡
¢¡
¢
x[−n] = x 1 [−n] x 2[−n] = − x 1 [n] − x 2 [n] = x[n]
c. Since x1 [n] is even and x2 [n] is odd, we have
x 1 [−n] = x 1 [n]
and
Therefore, x[n] is odd.
and
x 2 [−n] = −x 2 [n]
¡
¢
x[−n] = x 1 [−n] x 2[−n] = x 1 [n] − x 2 [n] = −x[n]
1.39.
The signal x[n]
1.5
Amplitude
1
0.5
0
−0.5
−1
−1.5
−15
−10
−5
0
5
10
15
10
15
10
15
Sample index n
The signal x[−n]
1.5
Amplitude
1
0.5
0
−0.5
−1
−1.5
−15
−10
−5
0
5
Sample index n
The signal xe [n]
1.5
Amplitude
1
0.5
0
−0.5
−1
−1.5
−15
−10
−5
0
Sample index n
5
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
32
The signal xo [n]
1.5
Amplitude
1
0.5
0
−0.5
−1
−1.5
−15
−10
−5
0
Sample index n
1.40.
a.
1
2
3
4
5
6
7
8
9
10
11
12
xa = @( t ) . . .
( 2 ) . * ( ( t >=0)&( t < 1 ) ) + . . .
( 1 ) . * ( ( t >=1)&( t < 2 ) ) + . . .
( t − 1 ) . * ( ( t >=2)&( t < 3 ) ) + . . .
( 2 ) . * ( ( t >=3)&( t < 4 ) ) ;
t = [ −1:0.01:6];
plot ( t , xa ( t ) ) ;
axis ( [ − 1 , 6 , − 1 , 3] ) ;
t i t l e ( ’ The s i g n a l x_ { a } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
b.
1
2
3
4
5
6
7
8
9
10
11
12
13
% Define the s i g n a l
xb = @( t ) . . .
( t ) . * ( ( t >=0)&( t < 2 ) ) + . . .
(− t + 4 ) . * ( ( t >=2)&( t < 3 ) ) + . . .
(−2 * t + 7 ) . * ( ( t >=3)&( t < 4 ) ) + . . .
( t − 5 ) . * ( ( t >=4)&( t < 5 ) ) ;
t = [ −1:0.01:6];
plot ( t , xb ( t ) ) ;
axis ( [ − 1 , 6 , − 2 , 3] ) ;
t i t l e ( ’ The s i g n a l x_ { b } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.41.
a. Compute and graph x a (t ):
5
10
15
33
1
2
3
4
5
6
7
8
9
10
11
12
% Define the s i g n a l xa ( t ) using data points in v e c t o r s tp and xp
tp = [ − 2 , − 1 , 0 , 1 , 2 , 3 , 4] ;
xp = [ 0 , 0 , 2 , 1 , 1 , 0 , 0 ] ;
% Obtain s i g n a l xa ( t ) through i n t e r p o la t i o n
t = [ −2:0.01:4];
xa = interp1 ( tp , xp , t , ’ l i n e a r ’ ) ;
plot ( t , xa , ’b− ’ , tp , xp , ’ ro ’ ) ;
axis ( [ − 2 , 4 , − 1 , 3] ) ;
t i t l e ( ’ The s i g n a l x_ { a } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph x b (t ):
1
2
3
4
5
6
7
8
9
10
11
12
% Define the s i g n a l xb ( t ) using data points in v e c t o r s tp and xp
tp = [ −2 , −1 ,0 ,2 ,3 ,4];
xp = [ 0 , 0 , 1 . 5 , − 1 . 5 , 0 , 0 ] ;
% Obtain s i g n a l xb ( t ) through i n t e r p o la t i o n
t = [ −2:0.01:4];
xb = interp1 ( tp , xp , t , ’ l i n e a r ’ ) ;
plot ( t , xb , ’b− ’ , tp , xp , ’ ro ’ ) ;
axis ( [ − 2 , 4 , − 2 , 2] ) ;
t i t l e ( ’ The s i g n a l x_ { b } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
b. Compute and graph g 1 (t ):
1
2
3
4
5
6
7
g1 = xa+xb ;
plot ( t , g1 ) ;
axis ( [ − 2 , 4 , − 1 , 4] ) ;
t i t l e ( ’ The s i g n a l g_ { 1 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 2 (t ):
1
2
3
4
5
6
7
g2 = xa . * xb ;
plot ( t , g2 ) ;
axis ( [ − 2 , 4 , − 2 , 4] ) ;
t i t l e ( ’ The s i g n a l g_ { 2 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 3 (t ):
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
34
1
2
3
4
5
6
7
g3 = 2 * xa−xb +3;
plot ( t , g3 ) ;
axis ( [ − 2 , 4 , 2 , 7 ] ) ;
t i t l e ( ’ The s i g n a l g_ { 3 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.42.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
t = [ −1:0.01:4];
g1 = ( 0 . 5 * t + 2 ) . * ( ( t >=0)&( t < 1 ) ) + . . .
( 0 . 5 * t + 1 ) . * ( ( t >=1)&( t < 2 ) ) + . . .
(−2 * t + 4 ) . * ( ( t >=2)&( t < 3 ) ) + . . .
( t − 4 ) . * ( ( t >=3)&( t < 4 ) ) ;
subplot ( 2 , 1 , 1 ) ;
plot ( t , g1 ) ;
t i t l e ( ’ g_ { 1 } ( t ) ’ ) ;
xlabel ( ’ t ’ ) ;
grid ;
subplot ( 2 , 1 , 2 ) ;
g2 = ( t ) . * ( ( t >=0)&( t < 1 ) ) + . . .
( 0 . 5 * t ) . * ( ( t >=1)&( t < 2 ) ) + . . .
( 2 * t − 5 ) . * ( ( t >=2)&( t < 3 ) ) ;
plot ( t , g2 ) ;
t i t l e ( ’ g_ { 2 } ( t ) ’ ) ;
xlabel ( ’ t ’ ) ;
grid ;
1.43.
a. Compute and graph x a (t ):
1
2
3
4
5
6
7
8
9
10
11
12
% Define the s i g n a l xa ( t ) using data points in v e c t o r s tp and xp
tp = [ −100 , −1.5 , −0.5 ,1 ,3 ,4 ,100];
xp = [ 0 , 0 , 1 , 1 , − 1 , 0 , 0 ] ;
x = @( t ) interp1 ( tp , xp , t , ’ l i n e a r ’ ) ;
% Obtain s i g n a l xa ( t ) through i n t e r p o la t i o n
t = [ −10:0.02:10];
plot ( t , x ( t ) , ’ b− ’ , tp , xp , ’ ro ’ ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
35
b.
1
2
3
4
5
6
7
Compute and graph g 1 (t ):
g1 = x(− t ) ;
plot ( t , g1 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 1 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 2 (t ):
1
2
3
4
5
6
7
g2 = x ( 2 * t ) ;
plot ( t , g2 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 2 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 3 (t ):
1
2
3
4
5
6
7
g3 = x ( t / 2 ) ;
plot ( t , g3 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 3 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 4 (t ):
1
2
3
4
5
6
7
g4 = x(− t + 3 ) ;
plot ( t , g4 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 4 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 5 (t ):
1
2
3
4
5
6
7
g5 = x ( ( t − 1) / 3) ;
plot ( t , g5 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 5 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
36
Compute and graph g 6 (t ):
1
2
3
4
5
6
7
g6 = x ( 4 * t −3);
plot ( t , g6 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 6 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 7 (t ):
1
2
3
4
5
6
7
g7 = x(1− t / 3 ) ;
plot ( t , g7 ) ;
axis ([ −10 ,10 , −2 ,2]);
t i t l e ( ’ The s i g n a l g_ { 7 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.44.
a. Compute and graph x (t ):
1
2
3
4
5
6
7
8
x = @( t ) ( exp(− t )−exp(−2 * t ) ) . * ( t >=0);
t = [ −10:0.02:10];
plot ( t , x ( t ) ) ;
axis ( [ − 10 , 10 , − 0. 2 , 0. 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
b.
Compute and graph g 1 (t ):
1
2
3
4
5
6
7
g1 = @( t ) x ( 2 * t −1);
plot ( t , g1 ( t ) ) ;
axis ( [ − 10 , 10 , − 0. 2 , 0. 5] ) ;
t i t l e ( ’ The s i g n a l g_ { 1 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 2 (t ):
37
1
2
3
4
5
6
7
g2 = @( t ) x(− t + 2 ) ;
plot ( t , g2 ( t ) ) ;
axis ( [ − 10 , 10 , − 0. 2 , 0. 5] ) ;
t i t l e ( ’ The s i g n a l g_ { 2 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 3 (t ):
1
2
3
4
5
6
7
g3 = @( t ) x(−3 * t + 5 ) ;
plot ( t , g3 ( t ) ) ;
axis ( [ − 10 , 10 , − 0. 2 , 0. 5] ) ;
t i t l e ( ’ The s i g n a l g_ { 3 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph g 4 (t ):
1
2
3
4
5
6
7
g4 = @( t ) x ( ( t − 1) / 3) ;
plot ( t , g4 ( t ) ) ;
axis ( [ − 10 , 10 , − 0. 2 , 0. 5] ) ;
t i t l e ( ’ The s i g n a l g_ { 4 } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.45.
a.
1
2
3
4
5
6
7
8
t = [ −2:0.01:8];
x = ss_step ( t )+ ss_step ( t −1)−3* ss_step ( t −2)+ ss_step ( t −3);
plot ( t , x ) ;
axis ( [ − 2 , 8 , − 5 , 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
b.
1
2
3
4
5
6
7
x = ss_ramp ( t )−2 * ss_ramp ( t −2)+ss_ramp ( t −3);
plot ( t , x ) ;
axis ( [ − 2 , 8 , − 5 , 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
38
c.
1
2
3
4
5
6
7
x = ss_step ( t )+ss_ramp( t −2)−ss_step ( t −3)−ss_ramp ( t −4);
plot ( t , x ) ;
axis ( [ − 2 , 8 , − 5 , 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
d.
1
2
3
4
5
6
7
x = ss_pulse ( ( t −1)/2)− ss_pulse ( ( t −2)/1.5)+2 * ss_pulse ( t −3);
plot ( t , x ) ;
axis ( [ − 2 , 8 , − 5 , 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
e.
1
2
3
4
5
6
7
x = s s _ t r i ( t )+2 * s s _ t r i ( t −1)+1.5 * s s _ t r i ( t −3)− s s _ t r i ( t −4);
plot ( t , x ) ;
axis ( [ − 2 , 8 , − 5 , 5] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.46.
a. Compute and graph x (t ) using unit-ramp functions.
1
2
3
4
5
6
7
8
9
x = @( t ) −ss_ramp ( t +1)+3.5 * ss_ramp( t ) . . .
−3*ss_ramp ( t −1) −0.5 * ss_ramp ( t −2)+ss_ramp ( t −3);
t = [ −5:0.01:5];
plot ( t , x ( t ) ) ;
axis ( [ − 5 , 5 , − 2 , 2] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
b. Compute and graph x (t ) using unit-triangle functions.
1
2
x = @( t ) −s s _ t r i ( t ) + 1 . 5 * s s _ t r i ( t −1)+ s s _ t r i ( t −2);
t = [ −5:0.01:5];
39
3
4
5
6
7
8
plot ( t , x ( t ) ) ;
axis ( [ − 5 , 5 , − 2 , 2] ) ;
t i t l e ( ’ The s i g n a l x ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
c. Compute and graph xe (t ), the even component of x (t ).
1
2
3
4
5
6
7
xe = 0 . 5 * x ( t ) + 0 . 5 * x(− t ) ;
plot ( t , xe ) ;
axis ( [ − 5 , 5 , − 2 , 2] ) ;
t i t l e ( ’ The s i g n a l x_ { e } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph x o (t ), the even component of x (t ).
1
2
3
4
5
6
7
xo = 0 . 5 * x ( t ) −0.5 * x(− t ) ;
plot ( t , xo ) ;
axis ( [ − 5 , 5 , − 2 , 2] ) ;
t i t l e ( ’ The s i g n a l x_ { o } ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
1.47.
a.
1
2
3
4
x = @(n) n . * ( ( n>=−4)&(n< = 4 ) ) ;
n = [ − 10: 10] ;
stem (n , x (n ) ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
b.
1
2
3
1
2
3
g1 = x (n−3);
stem (n , g1 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g2 = x ( 2 * n−3);
stem (n , g2 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
CHAPTER 1. SIGNAL REPRESENTATION AND MODELING
40
1
2
3
1
2
3
1
2
3
4
5
6
7
8
9
1
2
3
1
2
3
1
2
3
g3 = x(−n ) ;
stem (n , g3 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g4 = x(2−n ) ;
stem (n , g4 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g5 = zeros ( s i z e (n ) ) ;
for k = 1 : length ( g5 ) ,
nn = k−11;
i f (mod(nn, 2 ) = = 0 ) ,
g5 ( k ) = x (nn / 2 ) ;
end ;
end ;
stem (n , g5 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g6 = x (n ) . * ( n==0);
stem (n , g6 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g7 = x (n ) . * ( n==3);
stem (n , g7 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
g8 = x (n ) . * ( ss_step (n+2)− ss_step (n− 2) ) ;
stem (n , g8 ) ;
axis ( [ − 10. 5 , 10. 5 , − 5 , 5] ) ;
1.48.
a.
1
2
3
4
x = @(n) 0 . 2 * ss_ramp (n+5) −0.3 * ss_ramp (n) + 0 . 1 * ss_ramp(n−10);
n = [ − 10: 10] ;
stem (n , x (n ) ) ;
axis ( [ − 1 0 . 5 , 1 0 . 5 , − 1 . 5 , 1 . 5 ] ) ;
b.
1
2
3
4
5
6
7
8
xe = 0 . 5 * x (n) + 0 . 5 * x(−n ) ;
xo = 0 . 5 * x (n) −0.5 * x(−n ) ;
subplot ( 2 , 1 , 1 ) ;
stem (n , xe ) ;
axis ( [ − 1 0 . 5 , 1 0 . 5 , − 1 . 5 , 1 . 5 ] ) ;
subplot ( 2 , 1 , 2 ) ;
stem (n , xo ) ;
axis ( [ − 1 0 . 5 , 1 0 . 5 , − 1 . 5 , 1 . 5 ] ) ;
Chapter 2
Analyzing Continuous-Time Systems
in the Time Domain
2.1.
a.
¯
©
ª ¯
y 1 (t ) = Sys x 1 (t ) = ¯x 1 (t )¯ + x 1 (t )
¯
©
ª ¯
y 2 (t ) = Sys x 2 (t ) = ¯x 2 (t )¯ + x 2 (t )
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
¯
¯
= ¯α1 x 1 (t ) + α2 x 2 (t )¯ + α1 x 1 (t ) + α2 x 2 (t )
6=α1 y 1 (t ) + α2 y 2 (t )
The system is not linear.
¯
©
ª ¯
Sys x 1 (t − τ) = ¯x 1 (t − τ)¯ + x 1 (t − τ) = y 1 (t − τ)
The system is time-invariant.
b.
©
ª
y 1 (t ) = Sys x 1 (t ) = t x 1 (t )
©
ª
y 2 (t ) = Sys x 2 (t ) = t x 2 (t )
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
¤
£
=t α1 x 1 (t ) + α2 x 2 (t )
=α1 t x 1 (t ) + α2 t x 2 (t )
=α1 y 1 (t ) + α2 y 2 (t )
The system is linear.
The system is not time-invariant.
c.
©
ª
Sys x 1 (t − τ) = t x 1 (t − τ) 6= y 1 (t − τ)
©
ª
y 1 (t ) = Sys x 1 (t ) = e −t x 1 (t )
41
42
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
©
ª
y 2 (t ) = Sys x 2 (t ) = e −t x 2 (t )
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
¤
£
=e −t α1 x 1 (t ) + α2 x 2 (t )
=α1 e −t x 1 (t ) + α2 e −t x 2 (t )
=α1 y 1 (t ) + α2 y 2 (t )
The system is linear.
©
ª
Sys x 1 (t − τ) = e −t x 1 (t − τ) 6= y 1 (t − τ)
The system is not time-invariant.
d.
©
ª
y 1 (t ) = Sys x 1 (t ) =
©
ª
y 2 (t ) = Sys x 2 (t ) =
ˆ t
x 1 (λ) d λ
ˆ t
x 2 (λ) d λ
−∞
−∞
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
=
ˆ t
−∞
=α1
£
¤
α1 x 1 (λ) + α2 x 2 (λ) d λ
ˆ t
−∞
x 1 (λ) d λ + α2
ˆ t
x 2 (λ) d λ
−∞
=α1 y 1 (t ) + α2 y 2 (t )
The system is linear.
©
ª
Sys x 1 (t − τ) =
ˆ t
−∞
x 1 (λ − τ) d λ
Let γ = λ − τ. It follows that d γ = d λ. Substituting these into the integral and adjusting the limits yields
©
ª
Sys x 1 (t − τ) =
ˆ t −τ
−∞
¡ ¢
x 1 γ d γ = y 1 (t − τ)
The system is time-invariant.
e.
©
ª
y 1 (t ) = Sys x 1 (t ) =
©
ª
y 2 (t ) = Sys x 2 (t ) =
ˆ t
x 1 (λ) d λ
ˆ t
x 2 (λ) d λ
t −1
t −1
43
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
=
ˆ t
t −1
=α1
£
¤
α1 x 1 (λ) + α2 x 2 (λ) d λ
ˆ t
t −1
x 1 (λ) d λ + α2
ˆ t
x 2 (λ) d λ
t −1
=α1 y 1 (t ) + α2 y 2 (t )
The system is linear.
©
ª
Sys x 1 (t − τ) =
ˆ t
t −1
x 1 (λ − τ) d λ
Let γ = λ − τ. It follows that d γ = d λ. Substituting these into the integral and adjusting the limits yields
©
ª
Sys x 1 (t − τ) =
ˆ t −τ
t −τ−1
¡ ¢
x 1 γ d γ = y 1 (t − τ)
The system is time-invariant.
f.
©
ª
y 1 (t ) = Sys x 1 (t ) = (t + 1)
ª
y 2 (t ) = Sys x 2 (t ) = (t + 1)
©
ˆ t
x 1 (λ) d λ
ˆ t
x 2 (λ) d λ
−∞
−∞
Using x (t ) = α1 x 1 (t ) + α2 x 2 (t ) as input we obtain
ª
©
y (t ) = Sys α1 x 1 (t ) + α2 x 2 (t )
= (t + 1)
ˆ t
−∞
=α1 (t + 1)
£
¤
α1 x 1 (λ) + α2 x 2 (λ) d λ
ˆ t
−∞
x 1 (λ) d λ + α2 (t + 1)
ˆ t
x 2 (λ) d λ
−∞
=α1 y 1 (t ) + α2 y 2 (t )
The system is linear.
©
ª
Sys x 1 (t − τ) = (t + 1)
ˆ t
−∞
x 1 (λ − τ) d λ
Let γ = λ − τ. It follows that d γ = d λ. Substituting these into the integral and adjusting the limits yields
©
ª
Sys x 1 (t − τ) = (t + 1)
The system is not time-invariant.
ˆ t −τ
−∞
¡ ¢
x 1 γ d γ 6= y 1 (t − τ)
44
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
2.2.
a.
w (t ) =3 x (t )
y (t ) =w (t − 2) = 3 x (t − 2)
b.
w̄ (t ) =x (t − 2)
ȳ (t ) =3 w̄ (t ) = 3 x (t − 2)
Input-output relationship of the system does not change when the order of the two subsystems is
changed.
2.3.
a.
Using the first configuration:
w (t ) =3 x (t )
y (t ) =t w (t ) = 3t x (t )
Using the second configuration:
w̄ (t ) =t x (t )
ȳ (t ) =3 w̄ (t ) = 3t x (t )
Input-output relationship of the system does not change when the order of the two subsystems is
changed.
b.
Using the first configuration:
w (t ) =3 x (t )
y (t ) =w (t ) + 5 = 3 x (t ) + 5
Using the second configuration:
w̄ (t ) =x (t ) + 5
£
¤
ȳ (t ) =3 w̄ (t ) = 3 x (t ) + 5 = 3 x (t ) + 15
Input-output relationship of the system changes when the order of the two subsystems is changed.
2.4.
Writing the KVL around the loop on the left yields
¤
£
x (t ) =R i L (t ) + i C (t ) + y (t )
=R i L (t ) + R i C (t ) + y (t )
45
Recognizing that
i C (t ) = C
d v c (t )
d y (t )
=C
dt
dt
we have
x (t ) = R i L (t ) + RC
d y (t )
+ y (t )
dt
Differentiating both sides of this result and recognizing that
y (t ) = v L (t ) = L
d i L (t )
dt
we get
d x (t )
d i L (t )
d 2 y (t ) d y (t )
=R
+ RC
+
dt
dt
dt2
dt
=
R
d 2 y (t ) d y (t )
y (t ) + RC
+
L
dt2
dt
Thus the differential equation for the circuit is
1
1 d x (t )
1 d y (t )
d 2 y (t )
+
y (t ) =
+
2
dt
RC d t
LC
RC d t
Initial conditions are found through
y (0) = v C (0) = 2
and
¯
d y (t ) ¯¯
R i L (0) + RC
+ y (0) = x (0)
d t ¯t =0
⇒
¯
1
1
d y (t ) ¯¯
1
y (0) − i L (0) −
x (0)
=−
¯
d t t =0
RC
C
RC
=−
1
1
2
− −
x (0)
RC C RC
2.5. Let the currents of the two capacitors be i 1 (t ) and i 2 (t ). Begin by writing the nodal equations for
the circuit:
v 1 (t ) − x (t ) v 1 (t ) − v 2 (t )
+
+ i 1 (t ) = 0
R1
R2
v 2 (t ) − v 1 (t )
+ i 2 (t ) = 0
R2
Using the relationships
v 2 (t ) = y (t ) ,
i 1 (t ) = C 1
d v 2 (t )
d y (t )
d v 1 (t )
, and i 2 (t ) = C 2
= C2
dt
dt
dt
nodal equations become
d v 1 (t )
v 1 (t ) − x (t ) v 1 (t ) − y (t )
=0
+
+C 1
R1
R2
dt
(P.2.5.1)
y (t ) − v 1 (t )
d y (t )
+C 2
=0
R2
dt
(P.2.5.2)
46
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
Next, let us solve for v 1 (t ) from Eqn. (P2.5.2)
v 1 (t ) = y (t ) + R 2C 2
d y (t )
dt
and differentiate both sides to obtain
d v 1 (t ) d y (t )
d 2 y (t )
=
+ R 2C 2
dt
dt
dt2
Substituting the last two results into Eqn. ((P.2.5.1) and simplifying the differential equation obtained
yields
¤ d y (t )
d 2 y (t ) £
+ R 1 (C 1 +C 2 ) + R 2C 2
+ y (t ) = x (t )
R 1 R 2C 1 C 2
2
dt
dt
The initial conditions are
¯
¤
d y (t ) ¯¯
1 £
and
y () = v 2 (0) = 2 V
=
v 1 (0) − v 2 (0)
¯
d t t =0 R 2C 2
2.6.
Let w (t ) be the number of encounters between prey and predators at time t :
w (t ) = K x (t ) y (t )
The growth rate of prey is
d x (t )
=A x (t ) −C w (t )
dt
=A x (t ) − C̄ x (t ) y (t )
The growth rate of predators is
d y (t )
= − B y (t ) + D w (t )
dt
= − B y (t ) + D̄ x (t ) y (t )
The differential equations derived form a nonlinear system.
2.7.
Using Eqn. (2.57) with t 0 = 0 yields the solution
y (t ) =e
−4t
=e
−4t
y (0) +
ˆ t
y (0) + 4
e −4 (t −τ) r (τ) d t
0
ˆ t
e −4 (t −τ) u (τ) d t
0
=e −4t y (0) + 4 e −4t
=e
−4t
y (0) + 1 − e
ˆ t
−4t
0
e 4τ d t
47
a.
y (t ) = 1 − e −4t ,
b.
y (t ) = 1 + 4 e −4t ,
c.
y (t ) = 1 ,
d.
y (t ) = 1 − 2 e −4t ,
t ≥0
e.
y (t ) = 1 − 4 e −4t ,
t ≥0
t ≥0
t ≥0
t ≥0
2.8.
a.
y (t ) =e
−4t
=e
−4t
(−1) + e
−4t
(−1) + e
−4t
y (t ) = e
−2t
(2) + e
−2t
ˆ t
0
e 4τ u (τ) d τ
ˆ t
e 4τ d τ
0
0
1 5
= − e −4t ,
4 4
b.
ˆ t
t ≥0
e 2τ (2) [u (τ) − u (τ − 5)] d τ
If 0 < t < 5 then
y (t ) =2 e
−2t
=1 + e
+2e
−2t
ˆ t
e 2τ d τ
ˆ 5
e 2τ d τ
0
−2t
If t > 5, then
y (t ) =2 e
−2t
+2e
−2t
0
£
¤
= e 10 + 1 e −2t
Therefore, the complete solution is
y (t ) =
(
1 + e −2t ,
£ 10
¤
e + 1 e −2t ,
0<t <5
t >5
c.
y (t ) =e −5t (0.5) + e −5t
=0.5 e
−5t
+3e
−5t
ˆ t
3 e 5τ δ (τ) d τ
0
= 3.5 e −5t ,
t >0
48
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
d.
y (t ) =e −5t (−4) + e −5t
ˆ t
e 5τ 3τ u (τ) d τ
= − 4 e −5t + 3 e −5t
ˆ t
τ e 5τ d τ
0
0
Using Eqn. (B.16) from Appendix B.2 we get
ˆ t
0
τ e 5τ d τ =
and
y (t ) =
¤
1 £
5 t e 5t − e 5t + 1
25
3 97 −5t
3
t−
−
e
,
5
25 25
t ≥0
e.
y (t ) =e
−t
(−1) + e
−t
ˆ t
e τ 2 e −2τ u (τ) d τ
0
= − e −t + 2 e −t
ˆ t
=e −t − 2 e −2t ,
e −τ d τ
0
t ≥0
2.9.
a. Characteristic equation is
s 2 + 3s + 2 = 0
⇒
(s + 1) (s + 2) = 0
The solutions of the characteristic equation are s 1 = −1 and s 2 = −2. The homogeneous solution is in
the form
y (t ) = c 1 e −t + c 2 e −2t ,
t ≥0
In order to satisfy the initial conditions we need
y (0) = c 1 + c 2 = 3
and
¯
d y (t ) ¯¯
= −c 1 − 2 c 2 = 0
d t ¯t =0
which can be solved to yield c 1 = 6 and c 2 = −3. The homogeneous solution is
y (t ) = 6 e −t − 3 e −2t ,
t ≥0
b. Characteristic equation is
s 2 + 4s + 3 = 0
⇒
(s + 1) (s + 3) = 0
⇒
s 1,2 = −1 , −3
49
The solutions of the characteristic equation are s 1 = −1 and s 2 = −3. The homogeneous solution is in
the form
y (t ) = c 1 e −t + c 2 e −3t ,
t ≥0
In order to satisfy the initial conditions we need
y (0) = c 1 + c 2 = −2
and
¯
d y (t ) ¯¯
= −c 1 − 3 c 2 = 1
d t ¯t =0
which can be solved to yield c 1 = −5/2 and c 2 = 1/2. The homogeneous solution is
5
1
y (t ) = − e −t + e −3t ,
2
2
t ≥0
c. Characteristic equation is
s2 − 1 = 0
⇒
(s + 1) (s + 2) = 0
The solutions of the characteristic equation are s 1 = 1 and s 2 = −1. The homogeneous solution is in the
form
y (t ) = c 1 e t + c 2 e −t ,
t ≥0
In order to satisfy the initial conditions we need
y (0) = c 1 + c 2 = 1
and
¯
d y (t ) ¯¯
= c 1 − c 2 = −2
d t ¯t =0
which can be solved to yield c 1 = −1/2 and c 2 = 3/2. The homogeneous solution is
3
1
y (t ) = − e t + e −t ,
2
2
t ≥0
d. Characteristic equation is
s 3 + 6s 2 + 6s + 2 = 0
⇒
(s + 1) (s + 2) (s + 3) = 0
The solutions of the characteristic equation are s 1 = −1, s 2 = −2 and s 3 = −3. The homogeneous
solution is in the form
y (t ) = c 1 e −t + c 2 e −2t + c 3 e −3t ,
t ≥0
In order to satisfy the initial conditions we need
y (0) = c 1 + c 2 + c 3 = 2
¯
d y (t ) ¯¯
= −c 1 − 2 c 2 − 3 c 3 = −1
d t ¯t =0
50
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
and
¯
d 2 y (t ) ¯¯
= c1 + 4 c2 + 9 c3 = 1
d t 2 ¯t =0
which can be solved to yield c 1 = 4, c 2 = −3 and c 3 = 1. The homogeneous solution is
y (t ) = 4 e −t − 3 e −2t + e −3t ,
t ≥0
2.10.
a. The characteristic equation is
s2 + 3 = 0
⇒
³
s+j
p ´³
p ´
3 s−j 3 =0
Therefore the homogeneous solution is in the form
³p ´
³p ´
y (t ) = d 1 cos 3 t + d 2 sin 3 t ,
t ≥0
Coefficients d 1 and d 2 are determined through the initial conditions.
y (0) = d 1 = 2
³p ´ p
³p ´
p
d y (t )
= − 3 d 1 sin 3 t + 3 d 2 cos 3 t
dt
¯
p
d y (t ) ¯¯
= 3 d2 = 0
¯
d t t =0
Therefore
y (t ) = 2 cos
b. The characteristic equation is
s 2 + 2s + 2 = 0
⇒
³p
d2 = 0
⇒
´
3t ,
(s + 1)2 + 1 = 0
t ≥0
¡
⇒
Therefore the homogeneous solution is in the form
s +1+ j
y (t ) = d 1 e −t cos (t ) + d 2 e −t sin (t ) ,
¢¡
¢
s +1− j = 0
t ≥0
Coefficients d 1 and d 2 are determined through the initial conditions.
y (0) = d 1 = −2
d y (t )
= e −t (d 2 − d 1 ) cos (t ) + e −t (−d 1 − d 2 ) sin (t )
dt
Therefore
¯
d y (t ) ¯¯
= d 2 − d 1 = −1
d t ¯t =0
⇒
d 2 − 1 + d 1 = −3
y (t ) = −2 e −t cos (t ) − 3 e −t sin (t ) ,
t ≥0
51
c. The characteristic equation is
s 2 + 4s + 13 = 0
⇒
(s + 2)2 + 9 = 0
⇒
Therefore the homogeneous solution is in the form
¡
¢¡
¢
s +2+ j 3 s +2− j 3 = 0
y (t ) = d 1 e −2t cos (3t ) + d 2 e −2t sin (3t ) ,
t ≥0
Coefficients d 1 and d 2 are determined through the initial conditions.
y (0) = d 1 = 5
d y (t )
= e −2t (−2 d 1 + 3 d 2 ) cos (3t ) + e −2t (−3 d 1 − 3 d 2 ) sin (t )
dt
Therefore
¯
d y (t ) ¯¯
= −2 d 1 + 3 d 2 = 0
d t ¯t =0
y (t ) = 5 e −2t cos (3t ) +
⇒
d2 =
10 −2t
e
sin (3t ) ,
3
10
2
d1 =
3
3
t ≥0
d. The characteristic equation is
s 3 + 3 s 2 + 4s + 2 = 0
⇒
Therefore the homogeneous solution is in the form
¡
¢¡
¢
(s + 1) s + 1 + j s + 1 − j = 0
y (t ) = c 1 e −t + d 2 e −t cos (t ) + d 3 e −t sin (t ) ,
t ≥0
The derivatives are
d y (t )
= −c 1 e −t + (d 3 − d 2 ) e −t cos (t ) + (−d 3 − d 2 ) e −t sin (t )
dt
and
d 2 y (t )
= c 1 e −t − 2d 3 e −t cos (t ) + 2d 2 e −t sin (t )
dt2
Imposing the initial conditions yields
y (0) = c 1 + d 2 = 1
¯
d y (t ) ¯¯
= −c 1 − d 2 + d 3 = 0
d t ¯t =0
and
¯
d 2 y (t ) ¯¯
= c 1 − 2d 3 = −2
d t 2 ¯t =0
Coefficient values are c 1 = 0, d 2 = 1 and d 3 = 1. The homogeneous solution is
y (t ) = e −t cos (t ) + e −t sin (t ) ,
t ≥0
52
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
2.11.
a. The characteristic equation is
s 2 + 2s + 1 = 0
⇒
(s + 1)2 = 0
Therefore the homogeneous solution is in the form
y (t ) = c 1 e −t + c 2 t e −t ,
t ≥0
Coefficients c 1 and c 2 are determined through the initial conditions.
y (0) = c 1 = 1
d y (t )
= −c 1 e −t + c 2 e −t − c 2 t e −t
dt
Therefore
¯
d y (t ) ¯¯
= −c 1 + c 2 = 0
d t ¯t =0
⇒
y (t ) = e −t + t e −t ,
c2 = 1
t ≥0
b. The characteristic equation is
s 3 + 7s 2 + 16s + 12 = 0
⇒
(s + 2)2 (s + 3) = 0
Therefore the homogeneous solution is in the form
y (t ) = c 1 e −2t + c 2 t e −2t + c 3 e −3t ,
t ≥0
Coefficients c 1 , c 2 and c 3 are determined through the initial conditions.
y (0) = c 1 + c 3 = 1
(P.2.11.1)
d y (t )
= −2 c 1 e −2t + c 2 e −2t − 2 c 2 t e −2t − 3 c 3 e −3t
dt
¯
d y (t ) ¯¯
= −2 c 1 + c 2 − 3 c 3 = −2
d t ¯t =0
(P.2.11.2)
d 2 y (t )
= 4 c 1 e −2t − 2 c 2 e −2t − 2 c 2 e −2t + 4 c 2 t e −2t + 9 c 3 e −3t
dt2
¯
d 2 y (t ) ¯¯
= 4 c1 − 4 c2 + 9 c3 = 1
d t 2 ¯t =0
Solving Eqns. (P.2.11.1), (P.2.11.2) and (P.2.11.3) for the coefficients leads to
c1 = 4 ,
c 2 = −3 ,
c 3 = −3 ,
Therefore
y (t ) = 4 e −2t − 3 t e −2t − 3 e −3t ,
t ≥0
(P.2.11.3)
53
c. The characteristic equation is
s 3 + 6s 2 + 12s + 8 = 0
(s + 2)3 = 0
⇒
Therefore the homogeneous solution is in the form
y (t ) = c 1 e −2t + c 2 t e −2t + c 3 t 2 e −2t ,
t ≥0
Coefficients c 1 , c 2 and c 3 are determined through the initial conditions.
y (0) = c 1 = −1
(P.2.11.4)
d y (t )
= −2 c 1 e −2t + c 2 e −2t − 2 c 2 t e −2t + 2 c 3 t e −2t − 2 c 3 t 2 e −2t
dt
¯
d y (t ) ¯¯
= −2 c 1 + c 2 = 0
d t ¯t =0
c 2 = 2 c 1 = −2
⇒
(P.2.11.5)
d 2 y (t )
= 4 c 1 e −2t − 2 c 2 e −2t − 2 c 2 e −2t + 4 c 2 t e −2t + 2 c 3 e −2t − 4 c 3 t e −2t − 4 c 3 t e −2t + 4 c 3 t 2 e −2t
dt2
¯
d 2 y (t ) ¯¯
= 4 c1 − 4 c2 + 2 c3 = 1
d t 2 ¯t =0
(P.2.11.6)
Solving Eqns. (P.2.11.1), (P.2.11.2) and (P.2.11.3) for the coefficients leads to
c 1 = −1 ,
c 2 = −2 ,
c 3 = −1.5; ,
Therefore
y (t ) = −e −2t + −2 t e −2t − 1.5 e −3t ,
t ≥0
2.12.
The particular solution is in the form
y p = k1 t + k2
Since it must satisfy the differential equation, we have
¤
£
k 1 + 4 k 1 t + k 2 = 4t
which leads to coefficient values k 1 = 1 and k 2 = −1/4. The characteristic equation is
s +4 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −4t ,
t ≥0
and the total solution is in the form
y (t ) = c 1 e −4t + t −
1
,
4
t ≥0
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
54
The total solution must satisfy the initial conditions.
y (0) = c 1 −
1
=0
4
c1 =
⇒
1
4
Therefore
1
1
y (t ) = e −4t + t −
4
4
¤
1£
=t − 1 − e −4t ,
4
t ≥0
2.13.
a. The particular solution is in the form
y p = k1
Since it must satisfy the differential equation, we have
4 k1 = 1
k1 = −
⇒
1
4
The characteristic equation is
s +4 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −4t ,
t ≥0
1
,
4
t ≥0
and the total solution is in the form
y (t ) = c 1 e −4t +
The total solution must satisfy the initial condition.
y (0) = c 1 +
Therefore
1
= −1
4
5
y (t ) = − e −4t ,
4
c1 = −
⇒
5
4
t ≥0
b. The particular solution is in the form
y p = k 1 sin (2t ) + k 2 cos (2t ) + k 3 cos (t ) + k 4 sin (t )
(P.2.13.1)
The particular solution must satisfy the differential equation.
d y p (t )
dt
= 2 k 1 cos (2t ) − 2 k 2 sin (2t ) − k 3 sin (t ) + k 4 cos (t )
(P.2.13.2)
Using Eqns. (P.2.13.1) and (P.2.13.2) in the differential equation we have
d y p (t )
dt
£
¤
£
¤
£
¤
£
¤
+ 2 y p (t ) = 2 k 1 + 2 k 2 cos (2t ) + 2 k 1 − 2 k 2 sin (2t ) + 2 k 3 + k 4 cos (t ) + − k 3 + 2 k 4 sin (t )
=2 sin (2t ) + 4 cos (t )
55
which leads to the set of equations
2k 1 + 2k 2 =0
2 k 1 − 2 k 2 =2
2 k 3 + k 4 =4
−k 3 + 2 k 4 =0
and can be solved to yield
k1 =
1
,
2
k2 = −
1
,
2
k3 =
8
,
5
k4 =
4
5
The characteristic equation is
s +2 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −2t ,
t ≥0
and the total solution is in the form
y (t ) = c 1 e −2t +
1
1
8
4
sin (2t ) − cos (2t ) + cos (t ) + sin (t ) ,
2
2
5
5
t ≥0
The total solution must satisfy the initial condition.
y (0) = c 1 −
Therefore
y (t ) =
1 8
+ =2
2 5
c1 =
⇒
9
10
9 −2t 1
1
8
4
e
+ sin (2t ) − cos (2t ) + cos (t ) + sin (t ) ,
10
2
2
5
5
t ≥0
c. The particular solution is in the form
y p = k1 t + k2
The particular solution must satisfy the differential equation.
We obtain the set of equations
¤
£
k1 + 5 k1 t + k2 = 3 t
5 k 1 =3
k 1 + 5 k 2 =0
The coefficients of the particular solution are k 1 = 3/5 and k 2 = −3/25. The characteristic equation is
s +5 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −5t ,
t ≥0
56
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
and the total solution is in the form
y (t ) = c 1 e −5t +
3
3
t−
,
5
25
t ≥0
The total solution must satisfy the initial condition.
y (0) = c 1 −
Therefore
y (t ) = −
3
= −4
25
c1 = −
⇒
97 −5t 3
3
e
+ t−
,
25
5
25
97
25
t ≥0
d. The particular solution is in the form
y p = k 1 e −2t
(P.2.13.1)
The particular solution must satisfy the differential equation.
− 2 k 1 e −2t + k 1 e −2t − 2 e −2t
which leads to k 1 = −2. The characteristic equation is
s +1 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −t ,
t ≥0
and the total solution is in the form
y (t ) = c 1 e −t − 2 e −2t ,
t ≥0
The total solution must satisfy the initial condition.
y (0) = c 1 − 2 = −1
⇒
c1 = 1
Therefore
y (t ) = e −t − 2 e −2t ,
t ≥0
2.14.
a. The particular solution is in the form
y p = k1
Since it must satisfy the differential equation, we have k 1 = 1/2. The characteristic equation is
s 2 + 3s + 2 = 0
⇒
(s + 1) (s + 2) = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −t + c 2 e −2t ,
t ≥0
(P.2.13.1)
57
and the total solution is in the form
y (t ) = c 1 e −t + c 2 e −2t +
1
,
2
t ≥0
The total solution must satisfy the initial conditions.
y (0) = c 1 + c 2 +
1
=3
2
c1 + c2 =
⇒
5
2
d y (t )
= −c 1 e −t − 2 c 2 e −2t
dt
¯
d y (t ) ¯¯
= −c 1 − 2 c 2 = 0
d t ¯t =0
The coefficients are found as c 1 = 5 and c 2 = −5/2. Therefore
y (t ) = 5 e −t −
5 −2t 1
e
+ ,
2
2
t ≥0
b. The particular solution is in the form
y p = k1 t + k2
The particular solution must satisfy the differential equation.
d y p (t )
dt
4 k 1 + 3 (k 1 t + k 2 ) = t + 1
= k1
k1 =
⇒
1
,
3
k2 = −
1
9
The characteristic equation is
s 2 + 4s + 3 = 0
⇒
(s + 1) (s + 3) = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −t + c 2 e −3t ,
t ≥0
and the total solution is in the form
y (t ) = c 1 e −t + c 2 e −3t +
1
1
t− ,
3
9
t ≥0
The total solution must satisfy the initial conditions.
y (0) = c 1 + c 2 −
1
=2
9
c1 + c2 =
⇒
19
9
d y (t )
1
= −c 1 e −t − 3 c 2 e −3t +
dt
3
¯
1
d y (t ) ¯¯
= −c 1 − 3 c 2 + = 0
¯
d t t =0
3
⇒
−c 1 − 3 c 2 = −
1
3
58
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
The coefficients are found as c 1 = 3 and c 2 = −8/9. Therefore
y (t ) = 3 e −t −
1
8 −3t 1
e
+ t− ,
9
3
9
t ≥0
c. The particular solution is in the form
y p = k1
Since it must satisfy the differential equation, we have k 1 = 1/3. The characteristic equation is
³
p ´
p ´³
s2 + 3 = 0
⇒
s+j 3 s−j 3 =0
Therefore the homogeneous solution is in the form
³p ´
³p ´
y h (t ) = d 1 cos 3 t + d 2 sin 3 t ,
t ≥0
and the total solution is in the form
y (t ) = d 1 cos
³p ´ 1
³p ´
3 t + d 2 sin 3 t + ,
3
t ≥0
The total solution must satisfy the initial conditions.
y (0) = d 1 +
Therefore
1
=1
3
d1 =
⇒
2
3
³p ´ p
³p ´
p
d y (t )
= − 3 d 1 sin 3 t + 3 d 2 cos 3 t
dt
¯
p
d y (t ) ¯¯
= 3 d2 = 0
⇒
d2 = 0
¯
d t t =0
y (t ) =
d. The particular solution is in the form
³p ´ 1
2
cos 3 t + ,
3
3
t ≥0
y p = k 1 e −2t
Since it must satisfy the differential equation, we have
4 k 1 e −2t − 4 k 1 e −2t + k 1 e −2t = e −2t
leading to k 1 = 1. The characteristic equation is
s 2 + 2s + 1 = 0
⇒
(s + 1)2 = 0
Therefore the homogeneous solution is in the form
y h (t ) = c 1 e −t + c 2 t e −t ,
t ≥0
and the total solution is in the form
y (t ) = c 1 e −t + c 2 t e −t + e −2t ,
t ≥0
59
The total solution must satisfy the initial conditions.
y (0) = c 1 + 1 = 1
Therefore
⇒
c1 = 0
d y (t )
= −c 1 e −t + c 2 e −t − c 2 t e −t − 2 e −2t
dt
¯
d y (t ) ¯¯
= −c 1 + c 2 − 2 = 0
⇒
c2 = 2
d t ¯t =0
y (t ) = 2t e −t + e −2t ,
t ≥0
2.15. Using the intermediate variable w (t ) we have
d 2 w (t )
d w (t )
+4
+ 3 w (t ) = x (t )
dt2
dt
and the output signal y (t ) is computed as
y (t ) =
d w (t )
− 2 w (t )
dt
Using the output equation, the initial conditions can be expressed as
¯
d w (t ) ¯¯
y (0) =
− 2 w (0) = −2
d t ¯t =0
and
¯
¯
¯
d y (0) ¯¯
d w (t ) ¯¯
d 2 w (t ) ¯¯
−
2
=
=1
d t ¯t =0
d t 2 ¯t =0
d t ¯t =0
(P.2.15.1)
(P.2.15.2)
The second derivative in Eqn. (P.2.15.2) can be resolved as
¯
¯
d 2 w (t ) ¯¯
d w (t ) ¯¯
= −4
− 3 w (0) + x (0)
d t 2 ¯t =0
d t ¯t =0
which can be used in Eqn. (P.2.15.2) to yield
¯
d w (t ) ¯¯
− 3 w (0) = 1
−6
d t ¯t =0
where we have assumed that x (0) = 0. To simplify the notation, let
¯
d w (t ) ¯¯
a=
and
b = w (0)
d t ¯t =0
so that the Eqns. (P.2.15.1) and (P.2.15.3) become
a − 2b = −2
−6a − 3b = 1
with the solutions
¯
d w (t ) ¯¯
= −8/15
a=
d t ¯t =0
and
b = w (0) = 11/15
(P.2.15.3)
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
60
1
−8/15
11/15
b
b
R
x (t)
dt
b
R
b
b
t=0
dt
−2
b
y (t)
b
b
t=0
−4
−3
2.16. a.
R
x (t)
dt
R
b
dt
y (t)
b
−4
−13
b.
1
6
x (t)
R
R
dt
b
R
dt
b
3
dt
b
y (t)
−3
−4
−2
c.
1
4
x (t)
R
dt
b
R
dt
b
R
dt
−6
−12
−8
b
2
y (t)
61
2.17.
a.
w (t ) =h 1 (t ) ∗ x (t )
y (t ) =h 2 (t ) ∗ w (t )
£
¤
=h 2 (t ) ∗ h 1 (t ) ∗ x (t )
¤
£
= h 2 (t ) ∗ h 1 (t ) ∗ x (t )
Therefore
h eq (t ) = h 2 (t ) ∗ h 1 (t ) = h 1 (t ) ∗ h 2 (t )
b.
h eq (t ) =
Since
ˆ ∞
−∞
Π (τ − 0.5) Π (t − τ − 0.5) d τ
Π (τ − 0.5) =
½
1,
0,
0<τ<1
otherwise
½
1,
0,
t −1 < τ < t
otherwise
and
Π (t − τ − 0.5) =
the convolution integral can be written as follows:
t <0 :
h eq (t ) = 0
0<t <1 :
h eq (t ) =
ˆ t
1<t <2 :
h eq (t ) =
ˆ 1
t >2 :
0
(1) (1) d τ = t
(1) (1) d τ = 2 − t
t −1
h eq (t ) = 0
The signal heq (t)
1


t,
2
−
t,
h eq (t ) =

0,
Amplitude
The equivalent impulse response is
0<t <1
1<t <2
otherwise
0.5
0
−2
−1
0
1
t (sec)
c.
w (t ) =
ˆ ∞
−∞
h 1 (τ) u (t − τ) d τ
2
3
4
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
62
Since
u (t − τ) =
½
τ<t
τ>t
1,
0,
the convolution integral can be written as
w (t ) =
ˆ t
h 1 (τ) d τ
−∞
and can be evaluated as
t <0 :
w (t ) = 0
0<t <1 :
t >1 :
w (t ) =
ˆ t
0
(1) d τ = t
w (t ) =
ˆ 1
(1) d τ = 1
0
The signal w(t)
1

0,



t,
w (t ) =



1,
Amplitude
The signal w (t ) is
t <0
0<t <1
t >1
0.8
0.6
0.4
0.2
0
−0.2
−2
−1
0
1
t (sec)
Similarly
y (t ) =
ˆ ∞
−∞
h eq (τ) u (t − τ) d τ =
ˆ t
h eq (τ) d τ
−∞
which can be evaluated as
t <0 :
0<t <1 :
1<t <2 :
t >2 :
y (t ) = 0
y (t ) =
ˆ t
0
τdτ =
t2
2
y (t ) =
ˆ 1
τdτ+
ˆ t
1
(2 − τ) d τ = −
y (t ) =
ˆ 1
τdτ+
ˆ 2
(2 − τ) d τ = 1
0
0
1
t2
+ 2t − 1
2
2
3
4
63
The signal y(t)
1
t <0
Amplitude
The response of the system is


0,





t2



,
2
y (t ) =
2

 t


− + 2t − 1 ,


2



1,
0<t <1
1<t <2
0.8
0.6
0.4
0.2
0
−0.2
−2
t >2
−1
0
2
1
3
4
t (sec)
2.18.
a.
y 1 (t ) =h 1 (t ) ∗ x (t )
y 2 (t ) =h 2 (t ) ∗ x (t )
y (t ) =y 1 (t ) + y 2 (t )
=h 1 (t ) ∗ x (t ) + h 2 (t ) ∗ x (t )
¤
£
= h 1 (t ) + h 2 (t ) ∗ x (t )
Therefore
h eq
(t ) = h 1 (t ) + h 2 (t )
replacemen
b.
The signal heq (t)
Amplitude
1
¡
¢
h eq (t ) = e −t − e −3t u (t )
0.5
0
−0.5
−2
−1
0
1
t (sec)
c.
y 1 (t ) =
Since
ˆ ∞
−∞
u (t − τ) =
h 1 (τ) u (t − τ) d τ
½
1,
0,
τ<t
τ>t
2
3
4
5
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
64
the convolution integral can be written as
y 1 (t ) =
ˆ t
−∞
h 1 (τ) d τ =
ˆ t
−∞
e −τ d τ = 1 − e −t ,
t ≥0
Similarly for y 2 (t ) we obtain
y 2 (t ) =
ˆ t
−∞
h 2 (τ) d τ =
ˆ t
−∞
−e −3τ d τ = −
¤
1£
1 − e −3t ,
3
t ≥0
and the output signal is
·
replacemen
y 2 (t ) =
y (t ) =y 1 (t ) +replacemen
¸
1 −3t
2
−t
u (t )
−e + e
3
3
The signal y1 (t)
The signal y2 (t)
1
Amplitude
Amplitude
1
0.5
0
−0.5
−2
0
−1
replacemen
1
2
3
0
−0.5
−2
5
4
0.5
0
−1
1
t (sec)
The signal y (t)
1
Amplitude
2
t (sec)
0.5
0
−0.5
−2
−1
0
1
2
3
4
t (sec)
2.19.
a.
y 1 (t ) = h 1 (t ) ∗ x (t )
w (t ) = h 2 (t ) ∗ x (t )
y 3 (t ) = h 3 (t ) ∗ w (t ) = h 2 (t ) ∗ h 3 (t ) ∗ x (t )
5
3
4
5
65
The output signal is
y (t ) = y 1 (t ) + y 2 (t )
¤
£
= h 1 (t ) + h 2 (t ) ∗ h 3 (t ) ∗ x (t )
and the equivalent impulse response is
h eq (t ) = h 1 (t ) + h 2 (t ) ∗ h 3 (t )
b. Carrying out convolution operation we obtain
h 2 (t ) ∗ h 3 (t ) = h 3 (t − 2) = e −2(t −2) u (t − 2)
and the equivalent impulse response is
replacemen
h eq (t ) = e −t u (t ) + e −2(t −2) u (t − 2)
heq (t)
Amplitude
1
0.5
0
−0.5
−2
0
−1
1
2
3
5
4
t (sec)
c.
£
¤
y 1 (t ) = u (t ) ∗ h 1 (t ) = 1 − e −t u (t )
w (t ) = u (t ) ∗ h 2 (t ) = u (t − 2)
¤
£
y 2 (t ) = w (t ) ∗ h 3 (t ) = 1 − e −2(t −2) u (t − 2)
replacemen
replacemen
£
¤
£
¤
y (t ) = 1 − e −t u (t ) + 1 − e −2(t −2) u (t − 2)
y1 (t)
1
Amplitude
Amplitude
1
0.5
0
−0.5
−2
w (t)
−1
0
1
2
t (sec)
3
4
5
0.5
0
−0.5
−2
−1
0
1
2
t (sec)
3
4
5
replacemen
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
66
y2 (t)
y (t)
2
1
Amplitude
Amplitude
1.5
0.5
0
1
0.5
0
−0.5
−2
0
−1
1
2
3
4
5
−0.5
−2
−1
0
t (sec)
1
2
3
t (sec)
2.20.
a.
y 1 (t ) = h 1 (t ) ∗ x (t )
w (t ) = h 2 (t ) ∗ x (t )
y 3 (t ) = h 3 (t ) ∗ w (t ) = h 2 (t ) ∗ h 3 (t ) ∗ x (t )
y 4 (t ) = h 4 (t ) ∗ w (t ) = h 2 (t ) ∗ h 4 (t ) ∗ x (t )
The output signal is
y (t ) = y 1 (t ) + y 3 (t ) + y 4 (t )
¤
£
= h 1 (t ) + h 2 (t ) ∗ h 3 (t ) + h 2 (t ) ∗ h 4 (t ) ∗ x (t )
and the equivalent impulse response is
h eq (t ) = h 1 (t ) + h 2 (t ) ∗ h 3 (t ) + h 2 (t ) ∗ h 4 (t )
b. Carrying out convolution operations we obtain


and
t,
2
−
t,
h 2 (t ) ∗ h 3 (t ) =

0,
0≤t <1
1≤t <2
otherwise
h 2 (t ) ∗ h 4 (t ) = u (t − 1) − u (t − 2)
⇒
⇒
h 2 (t ) ∗ h 3 (t ) = Λ (t − 1)
h 2 (t ) ∗ h 4 (t ) = Π (t − 1.5)
The equivalent impulse response is
h eq (t ) = e −t u (t ) + Λ (t − 1) + Π (t − 1.5)
c.
¡
¢
y 1 (t ) = u (t ) ∗ h 1 (t ) = 1 − e −t u (t )
4
5
67

 t,
1,
w (t ) = u (t ) ∗ h 2 (t ) =

0,





0≤t <1
t ≥1
otherwise
t 2 /2 ,
−t /2 + 2t − 1 ,
y 3 (t ) = w (t ) ∗ h 3 (t ) =

1,



0,
replacemen
0≤t <1
1≤t <2
t ≥2
otherwise
2
replacemen
y 4 (t ) = h 4 (t ) ∗ w (t ) = w (t − 1)
y1 (t)
w (t)
0.5
0
−0.5
−2
replacemen
1
Amplitude
Amplitude
1
−1
0
1
2
3
t (sec)
0.5
0
−0.5
−2
5
4
replacemen
−1
0
1
y3 (t)
Amplitude
Amplitude
4
5
3
4
5
1
0.5
0
−1
0
1
2
3
4
0.5
0
−0.5
−2
5
−1
0
t (sec)
d x (t )
.
dt
1
2
t (sec)
2.21.
Let w (t ) =
3
y4 (t)
1
−0.5
−2
2
t (sec)
©
ª
y (t ) = Sys x (t ) =
½
ˆ ∞
−∞
h (τ) x (t − τ) d τ
ˆ ∞
¾
©
ª
d x (t )
Sys
= Sys w (t ) =
h (τ) w (t − τ) d τ
dt
−∞
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
68
d y (t ) d
=
dt
dt
·ˆ ∞
=
ˆ ∞
=
ˆ ∞
−∞
¸
h (τ) x (t − τ) d τ
¤
d £
h (τ) x (t − τ) d τ
−∞ d t
h (τ)
−∞
ˆ ∞
¤
d £
x (t − τ) d τ
dt
½
d x (t )
=
h (τ) w (t − τ) d τ = Sys
dt
−∞
¾
2.22.
´∞
a.
x (t ) ∗ δ (t ) =
b.
x (t ) ∗ δ (t − t 0 ) =
c.
−∞ x (τ) δ (t − τ) d τ = x (t )
´∞
−∞ x (τ) δ (t − t 0 − τ) d τ = x (t − t 0 )
x (t ) ∗ u (t − 2) =
ˆ ∞
x (τ) u (t − 2 − τ) d τ
½
τ < t −2
τ > t −2
−∞
Since
u (t − 2 − τ) =
1,
0,
the convolution integral can be written as
x (t ) ∗ u (t − 2) =
ˆ t −2
x (τ) d τ
−∞
d.
x (t ) ∗ u (t − t 0 ) =
ˆ ∞
x (τ) u (t − t 0 − τ) d τ
½
τ < t − t0
τ > t − t0
−∞
Since
u (t − t 0 − τ) =
1,
0,
the convolution integral can be written as
x (t ) ∗ u (t − t 0 ) =
e.
Since
ˆ t −t0
x (τ) d τ
−∞
µ
¶ ˆ ∞
¶
µ
t − t0
t − t0 − τ
x (t ) ∗ Π
=
dτ
x (τ) Π
T
T
−∞
Π (t − t 0 − τ) =
½
1,
0,
t − t 0 − T /2 < τ < t − t 0 + T /2
otherwise
the convolution integral can be written as
¶ ˆ t −t0 +T /2
t − t0
=
x (τ) d τ
x (t ) ∗ Π
T
t −t 0 −T /2
µ
69
2.23.
y (t ) =
=
=
ˆ ∞
h (τ) x (t − τ) d τ
−∞
ˆ ∞
£
−∞
ˆ ∞
−∞
¤
δ (τ) − δ (τ − 1) x (t − τ) d τ
δ (τ) x (t − τ) d τ −
ˆ ∞
−∞
δ (τ − 1) x (t − τ) d τ
y (t) = x (t) − x (t − 1)
1
Amplitude
Using the sifting property of the unit-impulse
function, we have
y (t ) = x (t ) − x (t − 1)
0.5
0
−0.5
−1
−2
−1
0
2
1
3
4
5
6
t (sec)
2.24.
y (t ) =
ˆ ∞
=
ˆ ∞
=
ˆ ∞
−∞
−∞
−∞
h (τ) x (t − τ) d τ
£
¤
δ (τ) + 0.5 δ (τ − 1) + 0.3 δ (τ − 2) + 0.2 δ (τ − 3) x (t − τ) d τ
δ (τ) x (t − τ) d τ + 0.5
+ 0.3
ˆ ∞
−∞
ˆ ∞
−∞
δ (τ − 1) x (t − τ) d τ
δ (τ − 2) x (t − τ) d τ + 0.2
ˆ ∞
−∞
δ (τ − 3) x (t − τ) d τ
y (t)
Using the sifting property of the unit-impulse
function, we have
y (t ) = x (t )+0.5 x (t − 1)+0.3 x (t − 2)+0.2 x (t − 3)
Amplitude
1
0.8
0.6
0.4
0.2
0
−0.2
−2
0
2
4
t (sec)
6
8
10
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
70
2.25.
x̃ (t ) =
=
·
∞
X
n=−∞
¸
δ (t − nT s ) ∗ x (t )
ˆ ∞·
∞
X
−∞ n=−∞
¸
δ (τ − nT s ) x (t − τ) d τ
Changing the order of integration and summation yields
∞
X
x̃ (t ) =
ˆ ∞
n=−∞ −∞
δ (τ − nT s ) x (t − τ) d τ
Using the sifting property of the unit-impulse function on each integral leads to the result
x̃ (t ) =
∞
X
n=−∞
x (t − nT s ) d τ
which is clearly a periodic extension of the signal x (t ).
2.26.
a.
y (t ) =
Case 1: t < 0
ˆ ∞
u (λ) e
−2(t −λ)
−∞
ˆ t
0
y (t ) =
Case 1: t < 0
ˆ ∞
−∞
¢
1¡
1 − e −2t
2
0
y (t ) = 0
y (t ) =
c.
y (t ) =
Case 2: t ≥ 2
e −2(t −λ) d λ =
ˆ ∞h
i
i
h
u (λ) e −(t −λ) − e −2(t −λ) u (t − λ) d λ =
e −(t −λ) − e −2(t −λ) u (t − λ) d λ
Case 2: t ≥ 0
Case 1: t < 2
0
e −2(t −λ) u (t − λ) d λ
y (t ) = 0
Case 2: t ≥ 0
b.
u (t − λ) d λ =
ˆ ∞
ˆ ∞
−∞
ˆ th
0
i
¢
1¡
e −(t −λ) − e −2(t −λ) d λ = 1 − e −t − 1 − e −2t
2
u (λ − 2) e
−2(t −λ)
u (t − λ) d λ =
ˆ ∞
2
e −2(t −λ) u (t − λ) d λ
y (t ) = 0
ˆ t
2
e −2(t −λ) d λ =
¢
1¡
1 − e −2(t −2)
2
71
d.
y (t ) =
Case 1: t < 0
ˆ ∞
−∞
£
¤
u (λ) − u (λ − 2) e
−2(t −λ)
u (t − λ) d λ =
ˆ 2
0
e −2(t −λ) u (t − λ) d λ
y (t ) = 0
Case 2: 0 ≤ t < 2
ˆ t
0
Case 3: t ≥ 2
ˆ 2
0
e.
y (t ) =
Case 1: t < 0
ˆ ∞
−∞
e −2(t −λ) d λ =
e −2(t −λ) d λ =
¢
1¡
1 − e −2t
2
¢
1¡ 4
e − 1 e −2t
2
e −λ u (λ) e −2(t −λ) u (t − λ) d λ =
ˆ ∞
0
e −2t +λ u (t − λ) d λ
y (t ) = 0
Case 2: t ≥ 0
y (t ) =
ˆ t
0
e −2t +λ d λ = e −t − e −2t
2.27.
a.
ˆ 4
¶
λ−2
u (t − λ) d λ =
u (t − λ) d λ
y (t ) =
Π
4
0
−∞
ˆ ∞
Case 1: t < 0
µ
y (t ) = 0
Case 2: 0 ≤ t < 4
y (t ) =
ˆ t
0
(1) d λ = t
y (t ) =
ˆ 4
(1) d λ = 4
Case 3: t > 4
b.
y (t ) =
Case 1: t < 0
ˆ ∞
−∞
3Π
µ
0
ˆ 4
¶
λ − 2 −(t −λ)
u (t − λ) d λ =
e
3 e −(t −λ) u (t − λ) d λ
4
0
y (t ) = 0
Case 2: 0 ≤ t < 4
y (t ) =
ˆ t
0
¡
¢
3 e −(t −λ) d λ = 3 1 − e −t
72
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
Case 3: t > 4
y (t ) =
c.
Case 1: t < 0
ˆ 4
0
¡
¢
3 e −(t −λ) d λ = 3 e −t e 4 − 1
ˆ 4 µ
¶ µ
¶
¶
t −λ−2
t −λ−2
λ−2
Π
dλ =
dλ
Π
y (t ) =
Π
4
4
4
0
−∞
ˆ ∞
µ
y (t ) = 0
Case 2: 0 ≤ t < 4
y (t ) =
Case 3: 4 ≤ t < 8
y (t ) =
Case 4: t > 8
d.
Case 1: t < 0
(1) d λ = t
0
ˆ 4
t −4
(1) d λ = 8 − t
y (t ) = 0
µ
ˆ 4 µ
¶ µ
¶
¶
λ−2
t −λ−3
t −λ−3
y (t ) =
Π
Π
dλ =
dλ
Π
4
6
6
−∞
0
ˆ ∞
y (t ) = 0
Case 2: 0 ≤ t < 4
y (t ) =
ˆ t
0
(1) d λ = t
y (t ) =
ˆ 4
(1) d λ = 4
Case 3: 4 ≤ t < 6
Case 4: 6 ≤ t < 10
y (t ) =
Case 5: t > 10
ˆ t
0
ˆ 4
t −6
(1) d λ = 10 − t
y (t ) = 0
2.28.
y (t ) = x (t ) ∗ h (t ) =
ˆ ∞
−∞
x (τ) h (t − τ) d τ
Since x (t ) = 0 outside the interval t 1 < t < t 2 , the convolution integral can be written as
y (t ) = x (t ) ∗ h (t ) =
ˆ t2
t1
x (τ) h (t − τ) d τ
73
The nonzero range of the signal h (t ) is specified to be t 3 < t < t 4 .
h (t ) = 0 except for t 3 < t < t 4
and
h (t − τ) = 0 except for t 3 < t − τ < t 4
Equivalently
h (t − τ) = 0 except for t − t 4 < τ < t − t 3
For the integrand to be nonzero, we need
t − t3 > t 1
and
t − t4 < t2
which can also be expressed as
t1 + t3 < t < t2 + t4
Therefore
t5 = t1 + t3
and
t6 = t2 + t4
2.29.
Using the convolution integral, the output signal is
y (t ) =
ˆ ∞
h (λ) x (t − λ) d λ
−∞
First, let us assume that |h (t )| < ∞ for all t , and we can select the input signal to be x (t ) = h ∗ (−t ) so that
y (t ) =
ˆ ∞
−∞
h (λ) h ∗ (λ − t ) d λ
At t = 0 the output signal is
y (0) =
If the integral
ˆ ∞
∗
−∞
h (λ) h (λ) d λ =
ˆ ∞
−∞
does not converge, then neither does the integral
y (0) =
ˆ ∞
−∞
¯
¯
¯h (λ)¯2 d λ
¯
¯
¯h (λ)¯ d λ
ˆ ∞
−∞
¯
¯
¯h (λ)¯2 d λ
If the assumption |h (t )| < ∞ is not valid, then there is at least one value of t for which h (t ) is infinitely
large. In that case choosing the input signal to be x (t ) = δ (t ) leads to the output signal
y (t ) = h (t )
which is also infinitely large for at least one value of t .
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
74
2.30.
a. Let the input signal to the system be x1 (t ).
©
ª
y 1 (t ) = Sys x 1 (t ) = x 1 (t ) + α1 x 1 (t − τ1 ) + α2 x 1 (t − τ2 )
Similarly, if the input signal is x 2 (t )
©
ª
y 2 (t ) = Sys x 2 (t ) = x 2 (t ) + α1 x 2 (t − τ1 ) + α2 x 2 (t − τ2 )
The response of the system to the input signal x (t ) = β1 x 1 (t ) + β2 x 2 (t ) is
¤
¤
£
ª
£
©
Sys β1 x 1 (t ) + β2 x 2 (t ) = β1 x 1 (t ) + α1 x 1 (t − τ1 ) + α2 x 1 (t − τ2 ) + β2 x 2 (t ) + α1 x 2 (t − τ1 ) + α2 x 2 (t − τ2 )
= β1 y 1 (t ) + β2 y 2 (t )
The system is linear.
b.
The response to x 1 (t − a) is
©
ª
Sys x 1 (t − a) = x 1 (t − a) + α1 x 1 (t − τ1 − a) + α2 x 1 (t − τ2 − a) = y 1 (t − a)
The system is time-invariant.
c.
The system is causal provided that τ1 > 0 and τ2 > 0.
d.
The system is stable provided that α1 , α2 < ∞.
2.31.
a. Let x (t ) = δ (t ).
Therefore
©
ª
h (t ) = Sys δ (t ) =
ˆ t
−∞
δ (λ) d λ =
½
1,
0,
t >0
otherwise
h (t ) = u (t )
Since h (t ) = 0 for t < 0, the system is causal. However, since h (t ) is not absolute summable, the system
is not stable.
b. Let x (t ) = δ (t ).
Therefore
©
ª
h (t ) = Sys δ (t ) =
ˆ t
t −T
δ (λ) d λ =
½
µ
¶
h (t ) = Π
t − T /2
T
1,
0,
0<t <T
otherwise
Since h (t ) = 0 for t < 0, the system is causal. Also, since h (t ) is absolute summable, the system is stable.
c. Let x (t ) = δ (t ).
©
ª
h (t ) = Sys δ (t ) =
ˆ t +T
t −T
δ (λ) d λ =
½
1,
0,
−T < t < T
otherwise
75
Therefore
h (t ) = Π
µ
t
2T
¶
Since h (t ) has nonzero values for some t < 0, the system is not causal. It is stable, however, since h (t ) is
absolute summable.
2.32.
a.
x = @( t ) exp(− t ) . * cos ( 2 * t ) . * ( t >=0);
t = [ −1:0.01:5];
b. Compute and graph w (t ):
w = @( t ) 3 * x ( t ) ;
plot ( t ,w( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 4] ) ;
t i t l e ( ’w( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph y (t ):
y = @( t ) w( t −2);
plot ( t , y ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 4] ) ;
title ( ’y( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
c. Compute and graph w̄ (t ):
wbar = @( t ) x ( t −2);
plot ( t , wbar ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 4] ) ;
t i t l e ( ’ wbar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph ȳ (t ):
ybar = @( t ) 3 * wbar ( t ) ;
plot ( t , ybar ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 4] ) ;
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
76
t i t l e ( ’ ybar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
2.33.
Create an anonymous function to compute the signal x (t ):
x = @( t ) exp(− t ) . * cos ( 2 * t ) . * ( t >=0);
t = [ −1:0.01:5];
a. Compute and graph w (t ):
w = @( t ) 3 * x ( t ) ;
plot ( t ,w( t ) ) ;
axis ( [ − 1 , 5 , − 2 , 4] ) ;
t i t l e ( ’w( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph y (t ):
y = @( t ) t . * w( t ) ;
plot ( t , y ( t ) ) ;
axis ( [ − 1 , 5 , − 2 , 4] ) ;
title ( ’y( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph w̄ (t ):
wbar = @( t ) t . * x ( t ) ;
plot ( t , wbar ( t ) ) ;
axis ( [ − 1 , 5 , − 2 , 4] ) ;
t i t l e ( ’ wbar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph ȳ (t ):
ybar = @( t ) 3 * wbar ( t ) ;
plot ( t , ybar ( t ) ) ;
axis ( [ − 1 , 5 , − 2 , 4] ) ;
t i t l e ( ’ ybar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
77
b. Compute and graph w (t ):
w = @( t ) 3 * x ( t ) ;
plot ( t ,w( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 20] ) ;
t i t l e ( ’w( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph y (t ):
y = @( t ) w( t ) + 5 ;
plot ( t , y ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 20] ) ;
title ( ’y( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph w̄ (t ):
wbar = @( t ) x ( t ) + 5 ;
plot ( t , wbar ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 20] ) ;
t i t l e ( ’ wbar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Compute and graph ȳ (t ):
ybar = @( t ) 3 * wbar ( t ) ;
plot ( t , ybar ( t ) ) ;
axis ( [ − 1 , 5 , − 1 , 20] ) ;
t i t l e ( ’ ybar ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
2.34.
a.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 0.25 −1.25 * exp(−4 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/40;
78
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
t2 = [ 0 : Ts : 2 ] ;
yhat = zeros ( s i z e ( t2 ) ) ;
yhat ( 1 ) = −1; % I n i t i a l value
for k =1: length ( yhat ) −1 ,
g = −4* yhat ( k ) + 1 ;
yhat ( k+1) = yhat ( k )+Ts * g ; % Eqn . ( 2 . 1 8 5 )
end ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
b.
t = [0:0.001:10];
% Compute the exact s o lu t i o n
y = (1+exp(−2 * t ) ) . * ( ( t >=0)&( t < 5 ) ) . . .
+(exp ( 10) + 1) * exp(−2 * t ) . * ( t >=5);
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/20;
t2 = [ 0 : Ts : 1 0 ] ;
yhat = zeros ( s i z e ( t2 ) ) ;
yhat ( 1 ) = 2 ; % I n i t i a l value
for k =1: length ( yhat ) −1 ,
i f ( ( k−1) * Ts <5) ,
g = −2* yhat ( k ) + 2 ;
else
g = −2* yhat ( k ) ;
end ;
yhat ( k+1) = yhat ( k )+Ts * g ; % Eqn . ( 2 . 1 8 5 )
end ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ NorthEast ’ ) ;
c.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 3 . 5 * exp(−5 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/50;
t2 = [ 0 : Ts : 2 ] ;
79
yhat = zeros ( s i z e ( t2 ) ) ;
yhat ( 1 ) = 0 . 5 ; % I n i t i a l value
for k =1: length ( yhat ) −1 ,
i f ( k ==1) ,
g = −5* yhat ( k)+3/ Ts ; % Approximate unit impulse with r e c t a n g le
% that has a width o f Ts and area o f 1 .
else
g = −5* yhat ( k ) ;
end ;
yhat ( k+1) = yhat ( k )+Ts * g ; % Eqn . ( 2 . 1 8 5 )
end ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ NorthEast ’ ) ;
d.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 0 . 6 * t −0.12 −3.88 * exp(−5 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/50;
t2 = [ 0 : Ts : 2 ] ;
yhat = zeros ( s i z e ( t2 ) ) ;
yhat ( 1 ) = −4; % I n i t i a l value
for k =1: length ( yhat ) −1 ,
x = ( k−1) * Ts ;
g = −5* yhat ( k )+3 * x ;
yhat ( k+1) = yhat ( k )+Ts * g ; % Eqn . ( 2 . 1 8 5 )
end ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
e.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = exp(− t )−2 * exp(−2 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/10;
t2 = [ 0 : Ts : 2 ] ;
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
80
yhat = zeros ( s i z e ( t2 ) ) ;
yhat ( 1 ) = −1; % I n i t i a l value
for k =1: length ( yhat ) −1 ,
x = exp( −2 * (k−1) * Ts ) ;
g = −yhat ( k )+2 * x ;
yhat ( k+1) = yhat ( k )+Ts * g ; % Eqn . ( 2 . 1 8 5 )
end ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
2.35.
a.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 0.25 −1.25 * exp(−4 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/40;
t2 = [ 0 : Ts : 2 ] ;
ga = @( t , yhat ) −4* yhat +1;
[ t2 , yhat ] = ode45( ga , t2 , − 1) ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
b.
t = [0:0.001:10];
% Compute the exact s o lu t i o n
y = (1+exp(−2 * t ) ) . * ( ( t >=0)&( t < 5 ) ) . . .
+(exp ( 10) + 1) * exp(−2 * t ) . * ( t >=5);
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/20;
t2 = [ 0 : Ts : 1 0 ] ;
gb = @( t , yhat ) −2* yhat + 2 * ( ( t >=0)&( t < = 5 ) ) ;
[ t2 , yhat ] = ode45( gb , t2 , 2 ) ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
81
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ NorthEast ’ ) ;
c.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 3 . 5 * exp(−5 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/50;
t2 = [ 0 : Ts : 2 ] ;
gc = @( t , yhat ) −5* yhat +3/Ts * ( t ==0);
[ t2 , yhat ] = ode45( gc , t2 , 3 . 5 ) ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ NorthEast ’ ) ;
d.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = 0 . 6 * t −0.12 −3.88 * exp(−5 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/50;
t2 = [ 0 : Ts : 2 ] ;
gd = @( t , yhat ) −5* yhat +3 * t . * ( t >=0);
[ t2 , yhat ] = ode45( gd , t2 , − 4) ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
e.
t = [0:0.001:2];
% Compute the exact s o lu t i o n
y = exp(− t )−2 * exp(−2 * t ) ;
% Compute the approximate s o lu t i o n using Euler method
Ts = 1/10;
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
82
t2 = [ 0 : Ts : 2 ] ;
ge = @( t , yhat ) −yhat +2 * exp(−2 * t ) . * ( t >=0);
[ t2 , yhat ] = ode45( ge , t2 , − 1) ;
% Graph exact and approximate s o lu t i o n s
clf ;
plot ( t , y , ’− ’ , t2 , yhat , ’ r . ’ ) ; grid ;
t i t l e ( ’ Exact and approximate s o l u t i o n s ’ ) ;
xlabel ( ’Time ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ Exact solution ’ , ’ Approximate solution ’ , . . .
’ Location ’ , ’ SouthEast ’ ) ;
2.36.
a. The unit-ramp function can be expressed with an anonymous function as
xr = @( t ) t . * ( t >=0);
b. The unit-ramp response of the circuit is
¸
·
1 1 −4t
u (t )
y r (t ) = t − + e
4 4
and can be expressed with an anonymous function as
yr = @( t ) ( t −0.25+0.25 * exp(−4 * t ) ) . * ( t >=0);
c. The input signal can be expressed using unit-ramp functions as
x (t ) = r (t − 1) − r (t − 1) − r (t − 2) + r (t − 3)
which can be produced with MATLAB statements
t = [ −1:0.001:5];
inp = xr ( t )− xr ( t −1)−xr ( t −2)+ xr ( t −3);
d. The response of the circuit to the signal x (t ) is
y (t ) = y r (t − 1) − y r (t − 1) − y r (t − 2) + y r (t − 3)
and can be computed in MATLAB using
t = [ −1:0.001:5];
out = yr ( t )−yr ( t −1)−yr ( t −2)+yr ( t −3);
e. The input and the output signals can be graphed with the following statements:
83
plot ( t , inp , ’b ’ , t , out , ’ r ’ ) ;
t i t l e ( ’ x ( t ) and y ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ x ( t ) ’ , ’ y ( t ) ’ ) ;
axis ( [ − 1 , 5 , − 0 . 2 , 1 . 2 ] ) ;
grid ;
2.37.
a. The input signal can be expressed as
x (t ) = r (t − 1) − 2 u (t − 1) + r (t − 2) − r (t − 3)
b. The unit-step response of the circuit is
The unit-ramp response of the circuit is
£
¤
y u (t ) = 1 − e −4t u (t )
·
¸
1 1 −4t
y r (t ) = t − + e
u (t )
4 4
The output signal y (t ) can be computed through the following statements:
xu = @( t ) 1 * ( t >=0);
xr = @( t ) t . * ( t >=0);
yu = @( t ) (1−exp(−4 * t ) ) . * ( t >=0);
yr = @( t ) ( 0 . 2 5 * exp(−4 * t )+ t − 0 . 2 5 ) . * ( t >=0);
t = [ −1:0.001:5];
inp = xr ( t )− xr ( t −1)−2* xu ( t −1)+ xr ( t −2)−xr ( t −3);
out = yr ( t )−yr ( t −1)−2* yu ( t −1)+yr ( t −2)−yr ( t −3);
c. Use the following statements to graph the input and the output signals:
plot ( t , inp , ’b ’ , t , out , ’ r ’ ) ;
axis ( [ − 1 , 5 , − 1 . 2 , 1 . 2 ] ) ;
t i t l e ( ’ x ( t ) and y ( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
legend ( ’ x ( t ) ’ , ’ y ( t ) ’ ) ;
grid ;
CHAPTER 2. ANALYZING CONTINUOUS-TIME SYSTEMS IN THE TIME DOMAIN
84
2.38.
1
2
3
4
5
6
7
8
9
x = @( t ) s s _ t r i ( t −1);
t = [ −1:0.01:5];
y = x ( t )−x ( t −1);
plot ( t , y ) ;
axis ( [ − 1 , 5 , − 1 . 2 , 1 . 2 ] ) ;
t i t l e ( ’ y ( t )= x ( t )−x ( t −1) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
2.39.
1
2
3
4
5
6
7
8
9
x = @( t ) exp(− t ) . * ( t >=0);
t = [ −1:0.01:7];
y = x ( t ) + 0 . 5 * x ( t −1)+0.3 * x ( t −2)+0.2 * x ( t −3);
plot ( t , y ) ;
axis ( [ − 1 , 7 , − 0 . 2 , 1 . 2 ] ) ;
title ( ’y( t ) ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
grid ;
Chapter 3
Analyzing Discrete-Time Systems
in the Time-Domain
3.1.
a.
©
ª
y 1 [n] = Sys x 1 [n] = x 1 [n] u[n]
©
ª
y 2 [n] = Sys x 2 [n] = x 2 [n] u[n]
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
= α1 x 1 [n] + α2 x 2 [n] u[n]
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
©
ª
Sys x 1 [n − m] = x 1 [n − m] u[n] 6= y 1 [n − m]
The system is not time-invariant.
b.
©
ª
y 1 [n] = Sys x 1 [n] = 3 x 1 [n] + 5
©
ª
y 2 [n] = Sys x 2 [n] = 3 x 2 [n] + 5
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
=3 α1 x 1 [n] + α2 x 2 [n] + 5
6=α1 y 1 [n] + α2 y 2 [n]
The system is not linear.
The system is time-invariant.
c.
©
ª
Sys x 1 [n − m] = 3 x 1 [n − m] + 5 = y 1 [n − m]
©
ª
y 1 [n] = Sys x 1 [n] = 3 x 1 [n] + 5 u[n]
©
ª
y 2 [n] = Sys x 2 [n] = 3 x 2 [n] + 5 u[n]
85
86
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
=3 α1 x 1 [n] + α2 x 2 [n] + 5 u[n]
6=α1 y 1 [n] + α2 y 2 [n]
The system is not linear.
©
ª
Sys x 1 [n − m] = 3 x 1 [n − m] + 5 u[n] 6= y 1 [n − m]
The system is not time-invariant.
d.
©
ª
y 1 [n] = Sys x 1 [n] = n x 1 [n]
©
ª
y 2 [n] = Sys x 2 [n] = n x 2 [n]
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
=n α1 x 1 [n] + α2 x 2 [n]
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
©
ª
Sys x 1 [n − m] = n x 1 [n − m] 6= y 1 [n − m]
The system is not time-invariant.
e.
©
ª
y 1 [n] = Sys x 1 [n] = cos (0.2πn) x 1 [n]
©
ª
y 2 [n] = Sys x 2 [n] = cos (0.2πn) x 2 [n]
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
= cos (0.2πn) α1 x 1 [n] + α2 x 2 [n]
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
©
ª
Sys x 1 [n − m] = cos (0.2πn) x 1 [n − m] 6= y 1 [n − m]
The system is not time-invariant.
f.
©
ª
y 1 [n] = Sys x 1 [n] = x 1 [n] + 3 x 1 [n − 1]
©
ª
y 2 [n] = Sys x 2 [n] = x 2 [n] + 3 x 2 [n − 1]
87
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
¡
¢
= α1 x 1 [n] + α2 x 2 [n] + 3 α1 x 1 [n − 1] + α2 x 2 [n − 1]
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
©
ª
Sys x 1 [n − m] = x 1 [n − m] + 3 x 1 [n − m − 1] = y 1 [n − m]
The system is not time-invariant.
g.
©
ª
y 1 [n] = Sys x 1 [n] = x 1 [n] + 3 x 1 [n − 1] x 1[n − 2]
©
ª
y 2 [n] = Sys x 2 [n] = x 2 [n] + 3 x 2 [n − 1] x 2[n − 2]
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
¡
¢
¡
¢¡
¢
= α1 x 1 [n] + α2 x 2 [n] + 3 α1 x 1 [n − 1] + α2 x 2 [n − 1] α1 x 1 [n − 2] + α2 x 2 [n − 2]
6=α1 y 1 [n] + α2 y 2 [n]
The system is not linear.
©
ª
Sys x 1 [n − m] = x 1 [n − m] + 3 x 1 [n − m − 1] x 1 [n − m − 2] = y 1 [n − m]
The system is time-invariant.
3.2.
a.
n
X
©
ª
y 1 [n] = Sys x 1 [n] =
x 1 [k]
k=−∞
n
X
©
ª
x 2 [k]
y 2 [n] = Sys x 2 [n] =
k=−∞
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
=
n ¡
X
k=−∞
=α1
α1 x 1 [k] + α2 x 2 [k]
n
X
k=−∞
x 1 [k] + α2
=α1 y 1 [n] + α2 y 2 [n]
n
X
k=−∞
¢
x 2 [k]
88
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
The system is linear.
n
X
©
ª
Sys x 1 [n − m] =
x 1 [k − m]
k=−∞
Let k − m = r . The last relationship can be written as
X
©
ª n−m
Sys x 1 [n − m] =
x 1 [r ] = y 1 [n − m]
r =−∞
The system is time-invariant.
b.
n
©
ª X
y 1 [n] = Sys x 1 [n] =
x 1 [k]
k=0
n
©
ª X
y 2 [n] = Sys x 2 [n] =
x 2 [k]
k=0
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
=
n ¡
X
k=0
=α1
α1 x 1 [k] + α2 x 2 [k]
n
X
k=0
x 1 [k] + α2
n
X
¢
x 2 [k]
k=0
=α1 y 1 [n] + α2 y 2 [n]
The system is linear. For testing time invariance, let us assume that n > 0.
n
©
ª X
x 1 [k − m]
Sys x 1 [n − m] =
k=0
Let k − m = r . The last relationship can be written as
X
©
ª n−m
Sys x 1 [n − m] =
x 1 [r ] 6= y 1 [n − m]
r =−m
The system is not time-invariant.
c.
n
X
©
ª
y 1 [n] = Sys x 1 [n] =
x 1 [k]
k=n−2
n
X
©
ª
y 2 [n] = Sys x 2 [n] =
x 2 [k]
k=n−2
89
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
=
n ¡
X
k=n−2
=α1
α1 x 1 [k] + α2 x 2 [k]
n
X
k=n−2
x 1 [k] + α2
n
X
¢
x 2 [k]
k=n−2
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
n
X
©
ª
Sys x 1 [n − m] =
x 1 [k − m]
k=n−2
Let k − m = r . The last relationship can be written as
©
ª
Sys x 1 [n − m] =
n−m
X
r =n−2−m
x 1 [r ] = y 1 [n − m]
The system is time-invariant.
d.
n+2
X
©
ª
x 1 [k]
y 1 [n] = Sys x 1 [n] =
k=n−2
n+2
X
©
ª
x 2 [k]
y 2 [n] = Sys x 2 [n] =
k=n−2
Using x[n] = α1 x 1 [n] + α2 x 2 [n] as input we obtain
©
ª
y[n] = Sys α1 x 1 [n] + α2 x 2 [n]
=
n+2
X
k=n−2
=α1
¡
α1 x 1 [k] + α2 x 2 [k]
n+2
X
k=n−2
x 1 [k] + α2
n+2
X
¢
x 2 [k]
k=n−2
=α1 y 1 [n] + α2 y 2 [n]
The system is linear.
n+2
X
©
ª
Sys x 1 [n − m] =
x 1 [k − m]
k=n−2
Let k − m = r . The last relationship can be written as
n+2−m
X
©
ª
Sys x 1 [n − m] =
x 1 [r ] = y 1 [n − m]
r =n−2−m
The system is time-invariant.
90
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
3.3.
a.
w [n] =3 x[n]
y[n] =w [n − 2] = 3 x[n − 2]
b.
w̄[n] =x[n − 2]
ȳ[n] =3 w̄[n] = 3 x[n − 2]
Input-output relationship of the system does not change when the order of the two subsystems is
changed.
3.4.
a. Using the first configuration:
w [n] =3 x[n]
y[n] =n w [n] = 3n x[n]
Using the second configuration:
w̄[n] =n x[n]
ȳ[n] =3 w̄[n] = 3n x[n]
Input-output relationship of the system does not change when the order of the two subsystems is
changed.
b. Using the first configuration:
w [n] =3 x[n]
y[n] =w [n] + 5 = 3 x[n] + 5
Using the second configuration:
w̄[n] =x[n] + 5
¡
¢
ȳ[n] =3 w̄[n] = 3 x[n] + 5 = 3 x[n] + 15
Input-output relationship of the system changes when the order of the two subsystems is changed.
91
3.5.
a. Since the system is linear
©
ª
©
ª
©
ª
y[n] = Sys δ[n] + δ[n − 1] = Sys δ[n] + Sys δ[n − 1]
The system is also time-invariant, therefore
©
ª
Sys δ[n] = { 2 , 1, −1 }
↑
n=0
©
ª
Sys δ[n − 1] = { 0 , 2, 1, −1 }
⇒
↑
n=0
and
y[n] = { 2 , 3, 0, −1 }
↑
n=0
b. Since the system is linear
©
ª
y[n] = Sys δ[n] − 2 δ[n − 1] + δ[n − 2]
©
ª
©
ª
©
ª
= Sys δ[n] − 2 Sys δ[n − 1] + Sys δ[n − 2]
The system is also time-invariant, therefore
©
ª
Sys δ[n − 1] = { 0 , 2, 1, −1 }
↑
n=0
and
©
ª
Sys δ[n − 2] = { 0 , 0, 2, 1, −1 }
↑
n=0
The response is
y[n] = { 2 , −3, −1, 3, −1 }
↑
n=0
c.
u[n] − u[n − 5] = δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3] + δ[n − 4]
Using the linearity of the system we have
©
ª
y[n] = Sys δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3] + δ[n − 4]
©
ª
©
ª
©
ª
©
ª
©
ª
= Sys δ[n] + Sys δ[n − 1] + Sys δ[n − 2] + Sys δ[n − 3] + Sys δ[n − 4]
Since the system is also time-invariant, we have
©
ª
Sys δ[n − 1] ={ 0 , 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 2] ={ 0 , 0, 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 3] ={ 0 , 0, 0, 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 4] ={ 0 , 0, 0, 0, 2, 1, −1 }
↑
n=0
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
92
The output signal is
y[n] = { 2 , 3, 2, 2, 2, 0, −1 }
↑
n=0
d.
¡
¢
n u[n] − u[n − 5] = δ[n − 1] + 2 δ[n − 2] + 3 δ[n − 3] + 4 δ[n − 4]
Using the linearity of the system we have
©
ª
y[n] = Sys δ[n − 1] + 2 δ[n − 2] + 3 δ[n − 3] + 4 δ[n − 4]
©
ª
©
ª
©
ª
©
ª
= Sys δ[n − 1] + 2 Sys δ[n − 2] + 3 Sys δ[n − 3] + 4 Sys δ[n − 4]
Since the system is also time-invariant, we have
©
ª
Sys δ[n − 1] ={ 0 , 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 2] ={ 0 , 0, 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 3] ={ 0 , 0, 0, 2, 1, −1 }
↑
n=0
©
ª
Sys δ[n − 4] ={ 0 , 0, 0, 0, 2, 1, −1 }
↑
n=0
The output signal is
y[n] = { 0 , 2, 5, 7, 9, 1, −4 }
↑
n=0
3.6.
a. Since the system is linear
©
ª
©
ª
y[n] = Sys 5 δ[n − 1] = 5 Sys δ[n − 1] = { 5 , 10, 15 }
↑
n=0
b. Since the system is linear
©
ª
y[n] = Sys 3 δ[n − 1] + 2 δ[n − 2]
©
ª
©
ª
=3 Sys δ[n − 1] + 2 Sys δ[n − 2]
={ 3 , 12, 15, 4 }
↑
n=0
93
c. Since the system is linear
©
ª
y[n] = Sys δ[n] − 2 δ[n − 1] + 4 δ[n − 2]
©
ª
©
ª
©
ª
= Sys δ[n] − 2 Sys δ[n − 1] + 4 Sys δ[n − 2]
={ 1 , −3, 9, 4, 4 }
↑
n=0
d. The input signal can be written as
x[n] = δ[n] + δ[n − 1] + δ[n − 2]
Since the system is linear
©
ª
y[n] = Sys δ[n] + δ[n − 1] + δ[n − 2]
©
ª
©
ª
©
ª
= Sys δ[n] + Sys δ[n − 1] + Sys δ[n − 2]
={ 1 , 5, 9, 4, 1 }
↑
n=0
e. The input signal can be written as
x[n] = δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3]
The response of the system to δ[n − 3] is not given, and can not be obtained from the information
provided since the system is not time-invariant (we reach this conclusion by looking at the relationship
between the responses to δ[n], δ[n − 1] and δ[n − 2]). Therefore, the response cannot be computed in
this case.
3.7.
x[0] + x[−1] 1.7 + 0
=
= 0.85
2
2
x[1] + x[0] 2.3 + 1.7
=
=2
y[1] =
2
2
y[0] =
x[2] + x[1] 3.1 + 2.3
=
= 2.7
2
2
x[3] + x[2] 3.3 + 3.1
=
= 3.2
y[3] =
2
2
y[2] =
x[4] + x[3] 3.7 + 3.3
=
= 3.5
2
2
x[5] + x[4] 2.9 + 3.7
=
= 3.3
y[5] =
2
2
y[4] =
94
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
y[6] =
x[6] + x[5] 2.2 + 2.9
=
= 2.55
2
2
x[7] + x[6] 1.4 + 2.2
=
= 1.8
2
2
x[8] + x[7] 0.6 + 1.4
y[8] =
=
=1
2
2
y[7] =
x[9] + x[8] −0.2 + 0.6
=
= 0.2
2
2
x[10] + x[9] 0.4 − 0.2
=
= 0.1
y[10] =
2
2
y[9] =
Therefore
y[n] = { 0.85, 2, 2.7, 3.2, 3.5, 3.3, 2.55, 1.8, 1, 0.2, 0.1 }
↑
n=0
3.8.
y[0] =
x[0] + x[−1] + x − 2] + x[−3] 1.7 + 0 + 0 + 0
=
= 0.425
4
4
y[1] =
x[1] + x[0] + x − 1] + x[−2] 2.3 + 1.7 + 0 + 0
=
=1
4
4
y[2] =
x[2] + x[1] + x0] + x[−1] 3.1 + 2.3 + 1.7 + 0
=
= 1.775
4
4
y[3] =
x[3] + x[2] + x1] + x[0] 3.3 + 3.1 + 2.3 + 1.7
=
= 2.6
4
4
y[4] =
x[4] + x[3] + x2] + x[1] 3.7 + 3.3 + 3.1 + 2.3
=
= 3.1
4
4
y[5] =
x[5] + x[4] + x3] + x[2] 2.9 + 3.7 + 3.3 + 3.1
=
= 3.25
4
4
y[6] =
x[6] + x[5] + x4] + x[3] 2.2 + 2.9 + 3.7 + 3.3
=
= 3.025
4
4
y[7] =
x[7] + x[6] + x5] + x[4] 1.4 + 2.2 + 2.9 + 3.7
=
= 2.55
4
4
x[8] + x[7] + x6] + x[5] 0.6 + 1.4 + 2.2 + 2.9
=
= 1.775
4
4
x[9] + x[8] + x7] + x[6] −0.2 + 0.6 + 1.4 + 2.2
=
=1
y[9] =
4
4
y[8] =
y[10] =
x[10] + x[9] + x8] + x[7] 0.4 − 0.2 + 0.6 + 1.4
=
= 0.55
4
4
Therefore
y[n] = { 0.425, 1, 1.775, 2.6, 3.1, 3.25, 3.025, 2.55, 1.775, 1, 0.55 }
↑
n=0
3.9. The exponential smoother with α = 0.2 has the difference equation
y[n] = 0.8 y[n − 1] + 0.2 x[n]
95
Writing the output signal iteratively for n = 0, . . . , 10 we obtain
y[0] =0.8 y[−1] + 0.2 x[0] = (0.8) (0) + (0.2) (1.7) = 0.34
y[1] =0.8 y[0] + 0.2 x[1] = (0.8) (0.34) + (0.2) (2.3) = 0.8
y[2] =0.8 y[1] + 0.2 x[2] = (0.8) (0.8) + (0.2) (3.1) = 1.42
y[3] =0.8 y[2] + 0.2 x[3] = (0.8) (1.42) + (0.2) (3.3) = 2.08
y[4] =0.8 y[3] + 0.2 x[4] = (0.8) (2.08) + (0.2) (3.7) = 2.82
y[5] =0.8 y[4] + 0.2 x[5] = (0.8) (2.82) + (0.2) (2.9) = 3.4
y[6] =0.8 y[5] + 0.2 x[6] = (0.8) (3.4) + (0.2) (2.2) = 3.84
y[7] =0.8 y[6] + 0.2 x[7] = (0.8) (3.84) + (0.2) (1.4) = 4.12
y[8] =0.8 y[7] + 0.2 x[8] = (0.8) (4.12) + (0.2) (0.6) = 4.24
y[9] =0.8 y[8] + 0.2 x[9] = (0.8) (4.24) + (0.2) (−0.2) = 4.2
y[10] =0.8 y[9] + 0.2 x[10] = (0.8) (4.2) + (0.2) (0.4) = 4.28
Therefore
y[n] = { 0.34, 0.8, 1.42, 2.08, 2.82, 3.4, 3.84, 4.12, 4.24, 4.2, 4.28 }
↑
n=0
3.10. The difference equation is
y[n] = 1.005 y[n − 1] − x[n]
Writing the output signal iteratively for n = 1, . . . , 12 we obtain
y[1] =1.005 y[0] − x[1] = (1.005) (10000) − 250 = 9800
y[2] =1.005 y[1] − x[2] = (1.005) (9800) − 250 = 9599
y[3] =1.005 y[2] − x[3] = (1.005) (9599) − 250 = 9396.99
y[4] =1.005 y[3] − x[4] = (1.005) (9396.99) − 250 = 9193.98
y[5] =1.005 y[4] − x[5] = (1.005) (9193.98) − 250 = 8989.95
y[6] =1.005 y[5] − x[6] = (1.005) (8989.95) − 250 = 8784.9
y[7] =1.005 y[6] − x[7] = (1.005) (8784.9) − 250 = 8578.82
y[8] =1.005 y[7] − x[8] = (1.005) (8578.82) − 250 = 8371.72
y[9] =1.005 y[8] − x[9] = (1.005) (8371.72) − 250 = 8163.58
y[10] =1.005 y[9] − x[10] = (1.005) (8163.58) − 250 = 7954.39
y[11] =1.005 y[10] − x[11] = (1.005) (7954.39) − 250 = 7744.17
y[12] =1.005 y[11] − x[12] = (1.005) (7744.17) − 250 = 7532.89
96
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
3.11. The difference equation given by Eqn. (3.25) is
µ
¶
A
1
y[n − 1] +
y[n] =
2
y[n − 1]
a. With A = 5 and y[0] = 1 we have
µ
¶
µ
¶
A
1
5
1
y[1] = y[0] +
=
1+
=3
2
y[0]
2
1
µ
¶
µ
¶
A
1
5
1
=
3+
= 2.33333
y[2] = y[1] +
2
y[1]
2
3
µ
¶
µ
¶
A
1
5
1
=
2.33333 +
= 2.2381
y[3] = y[2] +
2
y[2]
2
2.33333
µ
¶
µ
¶
1
A
1
5
y[4] = y[3] +
=
2.2381 +
= 2.23607
2
y[3]
2
2.2381
µ
¶
µ
¶
A
1
5
1
=
2.23607 +
= 2.23607
y[5] = y[4] +
2
y[4]
2
2.23607
b. With A = 17 and y[0] = 1 we have
µ
¶
µ
¶
A
1
17
1
y[0] +
=
1+
=9
2
y[0]
2
1
µ
¶
µ
¶
1
A
1
17
y[2] = y[1] +
=
9+
= 5.44444
2
y[1]
2
9
µ
¶
µ
¶
A
1
17
1
=
5.44444 +
= 4.28345
y[3] = y[2] +
2
y[2]
2
5.44444
µ
¶
µ
¶
A
1
17
1
=
4.28345 +
= 4.12611
y[4] = y[3] +
2
y[3]
2
4.28345
µ
¶
µ
¶
1
A
1
17
y[5] = y[4] +
=
4.12611 +
= 4.12311
2
y[4]
2
4.12611
µ
¶
µ
¶
A
1
17
1
=
4.12311 +
= 4.12311
y[6] = y[5] +
2
y[5]
2
4.12311
y[1] =
97
c. With A = 132 and y[0] = 1 we have
µ
¶
µ
¶
A
1
132
1
=
1+
= 66.5
y[1] = y[0] +
2
y[0]
2
1
µ
¶
µ
¶
1
A
1
132
y[2] = y[1] +
=
66.5 +
= 34.2425
2
y[1]
2
66.5
µ
¶
µ
¶
A
1
132
1
=
34.2425 +
= 19.0487
y[3] = y[2] +
2
y[2]
2
34.2425
µ
¶
µ
¶
1
A
1
132
y[4] = y[3] +
=
19.0487 +
= 12.9891
2
y[3]
2
19.0487
µ
¶
µ
¶
A
1
132
1
=
12.9891 +
= 11.5757
y[5] = y[4] +
2
y[4]
2
12.9891
µ
¶
µ
¶
A
1
132
1
=
11.5757 +
= 11.4894
y[6] = y[5] +
2
y[5]
2
11.5757
µ
¶
µ
¶
1
A
1
132
y[7] = y[6] +
=
11.4894 +
= 11.4891
2
y[6]
2
11.4894
µ
¶
µ
¶
A
1
132
1
=
11.4891 +
= 11.4891
y[8] = y[7] +
2
y[7]
2
11.4891
3.12.
a.
Characteristic equation:
z 2 + 0.2 z − 0.63 = 0
⇒
(z − 0.7) (z + 0.9) = 0
Homogeneous solution is in the form
y[n] = c 1 (0.7)n + c 2 (−0.9)n ,
n≥0
Imposing initial conditions leads to the equations
1.4286 c 1 − 1.1111 c 2 = 5
2.0408 c 1 + 1.2346 c 2 = −3
which can be solved to obtain c 1 = 0.7044 and c 2 = −3.5944. The homogeneous solution is
y[n] = 0.7044 (0.7)n − 3.5944 (−0.9)n ,
n≥0
b.
Characteristic equation:
z 2 + 1.3 z + 0.4 = 0
⇒
(z + 0.5) (z + 0.8) = 0
98
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
Homogeneous solution is in the form
y[n] = c 1 (−0.5)n + c 2 (−0.8)n ,
n≥0
Imposing initial conditions leads to the equations
−2 c 1 − 1.25 c 2 = 0
4 c 1 + 1.5625 c 2 = 5
which can be solved to obtain c 1 = 3.3333 and c 2 = −5.3333. The homogeneous solution is
y[n] = 3.3333 (−0.5)n − 5.3333 (−0.8)n ,
n≥0
c.
Characteristic equation:
z 2 − 1.7 z + 0.72 = 0
⇒
(z − 0.8) (z − 0.9) = 0
Homogeneous solution is in the form
y[n] = c 1 (0.8)n + c 2 (0.9)n ,
n≥0
Imposing initial conditions leads to the equations
1.25 c 1 + 1.1111 c 2 = 1
1.5625 c 1 + 1.2346 c 2 = 2
which can be solved to obtain c 1 = 5.12 and c 2 = −4.86. The homogeneous solution is
y[n] = 5.12 (0.8)n − 4.86 (0.9)n ,
n≥0
d.
Characteristic equation:
z 2 − 0.49 = 0
⇒
(z − 0.7) (z + 0.7) = 0
Homogeneous solution is in the form
y[n] = c 1 (0.7)n + c 2 (−0.7)n ,
n≥0
Imposing initial conditions leads to the equations
1.4286 c 1 − 1.4286 c 2 = −3
2.0408 c 1 + 2.0408 c 2 = −1
which can be solved to obtain c 1 = −1.295 and c 2 = 0.805. The homogeneous solution is
y[n] = −1.295 (0.7)n + 0.805 (−0.7)n ,
n ≥0
99
e. Characteristic equation:
z 3 + 0.6 z 2 − 0.51 z − 0.28 = 0;
⇒
(z + 0.8) (z + 0.5) (z − 0.7) = 0
Homogeneous solution is in the form
y[n] = c 1 (−0.8)n + c 2 (−0.5)n + c 3 (0.7)n ,
n≥0
Imposing initial conditions leads to the equations
−1.25 c 1 − 2 c 2 + 1.4286 c 3 = 3
1.5625 c 1 + 4 c 2 + 2.0408 c 3 = 2
−1.9531 c 1 − 8 c 2 + 2.9155 c 3 = 1
(3.1)
which can be solved to obtain c 1 = −2.56, c 2 = 0.9167 and c 3 = 1.1433. The homogeneous solution is
y[n] = −2.56 (−0.8)n + 0.9167 (−0.5)n + 1.1433 (0.7)n ,
n≥0
3.13.
a.
Characteristic equation:
z 2 − 1.4 z + 0.85 = 0
¡
⇒
The roots are
¢¡
¢
z − 0.7 − j 0.6 z − 0.7 + j 0.6 = 0
z 1,2 = 0.9220 e ± j 0.7086
Homogeneous solution is in the form
y[n] = d 1 (0.9220) n cos (0.7086n) + d 2 (0.9220) n sin (0.7086n) ,
n≥0
Imposing initial conditions leads to the equations
0.6923 d 1 − 0.5385 d 2 = 2
0.1893 d 1 − 0.7456 d 2 = −2
which can be solved to obtain d 1 = 4.5004 and d 2 = 2.4171. The homogeneous solution is
y[n] = 4.5004 (0.9220) n cos (0.7086n) + 2.4171 (0.9220)n sin (0.7086n) ,
b.
Characteristic equation:
z 2 − 1.6 z + 1 = 0
⇒
¡
¢¡
¢
z − 0.8 − j 0.6 z − 0.8 + j 0.6 = 0
n ≥0
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
100
The roots are
z 1,2 = 1, e ± j 0.6435
Homogeneous solution is in the form
y[n] = d 1 cos (0.6435n) + d 2 sin (0.6435n) ,
n≥0
Imposing initial conditions leads to the equations
0.8 d 1 − 0.6 d 2 = 0
0.28 d 1 − 0.96 d 2 = 3
which can be solved to obtain d 1 = −3 and d 2 = −4. The homogeneous solution is
y[n] = −3 cos (0.6435n) − 4 sin (0.6435n) ,
n≥0
c.
Characteristic equation:
z2 + 1 = 0
The roots are
Homogeneous solution is in the form
¡
⇒
¢¡
¢
z − j1 z + j1 = 0
z 1,2 = ± j = 1 e ± j π/2
y[n] = d 1 cos (πn/2) + d 2 sin (πn/2) ,
n ≥0
Imposing initial conditions leads to the equations
d 1 cos (−π/2) + d 2 sin (−π/2) = 3
d 1 cos (−π) + d 2 sin (−π) = 2
which can be solved to obtain d 1 = −2 and d 2 = −3. The homogeneous solution is
y[n] = −2 cos (πn/2) − 3 sin (πn/2) ,
n≥0
d.
Characteristic equation:
z 3 − 2.5 z 2 + 2.44 z − 0.9 = 0
⇒
The roots are
z 1,2 = 1, e ± j 0.6435
Homogeneous solution is in the form
¡
¢¡
¢
z − 0.8 − j 0.6 z − 0.8 + j 0.6 (z − 0.9) = 0
and
z 3 = 0.9
y[n] = d 1 cos (0.6435n) + d 2 sin (0.6435n) + c 3 (0.9)n ,
n≥0
Imposing initial conditions leads to the equations
0.8 d 1 − 0.6 d 2 + 1.1111 c 3 = 1
0.28 d 1 − 0.96 d 2 + 1.2346 c 3 = 2
−0.352 d 1 − 0.936 d 2 + 1.3717 c 3 = 3
which can be solved to obtain d 1 = −1.2562, d 2 = −0.4227, and c 3 = 1.5762. The homogeneous solution
is
n≥0
y[n] = −1.2562 cos (0.6435n) − 0.4227 sin (0.6435n) + 1.5762 (0.9)n ,
101
3.14.
a.
Characteristic equation:
z 2 − 1.4 z + 0.49 = 0
⇒
(z − 0.7)2 = 0
The roots are
z 1,2 = 0.7
Homogeneous solution is in the form
y[n] = c 1 (0.7)n + c 2 n (0.7)n ,
n ≥0
Imposing initial conditions leads to the equations
1.4286 c 1 − 1.4286 c 2 = 1
2.0408 c 1 − 4.0816 c 2 = 1
which can be solved to obtain c 1 = 0.91 and c 2 = 0.21. The homogeneous solution is
y[n] = 0.91 (0.7)n + 0.21 n (0.7)n ,
n ≥0
b.
Characteristic equation:
z 2 + 1.8 z + 0.81 = 0
⇒
(z + 0.9)2 = 0
The roots are
z 1,2 = −0.9
Homogeneous solution is in the form
y[n] = c 1 (−0.9)n + c 2 n (−0.9)n ,
n≥0
Imposing initial conditions leads to the equations
−1.1111 c 1 + 1.1111 c 2 = 0
1.2346 c 1 − 2.4691 c 2 = 2
which can be solved to obtain c 1 = −1.62 and c 2 = −1.62. The homogeneous solution is
y[n] = −1.62 (−0.9)n − 1.62 n (−0.9)n ,
n≥0
c.
Characteristic equation:
z 3 − 0.8 z 2 − 0.64 z + 0.512 = 0
⇒
(z + 0.8) (z − 0.8)2 = 0
The roots are
z 1 = −0.8
and
z 2,3 = 0.8
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
102
Homogeneous solution is in the form
y[n] = c 1 (−0.8)n + c 2 (0.8)n + c 3 n (0.8)n ,
n ≥0
Imposing initial conditions leads to the equations
−1.25 c 1 + 1.25 c 2 − 1.25 c 3 = 1
1.5625 c 1 + 1.5625 c 2 − 3.125 c 3 = 1
−1.9531 c 1 + 1.9531 c 2 − 5.8594 c 3 = 2
which can be solved to obtain c 1 = −0.136, c 2 = 0.552 and c 3 = −0.112. The homogeneous solution is
y[n] = −0.136 (−0.8)n + 0.552 (0.8)n − 0.112 n (0.8)n ,
n≥0
d.
Characteristic equation:
z 3 + 1.7 z 2 + 0.4 z − 0.3 = 0
⇒
(z + 1)2 (z − 0.3)2 = 0
The roots are
z 1,2 = −1
and
z 3 = 0.3
Homogeneous solution is in the form
y[n] = c 1 (−1)n + c 2 n (−1)n + c 3 (0.3)n ,
n≥0
Imposing initial conditions leads to the equations
−c 1 + c 2 + 3.3333 c 3 = 1
c 1 − 2 c 2 + 11.1111 c 3 = 2
−c 1 + 3 c 3 + 37.037 c 3 = 1
which can be solved to obtain c 1 = −2.2959, c 2 = −1.6154 and c 3 = 0.0959. The homogeneous solution is
y[n] = −2.2959 (−1)n − 1.6154 n (−1)n + 0.0959 (0.3)n ,
3.15.
a.
Characteristic equation is
z − 0.6 = 0
The homogeneous solution is in the form
y h [n] = c 1 (0.6)n ,
For the unit-step input the particular solution is
y p = k1
n≥0
n≥0
103
The particular solution must satisfy the difference equation, therefore
k 1 = 0.6 k 1 + 1
which leads to k 1 = 2.5. The complete solution is in the form
y[n] = c 1 (0.6)n + 2.5
In order to satisfy the specified initial condition we need
y[−1] = c 1 (0.6)−1 + 2.5 = 2
c 1 = −0.3
⇒
The signal y[n] is
y[n] = −0.3 (0.6)n + 2.5 ,
n≥0
b.
Characteristic equation is
z − 0.8 = 0
The homogeneous solution is in the form
y h [n] = c 1 (0.8)n ,
n≥0
For the unit-step input the particular solution is
y p = k 1 sin (0.2n) + k 2 cos (0.2n)
The particular solution must satisfy the difference equation, therefore
£
¤
k 1 sin (0.2n) + k 2 cos (0.2n) = 0.8 k 1 sin (0.2n − 0.2) + k 2 cos (0.2n − 0.2) + 2 si n (0.2n)
which leads to the equations
0.2159 k 1 − 0.1589 k 2 = −2
0.1589 k 1 + 0.2159 k 2 = 0
and yields the solutions k 1 = −6.0087 and k 2 = 4.4224. The forced solution of the difference equation is
in the form
y[n] = c 1 (0.8)n − 6.0087 sin (0.2n) + 4.4224 cos (0.2n)
In order to satisfy the specified initial condition we need
y[−1] = c 1 (0.8)−1 − 6.0087 sin (−0.2) + 4.4224 cos (−0.2) = 1
⇒
c 1 = −3.6224
The signal y[n] is
y[n] = −3.6224 (0.8)n − 6.0087 sin (0.2n) + 4.4224 cos (0.2n) ,
c.
Characteristic equation is
z 2 − 0.2 z − 0.63 = 0
⇒
(z + 0.7) (z − 0.9) = 0
n≥0
104
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
The homogeneous solution is in the form
y h [n] = c 1 (−0.7)n + c 2 (0.9)n
n≥0
For the exponential input signal the particular solution is
y p = k 1 e −0.2n
The particular solution must satisfy the difference equation, therefore
k 1 e −0.2n − 0.2 e −0.2(n−1) − 0.63 e −0.2(n−2) = e −0.2n
which leads to the equation
¡
¢
k 1 1 − 0.2 e 0.2 − 0.63 e 0.4 = 1
and yields the solution k 1 = −5.4309. The forced solution of the difference equation is in the form
y[n] = c 1 (−0.7)n + c 2 (0.9)n − 5.4309 e −0.2n
In order to satisfy the specified initial conditions we need
y[−1] = c 1 (−0.7)−1 + c 2 (0.9)−1 − 5.4309 e 0.2 = 0
and
y[−2] = c 1 (−0.7)−2 + c 2 (0.9)−2 − 5.4309 e 0.4 = 3
The solutions are found as
c 1 = 1.0285
and
c 2 = 7.2924
The signal y[n] is
y[n] = 1.0285 (−0.7)n + 7.2924 (0.9)n − 5.4309 e −0.2n ,
n≥0
d.
Characteristic equation is
z 2 + 1, 4 z + 0.85 = 0
The roots are
⇒
¡
¢¡
¢
z + 0.7 + j 0.6 z + 0.7 − j 0.6 = 0
z 1,2 = 0.922 e ± j 2.433
The homogeneous solution is in the form
y h [n] = d 1 (0.922)n cos (2.433n) + d 2 (0.922)n sin (2.433n)
For the unit-step input signal the particular solution is
y p = k1
The particular solution must satisfy the difference equation, therefore
k 1 + 1.4 k 1 + 0.85 k 1 = 1
⇒
k 1 = 0.3077
n≥0
105
The forced solution of the difference equation is in the form
y[n] = d 1 (0.922)n cos (2.433n) + d 2 (0.922)n sin (2.433n) + 0.3077
In order to satisfy the specified initial conditions we need
y[−1] = d 1 (0.922)−1 cos (−2.433) + d 2 (0.922)−1 sin (−2.433) + 0.3077 = −2
and
y[−2] = d 1 (0.922)−2 cos (−4.866) + d 2 (0.922)−2 sin (−4.866) + 0.3077 = 0
The solutions for d 1 and d 2 are found as
d 1 = 3.4926
and
d 2 = −0.8054
The signal y[n] is
y[n] = 3.4926 (0.922)n cos (2.433n) − 0.8054 (0.922)n sin (2.433n) + 0.3077 ,
e.
Characteristic equation is
z 2 + 1.6 z + 0.64 = 0
(z + 0.8)2 = 0
⇒
The homogeneous solution is in the form
y h [n] = c 1 (−0.8)n + c 2 n (−0.8)n
n≥0
For the unit-step input signal the particular solution is
y p = k1
The particular solution must satisfy the difference equation, therefore
k 1 + 1.6 k 1 + 0.64 k 1 = 1
⇒
k 1 = 0.3086
The forced solution of the difference equation is in the form
y[n] = c 1 (−0.8)n + c 2 n (−0.8)n + 0.3086
In order to satisfy the specified initial conditions we need
y[−1] = c 1 (−0.8)−1 − c 2 (−0.8)−1 + 0.3086 = 0
and
y[−2] = c 1 (−0.8)−2 − 2 c 2 (−0.8)−2 + 0.3086 = 1
The solutions are found as
c 1 = 0.0514
and
c 2 = −1.1956
The signal y[n] is
y[n] = 0.0514 (−0.8)n − 1.1956 n (−0.8)n + 0.3086 ,
n≥0
n ≥0
106
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
3.16.
a.
The characteristic equation is
z − (1 − α) = 0
and the homogeneous solution is in the form
y h [n] = c 1 (1 − α)n ,
n≥0
For a unit-step input signal the particular solution is in the form
y p [n] = k 1
Imposing the particular solution into the difference equation yields
k 1 = (1 − α) k 1 + α (1)
and subsequently k 1 = 1. Therefore, the complete solution is in the form
y[n] = c 1 (1 − α)n + 1 ,
n≥0
In order to satisfy the initial condition y[−1] = 0 we need
c 1 (1 − α)−1 + 1 = 0
c1 = α − 1
⇒
The output signal is
y[n] = (α − 1) (1 − α)n + 1
= 1 − (1 − α)n+1 ,
n≥0
b.
For a unit-ramp input signal the particular solution is in the form
y p [n] = k 1 n + k 2
Imposing the particular solution into the difference equation yields
¤
£
k 1 n + k 2 = (1 − α) k 1 (n − 1) + k 2 + α n
which requires k 1 = 1 and k 2 = (α − 1) /α. The particular solution is
y p [n] = n +
α−1
α
and the complete solution is in the form
y[n] = c 1 (1 − α)n + n +
α−1
,
α
n≥0
In order to satisfy the initial condition y[−1] = 0 we need
c 1 (1 − α)−1 − 1 +
α−1
=0
α
⇒
c1 =
1−α
α
107
The output signal is
y[n] =
=
1−α
α−1
(1 − α)n + n +
α
α
α−1
(1 − α)n+1
+n +
,
α
α
n≥0
3.17.
a.
y a (t ) − y a (t − T )
+ A y a (t ) = A x a (t )
T
Solving for y a (t ) we get
y a (t ) =
1
AT
y a (t − T ) +
x a (t )
1 + AT
1 + AT
b.
Let
t = nT
and
t − T = (n − 1) T
Defining
x[n] =x a (nT )
y[n] =y a (nT )
y[n − 1] =y a (nT − T )
we obtain
1
AT
y[n − 1] +
x[n]
1 + AT
1 + AT
which corresponds to an exponential smoother with
y[n] =
α=
AT
1 + AT
3.18.
a.
1
x[n]
b
D
b
y[n]
D
b
−0.2
0.63
1
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
108
b.
1
y[n]
−3
2
x[n]
D
b
D
b
D
b
b
2.5
−2.44
0.9
c.
1
y[n]
−1
x[n]
D
b
D
b
b
0.49
d.
1
y[n]
−2
x[n]
b
D
b
D
b
D
b
−0.6
0.51
0.28
3.19.
a.
w [n] = x[n] ∗ h 1[n]
y[n] = w [n] ∗ h 2[n]
= (x[n] ∗ h 1[n]) ∗ h 2 [n]
= x[n] ∗ (h 1 [n] ∗ h 2[n]) = x[n] ∗ h eq [n]
109
Therefore
h eq [n] = h 1 [n] ∗ h 2[n]
b.
h eq [n] =
We know that
X
k
h 1 [k] h 2[n − k] =
h 2 [n − k] = u[n − k] − u[n − k − 5] =
4
X
k=0
½
(1) h 2 [n − k]
1,
0,
−4 + n < k < n
otherwise
Therefore
h eq [n] = { 1 , 2, 3, 4, 5, 4, 3, 2, 1 }
↑
n=0
c.
y[n] = { 1 , 3, 6, 10, 15, 19, 22, 24, 25, . . . }
↑
n=0
3.20.
a.
It can easily be seen that
y 1 [n] = x[n] ∗ h 1[n]
y 2 [n] = x[n] ∗ h 2[n]
Combining these two relationships we have
y[n] = y 1 [n] + y 2[n]
= x[n] ∗ h 1[n] + x[n] ∗ h 2[n]
= x[n] ∗ (h 1 [n] + h 2 [n]) = x[n] ∗ h eq [n]
Therefore
h eq [n] = h 1 [n] + h 2[n]
b.
h[n] = (0.9)n u[n] + (−0.7)n u[n]
c.
y 1 [n] =
Similarly
y 2 [n] =
The output signal is
∞
X
h 1 [k] u[n − k] =
∞
X
h 2 [k] u[n − k] =
k=−∞
k=−∞
y[n] = y 1 [n] + y 2[n] =
∞
X
k=0
∞
X
k=0
1 − (0.9)n+1
,
1 − 0.9
n≥0
1 − (−0.7)n+1
,
1 + 0.7
n≥0
h 1 [k] =
h 2 [k] =
1 − (0.9)n+1 1 − (−0.7)n+1
+
,
1 − 0.9
1 + 0.7
n≥0
110
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
3.21.
a.
y 1 [n] = h 1 [n] ∗ x[n]
w [n] = h 2 [n] ∗ x[n]
y 3 [n] = h 3 [n] ∗ w [n] = h 2 [n] ∗ h 3[n] ∗ x[n]
The output signal is
y[n] = y 1 [n] + y 2[n]
¡
¢
= h 1 [n] + h 2 [n] ∗ h 3[n] ∗ x[n]
and the equivalent impulse response is
h eq [n] = h 1 [n] + h 2[n] ∗ h 3[n]
b.
h 2 [n] ∗ h 3[n] = δ[n − 2] ∗ e −0.2n u[n] = e −0.2(n−2) u[n − 2]
h eq [n] = e −0.1n u[n] + e −0.2(n−2) u[n − 2]
heq [n]
Amplitude
2
1.5
1
0.5
0
5
0
10
15
20
25
n
c.
y 1 [n] = h 1 [n] ∗ u[n] =
y 1 [n] =
n
X
k=0
e −0.1k =
∞
X
k=−∞
e −0.1k u[k] u[n − k]
1 − e −0.1(n+1)
,
1 − e −0.1
n≥0
w [n] = h 2 [n] ∗ u[n] = δ[n − 2] ∗ u[n] = u[n − 2]
y 2 [n] = h 3 [n] ∗ w [n] =
y 2 [n] =
n−2
X
k=0
e −0.2k =
y[n] = y 1 [n] + y 2[n] =
∞
X
k=−∞
e −0.2k u[k] u[n − k − 2]
1 − e −0.2(n−1)
,
1 − e −0.2
n≥2
1 − e −0.2(n−1)
1 − e −0.1(n+1)
u[n]
+
u[n − 2]
1 − e −0.1
1 − e −0.2
replacemen
111
w[n]
y1 [n]
10
1
Amplitude
Amplitude
8
6
4
2
0.5
0
0
0
5
10
15
20
0
25
10
n
y2 [n]
y[n]
6
15
20
25
15
20
25
15
4
Amplitude
Amplitude
5
n
2
10
5
0
0
0
5
10
15
20
25
0
5
10
n
n
3.22.
a.
y 1 [n] = h 1 [n] ∗ x[n]
w [n] = h 2 [n] ∗ x[n]
y 3 [n] = h 3 [n] ∗ w [n] = h 3 [n] ∗ h 2[n] ∗ x[n]
y 4 [n] = h 4 [n] ∗ w [n] = h 4 [n] ∗ h 2[n] ∗ x[n]
The output signal is
y[n] = y 1 [n] + y 3[n] + y 4[n]
= h 1 [n] ∗ x[n] + h 3[n] ∗ h 2[n] ∗ x[n] + h 4[n] ∗ h 2[n] ∗ x[n]
¡
¢
= h 1 [n] + h 3[n] ∗ h 2[n] + h 4[n] ∗ h 2[n] ∗ x[n]
and the equivalent impulse response is
h eq [n] = h 1 [n] + h 3[n] ∗ h 2[n] + h 4[n] ∗ h 2[n]
b.
¡
¢ ¡
¢
h 3 [n] ∗ h 2[n] = u[n] − u[n − 3] ∗ u[n] − u[n − 3] =
Using the unit ramp function r [n], the result can be written as
h 3 [n] ∗ h 2[n] = r [n] − 2 r [n − 3] + r [n − 6]
112
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
h 4 [n] ∗ h 2[n] = δ[n − 2] ∗ h 2 [n] = h 2 [n − 2] = u[n − 2] − u[n − 5]
The impulse response of the equivalent system is
h eq [n] = e −0.1n u[n] + r [n] − 2 r [n − 3] + r [n − 6] + u[n − 2] − u[n − 5]
c.
w [n] = h 2 n ∗ u[n] =
n
X
k=0
h 2 [k] = { 1 , 2, 3, 3, 3, 3, 3, 3, 3, . . . }
↑
n=0
y 3 [n] = h 3 [n] ∗ w [n] = { 1 , 3, 6, 8, 9, 9, 9, 9, 9, . . . }
↑
n=0
y 4 [n] = h 4 [n] ∗ w [n] = w [n − 2] = { 1 , 2, 3, 3, 3, 3, 3, 3, 3, . . . }
↑
n=2
y[n] = y 1 [n] + y 3[n] + y 4[n] = { 2 , 5, 10, 13, 15, 15, 15, 15, 15, . . . }
↑
n=0
3.23.
Convolution of the two signals x[n] and h[n] is
y[n] =
∞
X
k=−∞
x[k] h[n − k]
The limits of the summation can be adjusted based on the term h[n − k]. Since
h[n − k] = 0 if
(
n − k < Nh1
⇒ k > n − Nh1 , or
n − k > Nh2
⇒ k < n − Nh2
the convolution sum can be written as
y[n] =
n−N
Xh1
k=−n−Nh2
x[k] h[n − k]
We know that x[k] = 0 if k < N x1 or k > N x2 . If the lower limit of the summation is greater than N x2 all
terms of the summation are zero. Similarly if the upper limit of the summation is less than N x1 then all
terms of the summation are zero.
(
n − Nh2 > N x2 ⇒ n > Nh2 + N x2 , or
y[n] = 0 if
n − Nh1 < N x1 ⇒ n < Nh1 + N x1
Therefore, the nonzero samples of y[n] are in the index range
Nh1 + N x1 ≤ n ≤ Nh2 + N x2
3.24.
y[n] = x[n] ∗ h[n]
∞
X
k=−∞
x[k] h[n − k]
113
If the signal x[n] is time shifted by m samples, we have
x[n − m] ∗ h[n]
∞
X
k=−∞
x[k − m] h[n − k]
Let us use the variable change k − m = k̄.
x[n − m] ∗ h[n]
∞
X
k̄=−∞
x[k̄] h[n − m − k̄] = y[n − m]
If the signal h[n] is time shifted instead, then we have
x[n] ∗ h[n − m]
3.25.
a.
∞
X
k=−∞
x[k] h[n − m − k] = y[n − m]
n
X
©
ª
δ[k] =
Sys δ[n] =
k=−∞
Therefore, h[n] = u[n].
½
1,
0,
n ≥0
n <0
b.
n
©
ª X
Sys δ[n] =
e −0.1(n−k) δ[k]
−∞
= e −0.1n
=
½
n
X
e 0.1k δ[k]
−∞
e −0.1n ,
0,
n≥0
n<0
Therefore, h[n] = e −0.1n u[n].
c.
n
©
ª X
Sys δ[n] =
δ[k] = 1 for n ≥ 0
k=0
Therefore, h[n] = u[n].
d.
©
ª
Sys δ[n] =
n
X
k=n−10
δ[k] =
½
1,
0,
n − 10 ≤ 0 ≤ n
otherwise
⇒
0 ≤ n ≤ 10
n − 10 ≤ 0 ≤ n + 10
otherwise
⇒
−10 ≤ n ≤ 10
Therefore h[n] = u[n] − u[n − 11].
e.
©
ª
Sys δ[n] =
n+10
X
k=n−10
δ[k] =
Therefore h[n] = u[n + 10] − u[n − 11].
½
1,
0,
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
114
3.26.
a.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
n = [0:19];
ynm1 = 5 ;
ynm2 = −3;
y = [];
for i = 0: 19 ,
yn = −0.2 * ynm1+0.63 *ynm2;
y = [ y , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
end ;
% Display the r e s u l t s
[n’ , y ’ ]
% Compute the output from a n a l y t i c a l s o lu t i o n
y_anl = 0 . 7 0 4 4 * ( 0 . 7 ) . ^ n−3.5944 * ( −0.9).^n ;
[ n ’ , y_anl ’ ]
b.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
n = [0:19];
ynm1 = 0 ;
ynm2 = 5 ;
y = [];
for i = 0: 19 ,
yn = −1.3 * ynm1−0.4 * ynm2;
y = [ y , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
end ;
% Display the r e s u l t s
[n’ , y ’ ]
% Compute the output from a n a l y t i c a l s o lu t i o n
y_anl = 3. 3333 * ( − 0. 5) . ^n−5.3333 * ( −0.8).^n ;
[ n ’ , y_anl ’ ]
c.
1
2
3
4
5
6
7
8
9
10
11
12
n = [0:19];
ynm1 = 1 ;
ynm2 = 2 ;
y = [];
for i = 0: 19 ,
yn = 1 . 7 * ynm1−0.72 * ynm2;
y = [ y , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
end ;
% Display the r e s u l t s
[n’ , y ’ ]
115
13
14
15
% Compute the output from a n a l y t i c a l s o lu t i o n
y_anl = 5 . 1 2 * ( 0 . 8 ) . ^ n− 4 . 8 6 * ( 0 . 9 ) . ^n ;
[ n ’ , y_anl ’ ]
d.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
n = [0:19];
ynm1 = −3;
ynm2 = −1;
y = [];
for i = 0: 19 ,
yn = 0. 49 * ynm2;
y = [ y , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
end ;
% Display the r e s u l t s
[n’ , y ’ ]
% Compute the output from a n a l y t i c a l s o lu t i o n
y_anl = − 1. 295 * ( 0. 7) . ^ n+0.805 * ( −0.7).^ n ;
[ n ’ , y_anl ’ ]
e.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
n = [0:19];
ynm1 = 3 ;
ynm2 = 2 ;
ynm3 = 1 ;
y = [];
for i = 0: 19 ,
yn = −0.6 * ynm1+0.51 *ynm2+0.28 *ynm3;
y = [ y , yn ] ;
ynm3 = ynm2;
ynm2 = ynm1;
ynm1 = yn ;
end ;
% Display the r e s u l t s
[n’ , y ’ ]
% Compute the output from a n a l y t i c a l s o lu t i o n
y_anl = −2.56 * ( −0.8).^n+0.9167 * ( −0.5).^n + 1 . 1 4 3 3 * ( 0 . 7 ) . ^ n ;
[ n ’ , y_anl ’ ]
3.27.
a.
1
2
x = @(n) ( 0 . 9 5 ) . ^ n . * cos ( 0 . 1 * pi * n ) . * ( n>=0);
n = [0:29];
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
116
b.
1
2
3
4
w = @(n) 3 * x (n ) ;
[ n ’ ,w(n ) ’ ]
y = @(n) w(n−2);
[ n ’ , y (n ) ’ ]
c.
1
2
3
4
wtilde = @(n) x ( n−2);
[ n ’ , wtilde (n ) ’ ]
y t i l d e = @(n) 3 * wtilde (n ) ;
[ n ’ , y t i l d e (n ) ’ ]
3.28.
a.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
x = @(n) ( 0 . 9 5 ) . ^ n . * cos ( 0 . 1 * pi * n ) . * ( n>=0);
n = [0:29];
w = @(n) 3 * x (n ) ;
[ n ’ ,w(n ) ’ ]
stem (n ,w(n ) ) ;
y = @(n) n . * w(n ) ;
[ n ’ , y (n ) ’ ]
stem (n , y (n ) ) ;
wtilde = @(n) n . * x (n ) ;
[ n ’ , wtilde (n ) ’ ]
stem (n , wtilde (n ) ) ;
y t i l d e = @(n) 3 * wtilde (n ) ;
[ n ’ , y t i l d e (n ) ’ ]
stem (n , y t i l d e (n ) ) ;
b.
1
2
3
4
5
6
7
8
9
10
11
12
w = @(n) 3 * x (n ) ;
[ n ’ ,w(n ) ’ ]
stem (n ,w(n ) ) ;
y = @(n) w(n) + 5 ;
[ n ’ , y (n ) ’ ]
stem (n , y (n ) ) ;
wtilde = @(n) x (n) + 5 ;
[ n ’ , wtilde (n ) ’ ]
stem (n , wtilde (n ) ) ;
y t i l d e = @(n) 3 * wtilde (n ) ;
[ n ’ , y t i l d e (n ) ’ ]
stem (n , y t i l d e (n ) ) ;
117
3.29.
a. Anonymous functions for the two signals can be defined as follows:
1
2
b.
1
2
3
4
5
x1 = @(n ) n . * exp ( −0.2 * n ) . * ( ( n>=0)&(n < 2 0 ) ) ;
x2 = @(n ) cos ( 0 . 0 5 * pi * n ) . * ( ( n>=0)&(n < 2 0 ) ) ;
Anonymous functions for the systems can be defined as follows:
sys1 = @( x , n) x . * ( n>=0);
sys2 = @( x , n) 3 * x +5;
sys3 = @( x , n) 3 * x +5 * (n>=0);
sys4 = @( x , n) n . * x ;
sys5 = @( x , n) cos ( 0 . 2 * pi * n ) . * x ;
c.
Find the responses of each system to x 1 [n], x 2 [n], and x[n] = 5 x 1 [n] − 3 x 2[n]. The script below is
for the first system. It may be edited to test the remaining systems.
1
2
3
4
5
6
7
8
9
10
11
12
y1 = sys1 ( x1 (n ) , n ) ;
y2 = sys1 ( x2 (n ) , n ) ;
y = sys1 ( 5 * x1 (n)−3 * x2 (n ) , n ) ;
clf ;
subplot ( 2 , 1 , 1 ) ;
stem (n , y ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ Sys_ { 1 } \ { 5x_ { 1 } [ n]−3x_ { 2 } [ n ] \ } ’ ) ;
subplot ( 2 , 1 , 2 ) ;
stem (n , 5 * y1−3* y2 ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ 5y_ { 1 } [ n]−3y_ { 2 } [ n] ’ ) ;
d.
Find the responses of each system to x 1 [n − 1]. The script below is for the first system. It may be
edited to test the remaining systems.
1
2
3
4
5
6
7
8
9
clf ;
subplot ( 2 , 1 , 1 ) ;
stem (n , sys1 ( x1 (n ) , n ) ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ Sys_ { 1 } \ { x_ { 1 } [ n ] \ } ’ ) ;
subplot ( 2 , 1 , 2 ) ;
stem (n , sys1 ( x1 (n−1) ,n ) ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ Sys_ { 1 } \ { x_ { 1 } [ n−1] \ } ’ ) ;
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
118
3.30.
a.
1
2
3
4
5
6
function bal = s s _ l b a l (A , B , c , n)
bal = A ;
for k =1:n ,
bal = bal * (1+ c)−B ;
end ;
end
b.
1
2
3
4
5
6
A = 10000;
B = 150;
c = 0.01;
bal1 = s s _ l b a l (A , B , c , 1 2 )
bal2 = s s _ l b a l (A , B , c , 2 4 )
bal3 = s s _ l b a l (A , B , c , 3 6 )
3.31.
a.
1
2
3
4
5
6
7
8
9
10
function y = ss_sqrt (A , y _ i n i t , t o l )
ynm1 = 1 ;
for k =1:100 , % Quit i f i t doesn ’ t work in 100 i t e r a t i o n s .
yn = 0 . 5 * (ynm1+A/ynm1 ) ;
i f abs ( yn−ynm1) < t o l ,
break ;
end ;
ynm1 = yn ;
end ;
y = yn ;
b.
1
2
3
y1 = ss_sqrt ( 5 , 1 , 0. 00001)
y2 = ss_sqrt ( 17 , 1 , 0. 00001)
y3 = ss_sqrt ( 132 , 1 , 0. 00001)
3.32.
a. Unit-step response of the exponential smoother:
1
b.
yu = @(n , alpha ) (1−(1−alpha ) . ^ ( n + 1 ) ) . * ( n>=0);
Unit-ramp response of the exponential smoother:
119
1
c.
yr = @(n , alpha ) ((1 − alpha ) . ^ ( n+1)/ alpha+n+( alpha −1)/alpha ) . * ( n>=0);
The input signal is
x[n] = u[n] − 0.5 r [n − 3] + 0.8 r [n − 8] − 0.3 r [n − 18] − 1.5 u[n − 23]
which can be expressed in MATLAB and graphed as follows:
1
2
3
4
5
6
7
n = [ − 5: 28] ;
x = ss_step (n) −0.5 * ss_ramp (n−3)+0.8 * ss_ramp (n − 8 ) . . .
−0.3 * ss_ramp(n−18) −1.5 * ss_step (n−23);
stem (n , x ) ;
axis ( [ − 5 . 5 , 2 8 . 5 , − 2 , 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x [n] ’ ) ;
d.
The output of the exponential smoother is computed and graphed through the use of the
following:
1
2
3
4
5
6
7
alpha = 0 . 1 ;
y = yu (n , alpha ) −0.5 * yr (n−3, alpha ) + 0 . 8 * yr (n−8, alpha ) . . .
−0.3 * yr (n−18, alpha ) −1.5 * yu (n−23, alpha ) ;
stem (n , y ) ;
axis ( [ − 5 . 5 , 2 8 . 5 , − 2 , 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ y [ n ] fo r \ alpha =0.1 ’ ) ;
Edit the first line to try for different values of α.
120
CHAPTER 3. ANALYZING DISCRETE-TIME SYSTEMS IN THE TIME-DOMAIN
Chapter 4
Fourier Analysis for Continuous-Time Signals
and Systems
4.1.
a. The approximation error is
ε̃3 (t ) = x̃ (t ) − x̃ (3) (t ) = x̃ (t ) − b 1 sin (ω0 t ) − b 2 sin (2ω0 t ) − b 3 sin (3ω0 t )
and the cost function is
J=
ˆ T0
0
£
¤2
ε̃3 (t ) d t =
ˆ T0
0
£
¤2
x̃ (t ) − b 1 sin (ω0 t ) − b 2 sin (2ω0 t ) − b 3 sin (3ω0 t ) d t
The fundamental frequency is ω0 = 2π/T0 .
b.
For minimum J we need
dJ
=0
d b1
ˆ T0
⇒
2 ε̃3 (t )
0
·
¸
d ε̃3 (t )
dt = 0
d b1
Using orthogonality, this leads to
ˆ T0
2
T0
b1 =
0
x̃ (t ) sin (ω0 t ) d t =
4A
π
Similarly, differentiating J with respect to b 2 yields
dJ
=0
d b2
ˆ T0
⇒
and
b2 =
2
T0
0
ˆ T0
0
·
¸
d ε̃3 (t )
2 ε̃3 (t )
dt = 0
d b2
x̃ (t ) sin (2ω0 t ) d t = 0
Finally, differentiating J with respect to b 3 we get
dJ
=0
d b3
¸
d ε̃3 (t )
dt = 0
d b3
ˆ T0
ˆ T0
x̃ (t ) sin (3ω0 t ) d t =
and
b3 =
·
⇒
2
T0
2 ε̃3 (t )
0
0
121
4A
3π
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
122
4.2.
a. The fundamental period is T0 = 1 second which corresponds to a fundamental frequency of f 0 = 1
Hz or ω0 = 2π rad/s.
b.
The approximation error is
ε̃1 (t ) = x̃ (t ) − x̃ (1) (t )
= x̃ (t ) − a 0 − a 1 cos (2πt ) − b 1 sin (2πt )
and the cost function is
J=
ˆ 1
0
£
¤2
ε̃1 (t ) d t =
ˆ 1
0
£
¤2
x̃ (t ) − a 0 − a 1 cos (2πt ) − b 1 sin (2πt ) d t
Conditions for minimizing the cost function are found by differentiating the cost function with respect
to each unknown coefficient and setting the result equal to zero. The use of orthogonality properties
leads to the following results:
ˆ T0
ˆ 1
¸
·
d ε̃1 (t )
dJ
=0 ⇒
d t = 0 ⇒ a0 =
2 ε̃1 (t )
x̃ (t ) d t = 0.3 A
d a0
d a0
0
0
ˆ T0
ˆ 1
¸
·
dJ
A sin (0.6π)
d ε̃1 (t )
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ a1 = 2
x̃ (t ) cos (2πt ) d t =
d a1
d a1
π
0
−1
£
¤
ˆ T0
ˆ 1
¸
·
A 1 − cos (0.6π)
dJ
d ε̃1 (t )
=0 ⇒
d t = 0 ⇒ b1 = 2
2 ε̃1 (t )
x̃ (t ) sin (2πt ) d t =
d b1
d b1
π
0
0
4.3.
The approximation error is
ε̃2 (t ) = x̃ (t ) − x̃ (2) (t )
= x̃ (t ) − a 0 − a 1 cos (2πt ) − b 1 sin (2πt ) − a 2 cos (4πt ) − b 2 sin (4πt )
and the cost function is
ˆ 1
ˆ 1
£
¤2
£
¤2
J=
ε̃2 (t ) d t =
x̃ (t ) − a 0 − a 1 cos (2πt ) − b 1 sin (2πt ) − a 2 cos (4πt ) − b 2 sin (4πt ) d t
0
0
Conditions for minimizing the cost function are found by differentiating the cost function with respect
to each unknown coefficient and setting the result equal to zero. The use of orthogonality properties
leads to the following results:
ˆ T0
ˆ 1
·
¸
d ε̃1 (t )
dJ
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ a0 =
x̃ (t ) d t = 0.3 A
d a0
d a0
0
0
ˆ T0
ˆ 1
¸
·
A sin (0.6π)
d ε̃1 (t )
dJ
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ a1 = 2
x̃ (t ) cos (2πt ) d t =
d a1
d a1
π
0
0
£
¤
ˆ T0
ˆ 1
¸
·
A 1 − cos (0.6π)
dJ
d ε̃1 (t )
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ b1 = 2
x̃ (t ) sin (2πt ) d t =
d b1
d b1
π
0
0
ˆ T0
ˆ 1
¸
·
dJ
A sin (1.2π)
d ε̃1 (t )
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ a2 = 2
x̃ (t ) cos (4πt ) d t =
d a2
d a2
2π
0
0
£
¤
ˆ T0
ˆ 1
·
¸
A 1 − cos (1.2π)
d ε̃1 (t )
dJ
=0 ⇒
2 ε̃1 (t )
d t = 0 ⇒ b2 = 2
x̃ (t ) sin (4πt ) d t =
d b2
d b2
2π
0
0
123
4.4.
The discontinuities of the signal x̃ (t ) occur at time instants t = nT0 /2. The approximation using M
terms can be evaluated at these time instants to yield
x̃
(M)
µ
Since ω0 = 2π/T0 we have
x̃
(M)
µ
¶ M
¶
X
nT0
nT0
b k sin kω0
=
2
2
k=1
µ
¶ M
X
nT0
b k sin (knπ) = 0
=
2
k=1
The amplitude of the signal x̃ (t ) is equal to ±A right before or right after a discontinuity, therefore
µ
¶
µ
¶
¶
µ
nT0
(M) nT 0
(M) nT 0
= x̃
− x̃
= ±A
ε̃
2
2
2
4.5.
a. The fundamental period is T0 = 2 seconds which corresponds to a fundamental frequency of
f 0 = 1/2 Hz or ω0 = π rad/s.
b.
The approximation error is
ε̃2 (t ) = x̃ (t ) − x̃ (2) (t )
= x̃ (t ) − a 0 − a 1 cos (πt ) − a 2 cos (2πt )
and the cost function is
J=
ˆ 1
−1
£
¤2
ε̃2 (t ) d t =
ˆ 1
−1
£
¤2
x̃ (t ) − a 0 − a 1 cos (πt ) − a 2 cos (2πt ) d t
Conditions for minimizing the cost function are found by differentiating the cost function with respect
to each unknown coefficient and setting the result equal to zero. The use of orthogonality properties
leads to the following results:
dJ
=0 ⇒
d a0
dJ
=0 ⇒
d a1
dJ
=0 ⇒
d a2
¸
d ε̃2 (t )
2 ε̃2 (t )
dt = 0 ⇒
d a0
−1
ˆ 1
¸
·
d ε̃2 (t )
dt = 0 ⇒
2 ε̃2 (t )
d a1
−1
ˆ 1
·
¸
d ε̃1 (t )
2 ε̃1 (t )
dt = 0 ⇒
d a2
−1
ˆ 1
·
1
a0 =
2
ˆ 1
−1
x̃ (t ) d t =
1
2
a1 =
ˆ 1
x̃ (t ) cos (πt ) d t =
a2 =
ˆ 1
x̃ (t ) cos (2πt ) d t = 0
0
−1
2
π
4.6.
First change summation indices in Eqn. (4.26) from k to m, then multiply both sides of it with sin (kω0 t )
124
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
and integrate over one full period:
ˆ t0 +T0
t0
x̃ (t ) sin (kω0 t ) d t =
ˆ t0 +T0
a 0 sin (kω0 t ) d t
t0
+
ˆ t0 +T0 · ∞
a m cos (mω0 t ) sin (kω0 t ) d t
+
ˆ t0 +T0 · ∞
¸
b m sin (mω0 t ) sin (kω0 t ) d t
X
t0
¸
m=1
X
t0
m=1
Swapping the order of integration and summation we obtain
ˆ t0 +T0
t0
x̃ (t ) sin (kω0 t ) d t = a 0
+
+
∞
X
am
m=1
∞
X
bm
m=1
"ˆ
"ˆ
t 0 +T0
ˆ t0 +T0
sin (kω0 t ) d t
t0
cos (mω0 t ) sin (kω0 t ) d t
#
sin (mω0 t ) sin (kω0 t ) d t
#
t0
t 0 +T0
t0
Let us consider the three terms on the right side of the equal sign:
1. For k > 0, the first term on the right side of the equal sign evaluates to zero since it includes, as a
factor, the integral of a sine function over a full period.
2. In the second term we have a summation. Each term within the summation has a factor which is
the integral of the product of a sine function and a cosine function over a span of T0 . Because of
the orthogonality of basis functions, all terms of this summation disappear.
3. In the third term we have another summation. Each term within the summation has a factor
which is the integral of the product of two sine functions over a span of T0 . All terms of this
summation disappear with the exception of one term for which m = k.
Therefore, the last result simplifies to
ˆ t0 +T0
t0
ˆ t0 +T0
sin2 (kω0 t ) d t = b k
x̃ (t ) sin (kω0 t ) d t ,
for k = 1, . . . , ∞
x̃(t ) sin (kω0 t ) d t = b k
It follows that
2
bk =
T0
ˆ t0 +T0
t0
t0
T0
2
4.7.
The fundamental period is T0 = 3 seconds which corresponds to a fundamental frequency of f 0 = 1/3
Hz or ω0 = 2π/3 rad/s.
ˆ 2
ˆ
·ˆ 1
¸
1 3
1
a0 =
x̃ (t ) d t =
(1) d t = 1
(2) d t
3 0
3 0
1
125
¶
2πk t
dt
x̃ (t ) cos
3
0
ˆ 2
·ˆ 1
¶
¶ ¸
µ
µ
2
2πk t
2πk t
=
dt +
dt
(1) cos
(2) cos
3 0
3
3
1
· µ
¶
µ
¶¸
2πk
4πk
1
sin
+
sin
=0
=
πk
3
3
2
ak =
3
µ
ˆ 3
¶
2πk t
dt
3
0
ˆ 2
·ˆ 1
¶
¶ ¸
µ
µ
2
2πk t
2πk t
=
dt +
dt
(1) sin
(2) sin
3 0
3
3
1
·
µ
¶
µ
¶¸
2πk
4πk
1
2 − cos
− cos
=
πk
3
3
bk =
2
3
ˆ 3
x̃ (t ) sin
µ
4.8.
The fundamental period is T0 = 1 second which corresponds to a fundamental frequency of f 0 = 1 Hz or
ω0 = 2π rad/s.
ˆ
1
a0 =
0
ak = 2
ˆ 1
e −2t d t = 0.4323
The coefficients a k are found through
e −2t cos (2πk t ) d t
0
Using the entry (B.22) of the integral table in Appendix B we obtain
ak =
3.4587
4 + 4π2 k 2
Similarly, the coefficients b k are found through
bk = 2
ˆ 1
e −2t sin (2πk t ) d t
0
Using the entry (B.23) of the integral table in Appendix B we obtain
bk =
3.4587πk
4 + 4π2 k 2
4.9.
a. Using Eqns. (4.55), (4.60) and (4.61) we obtain
c0 = a0 = 1
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
126
ck =
·
µ
¶
µ
¶¸
¢
1¡
2πk
4πk
1
2 − cos
− cos
,
ak − j bk = − j
2
2πk
3
3
k = 1, . . . , ∞
·
µ
¶
µ
¶¸
¢
1¡
2πk
4πk
1
2 − cos
− cos
,
ak + j bk = j
2
2πk
3
3
k = 1, . . . , ∞
c −k =
b.
1
ck =
3
ˆ 3
1
=
3
·ˆ 1
x̃ (t ) e − j 2πk t /3 d t
0
(2) e
− j 2πk t /3
0
dt +
ˆ 2
(1) e
− j 2πk t /3
dt
1
·
µ
¶
µ
¶¸
2πk
4πk
1
2 − cos
− cos
,
= −j
2πk
3
3
¸
all k
For k = 0 the use of L’Hospital’s rule yields c 0 = 1.
c.
|ck |
Magnitude
1
0.8
0.6
0.4
0.2
0
−0.2
−20
−15
−10
−5
0
5
10
15
20
10
15
20
k
θk = ∡ ck
Phase (rad)
π
π
2
0
−
π
2
−π
−20
−15
−10
−5
0
5
k
4.10.
a. The EFS coefficients are determined from the TFS coefficients as follows:
c 0 = a 0 = 0.4323
127
ck =
¢
0.6647
1¡
ak − j bk =
,
2
2 + j 2πk
k >0
¢
1¡
0.6647
ak + j bk =
2
2 − j 2πk
k >0
c −k =
b.
The EFS coefficients can be found by direct application of the EFS integral as
ck =
ˆ 1
0
e −2t e − j 2πk t d t =
0.8647
,
2 + j 2πk
all k
c.
|ck |
Magnitude
0.5
0.4
0.3
0.2
0.1
0
−0.1
−20
−15
−10
−5
0
5
10
15
20
10
15
20
k
θk = ∡ ck
Phase (rad)
π
π
2
0
−
π
2
−π
−20
−15
−10
−5
0
5
k
4.11.
a. The fundamental period is T0 = 1 second which corresponds to a fundamental frequency of f 0 = 1
Hz or ω0 = 2π rad/s. The EFS coefficients are computed as
ck =
ˆ 1
0
= −a
(−at + a) e − j 2πk t d t
ˆ 1
0
te
− j 2πk t
dt +a
ˆ 1
0
e − j 2πk t d t
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
128
The first integral can be evaluated using entry (B.16) in Appendix B to yield
−a
ˆ 1
0
t e − j 2πk t d t =
¢ − j 2πk i
a h ¡
1
−
1
+
j
2πk
e
4π2 k 2
The second integral yields
a
ˆ 1
0
e − j 2πk t d t =
EFS coefficients are
ck = − j
i
a h
1 − e − j 2πk
j 2πk
a
,
2πk
all k 6= 0
The center coefficient c 0 is found separately as
c0 =
ˆ 1
0
(−at + a) d t =
a
2
b.
|ck |
Magnitude
1
0.8
0.6
0.4
0.2
0
−0.2
−20
−15
−10
−5
0
5
10
15
20
10
15
20
k
θk = ∡ ck
Phase (rad)
π
π
2
0
−
π
2
−π
−20
−15
−10
−5
0
5
k
c.
The TFS coefficients are computed from EFS coefficients as
a0 = c0
a k = c k + c −k ,
k = 1, . . . , ∞
b k = j (c k − c −k ) ,
k = 1, . . . , ∞
129
Thus
a
2
a
a
ak = − j
=0,
−j
2π (−k) ¶
µ 2π k
a
a
a
=
+j
,
bk = j − j
2π k
2π (−k)
πk
a0 =
k = 1, . . . , ∞
k = 1, . . . , ∞
4.12.
a.
The signals x̃ (t ) and g̃ (t ) can be written in terms of their EFS coefficients as
x̃ (t ) =
and
∞
X
k=−∞
∞
X
g̃ (t )
c k e j 2πk f 0 t
d k e j 2πk f 0 t
k=−∞
Using the relationship g̃ (t ) = −x̃ (t ) we have
∞
X
k=−∞
d k e j 2πk f 0 t =
∞
X
c k e − j 2πk f 0 t
k=−∞
which implies that
d0 = c0
b.
and
d k = c −k
For the signal x̃ (t ) used in Example 4-8 the EFS coefficients were
a
2πk
for k 6= 0
a
2πk
for k 6= 0
a
2
and
ck = j
a
2
and
dk = − j
c0 =
The EFS coefficients for g̃ (t ) are
d0 =
4.13.
Let x̃ (t ) and g̃ (t ) have the EFS coefficients c k and d k respectively, so that
x̃ (t ) =
and
g̃ (t ) =
Adding the two signals yields
x̃ (t ) + g̃ (t ) =
∞
X
c k e j 2πk f 0 t
k=−∞
∞
X
d k e j 2πk f 0 t
k=−∞
∞
X
k=−∞
(c k + d k ) e j 2πk f 0 t = a
130
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
Setting the coefficients of both sides equal to each other we get
c0 + d0 = a
⇒
d0 = a − c0
ck + dk = 0
⇒
d k = −c k ,
From Example 4-8
k 6= 0
a
,
k =0
2
ck =

 j a , k=
6 0
2πk
EFS coefficients d k for the signal g̃ (t ) are found as



a
,
k =0
2
dk =

 −j a , k =
6 0
2πk



1−
4.14.
The EFS coefficients are found as
1
ck =
T0
ˆ T0
1
=
T0
ˆ T0 /2
0
¯
¯
¯ sin (2πt /T0)¯ e − j 2πk t /T0 d t
sin (2πt /T0 ) e
− j 2πk t /T0
0
1
dt −
T0
ˆ T0
sin (2πt /T0 ) e − j 2πk t /T0 d t
T0 /2
The integrals can be evaluated using entry (B.23) of Appendix B. The result is

2

 ¡
¢ , k even
π
1
− k2
ck =


0,
k odd
4.15.
Let a periodic signal x̃ (t ) be written in terms of its EFS coefficients as
x̃ (t ) =
∞
X
c k e j kω0 t
−∞
Time shifting the signal by τ results in
x̃ (t − τ) =
∞
X
−∞
c k e j kω0 (t −τ) =
∞
X
−∞
c k e − j kω0 τ e j kω0 t =
It therefore follows that
d k = c k e − j kω0 τ
∞
X
−∞
d k e j kω0 t
131
4.16.
Let d k be the coefficients of the half-wave sinusoidal signal g̃ (t ) so that
g̃ (t ) =
∞
X
d k e j 2πk f 0 t
−∞
It was determined in Example 4-10 that the EFS coefficients for g̃ (t ) are


0,
k odd and k 6= ∓1





− j /4 ,
k =1

dk =
j /4 ,
k = −1



−1



¢ , k even
 ¡ 2
π k −1
The full-wave sinusoidal signal x̃ (t ) can be written in terms of the half-wave rectified sinusoidal signal
g̃ (t ) as
x̃ (t ) = g̃ (t ) + g̃ (t − T0 /2)
In terms of the EFS coefficients, the corresponding relationship is
c k = d k + e − j 2πk f 0 (T0 /2) d k
= d k + (−1)k d k
which results in
ck =
4.17.





2
¡
¢,
π 1 − k2
0,
k even
k odd
For the half-wave rectified sinusoidal signal x̃ (t ) the EFS coefficients were found as


0,
k odd and k 6= ∓1






− j /4 ,
k =1

ck =
j /4 ,
k = −1




−1



¢ , k even
 ¡ 2
π k −1
The CFS representation of the signal is
x̃ (t ) = d 0 +
k=1
with
dk =











¯ ¯
d k = 2 ¯c k ¯
∞
X
¡
¢
d k cos 2πk f 0 t + φk
and
0,
k odd and k 6= 1
1/2 ,
2
k =1
¡
¢ , k even
π k2 − 1
φk = tan−1
µ
¶
Im {c k }
= ∡ ck
Re {c k }
φk =







0,
k odd and k 6= 1
−π/2 , k = 1
π,
k even
132
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
4.18.
a.
¡ ¢
¡ ¢
X f = 3 sinc f
b.
¡ ¢
¡ ¢
X f = 3 sinc f e − j π f
c.
¡ ¢
¡ ¢
X f = 8 sinc 4 f
d.
¡ ¢
¡ ¢
X f = 4 sinc 2 f e − j 6π f
4.19.
Using the Fourier transform integral we have
¡ ¢
X f =
ˆ 1
0
e − j 2π f t d t −
ˆ 2
e − j 2π f t d t
1
¯1
¯2
e − j 2π f t ¯¯
e − j 2π f t ¯¯
=
¯ −
¯
− j 2π f ¯
− j 2π f ¯
0
1
− j 2π f
=
1
h
i
e − j 2π f − 1 − e − j 4π f + e − j 2π f
This result can be written as
h
i
h
i
1
1
e − j 2π f − e − j 4π f +
−1 + e − j 2π f
− j 2π f
− j 2π f
¡ ¢
X f =
i e−jπf h
i
e − j 3π f h j π f
e
− e−jπf +
−e j π f + e − j π f
− j 2π f
− j 2π f
h
i
¡ ¢
= sinc f e − j π f − e − j 3π f
=
4.20.
Using linearity of the Fourier transform we have
¡ ¢
©
ª
©
ª
©
ª
X f = F e −at u (t ) − e at u (−t ) = F e −at u (t ) − F e at u (−t )
Since
and
©
ª
F e −at u (t ) =
ˆ ∞
©
ª
F e at u (−t ) =
e
−at
u (t ) e
− j 2π f t
−∞
ˆ ∞
−∞
e
at
u (−t ) e
dt =
− j 2π f t
ˆ ∞
0
e −at e − j 2π f t d t =
1
a + j 2π f
ˆ 0
e at e − j 2π f t d t =
1
a − j 2π f
dt =
−∞
133
we obtain
¡ ¢
X f =
1
1
− j 4π f
−
= 2
a + j 2π f a − j 2π f
a + 4π2 f 2
4.21.
Using the unit-pulse function Π (t ) we have
©
ª
¡ ¢
F Π (t − 0.5) = sinc f e − j π f
and
©
ª
¡ ¢
F Π (t − 1.5) = sinc f e − j 3π f
Utilizing linearity of the Fourier transform
i
©
ª
¡ ¢h
F Π (t − 0.5) − Π (t − 1.5) = sinc f e − j π f − e − j 3π f
4.22.
Let the signal g (t ) be defined as
g (t ) = e −at u (t )
Its Fourier transform is
The signal x (t ) can be written as
¡ ¢
G f =
1
a + j 2π f
x (t ) = g (t ) − g (−t )
Since
¡ ¢
X f is found as
©
ª
¡ ¢
F g (−t ) = G − f =
¡ ¢
¡ ¢
¡ ¢
X f = G f +G − f =
1
a − j 2π f
1
1
− j 4π f
−
= 2
a + j 2π f a − j 2π f
a + 4π2 f 2
4.23.
The signal x (t ) can be written as
x (t ) = Λ (t − 0.5) − Λ (t − 1.5)
Using the time shifting property of the Fourier transform we obtain
and
Thus the transform of x (t ) is
³ω´
©
ª
e − j 0.5ω
F Λ (t − 0.5) = sinc2
2π
³ω´
©
ª
F Λ (t − 1.5) = sinc2
e − j 1.5ω
2π
X (ω) = sinc2
i
³ ω ´h
e − j 0.5ω − e − j 1.5ω
2π
134
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
4.24.
Using the duality property we have
F
X (t ) ←→ 2π x (−ω)
or equivalently
2a
F
←→ 2π e −a|−ω|
a2 + t 2
Multiplying both the numerator and the denominator of the time-domain signal by 4 yields
8a
F
4 a2 + 4 t 2
←→ 2π e −a|ω|
Let us choose
4 a2 = 1
so that
a=
⇒
1
2
4
F
←→ 2π e −|ω|/2
2
1+4t
Scaling both sides of the transform relationship by 1/2 we obtain the desired result:
2
F
←→ π e −|ω|/2
1+4t2
4.25.
Starting with the Fourier transform integral
X (ω) =
ˆ ∞
x (t ) e − j ωt d t
−∞
and conjugating both sides we get
∗
X (ω) =
·ˆ ∞
x (t ) e
− j ωt
−∞
dt
¸∗
=
ˆ ∞
x ∗ (t ) e j ωt d t
−∞
Since x (t ) is real-valued we have x ∗ (t ) = x (t ), and
∗
X (ω) =
ˆ ∞
−∞
x (t ) e j ωt d t = X (−ω)
Using the variable change λ = −t and d λ = −d t , the transform X (−ω) can be written as
X (−ω) =
ˆ −∞
∞
x (−λ) e − j ωλ (−d λ) =
ˆ ∞
x (−λ) e − j ωλ d λ
−∞
Since x (t ) is odd, we have x (−λ) = −x (λ), and
X (−ω) = −
ˆ ∞
−∞
x (λ) e − j ωλ d λ = −X (ω)
Therefore
X ∗ (ω) = −X (ω)
135
Since conjugating the transform causes it to be negated, the transform must be purely imaginary, that is
©
ª
Re X (ω) = 0
4.26.
a. Starting with
¡ ¢
F
Π (t ) ←→ sinc f
we obtain
Π
and
¡ ¢
Therefore, the transform X f is
b.
µ
µ ¶
¡ ¢
t
F
←→ 2 sinc 2 f
2
¶
¡ ¢
t −1 F
Π
←→ 2 sinc 2 f e − j 2π f
2
¡ ¢
¡ ¢
X f = 2 sinc 2 f e − j 2π f
The even component of x (t ) is
µ
µ
¶
¶
t −1
−t − 1
1
1
+ Π
x e (t ) = Π
2
2
2
2
and its odd component is
¶
¶
µ
µ
1
1
t −1
−t − 1
x o (t ) = Π
− Π
2
2
2
2
xo (t)
0.6
0.6
0.4
0.4
Amplitude
Amplitude
xe (t)
0.2
0
−0.2
−0.4
−0.6
−3
0.2
0
−0.2
−0.4
−0.6
−2
−1
0
1
2
3
−3
−2
−1
t
c.
The transform of the even component is
µ ¶¾
½
¡ ¢
¡ ¢
t
1
= 2 sinc 4 f
Π
Xe f = F
2
4
and the transform of the odd component is
½
¶
¶¾
µ
µ
¡ ¢
1
1
t +1
t −1
Xo f = F − Π
+ Π
2
2
2
2
¡ ¢ j 2π f
¡ ¢ − j 2π f
= − sinc 2 f e
+ sinc 2 f e
¡ ¢
¡
¢
= − j 2 sinc 2 f sin 2π f
0
t
1
2
3
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
136
¡ ¢
Using Euler’s formula on the transform X f we can write
© ¡ ¢ª
¡ ¢
¡
¢
Re X f = 2 sinc 2 f cos 2π f
¡
¢¸
·
¡
¢
sin 2π f
cos 2π f
=2
2π f
Using the trigonometric identity 2 si n (α) cos (α) = sin (2α) we get
¡ ¢
Similarly, the imaginary part of X f is
© ¡ ¢ª
¡ ¢
Re X f = 2 sinc 4 f
© ¡ ¢ª
¡ ¢
¡
¢
Im X f = −2 sinc 2 f sin 2π f
We have
¡ ¢
© ¡ ¢ª
X e f = Re X f
and
¡ ¢
© ¡ ¢ª
X o f = j Im X f
4.27.
X (f )
a.
Starting with
©
ª
¡ ¢
F Π (t ) = sinc f
and using the modulation property, we obtain
¡
¢ 1
¡
¢
¡ ¢ 1
X f = sinc f + 5 + sinc f − 5
2
2
1
0.8
0.6
0.4
0.2
0
−0.2
−10
−5
0
5
10
5
10
f (Hz)
X (f )
b.
Since
½ µ ¶¾
¡ ¢
t
= 2 sinc 2 f
F Π
2
1
0.8
0.6
0.4
the use of the modulation property yields
0.2
¡ ¢
¡
¢
¡
¢
X f = sinc 2 f + 10 + sinc 2 f − 10
−0.2
0
−10
−5
0
f (Hz)
137
X (f )
c.
Since
1
µ ¶
©
ª 1
f
F Π (2t ) = sinc
2
2
0.8
0.6
0.4
the use of the modulation property yields
¶
¶
µ
µ
¡ ¢ 1
1
f +5
f −5
X f = sinc
+ sinc
4
2
4
2
0.2
0
−0.2
−10
−5
0
5
10
5
10
f (Hz)
X (f )
d.
Since
1
µ ¶
©
ª 1
f
F Π (4t ) = sinc
4
4
0.8
0.6
0.4
the use of the modulation property yields
¶
¶
µ
µ
¡ ¢ 1
1
f +5
f −5
+ sinc
X f = sinc
8
4
8
4
0.2
0
−0.2
−10
−5
0
f (Hz)
4.28.
a. Starting with the inverse Fourier transform integral
x (t ) =
ˆ ∞
−∞
¡ ¢
X f e j 2π f t d f
and differentiating both sides with respect to t yields
d x (t )
d
=
dt
dt
Therefore
b.
·ˆ ∞
=
ˆ ∞
=
ˆ ∞
−∞
F
(
¸
d h ¡ ¢ j 2π f t i
df
X f e
−∞ d t
−∞
¡ ¢
j 2π f X f e j 2π f t d f
©
¡ ¢ª
d x (t )
= F −1 j 2π f X f
dt
Given that
¡ ¢
X f e j 2π f t d f
d k x (t )
dtk
⇒
)
¾
¡ ¢
d x (t )
= j 2π f X f
F
dt
½
¡
¢k ¡ ¢
= j 2π f X f
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
138
the use of the result in part (a) yields
"
)
(
#)
(
¡
¢k ¡ ¢ ¡
¢k+1 ¡ ¢
d d k x (t )
d k+1 x (t )
=
F
=
j
2π
f
j
2π
f
X
f
=
j
2π
f
X f
F
dt
d t k+1
dtk
4.29.
a. Starting with the Fourier transform integral
¡ ¢
X f =
ˆ ∞
x (t ) e − j 2π f t d t
−∞
and differentiating both sides with respect to f yields
¡ ¢
·ˆ ∞
¸
dX f
d
x (t ) e − j 2π f t d t
=
df
df
−∞
Therefore
b.
=
ˆ ∞
=
ˆ ∞
i
d h
x (t ) e − j 2π f t d t
−∞ d f
−∞
¡
¢
− j 2πt x (t ) e − j 2π f t d t
¡ ¢
©¡
¢
ª dX f
F − j 2πt x (t ) =
df
Given that
F
the use of the result in part (a) yields
F
n¡
n¡
− j 2πt
¢k
¡ ¢
o dk X f
x (t ) =
dfk
o
¢¡
¢k
d
− j 2πt − j 2πt x (t ) =
df
and therefore
F
n¡
− j 2πt
¢k+1
"
¡ ¢#
dk X f
dfk
¡ ¢
o d k+1 X f
x (t ) =
d f k+1
4.30.
The differentiation in time property states that
¡ ¢
F
x (t ) ←→ X f
implies
¢n ¡ ¢
d n x (t ) F ¡
←→ j 2π f X f
n
dt
Applying the duality property of the Fourier transform to both transform pairs leads to the statement
that
¡ ¢
¡
¢n
d n £ ¡ ¢¤
F
F
X (t ) ←→ x − f
implies
j 2πt X (t ) ←→ ¡ ¢n x − f
d −f
139
¡ ¢
¡ ¢
Defining two new functions as x̄ (t ) = X (t ) and X̄ f = x − f , the statement above becomes
¡ ¢
F
x̄ (t ) ←→ X̄ f
implies
¡
j 2πt
¢n
F
x̄ (t ) ←→
Multiplying both sides of the second transform pair by (−1)n we get
¡ ¢
F
x̄ (t ) ←→ X̄ f
implies
d n £ ¡ ¢¤
¡ ¢n X̄ f
d −f
¡
¢n
d n £ ¡ ¢¤
F
X̄ f
− j 2πt x̄ (t ) ←→
dfn
4.31.
Let w (t ) be the derivative of the signal x (t ), that is
¶
µ
d x (t )
t −4
w (t ) =
= Π (t − 0.5) − 0.5 Π
dt
2
¡ ¢
The transform W f is
¡ ¢
¡ ¢
¡ ¢
W f = sinc f e − j π f − sinc 2 f e − j 8π f
Using the differentiation property of the Fourier transform
¡ ¢
¡ ¢
W f = j 2π f X f
and
¡ ¢
i
¡ ¢
¡ ¢
¡ ¢ X f
1 h
sinc f e − j π f − sinc 2 f e − j 8π f
=
X f =
j 2π f
j 2π f
4.32.
Using duality we have
¡ ¢
F
X (t ) ←→ x − f
¡ ¢
¡ ¢
Since the signal x (t ) is even, we have x − f = x f . Let
τ = f 0 (1 + r ) − f 0 (1 − r ) = 2 f 0 r
and
λ = f0
The inverse transform we seek is
¡
¢
¡
¢
X (t ) = 2 f 0 sinc 2 f 0 r t sinc 2 f 0 t
4.33.
a.
The modulation property states that
¢
¢
©
¡
¢
ª 1 ¡
1 ¡
F sin 2π f 0 t p (t ) = P f − f 0 e − j π/2 + P f + f 0 e j π/2
2
2
Let the signal p (t ) be defined as
¶
µ
1
p (t ) = Π t −
2
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
140
with the corresponding transform
¡ ¢
¡ ¢
P f = sinc f e − j π f
The use of the modulation property with f 0 = 0.5 Hz yields
¢
¢
¡ ¢ 1 ¡
1 ¡
X f = P f − 0.5 e − j π/2 + P f + 0.5 e j π/2
2
2
¡
¢ − j π ( f −0.5) − j π/2 1
¡
¢
1
e
+ sinc f + 0.5 e − j π ( f +0.5) e j π/2
= sinc f − 0.5 e
2
2
£
¡
¢
¡
¢¤
1
=
sinc f − 0.5 + sinc f + 0.5 e − j π f
2
b.
The multiplication property states that
©
ª
¡ ¢
¡ ¢
F q (t ) p (t ) = Q f ∗ P f
Let the signals q (t ) and p (t ) be defined as
q (t ) = sin (πt )
and
¶
µ
1
p (t ) = Π t −
2
¡ ¢
¡ ¢
The transforms Q f and P f are
¢
¢
¡ ¢
1 ¡
1 ¡
Q f = − j δ f − 0.5 + j δ f + 0.5
2
2
¡ ¢
The transform X f is
⇒
¡ ¢
¡ ¢
P f = sinc f e − j π f
¡ ¢
¡ ¢
¡ ¢
X f =Q f ∗P f
=
¡
¢
¡
¢¤
1£
sinc f − 0.5 + sinc f + 0.5 e − j π f
2
4.34.
EFS coefficients for the signal x̃ (t ) are
µ ¶
k − j πk/3
1
e
c k = sinc
3
3
The Fourier transform of x̃ (t ) is
µ ¶
∞
∞
X
¡ ¢
¢ 1 X
¡
¢
k − j πk/3 ¡
X f =
ck δ f − k f 0 =
e
δ f − k/3
sinc
3 k=−∞
3
k=−∞
4.35.
a. The power in the pulse train is
P x = ⟨x 2 (t )⟩ =
1
T0
ˆ T0 /2
T0 /2
x 2 (t ) d t
1
T0
ˆ τ/2
τ/2
(1) d t
141
Since τ = d T0
Px =
b.
1
T0
ˆ d T0 /2
d T0 /2
(1) d t = d
EFS coefficients for the pulse train are
c k = d sinc (kd ) ,
k = −∞, . . . , ∞
The power spectral density is
∞
∞
X
X
¡
¢
¡
¢
¡ ¢
|c k |2 δ f − k f 0 =
d 2 sinc2 (kd ) δ f − k f 0
Sx f =
k=−∞
k=−∞
¡ ¢
where the fundamental frequency is f 0 = 1/T0 . S x f is shown below for d = 0.5.
Sx (f )
0.25
0.101
−9f0 −6f0 −3f0
0.101
0
3f0
6f0
9f0
f (Hz)
c. If EFS terms up to and including the M -th harmonic are retained, the normalized average power
of the signal would be
M
M
X
X
|c k |2 =
d 2 sinc2 (kd )
P x(M) =
k=−M
k=−M
and the percentage of this to the total average power in the signal x (t ) is
η=
M
X
P x(M) P x(M)
=
=
d sinc2 (kd )
Px
d
k=−M
It can be shown that, with d = 0.5 and M = 1 we get
1
X
k=−1
d.
Frequencies up to the third harmonic are needed to retain 95 percent of the signal power since
3
X
k=−3
e.
(0.5) sinc2 (0.5 k) = 0.9053
(0.5) sinc2 (0.5 k) = 0.9503
Frequencies up to the 21-st harmonic are needed to retain 99 percent of the signal power since
21
X
k=−21
(0.5) sinc2 (0.5 k) = 0.9908
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
142
4.36.
c. If EFS terms up to and including the M -th harmonic are retained, the percentage of this to the
total average power in the signal x (t ) is
η=
M
X
P x(M) P x(M)
=
=
d sinc2 (kd )
Px
d
k=−M
It can be shown that, with d = 0.2 and M = 4 we get
4
X
k=−4
d.
(0.2) sinc2 (0.2 k) = 0.9029
Frequencies up to the 11-th harmonic are needed to retain 95 percent of the signal power since
11
X
k=−11
e.
(0.2) sinc2 (0.2 k) = 0.9528
Frequencies up to the 51-st harmonic are needed to retain 99 percent of the signal power since
51
X
k=−51
(0.2) sinc2 (0.2 k) = 0.9900
ˆ ∞
ˆ ∞
4.37.
Parseval’s theorem states that
−∞
¯
¯
¯x (t )¯2 d t
−∞
¯ ¡ ¢¯2
¯X f ¯ d t
Let x (t ) = Π (t ). The normalized average power of x (t ) is
ˆ ∞
−∞
The Fourier transform of x (t ) is
Therefore
ˆ ∞
−∞
¯
¯
¯x (t )¯2 d t =
ˆ 1/2
−1/2
(1)2 d t = 1
¡ ¢
©
ª
¡ ¢
X f = F Π (t ) = sinc f
¯ ¡ ¢¯2
¯X f ¯ d t =
ˆ ∞
−∞
¯
¡ ¢¯
¯ sinc f ¯2 d f = 1
4.38.
a. For the signal x (t ) the fundamental frequency is f 0 = 10 Hz, and the EFS coefficients are
 3
 ,
2
ck =

0,
k = ±1
otherwise
The power spectral density is
¢ 9 ¡
¢
¡ ¢ 9 ¡
S x f = δ f + 10 + δ f − 10
4
4
Sx (f )
9
4
9
4
−10
10
f (Hz)
143
b.
For the signal x (t ) the fundamental frequency is f 0 = 10 Hz, and the EFS coefficients are


1,




3
ck =
,

2



 0,
k = ±2
k = ±3
Sx (f )
9
4
otherwise
9
4
1
1
¡ ¢ 9 ¡
¢ ¡
¢ ¡
¢ 9 ¡
¢−30 −20
S x f = δ f + 30 +δ f + 20 +δ f − 20 + δ f − 30
4
4
20
The power spectral density is
c.
30
f (Hz)
For the signal x (t ) the fundamental frequency is f 0 = 5 Hz, and the EFS coefficients are

5


,


4





 5,
2
ck =

 5



 4,




0,
Sx (f )
k = ±17
25
4
25
4
k = ±20
k = ±23
otherwise
25
16
25
16
−100
25
16
25
16
100
f (Hz)
The power spectral density is
¡ ¢ 25 ¡
¢ 25 ¡
¢ 25 ¡
¢ 25 ¡
¢ 25 ¡
¢ 25 ¡
¢
Sx f =
δ f + 230 +
δ f + 200 +
δ f + 170 +
δ f − 170 +
δ f − 200 +
δ f − 230
16
4
16
16
4
16
4.39.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
% Compute the s i g n a l x ( t )
A = 1;
T0 = 1 ;
xp = @( t ) A * ( t <=T0/2)−A * ( t >T0 / 2 ) ; % One period
t = [ −1.5:0.005:2.5];
x = xp (mod( t , T0 ) ) ;
% Compute the approximation
b1 = 4 * A/ pi ;
b2 = 0 ;
b3 = 4 * A/ ( 3 * pi ) ;
f0 = 1/T0 ;
xhat = b1 * sin ( 2 * pi * f0 * t )+b2 * sin ( 4 * pi * f0 * t )+b3 * sin ( 6 * pi * f0 * t ) ;
% Graph the s i g n a l and the approximation
plot ( t , x , t , xhat ) ;
axis ( [ − 1 . 5 , 2 . 5 , − 1 . 6 , 2 . 4 ] ) ;
t i t l e ( ’ The s i g n a l x ( t ) and i t s approximation ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
legend ( ’ O r i g i n a l s i g n a l ’ , ’ Approximation ’ ) ;
grid ;
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
144
4.40.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
% Compute the s i g n a l x ( t )
A = 1;
xp = @( t ) A * ( t < = 0 . 3 ) ; % One period
t = [ −1.5:0.005:2.5];
x = xp (mod( t , 1 ) ) ;
% Compute the approximation
a0 = 0 . 3 * A ;
a1 = A * sin ( 0 . 6 * pi ) / pi ;
b1 = A* (1−cos ( 0 . 6 * pi ) ) / pi ;
xhat = a0+a1 * cos ( 2 * pi * t )+b1 * sin ( 2 * pi * t ) ;
% Graph the s i g n a l and the approximation
plot ( t , x , t , xhat ) ;
axis ( [ − 1 . 5 , 2 . 5 , − 0 . 5 , 1 . 8 ] ) ;
t i t l e ( ’ The s i g n a l x ( t ) and i t s approximation ’ ) ;
xlabel ( ’ t ( sec ) ’ ) ;
legend ( ’ O r i g i n a l s i g n a l ’ , ’ Approximation ’ ) ;
grid ;
4.41. The code below computes and graphs the original signal, the approximation using M = 3
harmonics, and the approximation error.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
% Compute the s i g n a l x ( t )
xp = @( t ) 2 * ( t < 1) + 1 * ( ( t >=1)&( t < = 2 ) ) ; % One period
t = [ −4:0.005:6];
x = xp (mod( t , 3 ) ) ; % P e r i o d i c extension
% Compute the c o e f f i c i e n t s
k = [1:10];
a0 = 1 ;
a = ( sin ( 2 * pi * k /3)+ sin ( 4 * pi * k / 3 ) ) . / ( pi * k ) ;
b = (2−cos ( 2 * pi * k/3)− cos ( 4 * pi * k / 3 ) ) . / ( pi * k ) ;
% Approximation with M=3 harmonics
xhat = a0 * ones ( s i z e ( t ) ) ;
for m= 1: 3 ,
xhat = xhat+a (m) * cos ( 2 * pi *m* t /3)+b(m) * sin ( 2 * pi *m* t / 3 ) ;
end ;
plot ( t , x , t , xhat ) ;
axis ( [ − 3 , 3 , − 1 , 3. 5] ) ;
xlabel ( ’ t ’ ) ;
legend ( ’ O r i g i n a l s i g n a l ’ , ’ Approx . with M=3 ’ ) ;
grid ;
% Approximation e r r o r f o r M=3
plot ( t , x−xhat ) ;
axis ( [ − 3 , 3 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’ t ’ ) ;
grid ;
To use M = 4 harmonics, modify line 12 as
145
12
for m= 1: 4 ,
To use M = 5 harmonics, modify it as
12
for m= 1: 5 ,
4.42. The code below computes and graphs the original signal, the approximation using M = 3
harmonics, and the approximation error.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
% Compute the s i g n a l x ( t )
xp = @( t ) exp(−2 * t ) ; % One period
t = [ −1:0.005:2];
x = xp (mod( t , 1 ) ) ; % P e r i o d i c extension
% Compute the EFS c o e f f i c i e n t s
k = [1:10];
a0 = 0. 4323;
a = 3. 4587. / ( 4+ 4 * pi * pi * k . * k ) ;
b = 3.4587 * pi * k . / ( 4 + 4 * pi * pi * k . * k ) ;
% Approximation with M=3 harmonics
xhat = a0 * ones ( s i z e ( t ) ) ;
for m= 1: 3 ,
xhat = xhat+a (m) * cos ( 2 * pi *m* t )+b(m) * sin ( 2 * pi *m* t ) ;
end ;
plot ( t , x , t , real ( xhat ) ) ;
axis ( [ − 1 , 2 , − 0 . 2 , 1 . 8 ] ) ;
xlabel ( ’ t ’ ) ;
legend ( ’ O r i g i n a l s i g n a l ’ , ’ Approx . with M=3 ’ ) ;
grid ;
% Approximation e r r o r f o r M=3
plot ( t , x−real ( xhat ) ) ;
axis ( [ − 1 , 2 , − 0 . 5 , 0 . 5 ] ) ;
xlabel ( ’ t ’ ) ;
grid ;
To use M = 4 harmonics, modify line 12 as
12
for m= 1: 4 ,
To use M = 5 harmonics, modify it as
12
for m= 1: 5 ,
4.43. The code below computes and graphs the EFS line spectrum for duty cycle d = 0.4.
1
2
3
% Compute and graph the l i n e spectrum f o r d=0.4
k = [ − 20: 20] ;
d = 0.4;
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
146
4
5
6
7
8
9
c = d * sinc ( k * d ) ;
stem ( k , c ) ;
axis ( [ − 2 0 . 5 , 2 0 . 5 , − 0 . 3 , 1 . 2 ] ) ;
t i t l e ( ’ c_k ’ ) ;
xlabel ( ’ k ’ ) ;
grid ;
To repeat with other values of the duty cycle, modify the line
3
d = 0.4;
As the duty cycle is increased, the line spectrum becomes more concentrated around k = 0.
4.44. The code below computes and graphs the original signal and its approximation using M = 2
harmonics.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
% Compute the half −wave r e c t i f i e d sinusoid
t = [ −0.25:0.005:2.25];
xp = @( t ) sin ( 2 * pi * t ) . * ( t < = 0 . 5 ) ; % One period
x = xp (mod( t , 1 ) ) ;
% Compute the CFS c o e f f i c i e n t s f o r k = 0 , . . . , 1 0
k = [1:10];
dk = zeros ( s i z e ( k ) ) ;
d0 = 1/ pi ; % d0 = c0
dk ( 1 ) = 0 . 5 ;
for m= 2: 2: 10 ,
dk (m) = 2/( pi * (m*m− 1) ) ;
end ;
theta = zeros ( s i z e ( k ) ) ;
theta ( 1 ) = −pi / 2 ;
for m = 2 : 2 : 1 0 ,
theta (m) = pi ;
end ;
% Compute and graph the approximation using M=2 harmonics
omg0 = 2 * pi ;
xhat = d0 * ones ( s i z e ( t ) ) ;
for m= 1: 2 ,
xhat = xhat+dk (m) * cos (m* omg0 * t +theta (m) ) ;
end ;
plot ( t , x , t , xhat , ’ r ’ ) ;
axis ( [ − 0 . 2 5 , 2 . 2 5 , − 0 . 2 , 1 . 8 ] ) ;
xlabel ( ’ t ’ ) ;
legend ( ’ O r i g i n a l s i g n a l ’ , ’ Approx . fo r M=2 ’ ) ;
grid ;
To use M = 4 harmonics, modify line 21 as
21
for m= 1: 4 ,
To use M = 6 harmonics, modify it as
147
21
for m= 1: 6 ,
4.45. The code listed below defines an anonymous function and uses it to compute the transform of
with τ = 3 seconds and f 0 = 2 Hz.
1
2
3
4
5
6
7
8
9
10
11
12
µ ¶
¡
¢
t
cos 2π f 0 t Π
τ
% Anonymous function to compute the transform o f a pulse
P = @( f , tau ) tau * sinc ( f * tau ) ;
% Compute and graph the transform o f modulated pulse
tau = 3 ;
f0 = 2 ;
f = [ −4:0.005:4];
X = 0 . 5 * P( f +f0 , tau ) + 0 . 5 * P ( f−f0 , tau ) ;
plot ( f , X ) ;
axis ( [ − 4 , 4 , − 0. 6 , 2] ) ;
t i t l e ( ’X( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
grid ;
Parts (a) through (d) of Problem 4.27 can be verified using following code listings:
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
1
2
3
4
5
6
7
% V e r i f y part ( a ) o f Problem 4.27
f = [ −12:0.02:12];
X1 = 0 . 5 * sinc ( f +5)+0.5 * sinc ( f −5);
plot ( f , X1 ) ;
grid ;
axis ( [ − 12 , 12 , − 0. 3 , 1. 2] ) ;
t i t l e ( ’X( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
% V e r i f y part ( b ) o f Problem 4.27
X2 = sinc ( 2 * f +10)+ sinc ( 2 * f −10);
plot ( f , X2 ) ;
grid ;
axis ( [ − 12 , 12 , − 0. 3 , 1. 2] ) ;
t i t l e ( ’X( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
% V e r i f y part ( c ) o f Problem 4.27
X3 = 0. 25 * sinc ( ( f +5)/2)+0.25 * sinc ( ( f − 5) / 2) ;
plot ( f , X3 ) ;
grid ;
axis ( [ − 12 , 12 , − 0. 3 , 1. 2] ) ;
t i t l e ( ’X( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
CHAPTER 4. FOURIER ANALYSIS FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
148
1
2
3
4
5
6
7
% V e r i f y part (d ) o f Problem 4.27
X4 = 0.125 * sinc ( ( f +5)/4)+0.125 * sinc ( ( f − 5) / 4) ;
plot ( f , X4 ) ;
grid ;
axis ( [ − 12 , 12 , − 0. 3 , 1. 2] ) ;
t i t l e ( ’X( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
4.46. The script listed below computes the EFS coefficients, and then computes and graphs the signal
with only the dc component and the fundamental frequency retained. Recall from Problem 4.35 that,
for d 0 = 0.5, this corresponds to preserving 90 percent of the spectral power in the signal.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
% Compute the EFS c o e f f i c i e n t s
d = 0.5;
k = [ − 25: 25] ;
c = d * sinc ( k * d ) ;
o f f s e t = −k ( 1 ) + 1 ; % MATLAB index o f c e n t e r c o e f f i c i e n t " c0 "
% S e t f 0 = 1 Hz and c r e a t e a time v e c t o r .
f0 = 1 ;
t = [ −2:0.005:2];
% Compute the s i g n a l f o r M=1
m = 1;
xhat = zeros ( s i z e ( t ) ) ;
for i =−m:m,
xhat = xhat+c ( i + o f f s e t ) * exp ( j * 2 * pi * i * f0 * t ) ;
end ;
plot ( t , real ( xhat ) ) ;
axis ( [ − 2 , 2 , − 0 . 2 , 1 . 2 ] ) ;
grid ;
It was found in Problem 4.35 that, in order to preserve 95 percent of the spectral power, frequencies up
to and including the third harmonic are needed. The resulting signal can be computed and graphed by
modifying line 10 of the script as
10
m = 3;
Preserving 99 percent of the spectral power requires frequencies up to and including the 21-st
harmonic. The resulting signal can be computed and graphed by modifying line 10 of the script as
1
m = 21;
Chapter 5
Fourier Analysis for Discrete-Time Signals
and Systems
5.1.
Angular frequency:
Ω = 0.3π
Ω
= 0.15
Normalized frequency: F =
2π
3
Period: N =
= 20 samples
0.15
Normalized fundamental frequency:
x̃ (n) =
Therefore
c3 =
1
1
=
N 20
F0 =
1
,
2
,
Ω0 =
2π 2π
=
N
20
1 j (2π/20)3n 1 − j (2π/20)3n
e
+ e
2
2
c −3 = c 17 =
1
,
2
c k = 0 for all other k
x̃[n]
1.5
Amplitude
1
0.5
0
0
−0.5
19
−1
−1.5
5
0
10
15
20
Index n
c̃k
0.8
Amplitude
0.6
0.4
0.2
0
−0.2
0
2
4
6
10
8
Index k
149
12
14
16
18
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
150
5.2.
Angular frequencies of the two sinusoidal terms:
Normalized frequencies:
F 1 = 12 ,
F 2 = 0.28
Normalized fundamental frequency:
1
Period: N =
= 25 samples
0.04
x̃[n] = 1 +
Therefore
c3 =
1
,
2
Ω1 = 0.24π ,
Ω2 = 0.56π
F 0 = 0.04
1 j (2π/25)3n 1 − j (2π/25)3n 3 j (2π/25) 7n 3 − j (2π/25)7n
e
+ e
+
e
−
e
2
2
2j
2j
c −3 = c 22 =
1
,
2
c7 =
3
3
= −j ,
2j
2
c −7 = c 18 = −
3
3
=j
2j
2
|c̃k |
Amplitude
1.5
1
0.5
0
0
5
10
15
20
15
20
Index k
∠ c̃k
π
Radians
π
2
0
−
π
2
−π
0
5
10
Index k
5.3.
a.
c̃ k = 4 + 3 e − j (2π/8)k + 2 e − j (2π/8)2k + 1 e − j (2π/8)3k +
1 e − j (2π/8)5k + 2 e − j (2π/8)6k + 3 e − j (2π/8)7k ,
for k = 0, . . . , 7
151
Evaluating for k = 0, . . . , 7 yields
c̃ 0 = 2
c̃ 1 = 0.8536
c̃ 2 = 0
c̃ 3 = 0.1464
c̃ 4 = 0
c̃ 5 = 0.1464
c̃ 6 = 0
c̃ 7 = 0.8536
b.
c̃ k = 1 + e − j (2π/8)k + e − j (2π/8)2k
for k = 0, . . . , 7
Evaluating for k = 0, . . . , 7 yields
c̃ 0 = 0.3750
c̃ 1 = 0.2134 − j 0.2134
c̃ 2 = − j 0.1250
c̃ 3 = 0.0366 + j 0.0366
c̃ 4 = 0.1250
c̃ 5 = 0.0366 − j 0.0366
c̃ 6 = j 0.1250
c̃ 7 = 0.2134 + j 0.2134
c.
c̃ k = 1 + e − j (2π/8)k + e − j (2π/8)2k + e − j (2π/8)3k + e − j (2π/8)4k
Evaluating for k = 0, . . . , 7 yields
c̃ 0 = 0.6250
c̃ 1 = − j 0.3018
c̃ 2 = 0.1250
c̃ 3 = − j 0.0518
c̃ 4 = 0.1250
c̃ 5 = j 0.0518
c̃ 6 = 0.1250
c̃ 7 = j 0.3018
5.4.
a. Let
x̃[n] =
4
X
k=0
c̃ k e − j (2π/5) kn
for k = 0, . . . , 7
152
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
and
g̃ [n] = x̃[n − 1] =
Coefficients d˜k are found as
4
X
d˜k e − j (2π/5) kn
k=0
d˜k = 4 + e − j (2π/5)2k + 2 e − j (2π/5)3k + 3 e − j (2π/5)4k
= 4 + e − j 4πk/5 + 2 e − j 6πk/5 + 3 e − j 8πk/5 ,
for k = 0, . . . , 4
Evaluating for k = 0, . . . , 4 we get
d˜0 = 2
d˜1 = 0.5 + j 0.6882
d˜2 = 0.5 + j 0.1625
d˜3 = 0.5 − j 0.1625
d˜4 = 0.5 − j 0.6882
b.
Using the time shifting property of the DTFS, we have
d˜k = e − j 2πk/5 c̃ k
Table shown below illustrates the relationship between the coefficients c̃ k and d˜k :
k
0
1
2
3
4
c̃ k
2.0000+ j 0.0000
−0.5000+ j 0.6882
−0.5000+ j 0.1625
−0.5000− j 0.1625
−0.5000− j 0.6882
e − j 2πk/5
1.0000+ j 0.0000
0.3090− j 0.9511
−0.8090− j 0.5878
−0.8090+ j 0.5878
0.3090+ j 0.9511
d˜k
2.0000+ j 0.0000
0.5000+ j 0.6882
0.5000+ j 0.1625
0.5000− j 0.1625
0.5000− j 0.6882
5.5. For each set the following relationships hold true:
∗
= c̃ 7∗
c̃ 1 =c̃ −1
∗
= c̃ 6∗
c̃ 2 =c̃ −2
∗
= c̃ 5∗
c̃ 3 =c̃ −3
Furthermore, both c̃ 0 and c̃ 4 are real for each set of coefficients so that
c̃ 0 = c̃ 0∗
and
∗
= c̃ 4∗
c̃ 4 = c̃ −4
153
5.6.
a.
g̃ [n] + g̃ [−n]
2
g̃ [n] − g̃ [−n]
g̃ o [n] =
2
g̃ e [n] =
Tabular forms of the signals involved:
g̃ [n] = {. . . , 4, 0, 1, 2, 3, . . . }
↑
n=0
g̃ [−n] = {. . . , 4, 3, 2, 1, 0, . . . }
↑
n=0
g̃ e [n] = {. . . , 4, 1.5, 1.5, 1.5, 1.5, . . . }
↑
n=0
g̃ o [n] = {. . . , 0, − 1.5, −0.5, 0.5, 1.5, . . . }
↑
n=0
g̃[n]
g̃[−n]
4
4
n
0
(e)
(f )
g̃e [n]
g̃o [n]
4
4
0
n
0
(g)
b.
n
0
(h)
Let
g̃ e [n] =
and
g̃ o [n] =
Using Eqn. (5.22) we obtain
4
X
c̃ k e − j 2πk/5
k=0
4
X
d˜k e − j 2πk/5
k=0
c̃ 0 = 2
c̃ 1 = 0.5
c̃ 2 = 0.5
c̃ 3 = 0.5
c̃ 4 = 0.5
n
154
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
and
d˜0 = 0
d˜1 = j 0.6882
d˜2 = j 0.1625
d˜3 = − j 0.1625
d˜4 = − j 0.6882
5.7. Using the definition in Eqn. (5.47)
ỹ[n] = x̃[n] ⊗ h̃[n] =
NX
−1
k=0
x̃[k] h̃[n − k] ,
all n
and evaluating y[n + N ] we get
ỹ[n + N ] =
NX
−1
k=0
x̃[k] h̃[n + N − k] ,
all n
Since h[n] is periodic with period N we have h[n + N − k] = h[n − k], and
ỹ[n + N ] = ỹ[n]
5.8. Tabular form of the periodic convolution result is
ỹ[n] = {. . . , 5, 7, 9, 6, 3, . . . }
↑
n=0
If the DTFS cofficients for x̃[n] and h̃[n] are c̃ k and d˜k respectively, the DTFS coefficients of y[n] are
DTFS
ỹ[n] ←→ 5 c̃ k d˜k
The table below lists DTFS coefficients of the three signals:
k
0
1
2
3
4
Coeffs for x[n]
2.0000+ j 0.0000
0.5000− j 0.6882
0.5000− j 0.1625
0.5000+ j 0.1625
0.5000+ j 0.6882
Coeffs for h[n]
0.6000+ j 0.0000
0.1000− j 0.3078
0.1000+ j 0.0727
0.1000− j 0.0727
0.1000+ j 0.3078
Coeffs for y[n]
6.0000+ j 0.0000
−0.8090− j 1.1135
0.3090+ j 0.1004
0.3090− j 0.1004
−0.8090+ j 1.1135
5.9. Tabular form of the periodic convolution result is
ỹ[n] = {. . . , 1, 2, 3, 3, 3, 2, 1, 0, . . . }
↑
n=0
155
If the DTFS cofficients for x̃[n] and h̃[n] are c̃ k and d˜k respectively, the DTFS coefficients of y[n] are
DTFS
ỹ[n] ←→ 8 c̃ k d˜k
The table below lists DTFS coefficients of the three signals:
k
0
1
2
3
4
5
6
7
Coeffs for x[n]
0.3750+ j 0.0000
0.2134− j 0.2134
0.0000− j 0.1250
0.0366+ j 0.0366
0.1250+ j 0.0000
0.0366− j 0.0366
0.0000+ j 0.1250
0.2134+ j 0.2134
Coeffs for h[n]
0.6250+ j 0.0000
0.0000− j 0.3018
0.1250+ j 0.0000
0.0000− j 0.0518
0.1250+ j 0.0000
0.0000+ j 0.0518
0.1250+ j 0.0000
0.0000+ j 0.3018
Coeffs for y[n]
1.8750+ j 0.0000
−0.5152− j 0.5152
0.0000− j 0.1250
0.0152− j 0.0152
0.1250+ j 0.0000
0.0152+ j 0.0152
0.0000+ j 0.1250
−0.5152+ j 0.5152
5.10.
a. The transform is
X (Ω) = 1 + e − j Ω = 1 + cos (Ω) − j sin (Ω)
with magnitude
|X (Ω)| =
and phase
q
[1 + cos (Ω)]2 + sin2 (Ω) =
∡ X (Ω) = − tan
−1
·
q
2 + 2 cos (Ω)
sin (Ω)
1 + cos (Ω)
¸
|X (Ω)|
∡ X (Ω)
π
Phase (rad)
Magnitude
2
1.5
1
0.5
0
−π
−π/2
0
π/2
π
2
0
−
π
2
−π
−π
π
−π/2
Ω (rad)
b.
The transform is
X (Ω) = 1 − e − j Ω = 1 − cos (Ω) + j sin (Ω)
with magnitude
|X (Ω)| =
and phase
0
Ω (rad)
q
2
2
[1 − cos (Ω)] + sin (Ω) =
∡ X (Ω) = tan−1
·
q
sin (Ω)
1 − cos (Ω)
2 − 2 cos (Ω)
¸
π/2
π
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
156
|X (Ω)|
∡ X (Ω)
π
Phase (rad)
Magnitude
2
1.5
1
0.5
0
−π
0
−π/2
π/2
π
2
0
π
2
−
−π
−π
π
0
−π/2
Ω (rad)
c.
π/2
π
π/2
π
π/2
π
Ω (rad)
The transform is
X (Ω) = 1 + e − j Ω + e − j 2 Ω + e − j 3Ω
h
i
= e j 3Ω/2 + e j Ω/2 + e − j Ω/2 + e − j 3Ω/2 e − j 3Ω/2
= [2 cos (Ω/2) + 2 cos (3Ω/2)] e − j 3Ω/2
Magnitude and phase are shown below:
∡ X (Ω)
π
4
π
2
Phase (rad)
Magnitude
|X (Ω)|
5
3
2
1
0
−π
0
−π/2
π/2
0
−
π
2
−π
−π
π
−π/2
Ω (rad)
d.
0
Ω (rad)
The transform is
X (Ω) = e j 2Ω + e j Ω + 1 + e − j Ω + e − j 2 Ω
= 1 + 2 cos (Ω) + 2 cos (2Ω)
replacemen
Magnitude and phase are shown below:
|X (Ω)|
∡ X (Ω)
6
π
Phase (rad)
Magnitude
5
4
3
2
1
π
2
0
π
−
2
−π
0
−π
−π/2
0
Ω (rad)
π/2
π
−π
−π/2
0
Ω (rad)
157
e.
The transform is computed as
X (Ω) =
Using Euler’s formula:
∞
X
n=0
X (Ω) =
(0.7)n e − j Ωn =
1
1 − 0.7 e − j Ω
1
1 − 0.7 cos (Ω) + j 0.7 sin (Ω)
The magnitude of the transform is
¯
¯
1
1
¯ X (Ω)¯ = p
=p
1.49 − 1.4 cos (Ω)
[1 − 0.7 cos (Ω)]2 + [0.7 sin (Ω)]2
and the phase of the transform is
−1
∡ X (Ω) = − tan
µ
Magnitude and phase are shown below:
0.7 sin (Ω)
1 − 0.7 cos (Ω)
¶
∡ X (Ω)
π
3
π
2
Phase (rad)
Magnitude
|X (Ω)|
4
2
1
0
−π
0
−π/2
π/2
0
−
π
2
−π
−π
π
−π/2
f.
0
π/2
π
π/2
π
Ω (rad)
Ω (rad)
The transform is computed as
X (Ω) =
=
=
∞
X
(0.7)n cos (Ωn) e − j Ωn
n=0
∞
∞
1 X
1 X
(0.7)n e j 0.2πn e − j Ωn +
(0.7)n e − j 0.2πn e − j Ωn
2 n=0
2 n=0
1 − 0.7 cos (0.2π) e − j Ω
1 − 1.4 cos (0.2π) e − j Ω + 0.49 e − j 2Ω
Magnitude and phase are shown below:
∡ X (Ω)
|X (Ω)|
π
Phase (rad)
Magnitude
2
1.5
1
0.5
0
−π
−π/2
0
Ω (rad)
π/2
π
π
2
0
−
π
2
−π
−π
−π/2
0
Ω (rad)
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
158
5.11.
a.
Using the inverse DTFT relationship:
1
x[n] =
2π
1
=
2π
ˆ π
X (Ω) e j Ωn d Ω
−π
ˆ 0.2π
(1) e j Ωn d Ω
−0.2π
¯0.2π
e j Ωn ¯¯
sin (0.2πn)
=
= 0.2 sinc (0.2n)
=
j 2πn ¯−0.2π
πn
b.
x[n] =
1
2π
ˆ 0.4π
(1) e j Ωn d Ω
−0.4π
¯0.4π
e j Ωn ¯¯
sin (0.4πn)
=
= 0.4 sinc (0.4n)
=
¯
j 2πn −0.4π
πn
c.
1
x[n] =
2π
ˆ −0.2π
(1) e
−π
j Ωn
ˆ π
1
dΩ+
2π
(1) e j Ωn d Ω
0.2π
¯0.2π
¯π
e j Ωn ¯¯
e j Ωn ¯¯
sin (πn) sin (0.2πn)
=
−
+
=
¯
¯
j 2πn −π
j 2πn 0.2π
πn
πn
Using L’Hospital’s rule it can be shown that
sin (πn)
= δ[n]
πn
and therefore
x[n] = δ[n] −
sin (0.2πn)
= δ[n] − 0.2 sinc (0.2n)
πn
d.
x[n] =
1
2π
ˆ −0.1π
−0.2π
(1) e j Ωn d Ω +
1
2π
ˆ 0.2π
(1) e j Ωn d Ω
0.1π
¯−0.1π
¯0.2π
e j Ωn ¯¯
e j Ωn ¯¯
sin (0.2πn) sin (0.1πn)
−
=
+
=
¯
¯
j 2πn −0.2π j 2πn 0.1π
πn
πn
=0.2 sinc (0.2n) − 0.1 sinc (0.1n)
159
5.12.
a.
We know that
1
F
(0.5)n u[n] ←→
Using time shifting
1 − 0.5 e − j Ω
e − j 2Ω
F
(0.5)n−2 u[n − 2] ←→
Using linearity, and scaling both sides
1 − 0.5 e − j Ω
0.25 e − j 2Ω
F
(0.5)n u[n − 2] ←→
1 − 0.5 e − j Ω
b. Since
1
F
(0.8)n u[n] ←→
through the use of time shifting we obtain
1 − 0.8 e − j Ω
F
(0.8)n−10 u[n − 10] ←→
e − j 10Ω
1 − 0.8 e − j Ω
Linearity of the DTFT allows both sides to be scaled:
F
(0.8)n u[n − 10] ←→
Therefore
c.
(0.8)10 e − j 10Ω
1 − 0.8 e − j Ω
¡
¢ F 1 − (0.8)10 e − j 10Ω
(0.8)n u[n] − u[n − 10] ←→
1 − 0.8 e − j Ω
We know that
1
F
(0.8)n u[n] ←→
1 − 0.8 e − j Ω
Shifting the signal to the left by 5 samples and using the time shifting property of the DTFT we obtain
F
(0.8)n+5 u[n + 5] ←→
e j 5Ω
1 − 0.8 e − j Ω
Scaling both sides yields
F
(0.8)n u[n + 5] ←→
Shifting the signal to the right by 5 samples yields
F
(0.8)−5 e j 5Ω
1 − 0.8 e − j Ω
(0.8)n−5 u[n − 5] ←→
e − j 5Ω
1 − 0.8 e − j Ω
and with scaling we get
F
(0.8)n u[n − 5] ←→
Therefore
(0.8)5 e − j 5Ω
1 − 0.8 e − j Ω
¡
¢ F (0.8)−5 e j 5Ω − (0.8)5 e − j 5Ω
(0.8)n u[n + 5] − u[n − 5] ←→
1 − 0.8 e − j Ω
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
160
5.13.
a. We know that
µ ¶n
1
1
F
u[n] ←→
1 −jΩ
2
1− 2 e
Applying the time reversal property
µ ¶−n
1
1
F
u[−n] ←→
1
2
1− 2 e jΩ
or equivalently
1
F
(2)n u[−n] ←→
1 − 21 e j Ω
Using the time shifting property
e jΩ
F
(2)n+1 u[−n − 1] ←→
1 − 12 e j Ω
and scaling both sides we get
F
(2)n u[−n − 1] ←→
b.
1 jΩ
e
2
1 − 12 e j Ω
We know that
1
F
(0.8)n u[n] ←→
Applying the time reversal property
1 − 0.8 e − j Ω
1
F
(0.8)−n u[−n] ←→
1 − 0.8 e j Ω
or equivalently
1
F
(1.25)n u[−n] ←→
c.
1 − 0.8 e j Ω
The signal x[n] can be written as
x[n] = (0.8)−n u[−n − 1] + (0.8)n u[n]
We know that
1
F
(0.8)n u[n] ←→
Using time reversal property
1 − 0.8 e − j Ω
1
F
(0.8)−n u[−n] ←→
and using time shifting
1 − 0.8 e j Ω
F
(0.8)−n−1 u[−n − 1] ←→
Scaling both sides of the last relationship yields
F
(0.8)−n u[−n − 1] ←→
e jΩ
1 − 0.8 e j Ω
0.8 e j Ω
1 − 0.8 e j Ω
161
The desired transform is
X (Ω) =
d.
0.8 e j Ω
1 − 0.8 e j Ω
+
1
=
1 − 0.8 e − j Ω
0.36
1.64 − 1.6 cos (Ω)
The signal x[n] can be written as
x[n] =
Let
(
(0.8)−n ,
n = −5, . . . , −1
n
(0.8) ,
n = 0, . . . , 4
¡
¢
x 1 [n] = (0.8)n u[n] − u[n − 5]
and
¡
¢
x 2 [n] = (0.8)−n u[−n − 1] − u[−n − 6]
= 0.8 x 1 [−n − 1]
so that
x[n] = x 1 [n] + x 2 [n]
The transform of x 1 [n] is
X 1 (Ω) =
1 − (0.8)5 e − j 5Ω
=
1 − 0.3277 e − j 5Ω
1 − 0.8 e − j Ω
1 − 0.8 e − j Ω
Using the time shifting and time reversal properties we obtain
X 2 (Ω) = e
jΩ
X 1 (−Ω) =
and
X (Ω) = X 1 (Ω) + X 2 (Ω) =
=
¡
¢
e j Ω 1 − 0.3277 e j 5Ω
1 − 0.8 e j Ω
1 − 0.3277 e − j 5Ω
1 − 0.8 e − j Ω
+
¡
¢
e j Ω 1 − 0.3277 e j 5Ω
1 − 0.8 e j Ω
5.14.
X (Ω) =
X (Ω) =
·
∞
X
x[n] e − j Ωn
n=−∞
Conjugating both sides yields
∗
∞
X
x[n] e
n=−∞
− j Ωn
¸∗
=
∞
X
x ∗ [n] e j Ωn
n=−∞
Since x[n] is real-valued we have x ∗ [n] = x[n], and therefore
X ∗ (Ω) =
∞
X
n=−∞
x[n] e j Ωn = X (−Ω)
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
162
5.15.
a.
X (Ω) = 1 +
1 jΩ 1 −jΩ
e + e
= 1 + cos (Ω)
2
2
Magnitude of the transform is shown below:
|X (Ω)|
2.5
Magnitude
2
1.5
1
0.5
0
−π
−π/2
0
π/2
π
Ω (rad)
b.
h
i
h
i
h
i h
i
X (Ω) =5 + 4 e j Ω + e − j Ω + 3 e j 2Ω + e − j 2Ω + 2 e j 3Ω + e − j 3Ω + e j 4Ω + e − j 4Ω
=5 + 8 cos (Ω) + 6 cos (2Ω) + 4 cos (3Ω) + 2 cos (4Ω)
replacemen
Magnitude of the transform is shown below:
|X (Ω)|
30
Magnitude
25
20
15
10
5
0
−π
−π/2
0
π/2
π
Ω (rad)
c.
X (Ω) = cos (−0.8π) e j 4Ω + cos (−0.6π) e j 3Ω + cos (−0.4π) e j 2Ω + cos (−0.2π) e j Ω
+ 1 + cos (0.2π) e − j Ω + cos (0.4π) e − j 2Ω + cos (0.6π) e − j 3Ω + cos (0.8π) e − j 4Ω
=1 + 2 cos (0.2π) cos (Ω) + 2 cos (0.4π) cos (2Ω) + 2 cos (0.6π) cos (3Ω) + 2 cos (0.8π) cos (4Ω)
Magnitude of the transform is shown below:
163
|X (Ω)|
5
Magnitude
4
3
2
1
0
−π
0
−π/2
π
π/2
Ω (rad)
5.16.
a.
i 1h
i
1 h jΩ
e − e−jΩ +
e j 2Ω − e − j 2Ω
2
4
1
= j sin (Ω) + j sin (2Ω)
2
X (Ω) =
Magnitude and phase of the transform are shown below:
∡ X (Ω)
π
1
π
2
Phase (rad)
Magnitude
|X (Ω)|
1.5
0.5
0
−π
−π/2
0
π/2
π
0
−
π
2
−π
−π
−π/2
Ω (rad)
0
π/2
π
Ω (rad)
b.
h
i
h
i
h
i
h
i
h
i
X (Ω) = e j Ω − e − j Ω + 2 e j 2Ω − e − j 2Ω + 3 e j 3Ω − e − j 3Ω + 4 e j 4Ω − e − j 4Ω + 5 e j 5Ω − e − j 5Ω
= j 2 sin (Ω) + j 4 sin (2Ω) + j 6 sin (3Ω) + j 8 sin (4Ω) + j 10 sin (5Ω)
Magnitude and phase of the transform are shown below:
replacemen
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
164
∡ X (Ω)
π
25
π
2
Phase (rad)
Magnitude
|X (Ω)|
30
20
15
10
0
−
5
0
−π
−π/2
0
−π
−π
π
π/2
π
2
−π/2
Ω (rad)
0
π/2
π
Ω (rad)
c.
i
i
i
i
h
h
h
h
X (Ω) = sin (0.2π) e j Ω − e − j Ω + sin (0.4π) e j 2Ω − e − j 2Ω + sin (0.6π) e j 3Ω − e − j 3Ω + sin (0.8π) e j 4Ω − e − j 4Ω
replacemen
= j 2 sin (0.2π) sin (Ω) + j 2 sin (0.4π) sin (2Ω) + j 2 sin (0.6π) sin (3Ω) + j 2 sin (0.8π) sin (4Ω)
Magnitude and phase of the transform are shown below:
∡ X (Ω)
π
5
π
2
Phase (rad)
Magnitude
|X (Ω)|
6
4
3
2
0
−
1
0
−π
−π/2
0
−π
−π
π
π/2
π
2
−π/2
Ω (rad)
5.17.
a. Starting with
1
F
(0.7)n u[n] ←→
and using the modulation property we obtain
X (Ω) =
0
Ω (rad)
1 − 0.7 e − j Ω
1/2
1 − 0.7 e − j (Ω−0.2π)
Magnitude and phase characteristics are shown below.
+
1/2
1 − 0.7 e − j (Ω+0.2π)
π/2
π
165
∡ X (Ω)
π
2
π
2
Phase (rad)
Magnitude
|X (Ω)|
2.5
1.5
1
0.5
0
−π
−π/2
0
−
π
2
−π
−π
π
π/2
0
−π/2
Ω (rad)
b.
0
π/2
π
π/2
π
Ω (rad)
Starting with
1
F
(0.7)n u[n] ←→
and using the modulation property we obtain
X (Ω) =
1 − 0.7 e − j Ω
1/2 e − j π/2
1 − 0.7 e
+
− j (Ω−0.2π)
1/2 e j π/2
1 − 0.7 e − j (Ω+0.2π)
Magnitude and phase characteristics are shown below.
∡ X (Ω)
π
1
π
2
Phase (rad)
Magnitude
|X (Ω)|
1.5
0.5
0
−π
−π/2
0
π/2
0
−
π
2
−π
−π
π
−π/2
Ω (rad)
c.
0
Ω (rad)
Starting with
F
u[n] − u[n − 10] ←→
and applying the modulation property yields
1 − e − j 10Ω
1 − e−jΩ
µ ¶
µ ¶
1 1 − e − j 10(Ω−π/5)
1 1 − e − j 10(Ω+π/5)
X (Ω) =
+
2 1 − e − j (Ω−π/5)
2 1 − e − j (Ω+π/5)
Magnitude and phase characteristics are shown below.
replacemen
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
166
∡ X (Ω)
π
5
π
2
Phase (rad)
Magnitude
|X (Ω)|
6
4
3
2
0
−
1
0
−π
−π/2
0
−π
−π
π
π/2
π
2
−π/2
Ω (rad)
5.18.
a. Starting with
F
(0.7)n u[n] ←→
the use of the differentiation property yields
1
1 − 0.7 e − j Ω
·
¸
d
1
n (0.7) u[n] ←→ j
d Ω 1 − 0.7 e − j Ω
n
F
¸
·
i−1
d h
1
d
−jΩ
=
1
−
0.7
e
d Ω 1 − 0.7 e − j Ω
dΩ
i
ih
h
= − 2 1 − 0.7 e − j Ω j 0.7 e − j Ω
− j 1.4 e − j Ω
=¡
¢2
1 − 0.7 e − j Ω
The transform of x[n] is
b.
0
Ω (rad)
X (Ω) = ¡
1.4 e − j Ω
1 − 0.7 e − j Ω
¢2
The signal x[n] can be written as
x[n] = n 2 (0.7)n u[n] + n (0.7)n u[n]
From part (a) we have
Using the differentiation property
1.4 e − j Ω
F
n (0.7)n u[n] ←→ ¡
¢2
1 − 0.7 e − j Ω
"
#
−jΩ
1.4
e
d
n 2 (0.7)n u[n] ←→ j
¡
¢
d Ω 1 − 0.7 e − j Ω 2
"
#
¢
¡
− j 1.4 e − j Ω − 0.98 e − j 2Ω
1.4 e − j Ω
d
¡
¢ =
¡
¢3
d Ω 1 − 0.7 e − j Ω 2
1 − 0.7 e − j Ω
F
π/2
π
167
and
1.4 e − j Ω − 0.98 e − j 2Ω
¡
¢3
1 − 0.7 e − j Ω
F
n 2 (0.7)n u[n] ←→
Combining the results
c.
1.4 e − j Ω
1.4 e − j Ω − 0.98 e − j 2Ω
X (Ω) = ¡
¢2 +
¡
¢3
1 − 0.7 e − j Ω
1 − 0.7 e − j Ω
Let the signal x 1 [n] be defined as
¡
¢
x 1 [n] = (0.7)n u[n] − u[n − 10]
The transform X 1 (Ω) is
X 1 (Ω) =
9
X
n=0
(0.7)n e − j Ωn =
´
³
10
1 − 0.7 e − j Ω
1 − 0.7 e − j Ω
1 − 0.0282 e − j 10Ω
1 − 0.7 e − j Ω
d X 1 (Ω) j 0.282 e − j 10Ω − j 0.7e − j Ω
=
¡
¢2
dΩ
1 − 0.7 e − j Ω
X (Ω) = j
5.19.
a. The DTFT of x[n] are h[n] are
d X 1 (Ω) 0.7e − j Ω − 0.282 e − j 10Ω
=
¡
¢2
dΩ
1 − 0.7 e − j Ω
X (Ω) =
1 − e − j 10Ω
1 − e−jΩ
and
H (Ω) =
b.
y[n] =
Since
=
∞
X
k=−∞
1
1 − 0.8 e − j Ω
x[k] h[n − k] =
u[n − k] =
½
9
X
k=0
(0.8)n−k u[n − k]
1, n≥k
0, n<k
the convolution sum must be evaluated for three distinct possibilities:
n < 0:
y[n] = 0
0 ≤ n ≤ 9:
y[n] =
n
X
(0.8)n−k
k=0
= (0.8)n
= (0.8)n
n
X
(0.8)−k
k=0
·
1 − (0.8)−(n+1)
1 − (0.8)−1
=5 − 4 (0.8)n
¸
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
168
n > 9:
y[n] =
9
X
(0.8)n−k
k=0
= (0.8)n
= (0.8)
n
9
X
(0.8)−k
k=0
µ
1 − (0.8)−(10)
¶
1 − (0.8)−1
¢
=4 (1.25)10 − 1 (0.8)n
¡
Combining the results obtained, the signal y[n] is
¡
¢¡
¢
¡
¢
y[n] = 5 − 4 (0.8)n u[n] − u[n − 10] + 4 (1.25)10 − 1 (0.8)n u[n − 10]
¡
¢
=5 u[n] − u[n − 10] − 4 (0.8)n u[n] + 4 (1.25)10 (0.8)n u[n − 10]
c.
The DTFT of y[n] is
Y (Ω) =5
µ
1 − e − j 10Ω
¶
−4
µ
=5
µ
1 − e − j 10Ω
¶
−4
µ
=¡
1 − e−jΩ
1 − e−jΩ
1
1 − 0.8 e − j Ω
1 − e−jΩ
1 − 0.8 e − j Ω
¶
+ 4 (1.25)
10
µ
¶
(0.8)10 e − j 10Ω
1 − 0.8 e − j Ω
1 − e − j 10Ω
¢¡
¢
1 − e − j Ω 1 − 0.8 e − j Ω
=X (Ω) H (Ω)
5.20.
a.
X (Ω) = 2π
∞
X
m=−∞
δ (Ω − 0.3π − 2πm)
X (Ω)
2π
2π
2π
2π
2π
...
−3.7π
−1.7π
0.3π
2.3π
4.3π
...
b.
X (Ω) = 2π
Ω
∞ ¡
X
¢
δ (Ω − 0.2π − 2πm) + 3 δ (Ω − 0.4π − 2πm)
m=−∞
¶
169
X (Ω)
6π
6π
2π
6π
2π
2π
6π
2π
6π
2π
Ω
0.4π
c.
X (Ω) = π
∞ ¡
X
m=−∞
δ (Ω − π/5 − 2πm) + δ (Ω + π/5 − 2πm)
¢
X (Ω)
π
...
−4π
−2π
−
π
π
5
π
5
...
Ω
2π
4π
d.
X (Ω) = π
∞ ¡
X
2 δ (Ω − π/5 − 2πm) + 2 δ (Ω + π/5 − 2πm)
m=−∞
+ 3 δ (Ω − 2π/5 − 2πm) + 3 δ (Ω + 2π/5 − 2πm)
¢
X (Ω)
...
...
Ω
−4π
−2π
2π
4π
5.21.
a. Using Eqn. (5.181) with the DTFS coefficients found in Problem 5.3 we obtain
S x (Ω) =
∞ £
X
m=−∞
25.1327 δ (Ω − 2πm) + 4.5776 δ (Ω − 2π/8 − 2πm) + 0.1348 δ (Ω − 6π/8 − 2πm)
+ 0.1348 δ (Ω − 10π/8 − 2πm) + 4.5776 δ (Ω − 14π/8 − 2πm)
¤
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
170
b.
Using Eqn. (5.181) with the DTFS coefficients found in Problem 5.3 we obtain
S x (Ω) =
∞ £
X
m=−∞
0.8836 δ (Ω − 2πm) + 0.5722, δ (Ω − 2π/8 − 2πm) + 0.0982 δ (Ω − 4π/8 − 2πm)
+ 0.0168 δ (Ω − 6π/8 − 2πm) + 0.0982 δ (Ω − π − 2πm) + 0.0168 δ (Ω − 10π/8 − 2πm)
¤
+ 0.0982 δ (Ω − 12π/8 − 2πm) + 0.5722 δ (Ω − 14π/8 − 2πm)
c.
Using Eqn. (5.181) with the DTFS coefficients found in Problem 5.3 we obtain
S x (Ω) =
∞ £
X
m=−∞
2.4544 δ (Ω − 2πm) + 0.5722, δ (Ω − 2π/8 − 2πm) + 0.0982 δ (Ω − 4π/8 − 2πm)
+ 0.0168 δ (Ω − 6π/8 − 2πm) + 0.0982 δ (Ω − π − 2πm) + 0.0168 δ (Ω − 10π/8 − 2πm)
¤
+ 0.0982 δ (Ω − 12π/8 − 2πm) + 0.5722 δ (Ω − 14π/8 − 2πm)
5.22.
a. The DTFS coefficients are
µ
πk
sin
(2L + 1)
N
µ
¶
c̃ k =
πk
N sin
N
¶
and the power spectral density is
S x (Ω) =
∞
X
k=−∞
2π |c̃ k |2 δ (Ω − kΩ0 )
µ
¶ ¯2
¯
πk
¯
¯
¯
¶
sin
(2L + 1) ¯ µ
∞
X
¯
¯
2π
N
¯
¯
µ
¶ ¯ δ Ω−k
=
2π ¯
N
¯ N sin πk
¯
k=−∞
¯
¯
N
b.
The average power in the signal x[n] is
Px =
39 ¯
¯
1 X
¯x[n]¯2 = 7 = 0.1750
40 n=0
4
The DTFS coefficients up to the third harmonic are
c̃ 0 =0.1750
c̃ 1 = c̃ −1 =0.1665
c̃ 2 = c̃ −2 =0.1424
c̃ 3 = c̃ −3 =0.0114
171
The normalized average power in the first three harmonics is
P x,3 harmonics = (0.1750) 2 + 2 (0.1665)2 + 2 (0.1424)2 + 2 (0.0114)2 = 0.1494
which is 0.1494/0.1750 = 0.8538, or 85.38 percent of the total normalized average power.
c.
The average power in the signal x[n] is
Px =
39 ¯
¯
1 X
¯x[n]¯2 = 13 = 0.3250
40 n=0
4
The DTFS coefficients up to the third harmonic are
c̃ 0 =0.3250
c̃ 1 = c̃ −1 =0.2717
c̃ 2 = c̃ −2 =0.1424
c̃ 3 = c̃ −3 =0.0084
The normalized average power in the first three harmonics is
P x,3 harmonics = (0.3250) 2 + 2 (0.2717)2 + 2 (0.1424)2 + 2 (0.0084)2 = 0.2939
which is 0.2939/0.3250 = 0.9044, or 90.44 percent of the total normalized average power.
5.23.
a. The value of the system function at Ω0 = 0.2π is
H (0.2π) = 1.1159 + j 0.2039 = 1.1344 e j 0.1807
Therefore the steady-state response of the system is
y[n] =1.1344 e j 0.1807 x[n]
=1.1344 e j (0.2πn+0.1807)
b.
Using H (0.2π), the steady-state response of the system is
y[n] = 1.1344 cos (0.2πn + 0.1807)
c.
The value of the system function at Ω0 = 0.3π is
H (0.3π) = 1.2399 + j 0.3310 = 1.2833 e j 0.2609
Therefore the steady-state response of the system is
y[n] = (2) (1.2833) sin (0.3πn + 0.2609)
=2.5666 sin (0.3πn + 0.2609)
172
d.
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
The value of the system function at Ω0 = 0.1π is
H (0.1π) = 1.0561 + j 0.0976 = 1.0606 e j 0.0922
Using this result with H (0.2π) and from previous parts, the steady-state response of the system is
y[n] = (3) (1.0606) cos (0.1πn + 0.0922) − (5) (1.1344) sin (0.2πn + 0.1807)
=3.1818 cos (0.1πn + 0.0922) − 5.6720 sin (0.2πn + 0.1807)
5.24. The system function of a length-4 moving average filter is
H (Ω) =
i
1h
1 + e − j Ω + e − j 2Ω + e − j 3Ω
4
a. The value of the system function at Ω0 = 0.2π is
H (0.2π) = 0.4523 − j 0.6225 = 0.7694 e − j 0.9425
Therefore the steady-state response of the system is
y[n] =0.7694 e − j 0.9425 x[n]
=0.7694 e j (0.2πn−0.9425)
b. Using H (0.2π), the steady-state response of the system is
y[n] = 0.7694 cos (0.2πn − 0.9425)
c.
The value of the system function at Ω0 = 0.3π is
H (0.3π) = 0.0819 − j 0.5173 = 0.5237 e − j 1.4137
Therefore the steady-state response of the system is
y[n] = (2) (0.5237) sin (0.3πn − 1.4137)
=1.0474 sin (0.3πn − 1.4137)
d. The value of the system function at Ω0 = 0.1π is
H (0.1π) = 0.8370 − j 0.4265 = 0.9393 e − j 0.4712
Using this result with H (0.2π) and from previous parts, the steady-state response of the system is
y[n] = (3) (0.9393) cos (0.1πn − 0.4712) − (5) (0.7694) sin (0.2πn − 0.9425)
=2.8179 cos (0.1πn − 0.4712) − 3.8470 sin (0.2πn − 0.9425)
173
5.25.
The system function is
Y (Ω)
1 + 2e−jΩ
=
X (Ω) 1 + e − j Ω + 0.89 e − j 2Ω
All three periodic signals of Problem 5.3 have period N = 8 and the corresponding fundamental angular
frequency Ω0 = 2π/8 radians. Therefore DTFS coefficients d˜k of the output signal ỹ[n] are related to the
DTFS c̃ k coefficients of the input signal x̃[n] by
µ
¶
2πk
˜
dk = H
c̃ k
8
H (Ω) =
and the output signal is computed as
ỹ[n] =
a.
k
0
1
2
3
4
5
6
7
c̃ k
2.0000+ j 0.0000
0.8536+ j 0.0000
0.0000+ j 0.0000
0.1464+ j 0.0000
0.0000+ j 0.0000
0.1464+ j 0.0000
0.0000+ j 0.0000
0.8536− j 0.0000
7
X
d˜k e ( j 2π/8)kn
k=0
H (2πk/8)
1.0381+ j 0.0000
1.1674+ j 0.2638
2.0848+ j 0.7707
−3.1867− j 2.8385
−1.1236+ j 0.0000
−3.1867+ j 2.8385
2.0848− j 0.7707
1.1674− j 0.2638
d˜k
2.0761+ j 0.0000
0.9965+ j 0.2251
−0.0000+ j 0.0000
−0.4667− j 0.4157
−0.0000+ j 0.0000
−0.4667+ j 0.4157
−0.0000+ j 0.0000
0.9965− j 0.2251
The output signal is (one period shown)
ỹ[n] = {. . . , 3.1357, 4.4148, 0.7944, 0.2764, 1.0166, −0.2625, 3.3578, 3.8759, . . . }
↑
n=0
b.
k
0
1
2
3
4
5
6
7
c̃ k
0.3750+ j 0.0000
0.2134− j 0.2134
0.0000− j 0.1250
0.0366+ j 0.0366
0.1250+ j 0.0000
0.0366− j 0.0366
0.0000+ j 0.1250
0.2134+ j 0.2134
H (2πk/8)
1.0381+ j 0.0000
1.1674+ j 0.2638
2.0848+ j 0.7707
−3.1867− j 2.8385
−1.1236+ j 0.0000
−3.1867+ j 2.8385
2.0848− j 0.7707
1.1674− j 0.2638
d˜k
0.3893+ j 0.0000
0.3054− j 0.1928
0.0963− j 0.2606
−0.0127− j 0.2206
−0.1404+ j 0.0000
−0.0127+ j 0.2206
0.0963+ j 0.2606
0.3054+ j 0.1928
The output signal is (one period shown)
ỹ[n] = {. . . , 1.0268, 2.0855, 0.0006, 0.1433, −0.1438, 0.0163, 0.1117, −0.1262, . . . }
↑
n=0
174
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
c.
k
0
1
2
3
4
5
6
7
c̃ k
0.6250+ j 0.0000
0.0000− j 0.3018
0.1250+ j 0.0000
0.0000− j 0.0518
0.1250+ j 0.0000
0.0000+ j 0.0518
0.1250+ j 0.0000
0.0000+ j 0.3018
H (2πk/8)
1.0381+ j 0.0000
1.1674+ j 0.2638
2.0848+ j 0.7707
−3.1867− j 2.8385
−1.1236+ j 0.0000
−3.1867+ j 2.8385
2.0848− j 0.7707
1.1674− j 0.2638
d˜k
0.6488+ j 0.0000
0.0796− j 0.3523
0.2606+ j 0.0963
−0.1470+ j 0.1650
−0.1404+ j 0.0000
−0.1470− j 0.1650
0.2606− j 0.0963
0.0796+ j 0.3523
The output signal is (one period shown)
ỹ[n] = {. . . , 0.8948, 1.1819, 1.0217, 0.9264, 1.1643, 0.0113, −1.0475, 1.0374, . . . }
↑
n=0
5.26.
a.
X (k) = 1 + e − j 2πk/3 + e − j 4πk/3 ,
k = 0, 1, 2
X (0) = 3 + j 0
X (1) = 0 + j 0
X (2) = 0 + j 0
b.
X (k) = 1 + e − j 2πk/5 + e − j 4πk/5 ,
k = 0, . . . , 4
X (0) = 3 + j 0
X (1) = 0.5 − j 1.5388
X (2) = 0.5 + j 0.3633
X (3) = 0.5 − j 0.3633
X (4) = 0.5 + j 1.5388
c.
X (k) = 1 + e − j 2πk/7 + e − j 4πk/7 ,
k = 0, . . . , 6
175
X (0) = 3 + j 0
X (1) = 1.401 − j 1.7568
X (2) = −0.1235 − j 0.5410
X (3) = 0.7225 + j 0.3479
X (4) = 0.7225 − j 0.3479
X (5) = −0.1235 + j 0.5410
X (6) = 1.401 + j 1.7568
5.27.
a.
X (k) = 1 + e − j 2πk/8 + e − j 14πk/8 ,
Since e − j 14πk/8 = e j 2πk/8 and e − j 2πk/8 + e j 2πk/8 = 2 cos (2πk/8)
¶
µ
2πk
, k = 0, . . . , 7
X (k) = 1 + 2 cos
8
k = 0, . . . , 7
= { 3, 2.4142, 1, −0.4142, −1, −0.4142, 1, 2.4142 }
↑
k=0
b.
X (k) = 1 + e − j 2πk/8 + e − j 4πk/8 + e − j 12πk/8 + e − j 14πk/8 ,
Since e − j 12πk/8 = e j 4πk/8 and e − j 14πk/8 = e j 2πk/8
µ
¶
µ
¶
2πk
4πk
X (k) = 1 + 2 cos
+ 2 cos
,
8
8
k = 0, . . . , 7
k = 0, . . . , 7
= { 5, 2.4142, −1, −0.4142, 1, −0.4142, −1, 2.4142 }
↑
k=0
c.
X (k) = 1 + e − j 2πk/8 + e − j 4πk/8 + +e − j 6πk/8 + e − j 10πk/8 + e − j 12πk/8 + e − j 14πk/8 ,
Since e − j 10πk/8 = e j 6πk/8,
e − j 12πk/8 = e j 4πk/8 and e − j 14πk/8 = e j 2πk/8
µ
¶
µ
¶
µ
¶
2πk
4πk
6πk
X (k) = 1 + 2 cos
+ 2 cos
+ 2 cos
,
8
8
8
k = 0, . . . , 7
k = 0, . . . , 7
= { 7, 1, −1, 1, −1, 1, −1, 1 }
↑
k=0
5.28.
X [k] =
NX
−1
n=0
x[n] e − j 2πnk/N ,
k = 0, . . . , N − 1
The transform sample X [k] represents element i = k + 1 of the vector X. The coefficient matrix is
¤
£
W = w i j N ×N ,
w i j = e 2π(i −1) ( j −1)/N
176
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
which yields

1

 1
W=
 ..
 .
1
1
...
...
e − j 2π/N
e − j 2π(N −1)/N
1
e − j 2π(N −1)/N
. . . e − j 2π(N −1)(N −1)/N






5.29.
a.
x[n − 2]mod 4 = { 2, 1, 4, 3 }
↑
n=0
b.
x[n − 4]mod 5 = { 1, 1, 0, 0, 1 }
↑
n=0
c.
x[−n]mod 8 = { 1, 1, −3, −2, 1, 3, 2, 4 }
↑
n=0
d.
x[−n + 2]mod 8 = { 2, 4, 1, 1, −3, −2, 1, 3 }
↑
n=0
5.30.
a.
x E [n] =
x[n] + x ∗ [−n]mod 4
= { 4, 2, 2, 2 }
2
↑
n=0
xO [n] =
∗
x[n] − x [−n]mod 4
= { 0, 1, 0, −1 }
2
↑
n=0
b.
x E [n] =
x[n] + x ∗ [−n]mod 5
= { 1, 0.5, 0.5, 0.5, 0.5 }
2
↑
n=0
xO [n] =
∗
x[n] − x [−n]mod 5
= { 0, 0.5, 0.5, −0.5, −0.5 }
2
↑
n=0
c.
x E [n] =
x[n] + x ∗ [−n]mod 8
= { 1, 2.5, −0.5, 0.5, 1, 0.5, −0.5, 2.5 }
2
↑
n=0
xO [n] =
∗
x[n] − x [−n]mod 8
= { 0, 1.5, 2.5, 2.5, 0, −2.5, −2.5, −1.5 }
2
↑
n=0
177
5.31.
a.
X [k] = 1 + e − j 2πk/5 + e − j 4πk/5
= { 3, (0.5 − j 1.5388), (0.5 + j 0.3633), (0.5 − j 0.3633), (0.5 + j 1.5388) }
↑
k=0
b.
R[k] = e − j 2π/5 X [k]
= e − j 2πk/5 + e − j 4πk/5 + e − j 6πk/5
= { 3, (−1.3090 − j 0.9511), (−0.1910 − j 0.5878), (−0.1910 + j 0.5878), (−1.3090 + j 0.9511) }
↑
k=0
c.
r [n] = { 0, 1, 1, 1, 0 }
↑
n=0
d.
S[k] = e − j 4π/5 X [k]
= e − j 4πk/5 + e − j 6πk/5 + e − j 8πk/5
= { 3, (−1.3090 + j 0.9511), (−0.1910 + j 0.5878), (−0.1910 − j 0.5878), (−1.3090 − j 0.9511) }
↑
k=0
s[n] = { 0, 0, 1, 1, 1 }
↑
n=0
e.
Using the time shifting property
r [n] = x[n − 1]mod 5
s[n] = x[n − 2]mod 5
5.32.
a.
X [k] = 5 + 4 e − j 2πk/5 + 3 e − j 4πk/5 + 2 e − j 6πk/5 + e − j 8πk/5
= { 15, (2.5 − j 3.4410), (2.5 − j 0.8123), (2.5 + j 0.8123), (2.5 + j 3.4410) }
↑
k=0
178
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
b.
R[k] = 5 + 4 e j 2πk/5 + 3 e j 4πk/5 + 2 e j 6πk/5 + e j 8πk/5
= 5 + 4 e − j 8πk/5 + 3 e − j 6πk/5 + 2 e − j 4πk/5 + e − j 2πk/5
= { 15, (2.5 + j 3.4410), (2.5 + j 0.8123), (2.5 − j 0.8123), (2.5 − j 3.4410) }
↑
k=0
c.
r [n] = { 5, 1, 2, 3, 4 }
↑
n=0
It follows that r [n] = x[−n]mod 5 which is consistent with the time reversal property of the DFT.
5.33.
a. Let the transform X [k] be written as
X [k] = X E [k] + X O [k]
where X E [k] and X O [k] are the conjugate symmetric and conjugate antisymmetric components of X [k].
Using Eqns. (5.285) and (5.286) we have
DFT
x r [n] ←→ X E [k]
DFT
j x i [n] ←→ X O [k]
Details of computing X E [k] and X O [k] are in the table below:
k
0
1
2
3
4
5
X [k]
2+ j 3
1+ j 5
−2+ j 4
−1− j 3
2+ j 0
3+ j 1
X ∗ [−k]mod 6
2− j 3
3− j 1
2+ j 0
−1+ j 3
−2− j 4
1− j 5
X E [k]
2+ j 0
2+ j 2
0+ j 2
−1+ j 0
0− j 2
2− j 2
X O [k]
0+ j 3
−1+ j 3
−2+ j 2
0− j 3
2+ j 2
1+ j 3
©
ª
F x r [n] = X E [k] = { 2 , (2 + j 2), j 2, −1, − j 2, (2 − j 2) }
↑
k=0
b.
©
ª
F x i [n] = − j X O [k] = { 3 , (3 + j 1), (2 + j 2), −3, (2 − j 2), (3 − j 1) }
↑
k=0
179
5.34.
a.
k=0
1
2
3
4
k=0
1
2
3
4
x[k]
2
−3
4
1
6
x[k]
2
−3
4
1
6
h[0 − k]mod 5
1
0
0
1
1
h[1 − k]mod 5
1
1
0
0
1
x[k] h[0 − k]mod 5
2
0
0
1
6
x[k] h[1 − k]mod 5
2
−3
0
0
6
y[0] = 9
y[1] = 5
k=0
1
2
3
4
k=0
1
2
3
4
x[k]
2
−3
4
1
6
x[k]
2
−3
4
1
6
h[2 − k]mod 5
1
1
1
0
0
h[3 − k]mod 5
0
1
1
1
0
x[k] h[2 − k]mod 5
2
−3
4
0
0
x[k] h[3 − k]mod 5
0
−3
4
1
0
y[2] = 3
y[3] = 2
k=0
1
2
3
4
x[k]
2
−3
4
1
6
h[4 − k]mod 5
0
0
1
1
1
x[k] h[4 − k]mod 5
0
0
4
1
6
y[4] = 11
b.
The two transforms are
X [k] = 2 − 3 e − j 2πk/5 + 4 e − j 4πk/5 + e − j 6πk/5 + 6 e − j 8πk/5
and
H [k] = 1 + e − j 2πk/5 + e − j 4πk/5
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
180
c.
The product of the two transforms is
Y [k] =X [k] H [k]
=2 − e − j 2πk/5 + 3 e − j 4πk/5 + 2 e − j 6πk/5 + 11 e − j 8πk/5 + 7 e − j 10πk/5 + 6 e − j 12πk/5
Realizing that e − j 10πk/5 = 1 and e − j 12πk/5 = e − j 2πk/5 the transform Y [k] is
Y [k] =2 − e − j 2πk/5 + 3 e − j 4πk/5 + 2 e − j 6πk/5 + 11 e − j 8πk/5 + 7 + 6 e − j 2πk/5
=9 + 5 e − j 2πk/5 + 3 e − j 4πk/5 + 2 e − j 6πk/5 + 11 e − j 8πk/5
which is the DFT of the signal
y[n] = { 9, 5, 3, 2, 11 }
↑
n=0
5.35.
The signal x[n] has N x = 5 samples, and the signal h[n] has Nh = 3 samples (its last two samples have
zero amplitudes). Therefore, the length of the linear convolution result is N y = N x + Nh − 1 = 7 samples.
In other words, the significant samples of the linear convolution result is in the index range n = 0, . . . , 6.
The 7-point circular convolution of x[n] and h[n] would match the linear convolution of the two signals.
y[n] = { 2, − 1, 3, 2, 11, 7, 6 }
↑
n=0
5.36.
a. The DTFT of the signal x[n] is
X (Ω) =
Therefore
¯
¯
S[k] = X (Ω)¯
11
X
e − j Ωn
n=0
Ω=2πk/10
=
Splitting the summation into two parts
S[k] =
9
X
n=0
e − j (2πk/10)n +
11
X
e − j (2πk/10)n
n=0
11
X
e − j (2πk/10)n
n=10
Changing the variable of the second summation through m = n − 10
S[k] =
9
X
n=0
e − j (2πk/10)n +
1
X
e − j (2πk/10)(m+10)
m=0
and recognizing that
e − j (2πk/10)(m+10) = e − j (2πk/10)m
we obtain
S[k] =
9
X
n=0
e − j (2πk/10)n +
1
X
m=0
e − j (2πk/10)m
181
which can be written as
S[k] = 2 + 2 e − j 2πk/10 +
which is the 10-point DFT of the signal
9
X
e − j 2πkn/10
n=2
s[n] = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1 }
↑
n=0
b.
The DTFT evaluated at N equally-spaced frequencies is
¯
∞
X
¯
S[k] = X (Ω)¯
=
x[n] e − j (2πk/N )n
Ω=2πk/N
n=−∞
The infinite summation can be broken down into length-N summations as
#
"
∞
r NX
+N −1
X
− j (2πk/N )n
x[n] e
S[k] =
r =−∞
n=r N
Applying the variable change m = n − r N to the inner summation yields
#
"
∞
NX
−1
X
S[k] =
x[m + r N ] e − j (2πk/N )(m+r N )
r =−∞
m=0
Changing the order of summations and recognizing that
e − j (2πk/N )(m+r N ) = e − j (2πk/N )m
we obtain
S[k] =
NX
−1 · X
∞
m=0 r =−∞
¸
x[m + r N ] e − j (2πk/N )m
This result is the N -point DFT of the signal
s[n] =
∞
X
r =−∞
x[n + r N ]
5.37.
a. The DTFS coefficients c̃k of the signal x̃[n] are computed using the following:
>>
>>
>>
b.
The DTFS coefficients d˜k of the signal ỹ[n] are computed using the following:
>>
>>
c.
k = [0:7];
xn = [ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 ] ;
ck = s s _ d t f s ( xn , k )
gn = [ 5 , 6 , 7 , 0 , 1 , 2 , 3 , 4 ] ;
dk = s s _ d t f s ( gn , k )
The relationship d˜k = e − j 6πk/8 c̃ k can be verified using the following:
>>
>>
>>
>>
ek = exp(− j * 6 * pi * k / 8 ) ;
ck
ek
dk
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
182
5.38. Print the coefficients c̃k and d˜k and in the range k = −7, . . . , 7 and check symmetry properties
using the following code:
1
2
3
4
5
k = [ −7:7];
xn = [ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 ] ;
ck = s s _ d t f s ( xn , k )
gn = [ 5 , 6 , 7 , 0 , 1 , 2 , 3 , 4 ] ;
dk = s s _ d t f s ( gn , k )
5.39. The script “pr_05_39.m” listed below computes the DTFS coefficients of the periodic pulse
train:
1
2
3
4
5
6
7
k = [ 0 : N−1];
k ( 1 ) = eps ; % Avoid d i v i s i o n by zero
ck = sin ( pi * k/N* ( 2 * L + 1 ) ) . / (N* sin ( pi * k/N) ) ;
subplot (211)
stem ( k , abs ( ck ) ) ;
subplot (212)
stem ( k , angle ( ck ) ) ;
% DTFS c o e f f i c i e n t s
The graphs can be obtained by using the script as shown below:
a.
>>
>>
>>
N = 30;
L = 5;
pr_05_39 ;
>>
>>
>>
N = 30;
L = 8;
pr_05_39 ;
>>
>>
>>
N = 40;
L = 10;
pr_05_39 ;
>>
>>
>>
N = 40;
L = 15;
pr_05_39 ;
b.
c.
d.
183
5.40.
The script “pr_05_40.m” listed below computes the DTFS coefficients of the periodic pulse train. It is a
slightly modified version of the script used in Problem 5.39.
1
2
3
4
5
k = [−M:M]+ eps ;
N = 2 *M+1;
Omg = 2 * pi * k/N;
ck = sin ( pi * k/N* ( 2 * L + 1 ) ) . / (N* sin ( pi * k/N) ) ;
stem (Omg, ck ) ;
% DTFS c o e f f i c i e n t s
The graphs can be obtained by using the script as shown below:
a.
>> M = 20;
>> L = 3 ;
>> pr_05_40 ;
b.
>> M = 35;
>> L = 3 ;
>> pr_05_40 ;
c.
>> M = 60;
>> L = 3 ;
>> pr_05_40 ;
5.41.
a.
1
2
3
4
5
6
7
8
9
10
b.
k = [ −3:3]+ eps ;
n = [ − 50: 50] ;
ck = sin ( 7 * pi * k / 4 0 ) . / ( 4 0 * sin ( pi * k / 4 0 ) ) ; % DTFS c o e f f i c i e n t s
sum = 0 ;
for kk = 1 : 7 ,
sum = sum+ck ( kk ) * exp ( j * 2 * pi /40 * k ( kk ) * n ) ;
end ;
sum = real (sum ) ;
% Clean up imaginary part that i s
% due to roundoff e r r o r .
stem (n ,sum ) ;
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
184
1
2
3
4
5
6
7
8
9
10
k = [ −3:3]+ eps ;
n = [ − 50: 50] ;
ck = sin (13 * pi * k / 4 0 ) . / ( 4 0 * sin ( pi * k / 4 0 ) ) ; % DTFS c o e f f i c i e n t s
sum = 0 ;
for kk = 1 : 7 ,
sum = sum+ck ( kk ) * exp ( j * 2 * pi /40 * k ( kk ) * n ) ;
end ;
sum = real (sum ) ;
% Clean up imaginary part that i s
% due to roundoff e r r o r .
stem (n ,sum ) ;
5.42.
a. The output of the length-4 moving average filter can be computed using the following script:
>>
>>
>>
>>
>>
>>
>>
>>
>>
clear a l l
n = [0:49];
inp = exp ( j * 0 . 2 * pi * n ) ;
out = [ ] ;
for nn = 1: 50 ,
xn = inp (nn ) ;
yn = ss_movavg4 ( xn ) ;
out = [ out , yn ] ;
end ;
% Clear any p e r s i s t e n t v a r i a b le s
% Input stream
% Output stream
The steady-state response was found in Problem 5.24 and is computed using the following statement:
>>
yss = 0.7694 * exp ( j * ( 0 . 2 * pi * n− 0. 9425) ) ;
% Steady s t a t e response
The two responses may be printed on the screen for comparison using
>>
>>
out
yss
b.
1
2
3
4
5
6
7
8
9
10
11
12
clear a l l
% Clear any p e r s i s t e n t v a r i a b le s
n = [0:49];
inp = cos ( 0 . 2 * pi * n ) ;
% Input stream
out = [ ] ;
% Output stream
for nn = 1: 50 ,
xn = inp (nn ) ;
yn = ss_movavg4 ( xn ) ;
out = [ out , yn ] ;
end ;
yss = 0.7694 * cos ( 0 . 2 * pi * n−0.9425);
% Compare the r e s u l t s
[ n ’ , out ’ , yss ’ ]
185
c.
1
2
3
4
5
6
7
8
9
10
11
12
clear a l l
% Clear any p e r s i s t e n t v a r i a b le s
n = [0:49];
inp = 2 * sin ( 0 . 3 * pi * n ) ; % Input stream
out = [ ] ;
% Output stream
for nn = 1: 50 ,
xn = inp (nn ) ;
yn = ss_movavg4 ( xn ) ;
out = [ out , yn ] ;
end ;
yss = 1.0474 * sin ( 0 . 3 * pi * n−1.4137);
% Compare the r e s u l t s
[ n ’ , out ’ , yss ’ ]
d.
1
2
3
4
5
6
7
8
9
10
11
12
clear a l l
% Clear any p e r s i s t e n t v a r i a b le s
n = [0:49];
inp = 3 * cos ( 0 . 1 * pi * n)−5 * sin ( 0 . 2 * pi * n ) ; % Input stream
out = [ ] ;
% Output stream
for nn = 1: 50 ,
xn = inp (nn ) ;
yn = ss_movavg4 ( xn ) ;
out = [ out , yn ] ;
end ;
yss = 2.8179 * cos ( 0 . 1 * pi * n−0.9425) −3.8470 * sin ( 0 . 2 * pi * n−0.9425);
% Compare the r e s u l t s
[ n ’ , out ’ , yss ’ ]
5.43.
a. The script “pr_5_43a.m” listed below computes the output by iterating through the difference
equation for n = 0, . . . , 49.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
n = [0:49];
inp = ss_per ( xper , n ) ;
out = [ ] ;
ynm1 = 0 ;
ynm2 = 0 ;
xnm1 = 0 ;
for nn=1:50 ,
xn = inp (nn ) ;
yn = xn+2 *xnm1−ynm1−0.89 * ynm2;
xnm1 = xn ;
ynm2 = ynm1;
ynm1 = yn ;
out = [ out , yn ] ;
end ;
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
186
The variable “xper” should hold samples of one period prior to using this script. The variable “out”
holds the response of the system after the script completes.
b.
Compute the output of the system for the input signal in part (a) of Problem 5.3 using the
following statements:
>>
>>
>>
xper = [ 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 ] ;
pr_5_43a ;
out1 = out ( 2 6 : 5 0 ) ;
% One period o f the s i g n a l .
% Use the s c r i p t from part ( a ) .
% Discard the f i r s t 25 samples .
c.
Compute the steady-state output using DTFS coefficients. Us an anonymous function to
represent H (Ω) and then evaluate it at the frequencies Ωk = 2πk/8. Compute the DTFS coefficients of
the input and the output signals and construct the output signal.
>>
>>
>>
>>
>>
>>
>>
H = @(Omg) (1+2 * exp(− j *Omg) ) . / ( 1 + exp(− j *Omg) + 0. 89 * exp(− j * 2 *Omg) ) ;
k = [0:7];
Hk = H( 2 * pi * k / 8 ) ;
n = [25:49];
ck = s s _ d t f s ( xper , k ) ;
dk = Hk. * ck ;
out2 = real ( s s _ i n v d t f s ( dk , n ) ) ; % Clean imaginary part that i s
% due to roundoff e r r o r
The output computed from the difference equation and the correct steady-state output computed
through DTFS may be graphed simultaneously using the following statements:
>>
>>
>>
>>
stem (n , out1 ) ;
hold on ;
plot (n , out2 , ’ ro ’ ) ;
hold o f f ;
5.44.
a.
1
2
3
4
5
6
7
The function ss_dftmat(..) is listed below.
function W = ss_dftmat (N)
W = zeros (N,N) ;
for k =0:N−1,
for n=0:N−1,
W( k+1 ,n+1) = exp ( j * 2 * pi /N* n * k ) ;
end ;
end ;
b. Using ss_dftmat(..) the 10-point DFT of x[n] is computed as
>> W = ss_dftmat ( 1 0 ) ;
>> xn = [ 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 ] ’ ;
>> Xk=W* xn ;
187
c.
Using ss_dftmat(..) the 20-point DFT of x[n] is computed as
>> W = ss_dftmat ( 2 0 ) ;
>> xn = [ ones ( 1 , 1 0 ) , zeros ( 1 , 1 0 ) ] ’ ;
>> Xk=W* xn ;
5.45.
a.
One period of the signal x̃ (t ) is
x̃ (t ) =
½
sin (2πt ) , 0 ≤ t < 0.5
0,
0.5 ≤ t < 1
The following script computes estimated EFS coefficients using the function ss_efsapprox(..):
1
2
3
4
5
b.
t = [0:99]/100;
x = sin ( 2 * pi * t ) ;
x = x . * ( x >=0);
k = [ − 15: 15] ;
ck_est = ss_efsapprox ( x , k )
Actual EFS coefficients were determined in Example 4-10 of Chapter 4 as


0,
k odd and k 6= ∓1





− j /4 ,
k =1

ck =
j /4 ,
k = −1



−1



¢ , k even
 ¡ 2
π k −1
They can be computed with the following script:
>>
>>
kk = k+eps ;
% Avoid d i v i s i o n by zero .
ck_act = −0.25 * j * ( k ==1)+0.25 * j * ( k== −1) −1./( pi * ( kk . * kk − 1 ) ) . * (mod( k , 2) = = 0)
c.
The statement below allows the actual and estimated coefficients to be printed side by side for
comparison:
>>
conj ( [ k ’ , ck_act ’ , ck_est ’ ] )
Note: In the last MATLAB statement, vectors “ k_a t” and “ k_est” are transposed so that they can be
tabulated on the screen in column format. However, it must be remembered that they are complex
vectors. The transpose operation in MATLAB has an additional effect of conjugating complex vectors.
The conj(..) function is used for counteracting that.
5.46.
scriupt listed below defined an anonymous function to return the DTFT of the signal
a. The
¡
¢
x[n] = n u[n] − u[n − 12] .
CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS
188
1
2
3
Q = @(n ,Omg) n * exp(− j * n *Omg) ;
X = @(Omg) Q( 1 ,Omg)+Q( 2 ,Omg)+Q( 3 ,Omg)+Q( 4 ,Omg)+Q( 5 ,Omg)+Q( 6 ,Omg) + . . .
Q( 7 ,Omg)+Q( 8 ,Omg)+Q( 9 ,Omg)+Q( 10 ,Omg)+Q( 11 ,Omg) ;
b.
The transform X (Ω) may be evaluated at 10 equally spaced frequencies and the corresponding
inverse transform may be computed with the following code.
>>
>>
>>
k = [0:9];
Sk = X( 2 * pi * k / 1 0 ) ;
sn = i f f t ( Sk )
c.
The transform X (Ω) may be evaluated at 8 equally spaced frequencies and the corresponding
inverse transform may be computed with the following code.
>>
>>
>>
k = [0:7];
Sk = X( 2 * pi * k / 8 ) ;
sn = i f f t ( Sk )
Chapter 6
Sampling and Reconstruction
6.1.
a.
xs (t)
1
−1
t
1
b. Using A = 1 and τ = 1 the transform of the original signal x a (t ) is
¡ ¢
X a () = sinc2 f
The transform of the impulse sampled signal x s (t ) is
X s () =
∞
¡
¢
1 X
Xa f − k fs
T s k=−∞
=5
c.
∞
X
k=−∞
¡
¢
sinc2 f − 5k
Xs (f )
6
5
4
3
2
1
0
−1
−10
−5
0
f (Hz)
189
5
10
CHAPTER 6. SAMPLING AND RECONSTRUCTION
190
6.2.
X s () =
∞
¢
¡
1 X
Xa f − k fs
T s k=−∞
∞
X
= 100
k=−∞
¡
¢
X a f − 100k
Xs (f )
100
...
...
−135 −105
−65
−95
−35
−5 5
35
65
105
f (Hz)
135
95
6.3. Start with X a (ω) shown below.
Xa (ω)
1
−70π −10π 10π
ω (rad/s)
70π
The spectrum of the discrete-time signal x[n] is
µ
¶
∞
1 X
Ω − 2πk
X (Ω) =
Xa
T s k=−∞
Ts
= 100
∞
X
Xa
k=−∞
µ
Ω − 2πk
0.01
¶
X (Ω)
100
...
...
−2.7π
−2π
−1.3π −0.7π
0.7π
1.3π
2π
2.7π
Ω (rad)
6.4. If aliasing is to be avoided, the minimum sampling rate required is f s,mi n = 70 Hz. Let the
sampling rate be 90 percent of the required minimum, that is, f s = 63 Hz.
µ
¶
∞
Ω − 2πk
1 X
Xa
X (Ω) =
T s k=−∞
Ts
= 63
∞
X
k=−∞
Xa
µ
Ω − 2πk
1/63
¶
191
Xa (Ω/Ts )
1
−
10π
9
10π
9
Ω (rad)
X (Ω)
63
...
...
−2π
2π
−
10π
9
−
8π
9
8π
9
Ω (rad)
10π
9
X (Ω)
63
...
...
−2π
2π
−
10π
9
−
8π
9
8π
9
Ω (rad)
10π
9
6.5.
a. The continuous-time signal is
Using A = 1 and τ = 1 sec, we get

 1 − A |t | ,
τ
x a (t ) =

0,
x a (t ) =
(
−τ < t < τ
otherwise
1 − |t | ,
−τ < t < τ
0,
otherwise
The sampling interval is T s = 1/ f s = 1/12 sec. The sampled signal x[n] = x a (nT s ) is obtained as

 1 − |n| ,
12
x[n] =

0,
−12 ≤ n ≤ 12
otherwise
CHAPTER 6. SAMPLING AND RECONSTRUCTION
192
The signal x[n]
Amplitude
1
0.8
0.6
0.4
0.2
0
−0.2
−15
−10
−5
5
0
10
15
Sample index n
b. Let us use Eqn. (6.25). The spectrum of the continuous-time signal x a (t ) is
X a (ω) = Aτ sinc2
³ ωτ ´
2π
= sinc2
³ω´
2π
and X (Ω) is computed using Eqn. (6.25) as
µ
µ
¶
¶
∞
∞
Ω − 2πk
1 X
1 X
2 Ω − 2πk
Xa
sinc
=
X (Ω) =
T s k=−∞
Ts
T s k=−∞
2π T s
Using T s = 1/12 s, we get
X (Ω) = 12
∞
X
sinc
k=−∞
2
µ
¶
¶
µ
∞
X
Ω − 2πk
2 6Ω
= 12
− 12k
sinc
2π (1/12)
π
k=−∞
|X (Ω)|
Magnitude
15
10
5
0
−π
−3π/4
−π/2
0
−π/4
π/2
π/4
Ω (rad)
6.6.
The discrete-time signal is
or, equivalently

´
³
 sin πn ,
15
x[n] =

0,
x[n] = sin
Let
³ πn ´ ¡
15
n = 0, . . . , 14
otherwise
u[n] − u[n − 15]
p[n] = u[n] − u[n − 15]
¢
3π/4
π
193
so that
P (Ω) =
14
X
n=0
Using the modulation property, X (Ω) is
X (Ω) = − j
e − j Ωn =
sin (15 Ω/2) − j 7Ω
e
sin (Ω/2)
1
1
P (Ω − π/15) + j P (Ω + π/15)
2
2
|X (Ω)|
12
Magnitude
10
8
6
4
2
0
−π
−3π/4
−π/2
0
−π/4
π/4
π/2
π
3π/4
Ω (rad)
6.7.
a.
Not bandlimited. Cannot be sampled without loss of information.
b.
Not bandlimited. Cannot be sampled without loss of information.
c.
Bandlimited.
f max = 75 Hz,
f s ≥ 150 Hz.
d.
Bandlimited.
f max = 175 Hz,
f s ≥ 350 Hz.
e.
Not bandlimited. Cannot be sampled without loss of information.
6.8.
a, b.
xa (t) and x[n]
1
0.5
0
−0.5
−1
0
1
2
3
t (sec)
4
5
CHAPTER 6. SAMPLING AND RECONSTRUCTION
194
c.
¡
¢¯¯
x[n] = sin 2π f a t ¯
t =n/ f s
= sin
µ
¶
µ
¶
2πn
2πn
= sin
+ 2πr n
2.4
2.4
r, n : Integer
The normalized frequency is
F=
For r = 1 :
For r = 2 :
For r = 3 :
F = 1.4167
F = 2.4167
F = 3.4167
⇒
⇒
⇒
fa
+r
fs
f a = 3400 Hz
f a = 5800 Hz
f a = 8200 Hz
replacemen
6.9.
a.
Xa (f ) for fa = 1 kHz
j 0.5
−j 0.5
−1
replacemen
−8
−6
−4
−2
1
0
2
4
6
8
6
8
f (kHz)
Xa (f ) for fa = 3.4 kHz
j 0.5
−j 0.5
−3.4
replacemen
−8
−6
−4
3.4
−2
0
2
4
f (kHz)
Xa (f ) for fa = 5.8 kHz
j 0.5
−j 0.5
−5.8
−8
−6
5.8
−4
−2
0
f (kHz)
2
4
6
8
replacemen
195
Xa (f ) for fa = 8.2 kHz
j 0.5
−j 0.5
−8.2
−8
8.2
−6
−4
0
−2
2
4
6
8
f (kHz)
b.
Since all four continuous-time signals lead to the same discrete-time signal when sampled; they
all have the same DTFT spectrum.
µ
¶
2πn
x[n] = sin
= sin (0.8333πn)
2.4
X (Ω) =
∞ £
¤
π X
δ (Ω − 0.8333π − 2πm) − δ (Ω + 0.8333π − 2πm)
j m=−∞
X (Ω)
−2π
jπ
−jπ
−0.8333π
0.8333π
−π
0
π
2π
Ω (rad)
6.10.
a.
¯
¯
x[n] = x a (t )¯
µ
¶
500πn
= sin
= sin (0.4πn + 2πr n) ,
fs
t =n/ f s
The normalized frequency is
F=
f a 250
=
= 0.2 + r
fs
fs
Using r = 0 for proper sampling, the sampling rate must be
fs =
b.
For r = 1 :
250
= 1.2
fs
⇒
f s = 208.33 Hz
250
= 1250 Hz.
0.2
r, n : Integer
CHAPTER 6. SAMPLING AND RECONSTRUCTION
196
For r = 2 :
250
= 2.2
fs
⇒
f s = 113.64 Hz
c. Let the new signal frequency be f¯a .
For r = 1 :
For r = 2 :
f¯a
= 1.2
1250
f¯a
= 2.2
1250
f¯a
= 0.2 + r
fs
⇒
f¯a = 1500 Hz
⇒
f¯a = 2750 Hz
6.11.
a.
x[n] = 3 cos (πn) + 5 sin (2.5πn)
= 3 cos (πn) + 5 sin (0.5πn)
∞
X
©
ª
F 3 cos (πn) = 6π
δ (Ω − π − 2πm)
m=−∞
Let
∞ £
X
©
ª 5π
¤
F 5 sin (πn) =
δ (Ω − π/2 − 2πm) − δ (Ω + π/2 − 2πm)
j m=−∞
X̃ (Ω) = 3π δ (Ω − π) + 3π δ (Ω + π) − j 5π δ (Ω − π/2) + j 5π δ (Ω + π/2)
so that X (Ω) can be written as
X (Ω) =
∞
X
m=−∞
X̃ (Ω − 2πm)
X (Ω)
−j5π
6π
j5π
−j5π
−π/2
−3π/2
−π
6π
π/2
0
Ω (rad)
b.
j5π
The recovered signal would be
x̂ (t ) = 3 cos (100πt ) + 5 sin (50πt )
3π/2
π
197
6.12.
a.
¡
¢
¡
¢
¡
¢
¡
¢
X a (ω) = π δ f − 100π + π δ f + 100π + π δ f − 120π + π δ f + 120π
Xa (ω)
π
π
π
−120π −100π
π
100π 120π
ω (rad/s)
b.
The period is T s = 1/ f s = 1/400 = 0.0025 s, or equivalently T s = 2.5 ms. This corresponds to a duty
cycle of d = τ/T s = 0.2.
X̄ s (ω) = d
∞
X
k=−∞
= 0.2
c.
∞
X
sinc (kd ) X a (ω − kωs )
k=−∞
sinc (0.2k) X a (ω − 800πk)
X̄s (ω)
0.2π
0.2 sinc (0.2) π
−800π
0.2 sinc (0.2) π
0
800π
ω (rad/s)
6.13. The spectrum of the zero-order hold sampled signal is
µ
¶
∞
ωd T s − j ωdTs /2 X
X̄ s (ω) = d sinc
e
X a (ω − kωs )
2π
k=−∞
µ
¶
∞
0.2ω − j 0.1ω/800 X
= 0.2 sinc
e
X a (ω − 800πk)
800π
k=−∞
CHAPTER 6. SAMPLING AND RECONSTRUCTION
198
X̄s (ω)
−800π
0
800π
ω (rad/s)
6.14.
xr (t)
1
0.5
0
−0.5
−1
0
0.05
0.1
0.15
0.2
0.25
0.15
0.2
0.25
t (sec)
6.15.
xr (t)
1
0.5
0
−0.5
−1
0
0.05
0.1
t (sec)
6.16.
Using the geometric series formula
w [m] =
1 1 − e j 2πm
D 1 − e j 2πm/D
For m 6= nD then the numerator is equal to zero, and the denominator is nonzero, resulting in w [m] = 0.
If m = nD then both the numerator and the denominator are equal to zero, requiring the use of
199
L’Hospital’s rule:
w [nD] =
¯
¯
1
j 2π e j 2πm
¯
=1
¯
j
2πm/D
D j (2π/D) e
m=nD
6.17. Using the inverse DTFT relationship
1
h r [n] =
2π
ˆ π/L
(L) e
j Ωn
−π/L
The result can be simplified to
h r [n] =
¯π/L
L e j Ωn ¯¯
dΩ =
2π j n ¯−π/L
³ πn ´
L
= sinc (n/L)
sin
πn
L
6.18.
a. Not bandlimited. Cannot be downsampled without loss of information.
b.
Not bandlimited. Cannot be downsampled without loss of information.
c.
Ωmax = π/3. May be downsampled with D = 3.
d.
Ωmax = 2π/7. May be downsampled with D = 3.
6.19.
a.
H zoh (Ω) =
b.
2
X
n=0
Using Euler’s formula
£
¤
e − j Ωn = 1 + e − j Ω + e − j 2Ω = 1 + 2 cos (Ω) e − j Ω
x[n] =
The transform of the input signal is
replacemen
X (Ω) =
i
1 h j 0.1πn
e
− e − j 0.1πn
2j
∞ £
¤
π X
δ (Ω − 0.1π − 2πm) − δ (Ω + 0.1π − 2πm)
j m=−∞
X (Ω)
jπ
−jπ
−2π
−π
jπ
−jπ
−0.1π
0.1π
0
Ω (rad)
jπ
π
−jπ
2π
CHAPTER 6. SAMPLING AND RECONSTRUCTION
200
¶
µ
2
1 X
Ω − 2πk
X1 =
X
3 k=0
3
X1 (Ω)
j
π
3
−j
−2π
π
3
π
3
−j
−0.3π
0.3π
j
−π
π
3
0
j
π
3
−j
π
2π
π
2π
Ω (rad)
X 2 (Ω) = X 1 (3 Ω)
X2 (Ω)
j
π
3
−j
−0.1π
−2π
−π
π
3
0.1π
0
Ω (rad)
X 3 (Ω) = H zoh (Ω) X 2 (Ω)
X3 (Ω)
0.1π
−0.1π
−2π
−π
0
Ω (rad)
π
2π
π
3
201
6.20.
The impulse response of the first-order hold interpolation filter is
h f oh [n] = { 1/3, 2/3, 1, 2/3, 1/3 }
↑
n=−2
and its transform is
H f oh (Ω) =
1 j 2Ω 2 j Ω
2
1
e
+ e + 1 + e − j Ω + e − j 2Ω
3
3
3
3
= 1+
2
4
cos (Ω) + cos (2Ω)
3
3
The spectra X (Ω), X 1 (Ω) and X 2 (Ω) are as shown in the solution to Problem 6.19.
X 3 (Ω) = H f oh (Ω) X 2 (Ω)
X3 (Ω)
0.1π
−0.1π
−2π
−π
0
π
Ω (rad)
6.21.
a.
1
2
3
4
5
6
7
8
9
10
11
12
13
Xa = @( f ) 2. / ( 1+ 4 * pi * pi * f . * f ) ;
f = [ −7:0.01:7];
fs = 3;
Ts = 1/ f s ;
Xs = zeros ( s i z e ( f ) ) ;
for k= −5:5 ,
Xs = Xs+ f s * Xa ( f−k * f s ) ;
end ;
plot ( f , Xs ) ;
axis ( [ − 7 , 7 , − 1 , 10] ) ;
t i t l e ( ’ X_ { s } ( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
grid ;
% Original spectrum
2π
CHAPTER 6. SAMPLING AND RECONSTRUCTION
202
b.
1
2
3
4
5
6
7
8
9
10
11
12
13
c.
Xa = @( f ) 2. / ( 1+ 4 * pi * pi * f . * f ) ;
f = [ −7:0.01:7];
fs = 4;
Ts = 1/ f s ;
Xs = zeros ( s i z e ( f ) ) ;
for k= −5:5 ,
Xs = Xs+ f s * Xa ( f−k * f s ) ;
end ;
plot ( f , Xs ) ;
axis ( [ − 7 , 7 , − 1 , 10] ) ;
t i t l e ( ’ X_ { s } ( f ) ’ ) ;
xlabel ( ’ f (Hz) ’ ) ;
grid ;
% Original spectrum
Aliasing effect is present in both cases, but is less pronounced for f s = 4 Hz.
6.22.
¡ ¢
Compute and graph the spectrum X a f :
1
2
3
4
5
6
7
Xa = @( f , A , tau ) A * tau * sinc ( f * tau ) . * sinc ( f * tau ) ;
f = [ −10:0.02:10];
plot ( f , Xa ( f , 1 , 1 ) ) ;
axis ( [ − 10 , 10 , − 0. 25 , 1. 25] ) ;
xlabel ( ’ f (Hz) ’ ) ;
t i t l e ( ’ X_ { a } ( f ) ’ ) ;
grid ;
Compute and graph the spectrum X s (s) for the sampling rate f s = 5 Hz.
1
2
3
4
5
6
7
8
9
10
11
fs = 5;
% Edit t h i s f o r other sampling r a t e s .
Ts = 1/ f s ;
Xs = zeros ( s i z e ( f ) ) ;
for k= −15:15 ,
Xs = Xs+1/Ts * Xa ( f−k * fs , 1 , 1 ) ;
end ;
plot ( f , Xs ) ;
axis ([ −10 ,10 , −1 ,6]);
xlabel ( ’ f (Hz) ’ ) ;
t i t l e ( ’ X_ { s } ( f ) ’ ) ;
grid ;
6.23.
1
2
3
A = 1;
tau = 1 ;
Xa = @(omg) A * tau * ( sinc (omg* tau / ( 2 * pi ) ) ) . ^ 2 ;
203
4
5
6
7
8
9
10
11
12
13
14
15
16
17
f s = 12;
Omg = [ − 1 : 0 . 0 0 1 : 1 ] * pi ;
% Use Eqn . ( 6 . 2 5 )
X = zeros ( s i z e (Omg) ) ;
for k= −10:10 ,
X = X+12 * Xa ( (Omg−2* pi * k ) * f s ) ;
end ;
p1 = plot (Omg, X ) ;
axis ([ − pi , pi , − 2 , 1 5 ] ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
t i t l e ( ’ | X( \Omega ) | $ ’ ) ;
axis ([ − pi , pi , 0 , 1 5 ] ) ;
grid ;
6.24.
1
2
3
4
5
6
7
8
9
10
% Compute the transform o f p [ n ] .
Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi+eps ;
P = @(Omg) sin ( 7 . 5 *Omg) . / sin ( 0 . 5 *Omg) . * exp(− j * 7 *Omg) ;
% Use the modulation property o f the DTFT .
X = −0.5 * j * P (Omg−pi / 15) + 0. 5 * j * P (Omg+pi / 1 5 ) ;
plot (Omg, abs (X ) ) ;
axis ([ − pi , pi , 0 , 1 2 ] ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
t i t l e ( ’ | X( \Omega ) | ’ ) ;
grid ;
6.25. a.
1
2
3
4
5
6
7
8
9
10
11
xa = @( t ) sin ( 2 * pi * 1000 * t ) ;
t = [ 0 : 5 e −6:5e −3];
f s = 2400;
Ts = 1/ f s ;
n = [0:14];
plot (1000 * t , xa ( t ) ) ;
axis ( [ 0 , 5 , − 1 . 2 , 1 . 2 ] ) ;
hold on ;
stem(1000 * n * Ts , xa (n * Ts ) , ’ r ’ ) ;
hold o f f ;
grid ;
b.
1
2
3
xa = @( t ) sin ( 2 * pi * 1000 * t ) ;
xb = @( t ) sin ( 2 * pi * 3400 * t ) ;
xc = @( t ) sin ( 2 * pi * 5800 * t ) ;
CHAPTER 6. SAMPLING AND RECONSTRUCTION
204
4
5
6
7
8
9
10
11
12
13
14
xd = @( t ) sin ( 2 * pi * 8200 * t ) ;
t = [ 0 : 5 e −6:1e −3];
f s = 2400;
Ts = 1/ f s ;
n = [0:14];
plot (1000 * t , xa ( t ) , ’b ’ ,1000 * t , xb ( t ) , ’b−− ’ ,1000 * t , xc ( t ) , ’ g−− ’ ,1000 * t , xd ( t ) , ’ k−− ’ ) ;
axis ( [ 0 , 1 , − 1 . 2 , 1 . 2 ] ) ;
hold on ;
stem(1000 * n * Ts , xa (n * Ts ) , ’ r ’ ) ;
hold o f f ;
grid ;
6.26.
a. Lines 2 and 3 create a naturally sampled version of the signal rather then a zero-order hold
version. The vector “t1” contains 0s and 1s depending on whether the corresponding time instant in
vector “t” is within an active pulse of the signal p̃ (t ) or not (refer to Fig. 6.18).
The loop between lines 5 and 15 searches through this naturally sampled signal. When the left edge of a
pulse is encountered, its value is saved in the variable “value”, and the amplitude is adjusted to the
saved level for the duration of the pulse.
b.
A modified version of the function ss_zohsamp(..) is listed below as ss_zohsamp2(..). This function
is only for the case d = 1.
The function ss_zohsamp2(..) is listed below:
1
2
3
4
5
6
function xzoh = ss_zohsamp2( xa , Ts , t )
d e l t = t (2) − t ( 1 ) ;
% Time increment used in v e c t o r " t "
L = f l o o r ( Ts/ d e l t ) ; % Number o f samples per pulse ( must be i n t e g e r ! )
pulse = ones ( 1 , L ) ;
xzoh = downsample ( xa ( t ) , L ) ;
xzoh = kron ( xzoh , pulse ) ;
The script below can be used for testing.
1
2
3
4
5
x = @( t ) exp(−abs ( t ) ) ;
t = [ −4:0.001:3.999];
xzoh = ss_zohsamp2( x , 0 . 2 , t ) ;
plot ( t , xzoh ) ;
axis ( [ − 4 , 4 , − 0 . 2 , 1 . 2 ] ) ;
6.27.
a.
1
2
3
4
xa = @( t ) exp(−abs ( t ) ) ;
t = [ −4:0.001:4];
xzoh = ss_zohsamp( xa , 0 . 2 , 0 . 9 0 , t ) ;
plot ( t , xzoh ) ;
205
b.
1
2
3
4
1
2
3
4
1
2
3
4
% System with a=3
sys = t f ( [ 3 ] , [ 1 , 3 ] ) ;
y = lsim ( sys , xzoh , t ) ;
plot ( t , xzoh , t , y ) ;
% System with a=2
sys = t f ( [ 2 ] , [ 1 , 2 ] ) ;
y = lsim ( sys , xzoh , t ) ;
plot ( t , xzoh , t , y ) ;
% System with a=1
sys = t f ( [ 1 ] , [ 1 , 1 ] ) ;
y = lsim ( sys , xzoh , t ) ;
plot ( t , xzoh , t , y ) ;
6.28.
The system can be simulated with the following code:
1
2
3
4
5
6
n = [0:199];
xn = sin ( 0 . 1 * pi * n ) ; % Signal x [ n ] .
x1 = downsample ( xn , 3 ) ;
x2 = upsample( x1 , 3 ) ;
hzoh = ones ( 1 , 3 ) ; % Impulse response o f i n t e r p o la t i o n f i l t e r .
x3 = conv ( x2 , hzoh ) ;
Individual signals can be graphed using following code segments:
1
2
3
4
1
2
3
4
1
2
3
4
stem (n , xn ) ;
axis ( [ − 0 . 5 , 1 9 9 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x [n] ’ ) ;
stem ( [ 0 : 6 6 ] , x1 ) ;
axis ( [ − 0 . 5 , 6 6 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 1 } [ n ] ’ ) ;
stem ( [ 0 : 2 0 0 ] , x2 ) ;
axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 2 } [ n ] ’ ) ;
CHAPTER 6. SAMPLING AND RECONSTRUCTION
206
1
2
3
4
stem ( [ 0 : 2 0 0 ] , x3 ( 1 : 2 0 1 ) ) ;
axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 3 } [ n ] ’ ) ;
6.29.
The system can be simulated with the following code:
1
2
3
4
5
6
n = [0:199];
xn = sin ( 0 . 1 * pi * n ) ; % Signal x [ n ] .
x1 = downsample ( xn , 3 ) ;
x2 = upsample( x1 , 3 ) ;
hfoh = [ 1 / 3 , 2 / 3 , 1 , 2 / 3 , 1 / 3 ] ;
x3 = conv ( x2 , hfoh ) ;
Individual signals can be graphed using following code segments:
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
stem (n , xn ) ;
axis ( [ − 0 . 5 , 1 9 9 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x [n] ’ ) ;
stem ( [ 0 : 6 6 ] , x1 ) ;
axis ( [ − 0 . 5 , 6 6 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 1 } [ n ] ’ ) ;
stem ( [ 0 : 2 0 0 ] , x2 ) ;
axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 2 } [ n ] ’ ) ;
stem ( [ 0 : 2 0 0 ] , x3 ( 1 : 2 0 1 ) ) ;
axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ;
xlabel ( ’n ’ ) ;
t i t l e ( ’ x_ { 3 } [ n ] ’ ) ;
Chapter 7
Laplace Transform for Continuous-Time Signals
and Systems
7.1.
a.
jω
Using Laplace transform definition
X (s) =
ˆ ∞
e −2t u (t ) e −st d t
=
ˆ ∞
−2t −st
=
−∞
e
e
0
1
s +2
¯∞
e −(s+2)t ¯¯
dt =
− (s + 2) ¯0
σ
−2
For convergence of the integral, we need
Re {s + 2} > 0
⇒
|
Re {s} > −2
{z
}
Re {s} > −2
b.
jω
X (s) =
ˆ ∞
e −2t u (t ) e −st d t
=
ˆ ∞
e −2t e −st d t
−∞
1
σ
−2
¯∞
e −2 e −s
e −(s+2)t ¯¯
=
=
− (s + 2) ¯1
s +2
For convergence of the integral, we need
Re {s + 2} > 0
⇒
|
Re {s} > −2
{z
Re {s} > −2
207
}
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
208
c.
jω
X (s) =
ˆ ∞
e 2t u (−t ) e −st d t
=
ˆ 0
e 2t e −st d t =
−∞
−∞
¯0
e (2−s)t ¯¯
1
=
¯
(2 − s) −∞ 2 − s
2
σ
For convergence of the integral, we need
Re {2 − s} > 0
⇒
Re {s} < 2
|
{z
}
Re {s} < 2
d.
jω
X (s) =
ˆ ∞
=
ˆ 1
−∞
e 2t u (−t + 1) e −st d t
2t −st
e e
−∞
¯1
e 2 e −s
e (2−s)t ¯¯
=
dt =
2−s
(2 − s) ¯−∞
2
For convergence of the integral, we need
Re {2 − s} > 0
⇒
Re {s} < 2
|
{z
}
Re {s} < 2
e.
X (s) =
=
ˆ 1
0
(1) e −st d t +
−st ¯¯1
ˆ 2
(−1) e −st d t
1
−st ¯¯2
¤
£
e
e ¯
¯ = 1 1 − 2 e −s + e −2s
−
¯
¯
−s 0
−s 1 s
The ROC is the entire s-plane with the exception of Re {s} → −∞.
σ
209
f.
jω
X (s) =
=
ˆ 1
(1) e
−st
ˆ ∞
0
dt +
−st ¯¯1
−st ¯¯∞
(−1) e −st d t
1
¤
£
e ¯
e
¯ = a 1 − 2 e −s
+
¯
¯
−s 0
s 1
s
σ
For convergence of the second integral, we
need
Re {s} > 0
|
{z
} |
Re {s} < 0
{z
}
Re {s} > 0
7.2.
Using the definition of the Laplace transform given by Eqn. (7.1)
ˆ ∞·∞
¸
X −anT
X (s) =
e
δ (t − nT ) e −st d t
−∞
=
=
∞
X
e
n=0
−anT
−∞
n=0
∞
X
n=0
ˆ ∞
δ (t − nT ) e −st d t
e −anT e −snT =
∞
X
n=0
e −(s+a)T n =
The closed form expression is valid only if
¯ −(s+a)T ¯
©
ª
¯e
¯<1
⇒
Re − (s + a) < 0
⇒
1
1 − e −(s+a)T
© ª
© ª
Re s > − Re a
which establishes the ROC. The poles of X (s)
are found by solving
jω
j 4π/T
e −(s+a)T = 1
which, using the identity e j 2πk = 1, can also be
written as
e −(s+a)T = e j 2πk
j 2π/T
σ
−a
The poles are at
s k = −a − j
−j 2π/T
2πk
,
T
−j 4π/T
all integer k
as shown in the pole-zero diagram. (We are assuming that a is real-valued.)
|
{z
Re {s} > −a
}
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
210
7.3.
For the Fourier transform to exist, the ROC needs to include the j ω-axis.
a.
ROC:
b.
ROC:
c.
ROC:
d.
ROC:
© ª
− 1 < Re s < 1
© ª
Re s > −1
© ª
Re s < 1
© ª
Re s > −1
7.4.
For the signal to be causal the ROC needs to be to the right of a vertical line in the s-plane and include
s → ∞.
a.
ROC:
b.
ROC:
c.
ROC:
d.
ROC:
© ª
Re s > 1
© ª
Re s > −1
© ª
Re s > 2
© ª
Re s > −1
7.5.
The ROC of the transform an anti-causal signal is to the left of a vertical line in the s-plane.
a.
ROC:
b.
ROC:
c.
ROC:
d.
ROC:
7.6.
a.
© ª
Re s < −1
© ª
Re s < −2
© ª
Re s < 1
© ª
Re s < −1
¯
©
ª
¯
F x (t ) = X (s)¯
s= j ω
Since s = j ω is included in the ROC the Fourier transform exists, and is found as
©
ª
F x (t ) = ¡
b.
1 − ω
jω−1
¢
=¡
¢2
2 + ω2 + j ω
jω − jω−2
The desired Fourier transform is related to X (s) by
¯
ª
©
¯
F x (t ) e t = X (s)¯
s=−1+ j ω
211
Since the trajectory s = −1 + j ω is not included in the ROC, the Fourier transform of x (t ) e t does not
exist.
c.
The desired Fourier transform is related to X (s) by
¯
©
ª
¯
F x (t ) e −3t = X (s)¯
s=3+ j ω
Since the trajectory s = 3 + j ω is not included in the ROC, the Fourier transform of x (t ) e −3t does not
exist.
7.7.
a.
¯
©
ª
¯
F x (t ) = X (s)¯
b.
c.
¯
ª
©
¯
F x (t ) e −t = X (s)¯
s= j ω
=¡
1
1
¢
=¡
¢2
2
5 − ω + j 2ω
j ω + j 2ω + 5
1
1
¢
=¡
=¡
¢2
¡
¢
s=1+ j ω
8 − ω2 + j 4ω
1+ jω +2 1+ jω +5
The desired Fourier transform is related to X (s) by
¯
ª
©
¯
F x (t ) e t = X (s)¯
s=−1+ j ω
The trajectory s = −1 + j ω is not included in the ROC. Therefore the Fourier transform of x (t ) e t does
not exist.
d.
The desired Fourier transform is related to X (s) by
¯
ª
©
¯
F x (t ) e 3t = X (s)¯
s=−3+ j ω
The trajectory s = −3 + j ω is not included in the ROC. Therefore the Fourier transform of x (t ) e 3t does
not exist.
7.8.
jω
a.
The transform has a zero at s = 2 and
poles at s = −1, −2. Since x (t ) is causal, the
ROC is
© ª
Re s > −1
σ
−2 −1
2
The Fourier transform X (ω) is
¯
¯
X (ω) = X (s)¯
s= j ω
=¡
jω−2
jω−2
¢
=¡
¢2
2 − ω2 + j 3ω
j ω + j 3ω + 2
|
{z
Re {s} > −1
}
212
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
jω
b.
The transform has a zero at s = 0 and
poles at s = ∓1. Since x (t ) is causal, the ROC is
σ
© ª
Re s > 1
−1
1
The ROC does not include the j ω-axis of the
s-plane. Therefore the Fourier transform does
not exist.
{z
|
}
Re {s} > 1
jω
c.
The transform has a zero at s = −1 and
poles at s = 1, 3. Since x (t ) is specified to be
anti-causal, the ROC is
© ª
Re s < 1
σ
−1
1
3
The ROC includes the j ω-axis. The Fourier
transform X (ω) is
¯
¯
X (ω) = X (s)¯
s= j ω
=¡
jω+1
jω+1
¢
=¡
¢2
3 − ω2 − j 4ω
j ω − j 4ω + 3
|
{z
}
Re {s} < 1
jω
d.
The transform has zeros at s = 0, 1 and
poles at s = −2, 3. Since x (t ) is specified to be
anti-causal, the ROC must be
σ
© ª
Re s < −2
−2
The ROC does not include the j ω-axis. Consequently, the Fourier transform X (ω) does not
exist.
| {z }
Re {s} < −2
1
3
213
jω
e.
The transform has zeros at s = 0, 1 and
poles at s = −3, −5. Since the Fourier transform
of x (t ) e 4t exists, the ROC must include the trajectory s = −4 + j ω. Therefore the ROC is
© ª
− 5 < Re s < −3
σ
−5
−3
The ROC does not include the j ω-axis. Consequently, the Fourier transform X (ω) does not
exist.
|
{z
}
−5 < Re {s} < −3
7.9.
a.
1
,
s +1
©
ª
1
,
L e −3t u (t ) =
s +3
Using the linearity of the Laplace transform we obtain
©
ª
L e −t u (t ) =
X (s) =
© ª
Re s > −1
© ª
Re s > −3
3
5
−2 (s − 2)
,
−
=
s + 1 s + 3 (s + 1) (s + 3)
© ª
Re s > −1
b.
1
,
s +1
©
ª
−1
,
L e 3t u (−t ) =
s −3
Using the linearity of the Laplace transform we obtain
©
ª
L e −t u (t ) =
X (s) =
c.
© ª
Re s > −1
© ª
Re s < 3
3
2
s − 11
−
=
,
s + 1 s − 3 (s + 1) (s − 3)
©
ª
L δ (t ) = 1 ,
© ª
−1 < Re s < 3
all s
1
,
s +1
Using the linearity of the Laplace transform we obtain
© ª
Re s > −1
2
s +3
=
,
s +1 s +1
© ª
Re s > −1
©
ª
L e −t u (t ) =
X (s) = 1 +
d.
©
ª 1
L u (t ) = ,
s
© ª
Re s > 0
1
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
214
1
,
s +1
Using the linearity of the Laplace transform we obtain
©
ª
L e −t u (t ) =
X (s) =
e.
© ª
Re s > −1
1
1
1
,
−
=
s s + 1 s (s + 1)
©
ª
L cos (2t ) u (t ) =
s
s2 + 4
,
© ª
Re s > 0
© ª
Re s > 0
3
,
s2 + 9
Using the linearity of the Laplace transform we obtain
© ª
Re s > 0
s 3 + 6 s 2 + 9 s + 24
¢¡
¢ ,
X (s) = ¡ 2
s + 4 s2 + 9
© ª
Re s > 0
©
ª
L sin (3t ) u (t ) =
f.
The signal x (t ) can be written as
x (t ) =e −2t
·
¸
1 j 3t 1 − j 3t
e + e
u (t )
2
2
1
1
= e (−2+ j 3)t u (t ) + e (−2− j 3)t u (t )
2
2
n
o
© ª
1
,
Re s > −2
L e (−2+ j 3)t u (t ) =
s +2− j3
n
o
© ª
1
L e (−2+ j 3)t u (t ) =
,
Re s > −2
s +2+ j3
Using the linearity of the Laplace transform we obtain
X (s) =
1/2
1/2
s +2
+
=
,
s + 2 − j 3 s + 2 + j 3 s 2 + 4 s + 13
© ª
Re s > −2
7.10.
a.
© ª
1
,
Re s > −2
s +2
Using the time shifting property of the Laplace transform we get
©
ª
L e −2t u (t ) =
©
ª e −s
X (s) = L e −2(t −1) u (t − 1) =
,
s +2
b.
© ª
Re s > −2
Scaling the transform pair found in part (a) we obtain
©
ª
©
ª e −(s+2)
,
L e −2t u (t − 1) = e −2 L e −2(t −1) u (t − 1) =
s +2
© ª
Re s > −2
215
c.
Starting with the transform pair
©
ª
L e −2t u (t ) =
1
,
s +2
© ª
Re s > −2
and using the time scaling property of the Laplace transform we get
©
ª
L e 2t u (−t ) =
−1
1
−
,
−s + 2 s − 2
© ª
Re s < 2
Finally, using the time shifting property on this result leads to
©
ª −e s
X (s) = L e 2(t +1) u (−t − 1) =
,
s −2
d.
Using the transform pair found in part (c) with the linearity property of the Laplace transform we
obtain
e.
© ª
Re s < 2
©
ª
©
ª −e (s−2)
,
X (s) = L e 2t u (−t − 1) = e −2 L e 2(t +1) u (−t − 1) =
s −2
From earlier parts of the problem we have
©
ª
−1
L e 2t u (−t ) =
,
s −2
© ª
Re s < 2
Applying the time shifting property of the Laplace transform we get
ª −e −s
©
,
L e 2(t −1) u (−t + 1) =
s −2
and therefore
© ª
Re s < 2
©
ª −e −s
e −2 L e 2t u (−t + 1) =
,
s −2
© ª
Re s < 2
©
ª −e −(s−2)
,
X (s) = L e 2t u (−t + 1) =
s −2
© ª
Re s < 2
7.11.
a. Using the appropriate trigonometric identity, x (t ) is
x (t ) = cos (3t ) cos (π/6) u (t ) − sin (3t ) sin (π/6) u (t )
Transforms of individual terms are
s
©
ª
L cos (3t ) u (t ) =
s2 + 9
©
ª
L sin (3t ) u (t ) =
s2 + 9
Using the linearity of the Laplace transform we get
X (s) =
3
,
© ª
Re s > 0
,
© ª
Re s > 0
cos (π/6) s 3 sin (π/6) 0.866 s − 1.5
−
=
s2 + 9
s2 + 9
s2 + 9
© ª
Re s < 2
216
b.
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
Using Euler’s formula, x (t ) is
x (t ) =
1
1 j π/6 j 3t
e
e u (t ) + e − j π/6 e − j 3t u (t )
2
2
Transforms of individual terms are
n
o
L e j 3t u (t ) =
1
,
s − j3
n
o
L e − j 3t u (t ) =
1
,
s + j3
Using the linearity of the Laplace transform we get
© ª
Re s > 0
© ª
Re s > 0
1 j π/6
11 e − j π/6
e
X (s) = 2
+2
s − j3
s + j3
¡
¢
¡
¢
1 j π/6
s + j 3 + 12 e − j π/6 s − j 3
2e
=
s2 + 9
=
0.866 s − 1.5
,
s2 + 9
© ª
Re s > 0
©
ª 1
L u (t ) = ,
s
© ª
Re s > 0
=
7.12.
a. Using the relationships
and
cos (π/6) s − 3 sin (π/6)
s2 + 9
© ª
©
ª e −s
,
Re s > 0
L u (t − 1) =
s
the transform is found through the use of the linearity property as
X (s) =
1 − e −s
,
s
© ª
Re s > −∞
© ª
The ROC is the entire s-plane with the exception of Re s → −∞ since the pole at s = 0 is canceled when
¡
¢
the two terms are added. This is evident from the fact that X 0 + j 0 = 1 and that x (t ) is a finite length
signal.
x (t)
1
t
1
b.
Using the relationships
©
ª 1
L u (t ) = ,
s
© ª
Re s > 0
217
©
ª e −s
,
L u (t − 1) =
s
and
© ª
Re s > 0
©
ª e −2s
© ª
L u (t − 2) =
,
Re s > 0
s
the transform is found through the use of the linearity property as
© ª
1 − 2 e −s + e −2s
,
Re s > −∞
s
© ª
The ROC is the entire s-plane with the exception of Re s → −∞ since the pole at s = 0 is canceled when
¡
¢
1
the two terms are added. This is evident from the fact that X 0 + j 0 = 0 and that x (t ) is a finite length
signal.
X (s) =
x (t)
1
2
1
c.
and
t
Using the relationships
©
ª 1
L u (t ) = ,
s
©
ª e −s
L u (t − 1) =
,
s
© ª
Re s > 0
© ª
Re s > 0
the transform is found through the use of the linearity property as
X (s) =
1 − 2 e −s
,
s
© ª
Re s > 0
The ROC is the same as the ROCs of the two individual transforms that were added. Note that the pole
at s = 0 is not canceled in this case (the numerator of X (s) is nonzero for s = 0), and the signal x (t ) is of
infinite length.
x (t)
1
1
t
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
218
7.13.
a. The signal x (t ) can be written as
x (t ) = u (t ) − 2 u (t − 1) + u (t − 2)
which leads to the transform
X (s) =
b.
¢
1¡
1 − 2 e −s + e −2s ,
s
© ª
Re s > −∞
An expression for x (t ) using unit step functions is
x (t ) = u (t ) − 1.5 u (t − 1) + u (t − 2) − 0.5 u (t − 3)
It leads to the transform
X (s) =
¢
1¡
1 − 1.5 e −s + e −2s − 0.5 e −3s ,
s
© ª
Re s > −∞
7.14.
The Laplace transform of the unit ramp function r (t ) = t u (t ) was found in Example 7-18 as
©
ª 1
R (s) = L r (t ) = 2 ,
s
© ª
Re s > 0
©
ª 1
U (s) = L u (t ) = ,
s
© ª
Re s > 0
The transform of the unit step signal is
Writing a signal using these two functions allows computation of the Laplace transform in terms of the
corresponding transforms.
a.
Signal:
Transform:
x (t ) = r (t ) − r (t − 1) − r (t − 2) + r (t − 3)
X (s) =
¢
1 ¡
1 − e −s − e −2s + e −3s
2
s
b.
Signal:
Transform:
x (t ) = r (t ) − u (t − 1) − 2 r (t − 1) + u (t − 2) + r (t − 2)
X (s) =
¢
¢ 1¡
1 ¡
− e −s + e −2s
1 − 2 e −s + e −2s +
2
s
s
Signal:
x (t ) = r (t ) − 2 r (t − 1) + r (t − 2)
c.
Transform:
X (s) =
¢
1 ¡
1 − 2 e −s + e −2s
2
s
219
7.15.
a. Using Euler’s formula the signal x (t ) can be written as
£
¤
x (t ) = sin (πt ) u (t ) − u (t − 2)
´£
¤
1 ³ j πt
e
− e − j πt u (t ) − u (t − 2)
=
2j
=
1 j πt
1 − j πt
1 j πt
1 − j πt
e u (t ) −
e u (t − 2) −
e
u (t )
e
u (t − 2)
2j
2j
2j
2j
The transforms needed are
n
o
L e j πt u (t ) =
1
s − jπ
n
o
n
o
e −2s
L e j πt u (t − 2) = e j 2π L e j π(t −2) u (t − 2) =
s − jπ
n
o
1
L e − j πt u (t ) =
s + jπ
n
o
n
o
e −2s
L e − j πt u (t − 2) = e − j 2π L e − j π(t −2) u (t − 2) =
s + jπ
In the relationships above we have recognized that e ± j 2π = 0. The transform X (s) is
¡
¢
·
¸
π 1 − e −2s
1 1 − e −2s 1 − e −2s
X (s) =
=
−
2j
s − jπ
s + jπ
s 2 + π2
b.
Let us write x (t ) as
x (t ) = sin (πt ) u (t ) − sin (π (t − 2)) u (t − 2)
The transforms needed are
and
©
ª
L sin (πt ) u (t ) =
π
s 2 + π2
©
ª e −2s π
L sin (π (t − 2)) u (t − 2) = 2
s + π2
Using the linearity of the Laplace transform we obtain
¢
¡
π 1 − e −2s
π
e −2s π
X (s) = 2
−
=
s + π2 s 2 + π2
s 2 + π2
7.16.
a. Given that
1
s −a
through the use of the time differentiation property we get
©
ª
L e at u (t ) =
©
ª
L t e at u (t ) =
1
(s − a)2
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
220
Setting a = −2 yields
L
b.
−1
½
1
(s + 2)2
¾
= t e −2t u (t )
From the transform table in Appendix B.3
©
ª
L e at cos (ω0 t ) u (t ) =
Setting a = −3 and ω0 =
p
2 we have
L
c.
−1
½
s +3
(s + 3)2 + 2
¾
s −a
(s − a)2 + ω20
= e −3t cos
³p ´
2t u (t )
From the transform table in Appendix B.3
©
ª
L e at sin (ω0 t ) u (t ) =
Setting a = −1 and ω0 =
p
6 we have
n
L e
and
L −1
½
−t
ω0
(s − a)2 + ω20
³p ´
o
sin 6t u (t ) =
p
6
(s + 1)2 + 6
¾
³p ´
1 −t
=
e
sin
6t u (t )
p
(s + 1)2 + 6
6
1
7.17.
a. The transform X (s) has the partial fraction expansion
X (s) =
with residues
and
The partial fraction expansion for X (s) is
k2
k1
+
s +1 s +2
¯
1 ¯¯
=1
k1 =
s + 2 ¯s=−1
¯
1 ¯¯
k2 =
= −1
s + 1 ¯s=−2
X (s) =
1
1
−
s +1 s +2
Since the ROC is Re {s} > −1 both terms in the partial fraction expansion correspond to causal signal
components. Therefore
x (t ) = e −t u (t ) − e −2t u (t )
221
b.
The partial fraction expansion is the same as in part (a). Since the ROC is Re {s} < −2 both terms
in the partial fraction expansion correspond to anti-causal signal components. Therefore
x (t ) = −e −t u (−t ) + e −2t u (−t )
c.
The partial fraction expansion is the same as in part (a). Since the ROC is −2 < Re {s} < −1 the first
term in the partial fraction expansion corresponds to an anti-causal signal component, and teh second
term corresponds to a causal signal component, that is
X (s) =
1
1
−
s
+
1
s
+
| {z }
| {z2}
anti-causal
Therefore
causal
x (t ) = −e −t u (−t ) − e −2t u (t )
d.
The transform X (s) has the partial fraction expansion
X (s) =
with residues
and
k1
k2
k3
+
+
s +1 s +2 s +3
¯
(s − 1) (s − 2) ¯¯
=3
k1 =
(s + 2) (s + 3) ¯s=−1
¯
(s − 1) (s − 2) ¯¯
k2 =
= −12
(s + 1) (s + 3) ¯s=−2
¯
(s − 1) (s − 2) ¯¯
= 10
k3 =
(s + 1) (s + 2) ¯s=−3
The partial fraction expansion for X (s) is
X (s) =
3
12
10
−
+
s +1 s +2 s +3
Using the causal signal component for each term we have
x (t ) = 3 e −t u (t ) − 12 e −2t u (t ) + 10 e −3t u (t )
e.
The transform X (s) has the partial fraction expansion
X (s) =
with residues
k2
k3
k1
+
+
s −1 s +2 s +3
¯
1
(s + 1) (s − 2) ¯¯
=−
¯
6
(s + 2) (s + 3) s=1
¯
4
(s + 1) (s − 2) ¯¯
k2 =
=−
¯
3
(s − 1) (s + 3)
k1 =
s=−2
and
¯
5
(s + 1) (s − 2) ¯¯
=
k3 =
¯
(s − 1) (s + 2) s=−3 2
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
222
The partial fraction expansion for X (s) is
1/6
4/3
5/2
X (s) = −
−
+
| s{z− 1} | s + 2{z s + 3}
anti-causal
causal
Since the Fourier transform of x (t ) exists, the ROC of X (s) must include the j ω-axis of the s-plane.
Therefore, the term with the pole at s = 1 must correspond to an anti-causal signal component, and the
other two terms must correspond to causal signal components.
x (t ) =
1 t
4
5
e u (−t ) − e −2t u (t ) + e −3t u (t )
6
3
2
7.18.
a. Since the numerator order is the same as the denominator order, an constant term must be
extracted from X (s) to write it in the form
X (s) = 2 −
1
s +2
The ROC indicates a causal signal. Therefore
x (t ) = 2 δ (t ) − e −2t u (t )
b.
The transform X (s) has the partial fraction expansion
X (s) =
with residues
and
k2
k1
+
s +2 s +3
¯
s (s + 1) ¯¯
=2
k1 =
s + 3 ¯s=−2
¯
s (s + 1) ¯¯
k2 =
= −6
s + 2 ¯s=−3
The partial fraction expansion for X (s) is
X (s) =
6
2
−
s +2 s +3
The ROC for the transform is Re {s} < −3, and it indicates an anti-causal signal. Therefore
x (t ) = −2 e −2t u (−t ) + 6 e −3t u (−t )
c.
The transform X (s) has the partial fraction expansion
X (s) =
with residues
k1 =
k2
k1
+
s + j2 s − j2
¯
5
1
s + 5 ¯¯
= +j
s − j 2 ¯s=− j 2 2
4
223
and
k 2 = k 1∗ =
1
5
−j
2
4
The partial fraction expansion for X (s) is
¶
¶
µ
µ
1
1
5
5
1
1
+j
+ −j
X (s) =
2
4 s + j2
2
4 s − j2
The ROC indicates a causal signal. Therefore
¶
µ
¶
µ
1
5 − j 2t
5 j 2t
1
e
u (t ) + − j
e u (t )
x (t ) = + j
2
4
2
4
´
´
1³
5 ³ − j 2t
= e − j 2t + e j 2t u (t ) + j
e
− e j 2t u (t )
2
4
= cos (2t ) u (t ) +
d.
Factored form of X (s) is
and leads to the partial fraction form
X (s) = ¡
X (s) =
The residues are
and
5
sin (2t ) u (t )
2
s +6
¢¡
¢
s +1+ j2 s +1− j2
k1
k2
+
s +1+ j2 s +1− j2
¯
s + 6 ¯¯
5
1
k1 =
= +j
¯
s + 1 − j 2 s=−1− j 2 2
4
k 2 = k 1∗ =
5
1
−j
2
4
The ROC indicates a causal signal. Therefore
µ
¶
¶
µ
1
1
5 (−1− j 2)t
5 (−1+ j 2)t
x (t ) = + j
e
e
u (t ) + − j
u (t )
2
4
2
4
³
´
³
´
5
1
= e −t e − j 2t + e j 2t u (t ) + j e −t e − j 2t − e j 2t u (t )
2
4
=e −t cos (2t ) u (t ) +
e.
5 −t
e sin (2t ) u (t )
2
Partial fraction form of X (s) is
X (s) =
k 1,1
s +1
The residues are
k 1,2 =
+
k 1,2
2
(s + 1)
+
k2
s +2
¯
s (s − 1) ¯¯
=2
s + 2 ¯s=−1
224
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
¯
s (s − 1) ¯¯
=6
k2 =
¯
(s + 1)2 s=−2
and
¯
·
¸¯
d s (s − 1) ¯¯
s 2 + 4s − 2 ¯¯
k 1,1 =
=
= −5
¯
ds
s + 2 ¯s=−1
(s + 2)2 s=−1
The partial fraction expansion for X (s) is
X (s) = −
2
6
5
+
+
2
s + 1 (s + 1)
s +2
The ROC indicates a causal signal. Therefore
x (t ) = −5 e −t u (t ) + 2 t e −t u (t ) + 6 e −2t u (t )
7.19.
a. Pole-zero diagram is shown below:
jω
−1
1
|
{z
σ
}
−1 < Re {s} < 1
b.
Possible choices for the ROC are:
1.
2.
3.
4.
c.
© ª
Re s < −2
© ª
−2 < Re s < −1
© ª
−1 < Re s < 1
© ª
Re s > 1
The partial fraction form of X (s) is
X (s) =
The residues are
k1
k2
k3
+
+
s −1 s +1 s +2
¯
1
s (s − 3) ¯¯
=−
k1 =
3
(s + 1) (s + 2) ¯s=1
225
k2 =
and
k3 =
The partial fraction expansion for X (s) is
¯
s (s − 3) ¯¯
= −2
(s − 1) (s + 2) ¯s=−1
¯
10
s (s − 3) ¯¯
=
¯
3
(s − 1) (s + 1) s=−2
X (s) = −
1/3
2
10/3
−
+
s −1 s +1 s +2
The inverse transform is shown below for each choice of the ROC:
© ª
1
10 −2t
Re s < −2 : x (t ) = e t u (−t ) + 2 e −t u (−t ) −
e
u (−t )
3
3
© ª
10 −2t
1
e
u (t )
−2 < Re s < −1 : x (t ) = e t u (−t ) + 2 e −t u (−t ) +
3
3
© ª
10 −2t
1
e
u (t )
−1 < Re s < 1 : x (t ) = e t u (−t ) − 2 e −t u (t ) +
3
3
© ª
1
10 −2t
Re s > 1 : x (t ) = − e t u (t ) − 2 e −t u (t ) +
e
u (t )
3
3
The inverse transform x (t ) is square integrable if the ROC includes the j ω-axis. The only choice for ROC
that satisfies this condition is −1 < Re {s} < 1.
7.20.
a. The transform can be written as
X (s) = −
2
1
+
s +2 s +3
Possible choices for the ROC and the corresponding signals x (t ) are as follows:
© ª
Re s < −3 : x (t ) = e −2t u (−t ) − 2 e −3t u (−t )
© ª
−3 < Re s < −2 : x (t ) = e −2t u (−t ) + 2 e −3t u (t )
© ª
Re s > −2 : x (t ) = − e −2t u (t ) + 2 e −3t u (t )
b.
The transform can be written as
X (s) =
1
2
2
−
−
2
s + 2 (s + 2)
s +3
Possible choices for the ROC and the corresponding signals x (t ) are as follows:
© ª
Re s < −3 : x (t ) = −2 e −2t u (−t ) + t e −2t u (−t ) + 2 e −3t u (−t )
© ª
−3 < Re s < −2 : x (t ) = −2 e −2t u (−t ) + t e −2t u (−t ) − 2 e −3t u (t )
© ª
Re s > −2 : x (t ) = 2 e −2t u (t ) − t e −2t u (t ) − 2 e −3t u (t )
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
226
c.
The transform can be written as
X (s) = 1 −
12
7
+
s + 3 (s + 3)2
Possible choices for the ROC and the corresponding signals x (t ) are as follows:
© ª
Re s < −3 : x (t ) = δ (t ) + 7 e −3t u (−t ) − 12 t e −3t u (−t )
© ª
Re s > −3 : x (t ) = δ (t ) − 7 e −3t u (t ) + 12 t e −3t u (t )
d.
The transform can be written as
X (s) =
1
1
2−j3
s +2+ j3
+
1
1
2+j3
s +2− j3
Possible choices for the ROC and the corresponding signals x (t ) are as follows:
¶
¶
µ
µ
© ª
1
1
1 (−2− j 3)t
1 (−2+ j 3)t
e
e
Re s < −2 :
x (t ) = − − j
u (−t ) − + j
u (−t )
2
3
2
3
2
= −e −2t cos (3t ) u (−t ) + e −2t sin (3t ) u (−t )
3 µ
¶
¶
µ
© ª
1
1 (−2− j 3)t
1 (−2+ j 3)t
1
e
e
u (t ) + + j
u (t )
−j
Re s > −2 :
x (t ) =
2
3
2
3
2
= e −2t cos (3t ) u (t ) − e −2t sin (3t ) u (t )
3
7.21.
a. Let the transform X 1 (s) be defined as
X 1 (s) =
1
s +1
so that
x 1 (t ) = e −t u (t )
Since
we conclude that
¡
¢
X (s) = 1 − e −s X 1 (s)
x (t ) =x 1 (t ) − x 1 (t − 1)
=e −t u (t ) − e −(t −1) u (t − 1)
b.
Let the transform X 1 (s) be defined as
X 1 (s) =
s
1
= 1−
s +1
s +1
so that
x 1 (t ) = δ (t ) − e −t u (t )
227
Since
we conclude that
¡
¢
X (s) = 1 − e −s X 1 (s)
x (t ) =x 1 (t ) − x 1 (t − 1)
=δ (t ) − e −t u (t ) − δ (t − 1) + e −(t −1) u (t − 1)
c.
Let the transform X 1 (s) be defined as
X 1 (s) =
1
1
1
= −
s (s + 1) s s + 1
so that
x 1 (t ) = u (t ) − e −t u (t )
Since
we conclude that
¡
¢
X (s) = 1 − e −s X 1 (s)
x (t ) =x 1 (t ) − x 1 (t − 1)
=u (t ) − e −t u (t ) − u (t − 1) + e −(t −1) u (t − 1)
c.
Let the transform X 1 (s) be defined as
X 1 (s) =
1
s
so that
x 1 (t ) = u (t )
Since
we conclude that
¢
¡
X (s) = 1 − e −s + e −2s − e −3s X 1 (s)
x (t ) = u (t ) − u (t − 1) + u (t − 2) − u (t − 3)
7.22. Taking the Laplace transform of each side yields the following:
a.
3s Y (s) + 2 Y (s) = 7 X (s)
⇒
H (s) =
7
Y (s)
=
X (s) 3 s + 2
b.
s +1
Y (s)
= 2
X (s) s + 4s + 3
s 2 Y (s) + 4s Y (s) + 3 Y (s) = s X (s) + X (s)
⇒
H (s) =
s 2 Y (s) + 4 Y (s) = s 2 X (s) + s X (s) + 3 X (s)
⇒
H (s) =
c.
Y (s) s 2 + s + 3
=
X (s)
s2 + 4
228
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
7.23.
a.
H (s) =
Y (s)
1
=
s + 4 X (s)
⇒
s Y (s) + 4 Y (s) = X (s)
d y (t )
+ 4 y (t ) = x (t )
dt
b.
H (s) =
s
Y (s)
=
s + 4 X (s)
⇒
s Y (s) + 4 Y (s) = X (s)
d y (t )
+ 4 y (t ) = x (t )
dt
c.
H (s) =
s +1
s2 + 5 s + 6
=
Y (s)
X (s)
s 2 Y (s) + 5s Y (s) + 6 Y (s) = s X (s) + X (s)
d 2 y (t )
d x (t )
d y (t )
+ 6 y (t ) =
+ x (t )
+5
2
dt
dt
dt
d.
H (s) =
s +1
s2 + 5 s + 6
=
Y (s)
X (s)
s 2 Y (s) + 5s Y (s) + 6 Y (s) = s X (s) + X (s)
d 2 y (t )
d x (t )
d y (t )
+ 6 y (t ) =
+ x (t )
+5
2
dt
dt
dt
e.
H (s) =
Y (s)
s2 − s
=
s 3 + 5 s 2 + 8 s + 6 X (s)
s 3 Y (s) + 5s 2 Y (s) + 8s Y (s) + 6 Y (s) = s 2 X (s) − s X (s)
d 3 y (t )
d 2 y (t )
d y (t )
d 2 x (t ) d x (t )
+
5
+
8
+
6
y
=
−
(t
)
dt3
dt2
dt
dt2
dt
f.
H (s) =
s2 + s − 2
Y (s)
=
3
2
s + 7 s + 19 s + 13 X (s)
s 3 Y (s) + 7s 2 Y (s) + 19s Y (s) + 13 Y (s) = s 2 X (s) + s X (s) − 2 X (s)
d 2 y (t )
d y (t )
d 2 x (t ) d x (t )
d 3 y (t )
+
7
+
19
+
13
y
=
+
− 2 x (t )
(t
)
dt3
dt2
dt
dt2
dt
229
7.24.
Let the system function be H (s). For a unit-step input signal we have X (s) = 1/s, and the Laplace
transform of the unit-step response of the system is
Yu (s) =
H (s)
s
The solution method is based on determining H (s) from this relationship, and then finding a
differential equation from the knowledge of H (s).
a.
Yu (s) =
1
s +1
⇒
H (s) = s Yu (s) =
s
s +1
(s + 1) Y (s) = s X (s)
d x (t )
d y (t )
+ y (t ) =
dt
dt
b.
Yu (s) =
1
1
1
−
=
s s + 1 s (s + 1)
H (s) = s Yu (s) =
⇒
1
s +1
(s + 1) Y (s) = X (s)
d y (t )
+ y (t ) = x (t )
dt
c.
Yu (s) =
1
1
1
−
=
s + 1 s + 2 (s + 1) (s + 2)
H (s) = s Yu (s) =
⇒
s
(s + 1) (s + 2)
¡ 2
¢
s + 3 s + 2 Y (s) = s X (s)
d 2 y (t )
d y (t )
d x (t )
+3
+ 2 y (t ) =
2
dt
dt
dt
d.
Yu (s) =
1
1
2
2s 2 + 3s + 2
−
+
= 3
s s + 1 s + 2 s + 3s 2 + 2s
⇒
H (s) = s Yu (s) =
2s 2 + 3s + 2
s 2 + 3s + 2
¡ 2
¢
¡
¢
s + 3s + 2 Y (s) = 2s 2 + 3s + 2 X (s)
d y (t )
d 2 x (t )
d x (t )
d 2 y (t )
+
3
+
2
y
=
2
+3
+ 2 x (t )
(t
)
dt2
dt
dt2
dt
e.
Yu (s) =
1 0.3 (s + 1)
0.7 s 2 + 1.7 s + 5
¡
¢
−
=
s (s + 1)2 + 4
s s2 + 2 s + 5
⇒
H (s) = s Yu (s) =
0.7 s 2 + 1.7 s + 5
s2 + 2 s + 5
230
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
¡ 2
¢
¡
¢
s + 2 s + 5 Y (s) = 0.7 s 2 + 1.7 s + 5 X (s)
d y (t )
d 2 x (t )
d x (t )
d 2 y (t )
+
2
+
5
y
=
0.7
+ 1.7
+ 5 x (t )
(t
)
2
2
dt
dt
dt
dt
7.25.
Let the system function be H (s). For a unit-ramp input signal we have X (s) = 1/s 2 , and the Laplace
transform of the unit-step response of the system is
Yu (s) =
H (s)
s2
The solution method is based on determining H (s) from this relationship, and then finding a
differential equation from the knowledge of H (s).
a.
Yr (s) =
1
1
1
−
=
s s + 1 s (s + 1)
⇒
H (s) = s 2 Yr (s) =
s
Y (s)
=
s + 1 X (s)
(s + 1) Y (s) = s X (s)
d x (t )
d y (t )
+ y (t ) =
dt
dt
b.
Yr (s) =
2
4
−s 2 + s + 4
1
+
−
=
s + 1 s + 2 s + 3 (s + 1) (s + 2) (s + 3)
H (s) = s 2 Yr (s) =
−s 4 + s 3 + 4 s 2
Y (s)
=
3
2
s + 6 s + 11 s + 6 X (s)
¢
¡ 3
¢
¡
s + 6 s 2 + 11 s + 6 Y (s) = −s 4 + s 3 + 4 s 2 X (s)
d 3 y (t )
d 4 x (t ) d 3 x (t )
d 2 y (t )
d y (t )
d 2 x (t )
+
6
y
=
−
+
6
+
11
+
+
4
(t
)
dt3
dt2
dt
dt4
dt3
dt2
c.
Yr (s) =
0.7 s 2 + 1.7 s + 5
1 0.3 (s + 1)
¡
¢
−
=
s (s + 1)2 + 4
s s2 + 2 s + 5
H (s) = s 2 Yr (s) =
0.7 s 3 + 1.7 s 2 + 5 s Y (s)
=
s2 + 2 s + 5
X (s)
¡ 2
¢
¡
¢
s + 2 s + 5 Y (s) = 0.7 s 3 + 1.7 s 2 + 5 s X (s)
231
d y (t )
d 3 x (t )
d 2 x (t )
d x (t )
d 2 y (t )
+
2
+
5
y
=
0.7
+
1.7
+5
(t
)
dt2
dt
dt3
dt2
dt
7.26.
The response of the CTLTI system with system function H (s) to the signal x (t ) = e s0 t is
©
ª
Sys e s0 t = e s0 t H (s 0 )
a.
s 0 = −0.5 ,
H (−0.5) =3.3333 + j 0
=3.3333 e j 0
y (t ) = 3.3333 e −0.5t
b.
¡
¢
H −0.5 + j 2 = −0.0047 − j 0.6212
s 0 = −0.5 + j 2 ,
=0.6212 e − j 1.5783
y (t ) = 0.6212 e −0.5t − j 1.5783
c.
s0 = j 3 ,
¡ ¢
H j 3 =0.0462 − j 0.3692
=0.3721 e − j 1.4464
y (t ) = 0.3721 e j (3t −1.4464)
d.
The signal x (t ) is in the form
x (t ) = e s0 t + e s1 t
with s 0 = − j 3 and s 1 = j 3.
s0 = − j 3 ,
¡
¢
H − j 3 =0.0462 + j 0.3692
=0.3721 e j 1.4464
s1 = j 3 ,
¡ ¢
H j 3 =0.0462 − j 0.3692
=0.3721 e − j 1.4464
y (t ) =0.3721 e j (−3t +1.4464) + 0.3721 e j (3t −1.4464)
=0.7442 cos (3t − 1.4464)
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
232
e.
The signal x (t ) is in the form
x (t ) = e s0 t + e s1 t
with s 0 = j 2 and s 1 = j 3.
s0 = j 2 ,
¡ ¢
H j 2 =0.1500 − j 0.5500
=0.5701 e − j 1.3045
s1 = j 3 ,
¡ ¢
H j 3 =0.0462 − j 0.3692
=0.3721 e − j 1.4464
y (t ) = 0.5701 e j (2t −1.3045) + 0.3721 e j (3t −1.4464)
7.27.
Using the Laplace transform definition, H (s 0 ) is found as
ˆ ∞
h (t ) e −s0 t d t
·ˆ ∞
h (t ) e −s0 t d t
H (s 0 ) =
−∞
Conjugating both sides yields
H ∗ (s 0 ) =
=
−∞
ˆ ∞
¸∗
∗
h ∗ (t ) e −s0 t d t
−∞
Since h (t ) is real-valued we have h ∗ (t ) = h (t ) and
H ∗ (s 0 ) =
Therefore
ˆ ∞
−∞
¡ ¢
∗
h (t ) e −s0 t d t = H s 0∗
¡ ¢
H s 0∗ = H ∗ (s 0 ) = H0 e − j Θ0
7.28.
a.
s0 = j 2 ,
¡ ¢
H j 2 = − 0.1379 + 0.3448
=0.3714 e j 1.9513
y (t ) = 0.3714 cos (2t + 1.9513)
b.
s 0 = −0.5 + j 2 ,
¡
¢
H −0.5 + j 2 = − 0.3479 + 0.2879
=0.4516 e j 2.4503
233
y (t ) = 0.4516 e −0.5t cos (2t + 2.4503)
c.
¡
¢
H −1 + j 2 = − 0.6154 + j 0.0769
s 0 = −1 + j 2 ,
=0.6202 e j 3.0172
y (t ) = 0.6202 e −t sin (2t + 3.0172)
7.29.
The system function is
a.
©
ª
H (s) = L h (t ) =
1
s +1
Laplace transform of the input signal is
s
©
ª
X (s) = L cos (2t ) u (t ) =
s2 + 4
s
¡
¡
Laplace transform of the output signal is found as
Y (s) = H (s) X (s) =
(s + 1)
Partial fraction expansion of Y (s) is
Y (s) =
with the residues
¢=
s2 + 4
s
¢¡
¢
(s + 1) s + j 2 s − j 2
k2
k3
k1
+
+
s +1 s + j2 s − j2
s ¯¯
= −0.2
¯
s 2 + 4 s=−1
¯
¯
s
¡
¢ ¯¯
= 0.1 + j 0.2
+ 1) s − j 2
k1 =
k2 =
and
(s
s=− j 2
k 3 = k 2∗ = 0.1 − j 0.2
The output signal is found as
¡
¢
¡
¢
y (t ) = − 0.2 e −t u (t ) + 0.1 + j 0.2 e − j 2t u (t ) + 0.1 − j 0.2 e j 2t u (t )
³
´
³
´
= − 0.2 e −t u (t ) + 0.1 e − j 2t + e j 2t u (t ) + j 0.2 e − j 2t − e j 2t u (t )
= −0.2 e −t u (t ) + 0.2 cos (2t ) u (t ) + 0.4 sin (2t ) u (t )
{z
} |
{z
}
|
transient
steady-state
The transient component of the response is
y t (t ) = −0.2 e −t u (t )
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
234
and the steady-state component is
y ss (t ) = 0.2 cos (2t ) + 0.4 sin (2t )
b.
Laplace transform of the input signal is
©
ª
X (s) = L sin (3t ) u (t ) =
3
s2 + 9
Laplace transform of the output signal is found as
Y (s) = H (s) X (s) =
Partial fraction expansion of Y (s) is
3
3
¡
¢=
¡
¢¡
¢
2
(s + 1) s + 9
(s + 1) s + j 3 s − j 3
Y (s) =
with the residues
k2
k3
k1
+
+
s +1 s + j3 s − j3
¯
3 ¯¯
¢¯
k1 = ¡ 2
= 0.3
s + 9 s=−1
¯
¯
3
¡
¢ ¯¯
k2 =
= −0.15 + j 0.05
(s + 1) s − j 3 s=− j 3
and
k 3 = k 2∗ = −0.15 − j 0.05
The output signal is found as
¡
¢
¡
¢
y (t ) =0.3 e −t u (t ) + −0.15 + j 0.05 e − j 3t u (t ) + −0.15 − j 0.05 e j 3t u (t )
³
´
³
´
=0.3 e −t u (t ) − 0.15 e − j 3t + e + j 3t u (t ) + j 0.05 e − j 3t − e j 3t u (t )
= 0.3 e −t u (t ) − 0.3 cos (3t ) u (t ) + 0.1 sin (3t ) u (t )
{z
} |
{z
}
|
transient
steady-state
The transient component of the response is
y t (t ) = 0.3 e −t u (t )
and the steady-state component is
y ss (t ) = −0.3 cos (3t ) + 0.1 sin (3t )
7.30.
a.
H (s) = 1 −
2
s +3
⇒
h (t ) = δ (t ) − 2 e −3t u (t )
235
b.
H (s) = 1 +
Let
−3s − 1
s 2 + 3s + 2
= 1+
−3s − 1
(s + 1) (s + 2)
H1 (s) =
−3s − 1
(s + 1) (s + 2)
H1 (s) =
k2
k1
+
s +1 s +2
so that
The residues are
k1 =
and
k2 =
The system function is
¯
−3s − 1 ¯¯
=2
s + 2 ¯s=−1
¯
−3s − 1 ¯¯
= −5
s + 1 ¯s=−2
H (s) = 1 + H1 (s) = 1 +
and the impulse response is
5
2
−
s +1 s +2
h (t ) = δ (t ) + 2 e −t u (t ) − 5 e −2t u (t )
c.
H (s) =
=
The residues are
s2 − 1
¢¡
¢
(s + 2) s + 1 + j s + 1 − j
¡
k2
k3
k1
+
+
s +2 s +1+ j s +1− j
¯
s 2 − 1 ¯¯
k1 = 2
= 1.5
s + s + 2 ¯s=−2
¯
¯
s2 − 1
¡
¢ ¯¯
k2 =
= −0.25 − j 0.75
(s + 2) s + 1 − j s=−1− j
and
k 3 = k 2∗ = −0.25 + j 0.75
The impulse response is
¡
¢
¡
¢
h (t ) =1.5 e −2t u (t ) + −0.25 − j 0.75 e (−1− j )t u (t ) + −0.25 + j 0.75 e (−1+ j )t u (t )
´
³
´
³
=1.5 e −2t u (t ) − 0.25 e −t e − j t + e j t u (t ) − j 0.75 e −t e − j t − e j t u (t )
=1.5 e −2t u (t ) − 0.5 e −t cos (t ) u (t ) − 1.5 e −t sin (t ) u (t )
d.
H (s) =
−4
2
(s + 1)
+
1
s +1
236
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
The corresponding impulse response is
h (t ) = −4t e −t u (t ) + e −t u (t ) = (−4t + 1) e −t u (t )
7.31.
The system function is
H (s) =
The input signal can be written as
1
s +1
1 j 2t 1 − j 2t
e + e
2
2
Let s 0 = j 2. The system function evaluated at s = s 0 = j 2 is
x (t ) =
H (s 0 ) =
1
= 0.2 − j 0.4
j2+1
The system function evaluated at s = s 0∗ = − j 2 is
¡ ¢
H s 0∗ =
The response of the system is
1
= 0.2 + j 0.4
−j2+1
©
ª
y (t ) = Sys cos (2t )
n
o 1
n
o
1
= Sys e j 2t + Sys e − j 2t
2
2
¢ j 2t 1 ¡
¢
1¡
= 0.2 − j 0.4 e +
0.2 + j 0.4 e − j 2t
2
2
³
´
´
¢³
1
1¡
= (0.2) e j 2t + e − j 2t −
j 0.4 e j 2t − e − j 2t
2
2
=0.2 cos (2t ) + 0.4 sin (2t )
The result found matches the steady-state component of the result found in part (a) of Problem 7.29.
7.32.
The system function is found as
©
ª
H (s) = L e −t u (t ) − e −t u (t − 2)
©
ª
©
ª
= L e −t u (t ) − L e −t u (t − 2)
ª
©
ª
©
= L e −t u (t ) − e −2 L e −(t −2) u (t − 2)
£
¤
= 1 − e −2 (s+1)
1
s +1
Laplace transform of the input signal is
X (s) =
s + 0.5
2
(s + 0.5) + 4
=
s + 0.5
s 2 + s + 4.25
237
and the Laplace transform of the output signal is
Y (s) =H (s) X (s)
¤
£
= 1 − e −2 (s+1)
Let
Y1 (s) =
so that
s + 0.5
¡
¢
(s + 1) s 2 + s + 4.25
s + 0.5
¡
¢
(s + 1) s 2 + s + 4.25
£
¤
Y (s) = 1 − e −2 (s+1) Y1 (s)
The transform Y1 (s) can be expanded into partial fractions as
Y1 (s) =
with the residues
k2
k3
k1
+
+
s + 1 s + 0.5 + j 2 s + 0.5 − j 2
¯
s + 0.5 ¯¯
k1 = 2
= −0.1176
s + s + 4.25 ¯s=−1
¯
¯
s + 0.5
¡
¢ ¯¯
= 0.0588 + j 0.2353
k2 =
(s + 1) s + 0.5 − j 2
s=−0.5− j 2
and
k 3 = k 2∗ = 0.0588 − j 0.2353
Using the residues, y 1 (t ) is found as
¡
¢
¡
¢
y 1 (t ) = − 0.1176 e −t u (t ) + 0.0588 + j 0.2353 e (−0.5− j 2)t u (t ) + 0.0588 − j 0.2353 e (−0.5+ j 2)t u (t )
³
´
³
´
= − 0.1176 e −t u (t ) + 0.0588 e −0.5t e − j 2t + e j 2t u (t ) + j 0.2353 e −0.5t e − j 2t − e j 2t u (t )
= − 0.1176 e −t u (t ) + 0.1176 e −0.5t cos (2t ) u (t ) + 0.4706 e −0.5t sin (2t ) u (t )
The signal y (t ) is
y (t ) =y 1 (t ) − e −2 y 1 (t − 2)
£
¤
= −0.1176 e −t + 0.1176 e −0.5t cos (2t ) + 0.4706 e −0.5t sin (2t ) u (t )
£
¤
+ 0.1176 e −t − 0.1176 e −0.5t −1 cos (2t − 4) − 0.4706 e −0.5t −1 sin (2t − 4) u (t − 2)
7.33.
a.
jω
The system function has a zero at s = −1 and
poles at s = −2, −3.
σ
−3
−2
−1
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
238
b.
jω
H (s) =
¡
s−j
¢¡
s+j
¢
j1
(s + 2) (s + 3)
σ
The system function has zeros at s = ∓ j 1 and
poles at s = −2, −3.
−3
−2
−j1
c.
jω
j3
H (s) =
(s − 1) (s + 1)
¢¡
¢
(s + 2) s + 2 + j 3 s + 2 − j 3
¡
σ
The system function has zeros at s = ∓1 and
poles at s = −2 and s = −2 ± j 3.
−2
−1
1
−j3
7.34.
a.
H (s) = K
¡
s+j
¢¡
s−j
(s + 1) (s + 2)
µ ¶
1
=1
H (0) = K
2
H (s) =
b.
H (s) = K ¡
¢
=K
s2 + 1
s2 + 3 s + 2
⇒
K =2
¡
¢
2 s2 + 1
s2 + 3 s + 2
s 2 − 2.25
(s − 1.5) (s + 1.5)
¢¡
¢ =K 2
s + 2 s + 3.25
s + 1 − j 1.5 s + 1 + j 1.5
µ
¶
−2.25
13
=1
⇒
K =−
3.25
9
µ ¶
s 2 − 2.25
13
H (s) = −
9 s 2 + 2 s + 3.25
H (0) = K
239
c.
H (s) = K
¡
s−j
¢¡
H (s) =
H (s) = K
¢
(s − 1) (s + 1)
H (0) = −K = 1
d.
s+j
=K
s2 + 1
s2 − 1
⇒
K = −1
¡
¢
− s2 + 1
s2 − 1
s − 1.5
s − 1.5
¢¡
¢ =K 3
2
s
+
3.5
s
+ 6.25 s + 4.875
(s + 1.5) s + 1 − j 1.5 s + 1 + j 1.5
¶
µ
−1.5
=1
⇒
K = −3.25
H (0) = K
4.875
−3.25 (s − 1.5)
H (s) = 3
s + 3.5 s 2 + 6.25 s + 4.875
¡
7.35.
a.
replacemen
H (ω) =
jω−1
jω+2
H (ω)
|H (ω)|
6
π
Phase (rad)
Magnitude
1
0.8
0.6
0.4
π
2
0
−
0.2
0
−10
−5
0
5
π
2
−π
−10
10
−5
ω (rad/s)
b.
0
5
10
5
10
ω (rad/s)
¡
replacemen
|H (ω)|
¢¡
¢
jω− j1 jω+ j1
¢¡
¢
H (ω) = ¡
jω+2 jω+3
6
H (ω)
π
Phase (rad)
Magnitude
1
0.8
0.6
0.4
0.2
0
−10
−5
0
ω (rad/s)
5
10
π
2
0
−
π
2
−π
−10
−5
0
ω (rad/s)
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
240
c.
¡
¢¡
¢
jω−1 jω+1
¢¡
¢
H (ω) = ¡
jω+1+ j2 jω+1− j2
|H (ω)|
6
Phase (rad)
Magnitude
1.5
1
0.5
0
−10
−5
0
10
5
π
2
0
−
π
2
−π
−10
−5
ω (rad/s)
3
2
jω
1
2 poles
0
System is stable.
−1
−2
−3
−3
−2
−1
0
σ
1
0
ω (rad/s)
7.36.
a.
−4
H (ω)
π
2
3
5
10
241
b.
3
2
jω
1
0
System is unstable.
−1
−2
−3
−4
−3
−2
−1
0
1
2
3
σ
c.
3
2
jω
1
0
System is stable.
−1
−2
−3
−4
−3
−2
−1
0
σ
1
2
3
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
242
d.
3
2
jω
1
0
System is unstable.
−1
−2
−3
−4
−3
−2
−1
0
1
2
3
σ
7.37.
For a system function to represent a system that is both causal and stable, the following conditions
must be met:
• The order of the numerator of the system function must not be greater than the order of the
denominator.
• All poles of the system function must be in the left half s-plane. We need Re {s} < 0 for all poles.
a.
The system function can be written as
H (s) =
s −1
(s + 1) (s + 2)
Poles are at s = −1 and s = −2. This could be the system function of a causal and stable system.
b.
The system function can be written as
H (s) = ¡
s (s + 1)
¢¡
¢
s +1− j2 s +1+ j2
Poles of the system function are at s = −1 ± j 2. This could be the system function of a causal and stable
system.
c.
The system function can be written as
H (s) =
s2 + 1
s (s + 2) (s + 3)
Poles of the system function are at s = 0, s = −2 and s = −3. This could not be the system function of a
causal and stable system since there is a pole on the j ω-axis.
243
d.
The system function can be written as
H (s) = ¡
s +3
¢¡
¢
s −3− j1 s −3− j1
Poles of the system function are at s = 3 + j 1 and s = 3 − j 1. This could not be the system function of a
causal and stable system since the poles are in the right half of the s-plane.
7.38.
a. Taking the Laplace transform of both sides of the differential equation yields
s 2 Y (s) = −2s Y (s) − a Y (s) + X (s)
which can be rearranged as
¡ 2
¢
s + 2 s + a Y (s) = X (s)
and the system function is
H (s) =
1
Y (s)
= 2
X (s) s + 2 s + a
b.
For the system to be causal and stable, the poles of the system function must be in the left half of
the s-plane. The poles of H (s) are found as
s2 + 2 s + a = 0
p
s 1,2 = −1 ± 1 − a
⇒
If a > 1, the poles are complex-valued, and they are at
s 1,2 = −1 ± j
p
a −1
If a < 1, both poles are real-valued. The system cannot be stable if at least one is in the right half s plane
or on the j ω-axis.
p
−1+ 1−a ≥ 0
⇒
a≤0
Therefore, for stability we need a > 0.
7.39.
a.
Evaluating H (s) for s = j ω yields
The phase of the system function is
¯
¯
H (ω) = H (s)¯
s= j ω
=
jω−2
jω+2
∡ H (ω) = tan(−ω/2) − tan(ω/2) = −2 tan(ω/2)
244
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
∡H (ω)
π
Phase (rad)
π
2
0
−
π
2
−π
−10
−5
0
10
5
ω (rad/s)
b.
The transform of the input signal is
X (s) =
1
s +1
and the transform of the corresponding output signal is
Y (s) = H (s) X (s) =
s −2
(s + 1) (s + 2)
Using partial fraction expansion on Y (s) we get
Y (s) =
The residues are
k2
k1
+
s +1 s +2
¯
s − 2 ¯¯
= −3
k1 =
s + 2 ¯s=−1
and
and the response of the system is
¯
s − 2 ¯¯
k2 =
=4
s + 1 ¯s=−2
y (t ) = −3 e −t u (t ) + 4 e −2t u (t )
c.
For the input signal we obtain
1
X (ω) =
1+ jω
⇒
and for the output signal we obtain
−2 + j ω
Y (ω) =
2 − ω2 + j 3ω
⇒
¯
¯
¯ X (ω)¯ =
r
1
1 + ω2
v
u
¯
¯ u
4 + ω2
¯Y (ω)¯ = t ¡
¢2
2 − ω2 + 9ω2
245
|X (ω)|
|Y (ω)|
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
−10
0
−5
5
0
−10
10
0
−5
ω (rad/s)
5
ω (rad/s)
7.40.
a. The frequency response of the system is computed by evaluating H (s) for s = j ω:
¯
−ω2 + 5 − j 2ω
H (ω) = H (s)¯s= j ω =
−ω2 + 5 + j 2ω
The phase characteristic is
∡ H (ω) = tan−1
µ
¶
µ
¶
µ
¶
−2ω
2ω
2ω
−1
−1
−
tan
=
−2
tan
5 − ω2
5 − ω2
5 − ω2
∡H (ω)
π
Phase (rad)
π
2
0
−
π
2
−π
−10
−5
0
5
10
ω (rad/s)
b.
The transform of the input signal is
X (s) =
1
s +1
and the transform of the output signal is
Y (s) =
s2 − 2 s + 5
s2 − 2 s + 5
¡
¢= 3
s + 3 s2 + 7 s + 5
(s + 1) s 2 + 2 s + 5
10
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
246
which can be expressed in partial fraction form as
Y (s) =
The residues are
k2
k3
k1
+
+
s +1 s +1+ j2 s +1− j2
¯
s 2 − 2 s + 5 ¯¯
=2
s 2 + 2 s + 5 ¯s=−1
¯
¯
1
s2 − 2 s + 5
¡
¢ ¯¯
= − − j1
k2 =
2
(s + 1) s + 1 − j 2 s=−1− j 2
k1 =
and
1
k 3 = k 2∗ = − + j 1
2
The output signal is
µ
µ
¶ ¡
¶ ¡
¢
¢
1
1
1
1
y (t ) =2 e −t u (t ) + − − j 1 e − 2 + j 1 t u (t ) + − + j 1 e − 2 − j 1 t u (t )
2
2
1
1
=2 e −t u (t ) − e − 2 t cos (t ) u (t ) − 2 e − 2 t sin (t ) u (t )
c.
For the input signal
1
X (ω) =
1+ jω
⇒
and for the output signal
5 − ω2 − j 2ω
¡
¢
Y (ω) =
5 − 3ω2 + j ω 7 − ω2
⇒
¯
¯
¯ X (ω)¯ =
r
1
1 + ω2
v
¡
¢2
u
¯
¯ u
5 − ω2 + 4ω2
¯Y (ω)¯ = t ¡
¢2
¡
¢2
5 − 3ω2 + ω2 7 − ω2
|X (ω)|
|Y (ω)|
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
−10
−5
0
5
10
0
−10
ω (rad/s)
−5
0
ω (rad/s)
7.41.
a. The inverse of the system is
H −1 (s) =
(s + 3)(s + 4)
(s + 1)(s − 2)
5
10
247
The inverse system is causal but not stable since there is a pole at s = 2 in the right half s-plane.
b.
Let us express H (s) as
¶
µ
(s + 1)(s − 2) s + 2
H (s) =
(s + 3)(s + 4) s + 2
Let
H1 (s) =
and
(s + 1) (s + 2)
(s + 3) (s + 4)
s −2
s +2
so that H (s) = H1 (s) H2 (s) and the subsystem H2 (s) is an all pass system.
H2 (s) =
c.
The inverse system H1−1 (s) is
H1−1 (s) =
It is causal and stable.
(s + 3) (s + 4)
(s + 1) (s + 2)
d.
H (s) H1−1 (s) =
s −2
= H2 (s)
s +2
7.42.
a. Taking the Laplace transform of each side we obtain
(s + 5) Y (s) = (2s + 3) X (s)
which leads to the system function
µ ¶
1 (s + 5)
Y (s)
=
H (s) =
X (s)
2 s + 3/2
The inverse system has the system function
Hi (s) = H −1 (s) =
2 (s + 3/2)
s +5
The inverse system is both causal and stable. Its differential equation is
d x (t )
d y (t )
+ 5 y (t ) = 2
+ 3 x (t )
dt
dt
b.
Taking the Laplace transform of each side we obtain
which leads to the system function
¡ 2
¢
¡
¢
s + 7s + 12 Y (s) = s 2 + s X (s)
H (s) =
s2 + s
s (s + 1)
Y (s)
= 2
=
X (s) s + 7s + 12 (s + 3) (s + 4)
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
248
The system function for the inverse system is
Hi (s) = H −1 (s) =
(s + 3) (s + 4)
s (s + 1)
Since there is a pole at s = 0 the inverse system cannot be stable.
c.
Taking the Laplace transform of each side we obtain
which leads to the system function
¡ 2
¢
s + 3 Y (s) = (s + 5) X (s)
H (s) =
The system function for the inverse system is
Y (s)
s +5
= 2
X (s) s + 3
Hi (s) = H −1 (s) =
s2 + 3
s +5
The numerator order for the inverse system is 2, and the denominator order is 1. Therefore the inverse
system cannot be causal.
d.
Taking the Laplace transform of each side we obtain
which leads to the system function
¡
¢
(s − 3) Y (s) = s 2 + 2s + 1 X (s)
H (s) =
Y (s) s 2 + 2s + 1
=
X (s)
s −3
The numerator order is 2, and the denominator order is 1. Furthermore, there is a pole in the right half
of the s-plane. The original system cannot be causal and stable. The system function for the inverse
system is
s −3
s −3
=
Hi (s) = H −1 (s) = 2
s + 2s + 1 (s + 1)2
The inverse system is causal and stable. Its differential equation is
d x (t )
d y (t )
d 2 y (t )
+ y (t ) =
− 3 x (t )
+2
2
dt
dt
dt
7.43. Asymptotically we have
20 log10 |H (ω)| = 20 log10 |H1 (ω)| + 20 log10 |H2 (ω)| + 20 log10 |H3 (ω)| + 20 log10 |H4 (ω)|
At ω = 5 rad/s
20 log10 |H1 (5)| = 20 log10 (5) = 13.98 dB
and
20 log10 |H2 (ω)| = 20 log10 |H3 (ω)| = 20 log10 |H4 (ω)| = 0
resulting in
20 log10 |H (5)| = 13.98 dB
249
At ω = 300 rad/s we have
20 log10 |H (300)| = 20 log10 (300/1) − 20 log10 (300/5) − 20 log10 (300/40) + 20 log10 (300/300) = −3.52 dB
7.44.
a.
1 + 3 s −1
1 + 2 s −1
¢
¢
¡
¡
1 + 2 s −1 Y (s) = 1 + 3 s −1 X (s)
H (s) =
1
X(s)
1
W (s)
=
H1 (s) =
X (s) 1 + 2 s −1
H2 (s) =
1/s
b
b
3
Y (s)
−2
Y (s)
= 1 + 3 s −1
W (s)
b.
H (s) =
s −1 + s −3
1 + 9 s −1 + 26 s −2 + 24 s −3
1
X (s)
1/s
b
1/s
1/s
b
1
b
Y (s)
−9
−26
−24
c.
H (s) =
0.5 s + 2.5
0.5 s −2 + 2.5 s −3
=
s 3 + 2 s + 3.5 1 + 2 s −2 + 3.5 s −3
0.5
X (s)
1/s
1/s
b
1/s
−2
−3.5
b
2.5
Y (s)
250
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
7.45.
a.
H (s) =
Let
H1 (s) =
so that
s +1
s +3
(s + 1) (s + 2)
¡
¢
(s + 3) s 2 + 2s + 2
and
H2 (s) =
s +2
s 2 + 2s + 2
H (s) = H1 (s) H2 (s)
1
X(s)
1
1/s
b
1
b
1/s
−3
1/s
b
b
2
Y (s)
−2
−2
b.
H (s) =
s2 + 1
(s + 2)(s + 3) (s + 4)
and
H2 (s) =
Let
H1 (s) =
so that
1
s +2
s2 + 1
s2 + 1
= 2
(s + 3) (s + 4) s + 7s + 12
H (s) = H1 (s) H2 (s)
1
X(s)
1/s
b
1
−2
b
1/s
1/s
b
b
1
−7
−12
c.
Let
H1 (s) =
so that
H (s) =
(s − 1) (s − 2)
(s + 1) (s + 2)
s −1
s +1
and
H2 (s) =
H (s) = H1 (s) H2 (s)
s −2
s +2
Y (s)
251
1
X(s)
b
1
1/s
−1
b
1/s
b
−1
b
−2
Y (s)
−2
7.46.
a.
The system function can be written as
0.4
0.6 s + 0.4
+ 2
s + 3 s + 2s + 2
H (s) =
0.6
1/s
1/s
b
b
0.4
−2
−2
X(s)
Y (s)
1/s
b
0.4
−3
b.
The system function can be written as
H (s) =
10
8.5
2.5
−
+
s +2 s +3 s +4
1/s
b
2.5
−2
X(s)
−10
1/s
b
−3
1/s
−4
b
8.5
Y (s)
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
252
c.
The system function can be written as
H (s) = 1 +
12
6
−
s +1 s +2
1
X(s)
6
1/s
b
Y (s)
−1
1/s
−12
b
−2
7.47.
a.
ˆ ∞
0
x (t − 1) e −st d t
−s(λ+1)
−s
G u (s) =
Using the variable change t − 1 = λ we obtain
G u (s) =
ˆ ∞
x (λ) e
−1
If x (t ) = 0 for −1 < t < 0, then
dλ = e
ˆ ∞
−1
x (λ) e −sλ d λ =
x (t − 1) u (t ) = x (t − 1) u (t − 1)
and
G u (s) = e
−s
ˆ ∞
0
b.
x (λ) e −sλ d λ = e −s X u (s)
ˆ ∞
x (t + 2) e −st d t
−s(λ−2)
2s
G u (s) =
0
Using the variable change t + 2 = λ we obtain
G u (s) =
If x (t ) = 0 for 0 < t < 2, then
ˆ ∞
2
x (λ) e
dλ = e
ˆ ∞
2
x (λ) e −sλ d λ =
x (t + 2) u (t ) = x (t + 2) u (t + 2)
253
and
G u (s) = e 2s
ˆ ∞
0
c.
G u (s) =
x (λ) e −sλ d λ = e 2s X u (s)
ˆ ∞
x (2t ) e −st d t
0
Using the variable change 2t = λ we obtain
G u (s) =
ˆ ∞
d.
G u (s) =
x (λ) e
1
dλ =
2
2
0
ˆ ∞
e
−2t
ˆ ∞
−sλ/2 1
x (t ) e
−st
0
dt =
e.
X u (s) =
d X u (s)
=
ds
ˆ ∞
0
2
ˆ ∞
ˆ ∞
0
x (λ) e −sλ/2 d λ =
x (t ) e −(s+2)t d t = X u (s + 2)
x (t ) e −st d t
0
−t x (t ) e −st d t = −G u (s)
Therefore
G u (s) = −
d X u (s)
ds
7.48.
a. Writing KVL around the loop we obtain
x (t ) = R i (t ) + L
Recall that
i (t ) = C
d i (t )
+ y (t )
dt
d y (t )
dt
and
d 2 y (t )
d i (t )
=C
dt
dt2
Substituting these two relationships into the differential equation yields
x (t ) = RC
³s´
1
Xu
2
2
d y (t )
d 2 y (t )
+ LC
+ y (t )
dt
dt2
Rearranging terms we obtain
¶
µ
¶
µ
µ ¶
1
1
R d y (t )
d 2 y (t )
y
=
x (t )
+
+
(t
)
dt2
L
dt
LC
LC
Finally, substituting numerical values yields
d 2 y (t )
d y (t )
+2
+ 8 y (t ) = 8 x (t )
2
dt
dt
254
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
b.
Taking the Laplace transform of both sides of the differential equation using zero initial
conditions (a requirement for the system function) we get
¡ 2
¢
s + 2 s + 8 Y (s) = 8 X (s)
and
H (s) =
8
Y (s)
=
X (s) s 2 + 2 s + 8
c.
We can not use the system function found in part (b) to determine the transform of the output
signal as Y (s) = H (s) X (s) since initial conditions are specified. Recall that the system function is only
meaningful when the system is CTLTI which requires all initial conditions to be zero. In this case we
need to use the unilateral Laplace transform.
¯
¯
d y (t ) ¯¯
i (0) 0.5
d y (t ) ¯¯
=
=4
⇒
=
i (0) = C
d t ¯t =0
d t ¯t =0
C
1/8
L
½
¯
¾
d y (t ) ¯¯
d 2 y (t )
2
=s Y (s) − s y (0) −
dt2
d t ¯t =0
=s 2 Y (s) − 2s − 4
d y (t )
= s Y (s) − y (0) = s Y (s) − 2
dt
Computing the unilateral Laplace transform of both sides of the differential equation leads to
£
¤
s 2 Y (s) − 2s − 4 + 2 s Y (s) − 2 + 8 Y (s) = 8 X (s)
and
¡ 2
¢
s + 2 s + 8 Y (s) =8 X (s) + 2s + 8
8
= + 2s + 8
s
The transform Y (s) is
2 s2 + 8 s + 8
¢
Y (s) = ¡ 2
s s +2s +8
Its partial fraction expansion is
Y (s) =
k1
k2
k3
+
p +
p
s
s +1+ j 7 s +1− j 7
with residues
k1 = 1 ,
k 2 = 0.5 + j 0.945 ,
k 3 = 0.5 − j 0.945
The output signal is
p ¢
p ¢
¡
¢ ¡
¡
¢ ¡
y (t ) =u (t ) + 0.5 + j 0.945 e −1− j 7 t u (t ) + 0.5 − j 0.945 e −1+ j 7 t u (t )
³p ´
³p ´
=u (t ) + e −t cos 7t u (t ) + 1.89 e −t sin 7t u (t )
255
d.
Taking the unilateral Laplace transform of the differential equation leads to
¡ 2
¢
s + 2 s + 8 Y (s) =8 X (s) + 2s + 8
=
The transform Y (s) is
Y (s) =
Its partial fraction expansion is
Y (s) =
8
+ 2s + 8
s +2
2 s 2 + 12 s + 24
¡
¢
(s + 2) s 2 + 2 s + 8
k1
k2
k3
+
p +
p
s +2 s +1+ j 7 s +1− j 7
with residues
k1 = 1 ,
k 2 = 0.5 + j 1.323 ,
k 3 = 0.5 − j 1.323
The output signal is
p ¢
p ¢
¡
¢ ¡
¡
¢ ¡
y (t ) =u (t ) + 0.5 + j 1.323 e −1− j 7 t u (t ) + 0.5 − j 1.323 e −1+ j 7 t u (t )
³p ´
³p ´
=e −2t u (t ) + e −t cos 7t u (t ) + 2.646 e −t sin 7t u (t )
7.49.
a. The output voltage is
y (t ) = R 2 i 2 (t ) = R 2C 2
which leads to the relationship
¤
d £
v 0 (t ) − y (t )
dt
d v 0 (t )
1
d y (t )
=
y (t ) +
dt
R 2C 2
dt
(P.7.49.1)
Additionally, writing the KCL at the center node we get
which can be simplified to
¤
v 0 (t ) − x (t )
d v0
d £
+C 1
+C 2
v 0 (t ) − y (t ) = 0
R1
dt
dt
v 0 (t ) + R 1 (C 1 +C 2 )
d v 0 (t )
d y (t )
= x (t ) + R 1C 2
dt
dt
(P.7.49.2)
Substituting Eqn. (P.7.49.1) into Eqn. (P.7.49.2) and rearranging terms yields
v 0 (t ) = −R 1C 1
d y (t ) R 1 (C 1 +C 2 )
−
y (t ) + x (t )
dt
R 2C 2
(P.7.49.3)
Differentiating both sides of Eqn. (P.7.49.3)
d 2 y (t ) R 1 (C 1 +C 2 ) d y (t ) d x (t )
d v 0 (t )
= −R 1C 1
−
+
dt
dt2
R 2C 2
dt
dt
(P.7.49.4)
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
256
Equating the right sides of Eqns. (P.7.49.1) and (P.7.49.4) leads to the desired differential equation:
µ
¶
1
d x (t )
R 1C 1 + R 1C 2 + R 2C 2 d y (t )
d 2 y (t )
+
+
y (t ) =
R 1C 1
2
dt
R 2C 2
dt
R 2C 2
dt
Substituting numerical values into the differential equation we obtain
d 2 y (t )
d y (t )
d x (t )
+3
+ y (t ) =
2
dt
dt
dt
b.
Taking the Laplace transform of both sides of the differential equation using zero initial
conditions (a requirement for the system function) we get
¡ 2
¢
s + 3 s + 1 Y (s) = s X (s)
and
H (s) =
s
Y (s)
= 2
X (s) s + 3 s + 1
c.
y (0) = R 2 i 2 (0) = 2
Since
i 2 (t ) = C 2
it follows that
The current of the capacitor C 1 is
⇒
i 2 (0) =
y (0)
=2
R2
¤
d £
v 0 (t ) − y (t )
dt
¯
¯
d y (t ) ¯¯
d v 0 (t ) ¯¯
−C 2
i 2 (0) = C 2
d t ¯t =0
d t ¯t =0
¯
¯
d y (t ) ¯¯
d v 0 (t ) ¯¯
−
2=
¯
d t t =0
d t ¯t =0
C1
d v 0 (t ) x (t ) − v 0 (t )
=
− i 2 (t )
dt
R1
At time t = 0 we have
¯
1−3
x (0) − v 0 (0)
d v 0 (t ) ¯¯
− 2 = −4
− i 2 (0) =
=
C1
d t ¯t =0
R1
1
and we obtain
⇒
¯
d y (t ) ¯¯
= −6
d t ¯t =0
We are now ready to solve the differential equation.
L
½
¯
¾
d 2 y (t )
d y (t ) ¯¯
2
=s
Y
−
s
y
−
(s)
(0)
dt2
d t ¯t =0
=s 2 Y (s) − 2s + 6
d y (t )
= s Y (s) − y (0) = s Y (s) − 2
dt
¯
d v 0 (t ) ¯¯
= −4
d t ¯t =0
257
Computing the unilateral Laplace transform of both sides of the differential equation leads to
and
£ 2
¤
£
¤
s Y (s) − 2s + 6 + 3 s Y (s) − 2 + Y (s) = s X (s)
¡ 2
¢
s + 3 s + 1 Y (s) =s X (s) + 2s
=1 + 2s
The transform Y (s) is
Y (s) =
Its partial fraction expansion is
Y (s) = +
with residues
2s +1
s2 + 3 s + 1
k2
k1
+
s + 0.3820 s + 2.6180
k 1 = 0.1056 ,
k 2 = 1.8944
The output signal is
y (t ) =0.1056 e −0.3820 t u (t ) + 1.8944 e −2.6180 t u (t )
d. Taking the unilateral Laplace transform of the differential equation leads to
¡ 2
¢
s + 3 s + 1 Y (s) =s X (s) + 2s
=
The transform Y (s) is
Y (s) =
Its partial fraction expansion is
s
2 s2 + 5 s
+ 2s =
s +2
s +2
2 s2 + 5 s
¡
¢
(s + 2) s 2 + 3 s + 1
Y (s) = +
k1
k2
k1
+
+
s + 2 s + 0.3820 s + 2.6180
k1 = 2 ,
k 2 = −0.4472 ,
with residues
k 3 = 0.4472
The output signal is
y (t ) =e −2t u (t ) − 0.4472 e −0.3820 t u (t ) + 0.4472 e −2.6180 t u (t )
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
258
7.50.
a.
3
2
jω
1
0
−1
−2
−3
−4
−3
−2
−1
0
1
2
3
σ
b.
MATLAB script to evaluate the magnitude of X (s) at a grid of complex points in the s-plane:
[ sr , s i ] = meshgrid ( [ − 5 : 0 . 1 : 5 ] , [ − 5 : 0 . 1 : 5 ] ) ;
s = sr+ j * s i ;
Xs = @( s ) s . * ( s + 2 ) . / ( ( s + 1) . ^ 2+ 9) ;
XsMag = abs ( Xs ( s ) ) ;
XsMag = XsMag . * ( XsMag< = 2) + 2. * (XsMag> 2 ) ;
c.
1
2
3
4
5
6
7
8
9
10
11
12
¯
¯
Script to produce a three dimensional mesh plot of ¯ X (s)¯:
shading interp ;
% Shading method : I n t e r p o la t e d
colormap copper ;
% S p e c i f y the c o lo r map used .
m1 = mesh( sr , s i , XsMag ) ;
axis ( [ − 5 , 5 , − 5 , 5] ) ;
% Adjust transparency o f s u r f a c e l i n e s .
set (m1, ’ EdgeAlpha ’ , 0 . 6 ’ , ’ FaceAlpha ’ , 0 . 6 ) ;
% S p e c i f y x , y , z a xi s l a b e l s .
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
zlabel ( ’ | X( s ) | ’ ) ;
% S p e c i f y viewing angles .
view ( gca , [ 2 3 . 5 , 3 8 ] ) ;
d. Modify the script in part (c) to also evaluate the Laplace transform for s = j ω and draw it over the
three dimensional mesh plot:
1
% Define the t r a j e c t o r y s= j * omega
259
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
omega = [ − 5 : 0 . 0 1 : 5 ] ;
t r = j * omega ;
% Produce a mesh p lo t and hold i t .
shading interp ;
colormap copper ;
m1 = mesh( sr , s i , XsMag ) ;
hold on ;
% Superimpose a p lo t o f X( s ) magnitude values evaluated on the
% t r a j e c t o r y using ’ plot3 ’ function .
m2 = plot3 ( real ( t r ) , imag ( t r ) , abs ( Xs ( t r ) ) , ’b− ’ , ’ LineWidth ’ , 1 . 5 ) ;
hold o f f ;
axis ( [ − 5 , 5 , − 5 , 5] ) ;
% Adjust transparency o f s u r f a c e l i n e s .
set (m1, ’ EdgeAlpha ’ , 0 . 6 ’ , ’ FaceAlpha ’ , 0 . 6 ) ;
% S p e c i f y x , y , z a xi s l a b e l s .
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
zlabel ( ’ | X( s ) | ’ ) ;
% S p e c i f y viewing angles .
view ( gca , [ 2 3 . 5 , 3 8 ] ) ;
7.51.
a.
1
2
3
4
5
6
7
8
9
10
num = [ 1 , − 2] ;
den = [ 1 , 3 , 2 ] ;
pl s = roots ( den )
z r s = roots (num)
plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . .
real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ;
axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ;
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
grid ;
b.
1
2
3
4
5
6
7
8
9
10
c.
num = [ 1 , 0 ] ;
den = [ 1 , 0 , − 1] ;
pl s = roots ( den )
z r s = roots (num)
plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . .
real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ;
axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ;
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
grid ;
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
260
1
2
3
4
5
6
7
8
9
10
num = [ 1 , 1 ] ;
den = [ 1 , − 4 , 3] ;
pl s = roots ( den )
z r s = roots (num)
plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . .
real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ;
axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ;
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
grid ;
d.
1
2
3
4
5
6
7
8
9
10
num = [ 1 , − 1 , 0] ;
den = [1 , −1 , −6];
pl s = roots ( den )
z r s = roots (num)
plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . .
real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ;
axis ( [ − 3 , 4 , − 3 . 5 , 3 . 5 ] ) ;
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
grid ;
e.
1
2
3
4
5
6
7
8
9
10
num = [ 1 , − 1 , 0] ;
den = [ 1 , 8 , 1 5 ] ;
pl s = roots ( den )
z r s = roots (num)
plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . .
real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ;
axis ( [ − 6 , 1 , − 3 . 5 , 3 . 5 ] ) ;
xlabel ( ’ \sigma ’ ) ;
ylabel ( ’ j \omega ’ ) ;
grid ;
7.52.
a.
1
2
3
4
5
6
7
%%
X = @( s ) ( s − 2) . / ( s .^2+3 * s + 2 ) ;
omg = [ − 1 0 : 0 . 0 2 : 1 0 ] ;
Xomg = X( j *omg ) ;
plot (omg, abs (Xomg ) ) ;
axis ( [ − 1 0 , 1 0 , 0 , 1 . 2 ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
261
8
9
10
11
12
13
14
15
16
17
ylabel ( ’ Magnitude ’ ) ;
t i t l e ( ’ |H( \omega ) | ’ ) ;
grid ;
%%
plot (omg, angle (Xomg ) ) ;
axis ([ −10 ,10 , − pi , pi ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
t i t l e ( ’ \ angle H( \omega) ’ ) ;
grid ;
c.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
%%
X = @( s ) ( s + 1 ) . / ( s .^2−4 * s + 3 ) ;
omg = [ − 1 0 : 0 . 0 2 : 1 0 ] ;
Xomg = X( j *omg ) ;
plot (omg, abs (Xomg ) ) ;
axis ( [ − 1 0 , 1 0 , 0 , 0 . 4 ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
t i t l e ( ’ |H( \omega ) | ’ ) ;
grid ;
%%
plot (omg, angle (Xomg ) ) ;
axis ([ −10 ,10 , − pi , pi ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
t i t l e ( ’ \ angle H( \omega) ’ ) ;
grid ;
7.53.
a.
1
2
3
syms s t
Xs = 1/( s ^2+3 * s + 2 ) ;
x t = i l a p l a c e ( Xs )
b.
1
2
3
syms s t
Xs = ( s −1) * ( s −2)/(( s + 1) * ( s + 2) * ( s + 3 ) ) ;
x t = i l a p l a c e ( Xs )
c.
1
2
3
syms s t
Xs = s * ( s −1)/(( s +1)^2 * ( s + 2) )
x t = i l a p l a c e ( Xs )
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
262
7.54.
a.
1
sys = t f ( [ 1 , 1 , 0 ] , [ 1 , 4 , 1 3 ] )
b.
1
2
3
4
5
6
7
t = [ −1:0.01:10];
x = exp( −0.2 * t ) . * ( t >=0);
plot ( t , x ) ;
axis ( [ − 1 , 10 , − 0. 25 , 1. 25] ) ;
xlabel ( ’ t ( sec ) ’ ) ;
title ( ’x( t ) ’ ) ;
grid ;
c.
1
2
3
4
5
6
y = lsim ( sys , x , t ) ;
plot ( t , y ) ;
axis ( [ − 1 , 10 , − 0. 25 , 1. 25] ) ;
xlabel ( ’ t ( sec ) ’ ) ;
title ( ’y( t ) ’ ) ;
grid ;
7.55.
The code
1
2
3
4
5
% Enter numerator and denominator polynomials
num = [1 , −9 ,30 , −42 ,20];
den = [ 1 , 12 , 59 , 152 , 200 , 96] ;
% Compute the p o le s and the r e s i d u e s
[ r , p , k ] = residue (num, den )
results in the response
r =
62.5000 + 0.0000 i
−74.0000 + 0.0000 i
4.5500 − 7.1500 i
4.5500 + 7.1500 i
3.4000 + 0.0000 i
p =
−4.0000 + 0.0000 i
−3.0000 + 0.0000 i
−2.0000 + 2.0000 i
263
−2.0000 − 2.0000 i
−1.0000 + 0.0000 i
k =
[]
Correspondingly, the partial fraction expansion is
X (s) =
3.4
4.55 + j 7.15 4.55 − j 7.15 −74 62.5
+
+
+
+
s +1
s +2+ j2
s +2− j2
s +3 s +4
7.56.
a.
1
2
3
4
5
num1 = [ 1 , 1 ] ;
den1 = [ 1 , 5 , 6 ] ;
sys1 = t f (num1, den1 ) ;
bode( sys1 ) ;
grid ;
b.
1
2
3
4
5
num2 = [ 1 , − 1 , 0] ;
den2 = [ 1 , 5 , 8 , 6 ] ;
sys2 = t f (num2, den2 ) ;
bode( sys2 ) ;
grid ;
c.
1
2
3
4
5
num3 = conv ( [ 1 , − 1 ] , [ 1 , 2 ] ) ; % Use conv ( ) f o r polynomial m u lt i p li c a t i o n .
den3 = conv ( [ 1 , 1 ] , [ 1 , 6 , 1 3 ] ) ;
sys3 = t f (num3, den3 ) ;
bode( sys3 ) ;
grid ;
7.57. Use the following script to define H (s) as a function of parameters ζ and ω0 :
1
2
numH = @( zeta , omg0) [omg0 * omg0 ] ;
denH = @( zeta , omg0) [ 1 , 2 * zeta * omg0, omg0 * omg0 ] ;
Try with ζ = 0.01 and ω0 = 5 rad/s:
1
2
3
sys1 = t f (numH( 0 . 0 1 , 5 ) ,denH ( 0 . 0 1 , 5 ) ) ;
bode( sys1 , { 1 , 1 0 } ) ;
grid ;
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
264
Try with ζ = 0.1 and ω0 = 5 rad/s:
1
2
3
sys2 = t f (numH( 0 . 1 , 5 ) ,denH ( 0 . 1 , 5 ) ) ;
bode( sys2 , { 1 , 1 0 } ) ;
grid ;
Try with ζ = 0.5 and ω0 = 5 rad/s:
1
2
3
sys3 = t f (numH( 0 . 5 , 5 ) ,denH ( 0 . 5 , 5 ) ) ;
bode( sys3 , { 1 , 1 0 } ) ;
grid ;
Try with ζ = 1 and ω0 = 5 rad/s:
1
2
3
sys4 = t f (numH( 1 , 5 ) ,denH ( 1 , 5 ) ) ;
bode( sys4 , { 1 , 1 0 } ) ;
grid ;
Try with ζ = 2 and ω0 = 5 rad/s:
1
2
3
sys5 = t f (numH( 2 , 5 ) ,denH ( 2 , 5 ) ) ;
bode( sys5 , { 1 , 1 0 } ) ;
grid ;
Graph all five diagrams together for comparison.
1
2
3
4
5
6
7
8
9
10
bode( sys1 , { 1 , 1 0 } ) ;
hold on ;
bode( sys2 , { 1 , 1 0 } ) ;
bode( sys3 , { 1 , 1 0 } ) ;
bode( sys4 , { 1 , 1 0 } ) ;
bode( sys5 , { 1 , 1 0 } ) ;
hold o f f ;
legend ( ’ \ zeta =0.01 ’ , ’ \ zeta =0.1 ’ , ’ \ zeta =0.5 ’ , ’ \ zeta=1 ’ , ’ \ zeta=2 ’ , . . .
’ Location ’ , ’ SouthWest ’ )
grid ;
7.58.
a.
1
2
3
b.
Define H (s), H1 (s), and H1−1 (s).
H = zpk([ −1 ,2] ,[ −3 , −4] ,1)
H1 = zpk([ −1 , −2] ,[ −3 , −4] ,1)
H1inv = zpk([ −3 , −4] ,[ −1 , −2] ,1)
Graph |H (s)| and ∠ H (s).
265
1
2
3
4
5
6
7
8
omg = [ − 5 : 0 . 0 1 : 5 ] ;
Homg = freqresp (H,omg ) ;
subplot ( 1 , 2 , 1 ) ;
plot (omg, abs (Homg( 1 , : ) ) ) ;
grid ;
subplot ( 1 , 2 , 2 ) ;
plot (omg, angle (Homg( 1 , : ) ) ) ;
grid ;
Graph |H1 (s)| and ∠ H1 (s).
1
2
3
4
5
6
7
H1omg = freqresp (H1,omg ) ;
subplot ( 1 , 2 , 1 ) ;
plot (omg, abs (H1omg ( 1 , : ) ) ) ;
grid ;
subplot ( 1 , 2 , 2 ) ;
plot (omg, angle (H1omg ( 1 , : ) ) ) ;
grid ;
¯
¯
Graph ¯ H1−1 (s)¯ and ∠ H1−1 (s).
1
2
3
4
5
6
7
c.
1
2
3
H1invomg = freqresp ( H1inv ,omg ) ;
subplot ( 1 , 2 , 1 ) ;
plot (omg, abs (H1invomg ( 1 , : ) ) ) ;
grid ;
subplot ( 1 , 2 , 2 ) ;
plot (omg, angle (H1invomg ( 1 , : ) ) ) ;
grid ;
Graph unit step response of H (s).
clf ;
step (H, [ 0 : 0 . 0 1 : 5 ] )
grid ;
Graph unit step response of H (s) H1−1 (s).
1
2
3
clf ;
step (H* H1inv , [ 0 : 0 . 0 1 : 5 ] )
grid ;
7.59.
a. Recall that the differential equation is
d 2 y (t )
d y (t )
+ 8 y (t ) = 8 x (t )
+2
2
dt
dt
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
266
Using the initial conditions
y (0) = 2
and
¯
d y (t ) ¯¯
=4
d t ¯t =0
with unilateral Laplace transform leads to
£ 2
¤
£
¤
s Y (s) − 2s − 4 + 2 s Y (s) − 2 + 8 Y (s) = 8 X (s)
Solve for the output transform Y (s).
>>
>>
>>
>>
>>
>>
syms s t Ys
x t = heaviside ( t ) ;
% x ( t )=u( t )
Xs = laplace ( x t ) ;
% Laplace transform o f x ( t )
Y2 = s * s * Ys−2* s −4;
% Laplace transform o f d2y / dt2
Y1 = s * Ys −2;
% Laplace transform o f dy / dt
Ys = solve ( Y2+2 * Y1+8 * Ys−8* Xs , Ys )
% Solve f o r Y ( s )
Find y (t ) through inverse Laplace transform.
>>
y t = i l a p l a c e ( Ys )
% I n v e r s e Laplace transform o f Y ( s )
Graph the output signal.
>>
>>
b.
ezplot ( yt , [ 0 , 5 ] ) ; grid ;
axis ( [ 0 , 5 , − 1 , 3 ] ) ;
Solve for the output transform Y (s).
>>
>>
>>
>>
>>
>>
syms s t Ys
x t = exp(−2 * t ) * heaviside ( t ) ; % x ( t )= exp(−2 * t ) * u( t )
Xs = laplace ( x t ) ;
% Laplace transform o f x ( t )
Y2 = s * s * Ys−2* s −4;
% Laplace transform o f d2y / dt2
Y1 = s * Ys −2;
% Laplace transform o f dy / dt
Ys = solve ( Y2+2 * Y1+8 * Ys−8* Xs , Ys )
% Solve f o r Y ( s )
Find y (t ) through inverse Laplace transform.
>>
y t = i l a p l a c e ( Ys )
% I n v e r s e Laplace transform o f Y ( s )
Graph the output signal.
>>
>>
ezplot ( yt , [ 0 , 5 ] ) ; grid ;
axis ( [ 0 , 5 , − 1 , 3 ] ) ;
7.60.
a. Recall that the differential equation is
d 2 y (t )
d y (t )
d x (t )
+3
+ y (t ) =
2
dt
dt
dt
267
Using the initial conditions
y (0) = 2
with unilateral Laplace transform leads to
and
¯
d y (t ) ¯¯
= −6
d t ¯t =0
£ 2
¤
£
¤
s Y (s) − 2s + 6 + 3 s Y (s) − 2 + Y (s) = s X (s)
Solve for the output transform Y (s).
>>
>>
>>
>>
>>
>>
syms s t Ys
x t = heaviside ( t ) ;
% x ( t )=u( t )
Xs = laplace ( x t ) ;
% Laplace transform o f x ( t )
Y2 = s * s * Ys−2* s +6;
% Laplace transform o f d2y / dt2
Y1 = s * Ys −2;
% Laplace transform o f dy / dt
Ys = solve ( Y2+3 * Y1+Ys−s * Xs , Ys )
% Solve f o r Y ( s )
Find y (t ) through inverse Laplace transform.
>>
y t = i l a p l a c e ( Ys )
% I n v e r s e Laplace transform o f Y ( s )
Graph the output signal.
>>
>>
b.
ezplot ( yt , [ 0 , 5 ] ) ; grid ;
axis ( [ 0 , 5 , − 1 , 3 ] ) ;
Solve for the output transform Y (s).
>>
>>
>>
>>
>>
>>
syms s t Ys
x t = exp(−2 * t ) * heaviside ( t ) ; % x ( t )= exp(−2 * t ) * u( t )
Xs = laplace ( x t ) ;
% Laplace transform o f x ( t )
Y2 = s * s * Ys−2* s +6;
% Laplace transform o f d2y / dt2
Y1 = s * Ys −2;
% Laplace transform o f dy / dt
Ys = solve ( Y2+3 * Y1+Ys−s * Xs , Ys )
% Solve f o r Y ( s )
Find y (t ) through inverse Laplace transform.
>>
y t = i l a p l a c e ( Ys )
Graph the output signal.
>>
>>
ezplot ( yt , [ 0 , 5 ] ) ; grid ;
axis ( [ 0 , 5 , − 1 , 3 ] ) ;
% I n v e r s e Laplace transform o f Y ( s )
268
CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS
Chapter 8
z-Transform for Discrete-Time Signals and Systems
8.1.
a. Applying the z-transform definition:
X (z) = 1 + z −1 + z −2 =
Zeros:
Poles:
z 1 = −0.5 + j 0.866 ,
p1 = p2 = 0
z2 + z + 1
z2
z 2 = −0.5 − j 0.866
The transform converges at every point except the origin of the z-plane.
ROC:
|z| > 0
b.
Applying the z-transform definition:
X (z) = 1 + z −1 + z −2 + z −3 + z −4 =
Zeros:
Poles:
z 1 = 0.309 + j 0.9511 ,
p1 = p2 = p3 = p4 = 0
z 2 = 0.309 − j 0.9511 ,
z4 + z3 + z2 + z + 1
z4
z 3 = −0.809 + j 0.5878 ,
z 4 = −0.809 − j 0.5878
The transform converges at every point except the origin of the z-plane.
ROC:
|z| > 0
c.
Applying the z-transform definition:
X (z) = z 2 + z + 1 + z −1 + z −2 =
Zeros:
Poles:
z 1 = 0.309 + j 0.9511 ,
p1 = p2 = 0
z 2 = 0.309 − j 0.9511 ,
z4 + z3 + z2 + z + 1
z2
z 3 = −0.809 + j 0.5878 ,
z 4 = −0.809 − j 0.5878
In addition to the two poles at the origin of the z-plane, the transform also has two poles at infinity (due
to the z 2 term). Therefore the ROC must exclude both the origin of the z-plane and infinity.
ROC:
0 < |z| < ∞
d.
Applying the z-transform definition:
X (z) = z 4 + z 3 + z 2 + z + 1
Zeros:
z 1 = 0.309 + j 0.9511 ,
z 2 = 0.309 − j 0.9511 ,
269
z 3 = −0.809 + j 0.5878 ,
z 4 = −0.809 − j 0.5878
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
270
The transform X (z) has no finite poles. There are four poles at infinity. Therefore X (z) converges at all
points except where |z| → ∞.
ROC:
|z| < ∞
8.2.
a. No, since the ROC does not include the unit circle.
b.
Yes, since the ROC includes the unit circle.
¯
¯
X (Ω) = X (z)¯
c.
¡
¢
e jΩ e jΩ +2
¢¡
¢
= ¡ jΩ
z=e j Ω
e + 1/2 e j Ω + 3/2
Yes, since the ROC includes the unit circle.
¯
¯
X (Ω) = X (z)¯
d.
e j 2Ω + 5 e j Ω + 6
Yes, since the ROC includes the unit circle.
¯
¯
X (Ω) = X (z)¯
8.3.
a. ROC:
b.
z=e j Ω
e j 2Ω
=
¡ jΩ
¢¡
¢
e +1 e jΩ −1
¢¡
¢¡
¢
= ¡ jΩ
z=e j Ω
e +2 e jΩ −3 e jΩ −4
|z| > 0.75
1
Two complex poles are at 0.5 ∓ j 0.5 = p e ∓ j π/4
2
1
ROC: p < |z| < 1.25
2
c.
ROC:
|z| > 0.5
d.
ROC:
1
< |z| < 1.25
3
8.4.
a. Using z-transform definition
X (z) =
9
X
n z −n
n=0
Let A (z) be defined as
A (z) =
9
X
z −n
n=0
Using the closed form formula for the finite-length geometric series
A (z) =
1 − z −10
,
1 − z −1
¯ ¯
¯z ¯ > 0
271
Differentiating A (z) with respect to z yields
9
X
¤
d £
n z −n−1
A (z) = −
dz
n=0
and therefore
−z
9
¤ X
d £
A (z) =
n z −n = X (z)
dz
n=0
The derivative of the closed form expression for A (z) is
and X (z) is found as
¤ −z −2 + 10 z −11 − 9 z −12
d £
A (z) =
,
¡
¢2
dz
1 − z −1
X (z) = −z
b.
¯ ¯
¯z ¯ > 0
¤ z −1 − 10 z −10 + 9 z −11
d £
A (z) =
,
¡
¢2
dz
1 − z −1
¯ ¯
¯z ¯ > 0
The transform X (z) can be written as
X (z) = X 1 (z) + X 2 (z)
with
X 1 (z) =
and
X 2 (z) =
From part (a) we have
X 1 (z) =
9
X
n z −n
n=0
∞
X
10 z −n
n=10
z −1 − 10 z −10 + 9 z −11
,
¡
¢2
1 − z −1
X 2 (z) is found by using the variable change m = n − 10:
X 2 (z) = 10
∞
X
m=0
z −(m+10) =
10 z −10
,
1 − z −1
and
X (z) =
¯ ¯
¯z ¯ > 1
z −1 − 10 z −10 + 9 z −11 10 z −10
+
¡
¢2
1 − z −1
1 − z −1
z −1 − z −11
= ¡
¢2 ,
1 − z −1
c.
¯ ¯
¯z ¯ > 0
¯ ¯
¯z ¯ > 1
The transform X (z) can be written as
X (z) = X 1 (z) + X 2 (z)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
272
with
X 1 (z) =
and
X 2 (z) =
From part (a) we have
X 1 (z) =
19
X
n=10
9
X
n z −n
n=0
(−n + 20) z −n
z −1 − 10 z −10 + 9 z −11
,
¡
¢2
1 − z −1
X 2 (z) is found by using the variable change m = n − 10:
X 2 (z) =
9
X
m=0
(−m + 10) z −(m+10)
9
X
= −z −10
m=0
= −z −10
"
=
The transform X (z) is
m z −m + 10 z −10
9
X
z −m
m=0
#
·
−10 ¸
z −1 − 10 z −10 + 9 z −11
−10 1 − z
+ 10 z
¡
¢2
1 − z −1
1 − z −1
10 z −10 − 11 z −11 + z −21
,
¡
¢2
1 − z −1
X (z) = X 1 (z) + X 2 (z) =
8.5.
¯ ¯
¯z ¯ > 0
¯ ¯
¯z ¯ > 0
z −1 − 2 z −11 + z −21
,
¡
¢2
1 − z −1
¯ ¯
¯z ¯ > 0
Let X (z) be written as
X (z) = X 1 (z) + X 2 (z)
with
X 1 (z) =
and
X 2 (z) =
∞
X
z −n
n=0
n even
∞
X
(2/3)n z −n
n=1
n odd
Using n = 2m for the indices of the summation for X 1 (z) yields
X 1 (z) =
∞
X
m=0
z −2m =
1
,
1 − z −2
¯ −2 ¯
¯z ¯ < 1
⇒
¯ ¯
¯z ¯ > 1
273
Similarly, using n = 2m + 1 for the indices of the summation for X 2 (z) yields
X 2 (z) =
=
∞
X
(2/3)2m+1 z −(2m+1)
m=0
µ
¶ ∞ µ
¶
2 −1 X
4 −2 m
z
z
3
m=0 9
2 −1
6 z −1
3z
,
=
=
1 − 49 z −2 9 − 4 z −2
¯
¯
¯ 4 −2 ¯
¯ z ¯<1
¯9
¯
The transform X (z) is
X (z) =
⇒
¯ ¯ 2
¯z ¯ >
3
1
6 z −1
9 + 6 z −1 − 4 z −2 − 6 z −3
+
=
,
1 − z −2 9 − 4 z −2
9 − 13 z −2 + 4 z −4
or, using non-negative powers of z
X (z) =
9 z4 + 6 z3 − 4 z2 − 6 z
,
9 z 4 − 13 z 2 + 4
8.6.
a. Using Euler’s formula
g [n] =
The transform is
¯ ¯
¯z ¯ > 1
¯ ¯
¯z ¯ > 1
h
i
1
(0.9)n e j 0.3n + e − j 0.3n u[n]
2
´n 1 X
´n
∞ ³
∞ ³
1 X
0.9 e j 0.3 z −1
0.9 e − j 0.3 z −1
2 n=0
2 n=0
µ ¶
µ ¶
1
1
1
1
=
+
2 1 − 0.9 e j 0.3 z −1
2 1 − 0.9 e − j 0.3 z −1
G (z) =
=
=
1 − 0.9 cos (0.3) z −1
1 − 1.8 cos (0.3) z −1 + 0.81 z −2
£
¤
z z − 0.9 cos (0.3)
z 2 − 1.8 cos (0.3) z + 0.81
¡
¢
¯ ¯
z z − 0.8598
¯z ¯ > 0.9
= 2
,
z − 1.7196 z + 0.81
b.
Since x[n] = g [2n], the corresponding transform is
1 ¡p ¢ 1 ¡ p ¢
G z + G − z
2
2
¢
¢
µ ¶ p ¡p
µ ¶ p ¡p
z
z − 0.8598
z
z + 0.8598
1
1
=
+
p
p
2 z − 1.7196 z + 0.81
2 z + 1.7196 z + 0.81
X (z) =
=
z (z − 0.6685)
z 2 − 1.3370 z + 0.6561
,
¯ ¯
¯z ¯ > 0.81
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
274
c.
The signal x[n] is
x[n] = g [2n] = (0.9)2n cos ((0.3) 2n) u[2n]
= (0.81)n cos (0.6n) u[n]
and the corresponding transform is
X (z) =
£
¤
z z − 0.81 cos (0.6)
z 2 − 1.62 cos (0.6) z + 0.6561
¯ ¯
z (z − 0.6685)
¯z ¯ > 0.81
= 2
,
z − 1.3370 z + 0.6561
8.7.
a. Using z-transform definition
b.
c.
¯ ¯
¯z ¯ < ∞
X (z) = z −3 ,
¯ ¯
¯z ¯ > 0
Using z-transform definition
Using z-transform definition with the closed form formula for the infinite-length geometric series
X (z) =
d.
X (z) = z 2 ,
1
1 − 12 z −1
=
z
z − 21
¯ ¯ 1
¯z ¯ >
2
,
From part (c)
z
©
ª
Z (1/2)n u[n] =
z − 21
©
ª
Z (1/3)n u[n] =
z − 31
z
¯ ¯ 1
¯z ¯ >
2
,
¯ ¯ 1
¯z ¯ >
3
,
Using linearity of the z-transform
X (z) =
z
+
z
z − 31
¢
¡
z 2 z − 56
=
,
z 2 − 56 z + 61
e.
z − 21
From part (c)
ª
©
Z (1/2)n u[n] =
z
¯ ¯ 1
¯z ¯ >
2
¯ ¯ 1
¯z ¯ >
2
,
1
z−2
Using the time shifting property of the z-transform
¢
©
ª ¡
X (z) = Z (1/2)n−1 u[n − 1] = z −1
z
z − 21
=
1
z − 12
,
¯ ¯ 1
¯z ¯ >
2
275
f.
From part (c)
Using linearity of the z-transform
©
ª
Z (1/2)n u[n] =
z
,
1
z−2
¯ ¯ 1
¯z ¯ >
2
ª
©
ª
©
2z
Z (1/2)n−1 u[n] = (1/2)−1 Z (1/2)n u[n] =
,
z − 21
g.
From part (c)
Using linearity of the z-transform
ª
©
Z (1/2)n u[n] =
z
z − 21
,
¯ ¯ 1
¯z ¯ >
2
©
ª
©
ª
Z (1/2)n+1 u[n] = (1/2) Z (1/2)n u[n] =
h.
¯ ¯ 1
¯z ¯ >
2
From part (d)
ª
©
Z (1/3)n u[n] =
z
,
1
z−3
Applying the time reversal property of the z-transform
ª
ª
©
©
Z (1/3)−n u[−n] = Z (3)n u[−n] =
1
2z
,
z − 12
¯ ¯ 1
¯z ¯ >
2
¯ ¯ 1
¯z ¯ >
3
z −1
=
1
z −1 − 3
3
,
3−z
¯ ¯
¯z ¯ < 3
Replacing n with n + 1 and using time shifting property of the z-transform
and using linearity
i.
ª
©
Z (3)n+1 u[−n − 1] = (z)
3
3z
=
,
3−z 3−z
µ ¶
©
ª
1 3z
z
X (z) = Z (3)n u[−n − 1] =
=
,
3 3−z 3−z
¯ ¯
¯z ¯ < 3
¯ ¯
¯z ¯ < 3
From part (h)
¯ ¯
3
¯z ¯ < 3
,
3−z
Replacing n with n − 1 and using time shifting property of the z-transform
©
ª
Z (3)n u[−n] =
¢
©
ª ¡
Z (3)n−1 u[−n + 1] = z −1
3
3 z −1
3
,
=
=
3 − z 3 − z z (3 − z)
¯ ¯
0 < ¯z ¯ < 3
Note that the ROC for the new transform must also exclude the origin of the z-plane due to the pole
introduced at z = 0. Another way to explain the exclusion of the origin from the ROC is to realize that
the rightmost sample of x[n] with nonzero amplitude is now at n = 1 due to the right shift. Using
linearity we obtain
µ ¶
¯ ¯
©
ª
3
1
1
=
,
0 < ¯z ¯ < 3
X (z) = Z (3)n u[−n + 1] =
3 z (3 − z) z (3 − z)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
276
replacemen
8.8.
a.
xL [n]
1
Let
n
(1/2) , n ≥ 0
(1/2)n , n < 0
Amplitude
x[n] =
½
replacemen
µ ¶−n
1
x L [n] =
u[−n − 1] = (2)n u[−n − 1]
2
µ ¶n
1
u[n]
x R [n] =
2
0.5
0
−8
−6
−4
−2
0
Amplitude
x[n] = x L [n] + x R [n]
The two signal components are shown on the
right.
−6
−4
¯ ¯
©
ª
z
¯z ¯ < 2
X L (z) = Z (2)n u[−n − 1] = −
,
z −2
¾
½ µ ¶n
¯ ¯ 1
z
1
¯z ¯ >
u[n] =
,
X R (z) = Z
2
2
z − 12
Using linearity of the z-transform
X (z) = X L (z) + X R (z)
=
8.9.
a. We know that
Using the time reversal property
z
z
+
z − 2 z − 12
− 23 z
z 2 − 52 z + 1
©
ª
Z u[n] =
©
ª
X (z) = Z u[−n] =
8
2
4
6
8
0
−2
0
Index n
=−
6
0.5
−8
c.
4
1
so that
b.
2
Index n
xR [n]
,
z
,
z −1
1 ¯¯ ¯¯
< z <2
2
¯ ¯
¯z ¯ > 1
z1
1
=
,
z −1 − 1 1 − z
¯ ¯
¯z ¯ < 1
277
b.
Applying the time reversal property to the transform pair
©
ª
Z u[n − 1] =
yields the transform pair
©
ª
X (z) = Z u[−n − 1] =
c.
From part (a)
In addition we have
1
,
z −1
¯ ¯
¯z ¯ > 1
1
z
=
,
z −1 − 1 1 − z
©
ª
Z u[−n] =
1
,
1−z
ª
=
¡ −5 ¢
z
z
¯ ¯
¯z ¯ < 1
z −4
,
z −1
¯ ¯
¯z ¯ > 1
z5
z4
=
,
z −1 − 1 1 − z
¯ ¯
¯z ¯ < 1
1
z5
1 + z5
+
=
,
1−z 1−z
1−z
¯ ¯
¯z ¯ < 1
©
Z u[n − 5] =
Applying the time reversal property
z −1
©
ª
Z u[−n − 5] =
and, applying the linearity property
X (z) =
d.
¯ ¯
¯z ¯ < 1
The z transform of the unit-ramp function was found in Example 8-21 to be
©
ª
Z n u[n] =
z
(z − 1)
2
¯ ¯
¯z ¯ > 1
,
Applying the time reversal property leads to
©
ª
Z − n u[−n] = ¡
z −1
z
z
=
,
¢2 =
2
−1
(1 − z)
(z − 1)2
z −1
¯ ¯
¯z ¯ < 1
and, through the use of the linearity property, we get
©
ª
X (z) = Z n u[−n] =
e.
−z
(z − 1)
2
,
¯ ¯
¯z ¯ < 1
The transform of a cosine signal was found in Example 8-15 to be
©
ª
Z cos (Ω0 n) u[n] =
z [z − cos (Ω0 )]
,
z 2 − 2 cos (Ω0 ) z + 1
Applying the time reversal property yields
©
ª
Z cos (−Ω0 n) u[−n] =
=
¯ ¯
¯z ¯ > 1
£
¤
z −1 z −1 − cos (Ω0 )
z −2 − 2 cos (Ω0 ) z −1 + 1
1 − cos (Ω0 ) z
,
1 − 2 cos (Ω0 ) z + z 2
¯ ¯
¯z ¯ < 1
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
278
Since the cosine function is even, that is, cos (−Ω0 n) = cos (Ω0 n), we get
©
ª
X (z) = Z cos (Ω0 n) u[−n] =
1 − cos (Ω0 ) z
,
1 − 2 cos (Ω0 ) z + z 2
¯ ¯
¯z ¯ < 1
8.10.
a.
Starting with
¯ ¯
z
¯z ¯ > 1
,
z −1
the following relationships can be obtained through the use of the differentiation property of the
z-transform:
¯ ¯
©
ª
z
d h z i
¯z ¯ > 1
,
=
Z n u[n] = −z
2
dz z −1
(z − 1)
·
¸
¯ ¯
© 2
ª
d
z
z (z + 1)
¯z ¯ > 1
Z n u[n] = −z
=
,
2
3
d z (z − 1)
(z − 1)
©
ª
Z u[n] =
Using the relationships found, X (z) is obtained as
X (z) =
b.
z (z + 1)
(z − 1)
3
+3
z
(z − 1)
2
+5
z
5 z3 − 6 z2 + 3 z
,
=
z −1
(z − 1)3
¯ ¯
¯z ¯ > 1
Starting with
¯ ¯
©
ª
−z
¯z ¯ < 1
,
Z u[−n − 1] =
z −1
the following relationships can be obtained through the use of the differentiation property of the
z-transform:
¯ ¯
©
ª
−z
d h −z i
¯z ¯ < 1
=
,
Z n u[−n − 1] = −z
dz z −1
(z − 1)2
¸
·
¯ ¯
©
ª
−z (z + 1)
−z
d
¯z ¯ < 1
,
=
Z n 2 u[−n − 1] = −z
2
3
d z (z − 1)
(z − 1)
Using the relationships found, X (z) is obtained as
X (z) =
c.
−z (z + 1)
(z − 1)
3
−5
5 z 3 − 11 z 2 + 4 z
−z
=
z −1
(z − 1)3
¯ ¯
¯z ¯ < 1
The transform of a cosine signal was found in Example 8-15 to be
©
ª
Z cos (Ω0 n) u[n] =
z [z − cos (Ω0 )]
,
z 2 − 2 cos (Ω0 ) z + 1
Using the differentiation property of the z-transform
¯ ¯
¯z ¯ > 1
¸
·
©
ª
d
z [z − cos (Ω0 )]
X (z) = Z n cos (Ω0 n) u[n] = −z
,
d z z 2 − 2 cos (Ω0 ) z + 1
It can be shown that
¸
·
− cos (Ω0 ) z 2 + 2 z − cos (Ω0 )
z [z − cos (Ω0 )]
d
=
¢2
¡
d z z 2 − 2 cos (Ω0 ) z + 1
z 2 − 2 cos (Ω0 ) z + 1
¯ ¯
¯z ¯ > 1
279
and the transform X (z) is
¤
£
z cos (Ω0 ) z 2 − 2 z + cos (Ω0 )
,
X (z) =
¢2
¡
z 2 − 2 cos (Ω0 ) z + 1
d.
¯ ¯
¯z ¯ > 1
The transform of a sine signal was found in Example 8-16 to be
©
ª
Z sin (Ω0 n) u[n] =
sin (Ω0 ) z
,
z 2 − 2 cos (Ω0 ) z + 1
¯ ¯
¯z ¯ > 1
Using the differentiation property of the z-transform along with the linearity property
©
ª
©
ª
X (z) = Z n sin (Ω0 n) u[n] + Z sin (Ω0 n) u[n]
¸
·
d
sin (Ω0 ) z
sin (Ω0 ) z
= −z
+ 2
,
d z z 2 − 2 cos (Ω0 ) z + 1
z − 2 cos (Ω0 ) z + 1
It can be shown that
and the transform X (z) is
¡
¢
¸
·
− sin (Ω0 ) z 2 − 1
d
sin (Ω0 ) z
=¡
¢2
d z z 2 − 2 cos (Ω0 ) z + 1
z 2 − 2 cos (Ω0 ) z + 1
X (z) = ¡
=
¡
¢
sin (Ω0 ) z z 2 − 1
sin (Ω0 ) z
¢2 + 2
z − 2 cos (Ω0 ) z + 1
z 2 − 2 cos (Ω0 ) z + 1
2 sin (Ω0 ) z 2 [z − cos (Ω0 )]
¢2 ,
¡
z 2 − 2 cos (Ω0 ) z + 1
8.11.
a.
£
¯ ¯
¯z ¯ > 1
· 2¸
z
=1
z→∞ z 2
¤
x[0] = lim X (z) = lim
z→∞
b.
· 2¸
z
=0
z→∞ z 3
£
¤
x[0] = lim X (z) = lim
z→∞
c.
The first step is to write X (z) using non-negative powers of z:
X (z) =
z2 + z − 1
z 3 + 0.7 z 2 + 1.2 z − 1.5
Applying the initial value property
£
· 2¸
z
=0
z→∞ z 3
¤
x[0] = lim X (z) = lim
z→∞
¯ ¯
¯z ¯ > 1
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
280
8.12.
a. Using the correlation property, the transform of the autocorrelation function is
¡
¢
R xx (z) = X (z) X z −1
The transform of x[n] is
X (z) = 1 + z −1 + z −2
Therefore
¡
¢¡
¢
R xx (z) = 1 + z −1 + z −2 1 + z + z 2 = z + 2 + z −1
and
r xx [m] = δ[m + 1] + 2 δ[m] + δ[m − 1]
b.
The transform of x[n] is
X (z) =
Replacing z with z −1 leads to
¡
¢
1
1 − z −N ,
−1
1−z
¯ ¯
¯z ¯ > 0
1
(1 − z) ,
1−z
¯ ¯
¯z ¯ < ∞
¢
¡
X z −1 =
Therefore
¶
µ
¶µ
¡ −1 ¢
1 − z −N 1 − z N
=
R xx (z) = X (z) X z
1 − z −1
1−z
=
z N +1 − 2 z + z −N +1
(z − 1)
R xx (z) can be written as
R xx (z) =
which leads to the autocorrelation function
2
z
(z − 1)
¯ ¯
0 < ¯z ¯ < ∞
,
£ N
¤
−N
z
−
2
+
z
2
r xx [m] = (m + N ) u[m + N ] − 2m u[m] + (m − N ) u[m − N ]
8.13.
The transforms of the two signals are
X (z) = 1 + z −1 + z −2
and
Y (z) = 1 + z −1 + z −2 + z −3 + z −4
The transform of the cross correlation function is
¢
¢¡
¢ ¡
¡
R X Y (z) = X (z) Y z −1 = 1 + z −1 + z −2 1 + z + z 2 + z 3 + z 4
= z 4 + 2 z 3 + 3 z 2 + 3 z + 3 + 2 z −1 + z −2
and the cross correlation function is found as
r X Y [m] = { 1, 2, 3, 3, 3 , 2, 1 }
↑
m=0
281
8.14.
a. The convolution of x[m] and y[m] is
x[m] ∗ y[m] =
n=−∞
Replacing y[m] with y[−m] gives
x[m] ∗ y[−m] =
b.
∞
X
n=−∞
x[n] y[m − n]
x[n] y[n − m] = r x y [m]
Using correlation property of the z-transform
¡
¢
R x y (z) = X (z) Y z −1
Since
it follows that
©
ª
Z −1 X (z) = x[m]
and
© ¡
¢ª
Z −1 Y z −1 = y[−m]
©
¡
¢ª
r x y [m] = Z −1 X (z) Y z −1 = x[m] ∗ y[−m]
8.15.
a. The transform of X (z) is
X (z) =
Using the summation property
W (z) =
b.
∞
X
z
,
z −a
¯ ¯ ¯ ¯
¯ z ¯ > ¯a ¯
z
z2
,
X (z) =
z −1
(z − 1) (z − a)
¯ ¯
¡ ¯ ¯¢
¯z ¯ > max 1, ¯a ¯
Partial fraction form of W (z) is
with the residues
k1 =
Inverse transform is
W (z) =
k1 z
k2 z
+
z −1 z −a
1
,
1−a
and
k2 =
a
a −1
1
a
u[n] +
a n u[n]
1−a
a −1
¢
1 ¡
1 − a n+1 u[n]
=
1−a
w [n] =
c.
Using the closed form formula for finite-length geometric series
w [n] =
for n ≥ 0.
n
X
k=−∞
a k u[k] =
n
X
k=0
ak =
1 − a n+1
1−a
282
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
8.16. Let
x[n] = n u[n]
so that
w [n] =
The transform of x[n] is
n
X
k=−∞
x[k] =
X (z) =
n
X
k=−∞
k u[k] =
n
X
k
k=0
z
(z − 1)2
Using the differentiation property of the z-transform
W (z) =
z2
z
X (z) =
z −1
(z − 1)3
Following transform relationships were derived earlier (see Example 8-22)
Z
n u[n] ←→
Z
n 2 u[n] ←→
z2 − z
=
(z − 1)2 (z − 1)3
z
z2 + z
(z − 1)3
Combining these transform pairs using linearity we get
¡ 2
¢
Z
n + n u[n] ←→
Therefore
w [n] =
8.17.
a.
2 z2
(z − 1)3
n (n + 1)
u[n]
2
1
X (z)
k1
k2
=
=
+
z
(z + 1) (z + 2) (z + 1) (z + 2)
Next step is to determine the residues:
Partial fraction expansion for X (z) is
¯
1 ¯¯
=1
k1 =
(z + 2) ¯z=−1
¯
1 ¯¯
k2 =
= −1
(z + 1) ¯z=−2
z
z
−
z +1 z +2
Based on the ROC specified, x[n] is an anti-causal signal. Therefore
X (z) =
x[n] = − (−1)n u[−n − 1] + (−2)n u[−n − 1]
283
b.
The residues are
k1
z +1
k2
X (z)
k3
=
=
+
+
z
z (z + 1/2) (z + 2/3)
z
(z + 1/2) (z + 2/3)
¯
¯
z +1
¯
=3
k1 =
(z + 1/2) (z + 2/3) ¯z=0
¯
z + 1 ¯¯
k2 =
= −6
z (z + 2/3) ¯z=−1/2
and
k3 =
Partial fraction expansion for X (z) is
¯
z + 1 ¯¯
=3
z (z + 1/2) ¯z=−2/3
X (z) = 3 −
6z
3z
+
z + 1/2 z + 2/3
The specified ROC indicates a causal signal, therefore
x[n] = 3 δ[n] − 6 (−1/2)n u[n] + 3 (−2/3)n u[n]
c.
The residues are
and
z +1
X (z)
k1
k2
=
=
+
z
(z − 0.4) (z + 0.7) (z − 0.4) (z + 0.7)
¯
¯
14
z +1
¯
=
k1 =
¯
(z − 0.4) (z + 0.7) z=0.4 11
k2 =
Partial fraction expansion for X (z) is
¯
¯
3
z +1
¯
=−
¯
11
(z − 0.4) (z + 0.7) z=−0.7
µ ¶
¶
3
14
z
z
−
X (z) =
11 (z − 0.4)
11 (z + 0.7)
µ
The specified ROC indicates a causal signal, therefore
µ ¶
µ ¶
14
3
x[n] =
(0.4)n u[n] −
(−0.7)n u[n]
11
11
d.
The residues are
z +1
X (z)
k1
k2
k3
=
=
+
+
z
(z + 3/4) (z − 1/2) (z − 3/2) (z + 3/4) (z − 1/2) (z − 3/2)
¯
¯
z +1
4
¯
k1 =
=
(z − 1/2) (z − 3/2) ¯z=−3/4 45
¯
¯
6
z +1
¯
=−
k2 =
¯
+
3/4)
−
3/2)
5
(z
(z
z=1/2
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
284
and
k3 =
Partial fraction expansion for X (z) is
¯
¯
z +1
10
=
=¯¯
9
(z + 3/4) (z − 1/2) z=3/2
X (z) =
4
45 z
(z + 3/4)
−
6
5z
(z − 1/2)
+
10
9 z
(z − 3/2)
Based on the ROC specified, the terms with poles at z = 1/2 and z = −3/4 correspond to causal signal
components whereas the term with pole at z = 3/2 corresponds to an anti-causal signal component:
X (z) =
10
6
5z
9 z
+
−
(z + 3/4) (z − 1/2) (z − 3/2)
{z
} | {z }
|
4
45 z
causal
The signal x[n] is
x[n] =
µ
anti-causal
µ ¶
µ ¶
¶
6
10
4
(−3/4)n u[n] −
(1/2)n u[n] −
(3/2)n u[−n − 1]
45
5
9
e.
Partial fraction expansion for X (z) is as found in part (d). However, since the ROC is different
than that of part (d), the terms with poles at z = −3/4 and z = 3/2 correspond to anti-causal signal
components whereas the term with pole at z = 1/2 corresponds to a causal signal component:
X (z) =
4
45 z
−
x[n] = −
µ
10
9 z
+
(z + 3/4) (z − 1/2) (z − 3/2)
| {z } | {z } | {z }
anti-causal
The signal x[n] is
6
5z
causal
anti-causal
µ ¶
µ ¶
¶
6
10
4
(−3/4)n u[−n − 1] −
(1/2)n u[n] −
(3/2)n u[−n − 1]
45
5
9
8.18.
a.
Im {z}
Unit circle
−2
−1
−1/2
1
2
Re {z}
285
b.
k1
k2
X (z)
k3
k4
(z + 1) (z − 2)
=
=
+
+
+
z
z (z + 1/2) (z − 1) (z + 2)
z
(z + 1/2) (z − 1) (z + 2)
Next step is to determine the residues:
¯
¯
(z + 1) (z − 2)
¯
=2
k1 =
(z + 1/2) (z − 1) (z + 2) ¯z=0
¯
10
(z + 1) (z − 2) ¯¯
k2 =
=−
z (z − 1) (z + 2) ¯z=−1/2
9
¯
4
(z + 1) (z − 2) ¯¯
k3 =
=−
z (z + 1/2) (z + 2) ¯z=1
9
¯
4
(z + 1) (z − 2) ¯¯
k4 =
=−
¯
z (z + 1/2) (z − 1) z=−2
9
Partial fraction expansion for X (z) is
X (z) = 2 −
c.
10
9 z
(z + 1/2)
−
4
9z
(z − 1)
−
4
9z
(z + 2)
There are four possible choices for the ROC:
Case 1:
|z| < 1/2
x[n] = 2 δ[n] +
Case 2:
µ
µ ¶
¶
µ ¶
10
4
4
u[−n − 1] +
(−1/2)n u[−n − 1] +
(−2)n u[−n − 1]
9
9
9
1/2 < |z| < 1
x[n] = 2 δ[n] −
Case 3:
1 < |z| < 2
µ
µ ¶
¶
µ ¶
4
4
10
u[−n − 1] +
(−1/2)n u[n] +
(−2)n u[−n − 1]
9
9
9
µ ¶
¶
µ ¶
4
10
4
n
x[n] = 2 δ[n] −
u[n] +
(−1/2) u[n] −
(−2)n u[−n − 1]
9
9
9
Case 4:
|z| > 2
µ
µ
µ ¶
¶
µ ¶
10
4
4
n
x[n] = 2 δ[n] −
u[n] −
(−1/2) u[n] −
(−2)n u[n]
9
9
9
8.19.
a. Factored form of X (z) is
and
X (z) = ¡
z (z + 3)
¢¡
¢
z − 0.7 − j 0.6 z − 0.7 + j 0.6
X (z)
k1
k2
(z + 3)
¢¡
¢=
=¡
+
z
z − 0.7 − j 0.6 z − 0.7 + j 0.6
z − 0.7 − j 0.6 z − 0.7 + j 0.6
Residues are found as
¯
¯
¢ ¯¯
= 0.5 − j 3.0833
k1 = ¡
z − 0.7 + j 0.6 z=0.7+ j 0.6
(z + 3)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
286
and
k 2 = k 1∗ = 0.5 + j 3.0833
Partial fraction expansion of X (z) is
X (z) =
¡
¢
0.5 − j 3.0833 z
z − 0.7 − j 0.6
+
¡
¢
0.5 + j 3.0833 z
z − 0.7 + j 0.6
and the inverse transform is
¡
¢¡
¢n
¡
¢¡
¢n
x[n] = 0.5 − j 3.0833 0.7 + j 0.6 u[n] + 0.5 + j 3.0833 0.7 − j 0.6 u[n]
Since
0.5 − j 3.0833 = 3.1236 e − j 1.41
0.7 + j 0.6 = 0.922 e j 0.7086
and
the inverse transform can be written as
´n
´³
´n
³
´³
³
x[n] = 3.1236 e − j 1.41 0.922 e j 0.7086 u[n] + 3.1236 e j 1.41 0.922 e − j 0.7086 u[n]
= 6.2472 (0.922)n cos (0.7086n − 1.41) u[n]
b.
and
Factored form of X (z) is
X (z) = ¡
z2
¢¡
¢
z − 0.8 − j 0.6 z − 0.8 + j 0.6
X (z)
k1
z
k2
¢¡
¢=
=¡
+
z
z − 0.8 − j 0.6 z − 0.8 + j 0.6
z − 0.8 − j 0.6 z − 0.8 + j 0.6
Residues are found as
k1 = ¡
and
¯
¯
2
1
¢ ¯¯
= −j
3
z − 0.8 + j 0.6 z=0.8+ j 0.6 2
z
k 2 = k 1∗ =
1
2
+j
2
3
¢
− j 23 z
¡1
Partial fraction expansion of X (z) is
X (z) =
¡1
2
z − 0.8 − j 0.6
+
2
¢
+ j 23 z
z − 0.8 + j 0.6
and the inverse transform is
¶
µ
¶
µ
¢n
¢n
1
2 ¡
2 ¡
1
0.8 + j 0.6 u[n] + + j
0.8 − j 0.6 u[n]
−j
x[n] =
2
3
2
3
Since
2 5
1
− j = e − j 0.9273
and
0.8 + j 0.6 = e j 0.6436
2
3 6
the inverse transform can be written as
µ
¶
µ
¶
5 − j 0.9273 j 0.7086n
5
x[n] =
e
e
u[n] + e j 0.9273 e − j 0.7086n u[n]
6
6
=
5
cos (0.9273n − 0.7086) u[n]
3
287
c.
Factored form of X (z) is
X (z) = ³
and
X (z)
z +3
k2
k1
´³
´=
p
=³
+
p
p
p
3
3
3
1
1
1
z
z−
z−
z− 3+j 1
−
+j
−
z−
2
Residues are found as
z (z + 3)
´³
´
p
p
z − 23 − 21 z − 23 + j 12
2
2
2
¯
¯
¯
´ ¯¯
k1 = ³
p
z− 3+j 1 ¯
2
z +3
2
and
2
p
z= 23 + j 12
2
2
2
= 0.5 − j 3.866
k 2 = k 1∗ = 0.5 + j 3.866
Partial fraction expansion of X (z) is
X (z) =
¡
¢
0.5 − j 3.866 z
p
z − 23 − j 12
+
¡
¢
0.5 + j 3.866 z
p
z − 23 + j 12
and the inverse transform is
¡
¢
x[n] = 0.5 − j 3.866
Ãp
!n
!n
Ãp
¡
¢
1
1
3
3
+j
−j
u[n] + 0.5 + j 3.866
u[n]
2
2
2
2
Since
0.5 − j 3.866 = 3.8982 e
− j 1.4422
and
p
1
3
+ j = e j 0.5236
2
2
the inverse transform can be written as
´
³
´
³
x[n] = 3.8982 e − j 1.4422 e j 0.5236n u[n] + 3.8982 e j 1.4422 e − j 0.5236n u[n]
= 7.7964 cos (0.5236n − 1.4422) u[n]
8.20.
a. Factored form of X (z) is
X (z) =
and
Residues are found as
(z + 1) (z + 2)
(z − 1)2
k 2,2
X (z) (z + 1) (z + 2) k 1 k 2,1
=
+
+
=
2
z
z
z − 1 (z − 1)2
z (z − 1)
¯
(z + 1) (z + 2) ¯¯
k1 =
=2
¯
(z − 1)2
z=0
¯
(z + 1) (z + 2) ¯¯
k 2,2 =
=6
¯
z
z=1
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
288
and
¯
·
¸¯
z 2 − 2 ¯¯
d (z + 1) (z + 2) ¯¯
= −1
=
k 2,1 =
¯
dz
z
z 2 ¯z=1
z=1
Partial fraction expansion of X (z) is
X (z) = 2 −
b.
z
6z
+
z − 1 (z − 1)2
Dividing X (z) by z yields
k 1,2
k 1,1
z +1
k2
X (z)
=
+
=
+
2
2
z
z
+
0.9
z
−1
(z + 0.9) (z − 1)
(z + 0.9)
Residues are found as
¯
z + 1 ¯¯
= −0.0526
z − 1 ¯z=−0.9
¯
·
¸¯
d z + 1 ¯¯
−2 ¯¯
k 1,1 =
=
= −0.5540
d z z − 1 ¯z=−0.9 (z − 1)2 ¯z=−0.9
k 1,2 =
and
k2 =
Partial fraction expansion of X (z) is
X (z) = −
c.
(z
¯
¯
¯
2¯
+ 0.9)
z +1
z=1
= 0.5540
0.0526 z
0.5540 z
0.5540 z
−
+
2
z + 0.9
z −1
(z + 0.9)
Dividing X (z) by z yields
k 2,1
k 2,2
k 3,2
k 3,1
X (z)
k1
z2 + 4 z − 7
=
=
+
+
+
+
2
2
2
z
z
z + 0.9 (z + 0.9)
z − 1.2 (z − 1.2)2
z (z + 0.9) (z − 1.2)
Residues are found as
k1 =
¯
¯
¯
2¯
− 1.2)
z2 + 4 z − 7
(z + 0.9)2 (z
z=0
= −6.0014
¯
z 2 + 4 z − 7 ¯¯
= 2.4666
z (z − 1.2)2 ¯z=−0.9
¯
z 2 + 4 z − 7 ¯¯
= −0.1436
k 3,2 =
z (z + 0.9)2 ¯z=1.2
k 2,2 =
¯
·
¸¯
d z 2 + 4 z − 7 ¯¯
−z 4 − 8 z 3 + 32.04 z 2 − 33.6 z + 10.08 ¯¯
k 2,1 =
= 4.5355
=
¯
d z z (z − 1.2)2 ¯z=−0.9
z 2 (z − 1.2)4
z=−0.9
¯
·
¸¯
−z 4 − 8 z 3 + 14.61 z 2 + 25.2 z + 5.67 ¯¯
d z 2 + 4 z − 7 ¯¯
= 1.4658
=
k 3,1 =
¯
d z z (z + 0.9)2 ¯z=1.2
z 2 (z + 0.9)4
z=1.2
Partial fraction expansion of X (z) is
289
X (z) = −6.0014 +
2.4666 z
0.1436 z
4.5355 z
1.4658 z
+
−
+
2
z + 0.9
z − 1.2
(z + 0.9)
(z − 1.2)2
8.21.
a. Taking the z-transform of the difference equation
Y (z) = (1 + c) z −1 Y (z) − X (z)
and
H (z) =
b.
X (z) = −A + B z −1
c.
−z
z − (1 + c)
³ z ´
B
−A z + A + B
= −A +
=
z −1
z −1
z −1
µ
d.
¶
A +B
Az z−
A
Y (z) = H (z) X (z) =
(z − 1 − c) (z − 1)
Partial fraction expansion for Y (z) is found through
µ
¶
A +B
A z−
k1
k2
Y (z)
A
=
=
+
z
(z − 1 − c) (z − 1) z − 1 − c z − 1
The residues are
k1 = A −
The solution is
B
,
c
and
k2 =
µ
B
c
¶
B
B
y[n] = A −
(1 + c)n u[n] + u[n]
c
c
Setting y[N ] = 0 we obtain
B=
8.22.
a.
X (z) =
A c (1 + c)N
(1 + c)N − 1
z2 + 3 z
z 2 − 1.4 z + 0.85
290
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
Long division is carried out as follows:
1 + 4.4z −1 + 5.31z −2 + 3.694z −3 + 0.6581z −4
z 2 −1.4z +0.85 | z 2
2
+3z
z −1.4z +0.85
4.4z −0.85
4.4z −6.16 +3.74z −1
5.31 −3.74z −1
5.31 −7.434z −1
+4.5135z −2
3.694z −1
−4.5135z −2
3.694z −1
−5.1716z −2 +3.1399z −3
0.6581z −2 −3.1399z −3
The inverse transform x[n] is
x[n] = { 1 , 4.4, 5.31, 3.694, 0.6581, −2.2186; −3.6654, . . . }
↑
n=0
b.
X (z) =
Long division is carried out as follows:
z2
z 2 − 1.6 z + 1
1 + 1.6z −1 + 1.56z −2 + 0.896z −3 − 0.1264z −4
z 2 −1.6z +1 | z 2
z 2 −1.6z
+1
1.6z
−1
1.6z −2.56
+1.6z −1
1.56
−1.6z −1
1.56 −2.496z −1
+1.56z −2
0.896z −1
−1.56z −2
0.896z −1
−1.4336z −2 +0.896z −3
−0.1264z −2 −0.896z −3
The inverse transform x[n] is
x[n] = { 1 , 1.6, 1.56, 0.896, −0.1264, −1.0982; −1.6308, . . . }
↑
n=0
c.
X (z) =
z2 + 3 z
z 2 − 1.7321 z + 1
291
Long division is carried out as follows:
1 + 4.7321z −1 + 7.1962z −2 + 7.7324z −3 + 6.1971z −4
z 2 −1.7321z +1 | z 2
+3z
z 2 −1.7321z
+1
4.7321z
−1
4.7321z −8.1962 +4.7321z −1
7.1962 −4.7321z −1
7.1962 −12.4645z −1
+7.1962z −2
7.7324z −1
−7.1962z −2
7.7324z −1 −13.3933z −2 +7.7324z −3
6.1971z −2 −7.7324z −3
The inverse transform x[n] is
x[n] = { 1 , 4.7321, 7.1962, 7.7324, 6.1971, 3, −1, . . . }
↑
n=0
8.23.
There are four possible choices for the ROC:
Case 1:
|z| < 1/2
z2 − z − 2
−2 − z + z 2
=
z 3 + 1.5 z 2 − 1.5 z − 1 −1 − 1.5 z + 1.5 z 2 + z 3
Long division is carried out as follows:
X (z) =
2 − 2z − 5z 2 − 8.5z 3 + 18.25z 4
− 1 −1.5z +1.5z 2 + z 3 | −2 −z
+z 2
−2 −3z +3z 2
+2z 3
2z −2z 2
−2z 3
−3z 3
−2z 4
−5z 2
+z 3
+2z 4
2z +3z 2
−5z 2 −7.5z 3
+7.5z 4
5z 5
8.5z 3
−5.5z 4
−5z 5
8.5z 3 +12.75z 4 −12.75z 5 −8.5z 6
−18.25z 4 +7.75z 5 +8.5z 6
The inverse transform x[n] is
x[n] = { . . . , −35.125, 18.25, −8.5, 5, −2, 2 , 0, 0, . . . }
↑
n=0
Case 2:
1/2 < |z| < 1
X (z) = 2 −
4
4
10
z
z
z
9
9
9
−
= X L (z) + X R (z)
−
1
z + 2 z −1 z +2
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
292
X R (z) = 2 −
10
z
0.8889 z + 1
9
=
1
z + 0.5
z+2
4
4
z
z
−0.4444 z − 0.8889 z 2
X L (z) = − 9 − 9 =
z −1 z +2
−2 + z + z 2
For X R (z), long division is carried out as follows:
0.8889 + 0.5556z −1 − 0.2778z −2 + 0.1389z −3 − 0.0694z −4
z +0.5 | 0.8889z
+1
0.8889z +0.4444
0.5556
0.5556 +0.2778z −1
−0.2778z −1
−0.2778z −1 −0.1389z −2
0.1389z −2
0.1389z −2 +0.0694z −3
−0.0694z −3
For X L (z), long division is carried out as follows:
0.2222z + 0.5556z 2 + 0.3889z 3 + 0.4722z 4 + 0.4306z 5
− 2 +z +z 2 | −0.4444z −0.8889z 2
−0.4444z +0.2222z 2 +0.2222z 3
−1.3333z 2 −0.2222z 3
−1.3333z 2 +0.5556z 3 +0.5556z 4
−0.7778z 3 −0.5556z 4
−0.7778z 3 +0.3889z 4 +0.3889z 5
−0.9445z 4 −0.3889z 5
−0.9445z 4 +0.4722z 5 +0.4722z 6
−0.8611z 5 −0.4722z 6
The inverse transform x[n] is
x[n] = { . . . , 0.4306, 0.4722, 0.3889, 0.5556, 0.2222, 0.8889, 0.5556, −0.2778, 0.1389, −0.0694, . . . }
↑
n=0
Case 3:
1 < |z| < 2
X (z) = 2 −
10
4
4
9 z
9z
9z
−
= X L (z) + X R (z)
−
1
z + 2 z −1 z +2
X R (z) = 2 −
4
10
0.4444 z 2 − 0.1111 z − 1
9 z
9z
=
−
1
z 2 − 0.5 z − 0.5
z + 2 z −1
293
4
z
−0.4444 z
X L (z) = − 9 =
z +2
2+z
For X R (z), long division is carried out as follows:
0.4444 + 0.1111z −1 − 0.7222z −2 − 0.3056z −3 − 0.5139z −4
z 2 −0.5z −0.5 | 0.4444z 2 −0.1111z
0.4444z 2 −0.2222z −0.2222
0.1111z −0.7778
0.1111z −0.0556 −0.0556z −1
−0.7222 +0.0556z −1
−0.7222 +0.3611z −1
+0.3611z −2
−0.3056z −1
−0.3611z −2
−0.3056z −1
+0.1528z −2 +0.1528z −3
−0.5139z −2 −0.1528z −3
For X L (z), long division is carried out as follows:
−0.2222z + 0.1111z 2 − 0.0556z 3 + 0.0278z 4
2 +z | −0.4444z
−0.4444z −0.2222z 2
0.2222z 2
0.2222z 2 +0.1111z 3
−0.1111z 3
−0.1111z 3 −0.0556z 4
0.0556z 4
0.0556z 4 +0.0278z 5
−0.0278z 5
The inverse transform x[n] is
x[n] = { . . . , 0.0278, −0.0556, 0.1111, −0.2222, 0.4444, 0.1111, −0.7222, −0.3056, −0.5139 . . . }
↑
n=0
Case 4:
|z| > 2
X (z) =
z2 − z − 2
z 3 + 1.5 z 2 − 1.5 z − 1
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
294
Long division is carried out as follows:
z −1 − 2.5z −2 + 3.25z −3 − 7.625z −4 + 13.8125z −5
z 3 +1.5z 2 −1.5z − 1 | z 2
−z
2
−2
z +1.5z
−1.5
−z −1
−2.5z
−0.5
+z −1
−2.5z −3.75 +3.75z −1
+2.5z −2
3.25 −2.75z −1
−2.5z −2
3.25 +4.875z −1
−4.875z −2
−3.25z −3
−7.625z −1
+2.375z −2
+3.25z −3
−7.625z −1 −11.4375z −2 +11.4375z −3 +7.625z −4
13.8125z −2 −8.1875z −3 −7.625z −4
The inverse transform x[n] is
x[n] = { 0 , 1, −2.5, 3.25, −7.625, 13.8125, −28.9063, . . . }
↑
n=0
8.24.
a.
X (z) =
Long division:
z
z2 + 3 z + 2
z −1 − 3z −2 + 7z −3 − 15z −4 + 31z −5
z 2 +3z +2 | z
z +3 +2z −1
−3 −2z −1
−3 −9z −1 −6z −2
7z −1 +6z −2
7z −1 +21z −2 +14z −3
15z −2 −14z −3
15z −2 −45z −3 −30z −4
31z −3 +30z −4
The inverse transform x[n] is
x[n] = { 0 , 1, −3, 7, −15, 31, −63, . . . }
↑
n=0
b.
X (z) =
z +1
z 2 + 1.1667 z + 0.3333
295
Long division:
z −1 − 0.1667z −2 − 0.1388z −3 + 0.2175z −4 − 0.2075z −5
z 2 +1.1667z +0.3333 | z
+1
z +1.1667 +0.3333z −1
−0.1667 −0.3333z −1
−0.1667 −0.1945z −1 −0.0556z −2
−0.1388z −1 +0.0556z −2
−0.1388z −1 −0.1619z −2 −0.0463z −3
0.2175z −2 +0.0463z −3
0.2175z −2 +0.2538z −3 +0.0725z −4
0.2075z −3 −0.0725z −4
The inverse transform x[n] is
x[n] = { 0 , 1, −0.1667, −0.1388, 0.2175, −0.2075, 0.1696, . . . }
↑
n=0
c.
X (z) =
Long division:
z2 + z
z 2 + 0.3 z − 0.28
1 + 0.7z −1 + 0.07z −2 + 0.175z −3 − 0.0329z −4
z 2 +0.3z −0.28 | z 2
2
+z
z +0.3z −0.28
0.7z +0.28
0.7z +0.21 −0.196z −1
0.07 +0.196z −1
0.07 +0.021z −1 −0.0196z −2
0.175z −1 +0.0196z −2
0.175z −1 +0.0525z −2 −0.049z −3
−0.0329z −2 +0.049z −3
The inverse transform x[n] is
x[n] = { 1 , 0.7, 0.07, 0.175, −0.0329, 0.0589, −0.0269, . . . }
↑
n=0
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
296
8.25.
The partial fraction for X (z) is
X (z) =
4
10
45 z
9 z
+
+
1
3
z − 2 z + 4 z − 32
− 65 z
a.
For the specified ROC, the poles at z = 1/2 and z = −3/4 correspond to causal terms, and the pole
at z = 3/2 corresponds to an anti-causal term. The causal part of the transform is
X R (z) =
4
−1.1111 z 2 − 0.9444 z
45 z
+
=
z 2 + 0.25 z − 0.375
z − 21 z + 43
− 56 z
Long division:
−1.1111 − 0.6667z −1 − 0.25z −2 − 0.1875z −3 − 0.0329z −4
z 2 +0.25z −0.375 | −0.1111z 2 −0.9444z
−0.1111z 2 −0.2778z +0.4167
−0.6667z +0.4167
−0.6667z +0.1667
+0.25z −1
−0.25
−0.25z −1
−0.25 −0.0625z −1
0.0938z −2
−0.1875z −1 −0.0938z −2
−0.1875z −1 −0.0469z −2 +0.0703z −3
−0.0329z −2 +0.049z −3
The anti-causal part of the transform is
X L (z) =
10
1.1111 z
9 z
=
3
z − 2 −1.5 + z
Long division:
−0.7407z − 0.4938z 2 − 0.3292z 3 − 0.2195z 4
−1.5 +z | 0.1111z
0.1111z −0.7407z 2
0.7407z 2
0.7407z 2 −0.4938z 3
0.4938z 3
0.4938z 3 −0.3292z 4
0.3292z 4
0.3292z 4 −0.2195z 5
0.2195z 5
297
The inverse transform x[n] is
x[n] = { . . . , −0.1463, −0.2195, −0.3292, −0.4938, −0.7407, −1.1111, −0.6667, −0.25, −0.1875, −0.0469, −0.0586, . . . }
↑
n=0
b.
For the specified ROC, the pole at z = 1/2 corresponds to a causal term. The poles at and z = −3/4
and z = 3/2 correspond to anti-causal terms. The causal part of the transform is
X R (z) =
− 56 z
z − 12
=
−1.2 z
−0.5 + z
Long division:
−1.2 − 0.6z −1 − 0.3z −2 − 0.15z −3 − 0.075z −4
z −0.5 | −1.2z
−1.2z +0.6
−0.6
−0.6 +0.3z −1
−0.3z −1
−0.3z −1 +0.15z −2
−0.15z −2
−0.15z −2 +0.075z −3
−0.075z −3
The anti-causal part of the transform is
10
4
1.2 z 2 + 0.7 z
45 z
9 z
+
=
X L (z) =
z + 34 z − 32 z 2 − 0.75 z − 1.125
Long division:
−0.6222z − 0.6519z 2 − 0.1185z 3 − 0.5004z 4 + 0.2283z 5
− 1.125 −0.75z +z 2 | 0.7z
+1.2z 2
0.7z +0.4667z 2 −0.6222z 3
0.7333z 2 +0.6222z 3
0.7333z 2 +0.4889z 3 −0.6519z 4
0.1333z 3 +0.6519z 4
0.1333z 3 +0.0889z 4 −0.1185z 5
0.5630z 4 +0.1185z 5
0.5630z 4 +0.3753z 5 −0.5004z 6
−0.2568z 5 +0.5004z 6
The inverse transform x[n] is
x[n] = { . . . , 0.2283, −0.5004, −0.1185, −0.6519, −0.6222, −1.2, −0.6, −0.3, −0.15, −0.075, −0.0375, . . . }
↑
n=0
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
298
8.26.
a. Take the z-transform of both sides of the difference equation:
Y (z) = 0.9 z −1 Y (z) + X (z) + z −1 X (z)
The system function is
H (z) =
and its partial fraction form is
The corresponding impulse response is
Y (z)
1 − z −1
z −1
=
=
X (z) 1 − 0.9 z −1 z − 0.9
1
10
9z
H (z) =
−
9
z − 0.9
h[n] =
b.
10
1
δ[n] − (0.9)n u[n]
9
9
Take the z-transform of both sides of the difference equation:
Y (z) = 1.7 z −1 Y (z) − 0.72 z −2 Y (z) + X (z) − 2 z −1 X (z)
The system function is
H (z) =
1 − 2 z −1
z (z − 2)
Y (z)
=
=
−1
−2
X (z) 1 − 1.7 z + 0.72 z
(z − 0.9) (z − 0.8)
H (z)
z −2
k2
k1
=
+
=
z
(z − 0.9) (z − 0.8) z − 0.9 z − 0.8
The residues are
¯
z − 2 ¯¯
= −11
k1 =
z − 0.8 ¯z=0.9
and
¯
z − 2 ¯¯
k2 =
= 12
z − 0.9 ¯z=0.8
and the partial fraction form of the system function is
H (z) = −
12 z
11 z
+
z − 0.9 z − 0.8
and the impulse response is found as
h[n] = −11 (0.9)n u[n] + 12 (0.8)n u[n]
c.
Take the z-transform of both sides of the difference equation:
Y (z) = 1.7 z −1 Y (z) − 0.72 z −2 Y (z) + X (z) + z −1 X (z) + z −2 X (z)
The system function is
H (z) =
1 + z −1 + z −2
z2 + z + 1
Y (z)
=
=
X (z) 1 − 1.7 z −1 + 0.72 z −2 (z − 0.9) (z − 0.8)
299
Division of H (z) by z yields
H (z)
k1
z2 + z + 1
k2
k3
=
=
+
+
z
z (z − 0.9) (z − 0.8)
z
z − 0.9 z − 0.8
The residues are
¯
¯
z2 + z + 1
¯
= 1.3889
k1 =
(z − 0.9) (z − 0.8) ¯z=0
¯
z 2 + z + 1 ¯¯
= 30.111
k2 =
z (z − 0.8) ¯z=0.9
and
¯
z 2 + z + 1 ¯¯
k3 =
= −30.5
z (z − 0.9) ¯z=0.8
Partial fraction form of the system function is
H (z) = 1.3889 +
30.111 z 30.5 z
−
z − 0.9
z − 0.8
The impulse response is found as
h[n] = 1.3889 δ[n] + 30.1111 (0.9)n u[n] − 30.5 (0.8)n u[n]
d.
Take the z-transform of both sides of the difference equation:
Y (z) = z −1 Y (z) − 0.11 z −2 Y (z) − 0.07 z −3 Y (z) + z −1 X (z)
The system function is
H (z) =
z −1
z2
Y (z)
=
=
X (z) 1 − z −1 + 0.11 z −2 + 0.07 z −3 z 3 − z 2 + 0.11 z + 0.07
In factored form the system function is
H (z) =
z2
(z − 0.7) (z − 0.5) (z + 0.2)
Dividing H (z) by z we get
The residues are
and
H (z)
z
=
z
(z − 0.7) (z − 0.5) (z + 0.2)
¯
¯
z
¯
k1 =
= 2.7222
(z − 0.5) (z + 0.2) ¯z=0.7
¯
¯
z
¯
= −1.7857
k2 =
(z − 0.7) (z + 0.2) ¯z=0.5
¯
¯
z
¯
k3 =
= 0.0635
(z − 0.7) (z − 0.5) ¯z=−0.2
Partial fraction form of the system function is
H (z) =
2.7222 z 1.7857 z 0.0635 z
−
+
z − 0.7
z − 0.5
z + 0.2
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
300
The impulse response is
h[n] = 2.7222 (0.7)n u[n] − 1.7857 (0.5)n u[n] + 0.0635 (−0.2)n u[n]
8.27.
a. Begin by writing H (z) using negative powers of z:
H (z) =
z −1 + z −2
Y (z)
=
X (z) 1 + 5 z −1 + 6 z −2
The relationship between X (z) and Y (z) is
¤
¤
£
£
Y (z) 1 + 5 z −1 + 6 z −2 = X (z) z −1 + z −2
or, equivalently
Y (z) + 5 z −1 Y (z) + 6 z −2 Y (z) = z −1 X (z) + z −2 X (z)
Taking inverse transforms of both sides leads to
y[n] + 5 y[n − 1] + 6 y[n − 2] = x[n − 1] + x[n − 2]
b.
Multiply out numerator and denominator factors, and write the result using negative powers of z
to obtain
1 − 2 z −1 + z −2
Y (z)
=
H (z) =
11 −1 3 −2 4 −3
X (z)
z − z −
z
1+
30
5
15
Using the same steps as in part (a)
¸
·
£
¤
11 −1 3 −2 4 −3
Y (z) 1 +
= X (z) 1 − 2 z −1 + z −2
z − z −
z
30
5
15
Y (z) +
3
4 −3
11 −1
z Y (z) − z −2 Y (z) −
z Y (z) = X (z) − 2 z −1 X (z) + z −2 X (z)
30
5
15
and
y[n] +
c.
11
3
4
y[n − 1] − y[n − 2] −
y[n − 3] = x[n] − 2 x[n − 1] + x[n − 2]
30
5
15
The system function is
H (z) =
which leads to
z −1 + z −3
Y (z)
=
X (z) 1 + 1.2 z −1 − 1.8 z −3
£
¤
£
¤
Y (z) 1 + 1.2 z −1 − 1.8 z −3 = X (z) z −1 + z −3
and to the corresponding difference equation
y[n] + 1.2 y[n − 1] − 1.8 y[n − 3] = x[n − 1] + x[n − 3]
301
8.28.
a. Take the z-transform of both sides of the difference equation:
Y (z) = −0.1 z −1 Y (z) + 0.56 z −2 Y (z) + X (z) − 2 z −1 X (z)
The system function is
H (z) =
b.
Y (z)
1 − 2 z −1
z (z − 2)
=
=
X (z) 1 + 0.1 z −1 − 0.56 z −2 (z + 0.8) (z − 0.7)
Division of H (z) by z yields
H (z)
k2
k1
(z − 2)
=
+
=
z
(z + 0.8) (z − 0.7) z + 0.8 z − 0.7
The residues are found as
¯
z − 2 ¯¯
= 1.8667
k1 =
z − 0.7 ¯z=−0.8
and
k2 =
¯
z − 2 ¯¯
= −0.8667
z + 0.8 ¯z=0.7
Partial fraction form of the system function is
H (z) =
1.8667 z 0.8667 z
−
,
z + 0.8
z − 0.7
ROC:
and the impulse response is found as
¯ ¯
¯z ¯ > 0.8
h[n] = 1.8667 (−0.8)n u[n] − 0.8667 (0.7)n u[n]
c.
If x[n] = u[n] then
X (z) =
and the transform of the output signal is
Y (z) = H (z) X (z) =
z
,
z −1
ROC:
¯ ¯
¯z ¯ > 1
z 2 (z − 2)
(z + 0.8) (z − 0.7) (z − 1)
ROC:
Partial fraction form of Y (z) is
Y (z) =
¯ ¯
¯z ¯ > 1
0.8296 z 2.0222 z 1.8519 z
+
−
z + 0.8
z − 0.7
z −1
which leads to the output signal
£
¤
y[n] = 0.8296 (−0.8)n + 2.0222 (0.7)n − 1.8519 u[n]
d.
If x[n] = u[−n] then
X (z) =
−1
,
z −1
ROC:
¯ ¯
¯z ¯ < 1
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
302
and the transform of the output signal is
Y (z) = H (z) X (z) =
−z (z − 2)
(z + 0.8) (z − 0.7) (z − 1)
¯ ¯
ROC: 0.8 < ¯z ¯ < 1
Partial fraction form of Y (z) is
Y (z) =
1.0370 z 2.8889 z 1.8519 z
−
+
z + 0.8
z − 0.7
z −1
which leads to the output signal
y[n] = 1.0370 (−0.8)n u[n] − 2.8889 (0.7)n u[n] − 1.8519 u[−n − 1]
8.29.
a. Take the z-transform of both sides of the difference equation:
1
1
1
5
Y (z) = − z Y (z) − z 2 Y (z) + z X (z) + z 2 X (z)
6
6
6
6
The system function is
H (z) =
b.
Y (z)
z (z + 1)
,
=
X (z) (z + 2) (z + 3)
ROC:
¯ ¯
¯z ¯ < 2
Division of H (z) by z yields
k2
k1
H (z)
(z + 1)
=
+
=
z
(z + 2) (z + 3) z + 2 z + 3
The residues are found as
¯
z + 1 ¯¯
k1 =
= −1
z + 3 ¯z=−2
and
¯
z + 1 ¯¯
k2 =
=2
z + 2 ¯z=−3
Partial fraction form of the system function is
H (z) =
2z
−z
+
,
z +2 z +3
ROC:
and the impulse response is found as
¯ ¯
¯z ¯ < 2
h[n] = (−2)n u[−n − 1] − 2 (−3)n u[−n − 1]
c.
If x[n] = u[n] then
X (z) =
and the transform of the output signal is
Y (z) = H (z) X (z) =
z
,
z −1
ROC:
z 2 (z + 1)
(z − 1) (z + 2) (z + 3)
¯ ¯
¯z ¯ > 1
¯ ¯
ROC: 1 < ¯z ¯ < 2
303
Partial fraction form of Y (z) is
Y (z) =
which leads to the output signal
y[n] =
d.
If x[n] = u[−n] then
z −1
+
− 23 z
z +2
+
3
2z
z +3
2
3
1
u[n] + (−2)n u[−n − 1] − (−3)n u[−n − 1]
6
3
2
X (z) =
and the transform of the output signal is
Y (z) = H (z) X (z) =
−1
,
z −1
ROC:
¯ ¯
¯z ¯ < 1
−z (z + 1)
(z − 1) (z + 2) (z + 3)
Partial fraction form of Y (z) is
Y (z) =
which leads to the output signal
y[n] =
1
z
6
− 16 z
z −1
+
− 13 z
z +2
+
ROC:
¯ ¯
¯z ¯ < 1
1
2z
z +3
1
1
1
u[−n − 1] + (−2)n u[−n − 1] − (−3)n u[−n − 1]
6
3
2
8.30.
a.
The transform of the causal sinusoidal signal is
X (z) =
sin (0.01) z
sin (0.01) z
¢¡
¢
=¡
j
z 2 − 2 cos (0.01) z + 1
z − e 0.01 z − e − j 0.01
and the transform of the output signal is
Y (z) = H (z) X (z) =
=
0.04 sin (0.01) z 2
¡
¢¡
¢
(z − 0.96) z − e j 0.01 z − e − j 0.01
0.2264 z 0.4856 e − j 1.8061 z 0.4856 e j 1.8061 z
+
+
z − 0.96
z − e j 0.01
z − e − j 0.01
The output signal is
y[n] =0.2264 (0.96)n u[n] + 0.4856 e − j 1.8061 e j 0.01n + 0.4856 e j 1.8061 e − j 0.01n
=0.2264 (0.96)n u[n] + 0.9713 cos (0.01n − 1.8061) u[n]
b.
The system function evaluated at ω0 = 0.01 rad/s is
¯
¯
H (0.01) = H (z)¯
= 0.9445 − j 0.2264 = 0.9713 e − j 0.2353
z=e j 0.01
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
304
and the steady-state response of the system is
y ss [n] =0.9713 sin (0.01n − 0.2353)
=0.9713 cos (0.01n − 0.2353 − π/2)
=0.9713 cos (0.01n − 1.8061)
c.
The difference between the two responses is the transient term
y t [n] = 0.2264 (0.96)n u[n]
It can be shown that y t [140] ≈ 0.0007 which is less than 0.1 percent of the amplitude of the steady-state
response.
8.31.
a. Starting with
©
ª
z [z − cos (0.2 π)]
Z cos (0.2 πn) u[n] = 2
z − 2 cos (0.2 π) z + 1
z (z − 0.8090)
= 2
z − 1.6180 z + 1
and using the multiplication by an exponential property of the z-transform, the system function is
obtained as
³ z ´ h³ z ´
i
−
0.8090
ª
©
Z (0.8)n cos (0.2 πn) u[n] = ³ 0.8´2 0.8
³ z ´
z
+1
− 1.6180
0.8
0.8
z (z − 0.6472)
= 2
z − 1.2944 z + 0.64
b.
Im {z}
The zeros of the system function are at
Unit circle
z 1 = 0 , z 2 = 0.6472
and its poles are at
Re {z}
p 1,2 = 0.6472 ± 0.4702
Pole zero plot is shown. Poles are inside the
unit circle. Since h[n] is causal, it is also stable.
305
c.
The difference equation for the system is
y[n] − 1.2944 y[n − 1] + 0.64 y[n − 2] = x[n] − 0.5178 x[n − 1]
d.
The z-transform of the output signal is
Y (z) =H (z) X (z)
z 2 (z − 0.6472)
= 3
z − 2.2944 z 2 + 1.9344 z − 0.64
¡
¢
¡
¢
−0.0104 − j 0.6804 z
−0.0104 + j 0.6804 z
1.0209 z
+
+
=
z −1
z − 0.6472 − j 0.4702
z − 0.6472 + j 0.4702
=
1.0209 z 0.6804 e − j 1.5861 z 0.6804 e j 1.5861 z
+
+
z −1
z − 0.8 e j 0.6283
z − 0.8 e − j 0.6283
The output signal is
´n
´n
³
³
y[n] =1.0209 u[n] + 0.6804 e − j 1.5861 0.8 e j 0.6283 u[n] + 0.6804 e j 1.5861 0.8 e j 0.6283 u[n]
=1.0209 u[n] + 1.3608 (0.8)n cos (0.6283n − 1.5861)
8.32.
a.
X (z) =
Y (z) =
The system function is
z
z − 12
3z
z − 12
+
,
ROC:
|z| >
1
2
2z
z − 34
¡
¢
5 z z − 13
20
=¡
¢¡
¢,
z − 12 z − 43
ROC:
¡
¢
13
Y (z) 5 z − 20
=
H (z) =
,
X (z)
z − 34
|z| >
3
4
ROC:
|z| >
|z| >
1
2
3
4
The system is both causal and stable.
b.
X (z) =
z
z − 12
,
ROC:
The output signal may be written as
y[n] =
µ ¶ µ ¶n+2
µ ¶ µ ¶n+1
1 1
1 1
u[n + 2] +
u[n + 1]
2 2
2 2
306
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
and leads to the transform
! µ ¶ Ã
!
µ ¶ Ã
z
1
z
1 2
z
z
+
Y (z) =
2
2
z − 21
z − 21
¡1¢ 2
z (z + 1)
1
= 2
,
ROC:
< |z| < ∞
1
2
z−2
The system function is
µ ¶
1
Y (z)
z (z + 1) ,
=
H (z) =
X (z)
2
ROC:
|z| < ∞
and the impulse response is
h[n] = { 0.5 , 0.5 }
↑
n=−1
The system is stable but not causal.
c.
z
,
ROC:
z −1
·
¸
z
1
−1
Y (z) = z
,
=
2
(z − 1)
(z − 1)2
X (z) =
The system function is
H (z) =
The system is causal but not stable.
d.
Y (z)
1
=
,
X (z) z (z − 1)
|z| > 1
ROC:
ROC:
z −1
0.25 z
=
,
z − 0.8 z − 0.8
z
Y (z) =
,
ROC:
z − 0.8
X (z) = 1.25 −
The system function is
H (z) =
The system is causal but not stable.
Y (z)
z
=
,
X (z) z − 1
|z| > 1
ROC:
ROC:
|z| > 1
Y (z) = H1 (z) E (z)
E (z) = X (z) + H2 (z) Y (z)
Combining the two relationships we get
Y (z) = H1 (z) [X (z) + H2 (z) Y (z)]
H (z) =
|z| > 0.8
|z| > 0.8
8.33.
a.
Therefore
|z| > 1
H1 (z)
Y (z)
=
X (z) 1 − H1 (z) H2 (z)
307
b.
⇒
H1 (z) =
h 2 [n] = K δ[n − 1]
⇒
H2 (z) = K z −1 =
H1 (z)
=
H (z) =
1 − H1 (z) H2 (z)
c.
z
z −1
h 1 [n] = u[n]
K
z
z
z
z −1
³ z ´ µK ¶ = z −1−K
1−
z −1 z
For a causal system, the ROC of the system function found in part (b) is
¯ ¯ ¯
¯
¯ z ¯ > ¯1 + K ¯
ROC:
For the system to be stable, the ROC must include the unit circle of the z-plane. This requires
d.
¯
¯
¯1 + K ¯ < 1
−1 < 1 + K < 1
⇒
⇒
−2 < K < 0
With K = 3/2 the system function is
z
H (z) =
The transform of the unit step response is
Ã
Y (z) =
z
z − 52
Partial fraction form of Y (z) is
Y (z) =
and the unit-step response is
!
z − 25
³ z ´
z2
=¡
¢
z −1
z − 25 (z − 1)
1.6667 z
z − 25
−
0.6667 z
z −1
·
µ ¶n ¸
5
y[n] = − 0.6667 + 1.6667
u[n]
2
8.34.
Using the z-transform definition we have
H (z 0 ) =
and
∞
X
n=−∞
h[n] z 0−n
∞
X
¡ ¢
¡ ¢−n
H z 0∗ =
h[n] z 0∗
n=−∞
Conjugating both sides of the transform relationship for H (z 0 ) yields
£
¤∗
H (z 0 ) =
·
∞
X
n=−∞
h[n] z 0−n
¸∗
=
∞
X
n=−∞
¡ ¢−n
(h[n])∗ z 0∗
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
308
Since h[n] is specified to be real, (h[n])∗ = h[n], and therefore
£
H (z 0 )
¤∗
=
∞
X
n=−∞
¡ ¢
¡ ¢−n
= H z 0∗
h[n] z 0∗
which implies that
¯
¯ ¯ ¡ ¢¯
¯H (z 0 )¯ = ¯ H z ∗ ¯ = H0
and
0
¡
¢
∡ H z 0∗ = −∡ H (z 0 ) = −Θ0
8.35.
a.
The input signal can be expressed as x[n] = z 0n with
z 0 = 0.8 e j 0.4π
The steady-state response of the system for a complex exponential signal is
y[n] = H (z 0 ) x[n] = H (z 0 ) z 0n
The system function can be evaluated at z = z 0 to yield
Therefore
¡
¢2
¡
¢
³
´
0.8 e j 0.4π + 3 0.8 e j 0.4π
j 0.4π
=¡
H (z 0 ) = H 0.8 e
= 3.8717 e j 3.0777
¢2
¡
¢
0.8 e j 0.4π − 1.4 0.8 e j 0.4π + 0.85
y[n] = 3.8717 (0.8)n e j (0.4 πn+3.0777)
b.
Let z 1 = (0.9)n e 0.3π . The input signal can be expressed as
x[n] =
1 n 1 ¡ ∗ ¢n
z
z +
2 1 2 1
The system function can be evaluated at z = z 1 and at z = z 1∗ to yield
and
¡
¢2
¡
¢
³
´
0.9 e j 0.3π + 3 0.9 e j 0.3π
j 0.3π
= 11.3347 e − j 3.0515
=¡
H (z 1 ) = H 0.9 e
¢2
¡
¢
0.9 e j 0.3π − 1.4 0.9 e j 0.3π + 0.85
´
³
´
³
¡ ¢
H z 1∗ = H 0.9 e − j 0.3π = H ∗ 0.9 e j 0.3π = 11.3347 e j 3.0515
¡ ¢
In the last step we recognized that H (z) has only real coefficients and therefore H z 1∗ = H ∗ (z 1 ) (see
Problem 8.34). The output signal is
µ ¶
³
´n µ 1 ¶
³
´n
1
− j 3.0515
j 0.3π
y[n] =
11.3347 e
0.9 e
+
11.3347 e j 3.0515 0.9 e − j 0.3π
2
2
= 11.3347 (0.9)n cos (0.3πn − 3.0515)
309
8.36.
a.
Im {z}
Unit circle
The zero of the system function is at
z 1 = 0.4
and its poles are at
p1 = 0 ,
Re {z}
p 2,3 = 0.7 ± 0.6
Pole zero plot is shown.
b.
Im {z}
¯ ¯
¯ A1¯ =
¯ ¯
¯B 1 ¯ =
¯ ¯
¯B 2 ¯ =
¯ ¯
¯B 3 ¯ =
−→
B2
p2
−→
B1
π/6
−→
A1
p1
Re {z}
z1
0.6835 ,
∡ A 1 = 0.8206 rad
1,
∡ B 1 = 0.5236 rad
0.1938 ,
∡ B 2 = −0.5421 rad
1.1125 ,
∡ B 3 = 1.4210 rad
The magnitude and the phase of the system function are
−→
B3
p3
and
¯
¯
¯ H (π/6)¯ =
0.6835
= 3.1701
(1) (0.1938) (1.1125)
∡ H (π/6) = 0.8206−0.5236+0.5421−1.4210 = −0.5819 rad
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
310
c.
Im {z}
¯ ¯
¯ A1¯ =
¯ ¯
¯B 1 ¯ =
¯ ¯
¯B 2 ¯ =
¯ ¯
¯B 3 ¯ =
−→
B2
p2
−→
B1
π/4
−→
A1
p1
Re {z}
z1
0.7709 ,
∡ A 1 = 1.1611 rad
1,
∡ B 1 = 0.7854 rad
0.1073 ,
∡ B 2 = 1.5045 rad
1.3071 ,
∡ B 3 = 1.5654 rad
The magnitude and the phase of the system function are
−→
B3
p3
and
¯
¯
¯ H (π/4)¯ =
0.7709
= 5.4944
(1) (0.1073) (1.3071)
∡ H (π/4) = 1.1611−0.7854−1.5045−1.5654 = −2.6942 rad
8.37.
a. The system has zeros at z = ±2 and a pole at z = −1/2. The system function has a second-order
numerator and a first-order denominator, and therefore does not converge at |z| → ∞. The ROC may be
one of the following:
¯ ¯ 1
¯z ¯ <
: Neither stable nor causal
2
1 ¯¯ ¯¯
< z < ∞ : Stable, but not causal
2
In the first case the ROC includes neither the unit circle nor infinity. In the second case it includes the
unit circle but excludes infinity. Consequently, this cannot be the system function of a system that is
both stable and causal.
b. The system has two zeros at z = −1 and poles at z = −1/3 and z = 1/2. The ROC may be one of the
following:
¯ ¯ 1
¯z ¯ <
3
1 ¯¯ ¯¯ 1
< z <
3
2
¯ ¯ 1
¯z ¯ >
2
: Neither stable nor causal
: Neither stable nor causal
: Stable and causal
This system function could correspond to a system that is both stable and causal.
c.
The system has zeros at z = 0 and z = −1. Its poles are at z = −1/2 and z = −3/2. The ROC may be
311
one of the following:
¯ ¯ 1
¯z ¯ <
2
1 ¯¯ ¯¯ 3
< z <
2
2
¯ ¯ 3
¯z ¯ >
2
: Neither stable nor causal
: Stable, but not causal
: Causal, but not stable
This cannot be the system function of a system that is both stable and causal.
8.38.
a. Taking the z-transform of the difference equation leads to
Y (z) = 1.5 z −1 Y (z) − 0.54 z −2 Y (z) + X (z) + 3 z −1 X (z)
from which the system function can be obtained as
H (z) =
Y (z)
1 + 3 z −1
=
X (z) 1 − 1.5 z −1 + 0.54 z −2
The factored form of the system function is
H (z) =
z (z + 3)
(z − 0.6) (z − 0.9)
and the pole-zero diagram is shown below.
Im {z}
Unit circle
Re {z}
−3
0
0.6 0.9
Since the system is specified to be causal, the ROC of the system function is
¯ ¯
¯z ¯ > 0.9
The ROC includes the unit circle, therefore the system is stable.
b.
Taking the z-transform of the difference equation leads to
Y (z) = −0.64 z −2 Y (z) + 2 X (z)
312
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
from which the system function can be obtained as
H (z) =
2
2 z2
Y (z)
¡
¢¡
¢
=
=
X (z) 1 + 0.64 z −2
z − j 0.8 z + j 0.8
The pole-zero diagram is shown below.
Im {z}
Unit circle
j0.8
Two zeros
Re {z}
0
−j0.8
Since the system is specified to be causal, the ROC of the system function is
¯ ¯
¯z ¯ > 0.8
The ROC includes the unit circle, therefore the system is stable.
c.
Taking the z-transform of the difference equation leads to
Y (z) = 0.25 z −1 Y (z) − 0.125 z −2 Y (z) − 0.5 z −3 Y (z) + X (z)
from which the system function can be obtained as
H (z) =
or, using non-negative powers of z
1
Y (z)
=
−1
X (z) 1 − 0.25 z + 0.125 z −2 + 0.5 z −3
H (z) =
z3
z 3 − 0.25 z 2 + 0.125 z + 0.5
There are three zeros at z = 0. Poles are at z = −0.6718 and z = 0.4609 ± 0.7293. The pole-zero diagram is
shown below.
Im {z}
Unit circle
Three zeros
Re {z}
−0.67 0
313
Since the system is specified to be causal, the ROC of the system function is
¯ ¯
¯z ¯ > 0.8627
The ROC includes the unit circle, therefore the system is stable.
d.
Taking the z-transform of the difference equation leads to
Y (z) = 0.25 z −1 Y (z) − 0.5 z −2 Y (z) − 0.75 z −3 Y (z) + X (z) + z −1 X (z)
from which the system function can be obtained as
H (z) =
1 + z −1
Y (z)
=
X (z) 1 − 0.25 z −1 + 0.5 z −2 + 0.75 z −3
or, using non-negative powers of z
H (z) =
z 2 (z + 1)
z 3 − 0.25 z 2 + 0.5 z + 0.75
The factored form of the system function is
H (z) =
z 2 (z + 1)
¡
¢¡
¢
(z + 0.6709) z − 0.4605 − j 0.9518 z − 0.4605 + j 0.9518
and the pole-zero diagram is shown below.
Im {z}
Unit circle
Two zeros
Re {z}
−1 −0.67 0
Since the system is specified to be causal, the ROC of the system function is
¯ ¯
¯z ¯ > 1.0573
The ROC does not include the unit circle, therefore the system is not stable.
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
314
8.39.
a.
Im {z}
The system function is
H (z) =
Unit circle
−z
z − (1 + c)
It has a zero at z 1 = 0 and a pole at p 1 = 1 + c.
The ROC of the system function is |z| > 1 + c
which excludes the unit circle for any positive
interest rate c. Therefore, the system is unstable for any positive interest rate c.
b.
1+c
Re {z}
Consider the input signal
x[n] = −A δ[n]
which corresponds to borrowing the amount A initially and never making a payment. The transform of
the input signal is X (z) = −A, and the transform of the output signal is
Y (z) = H (z) X (z) =
Az
z − (1 + c)
which corresponds to the output signal
y[n] = A (1 + c)n u[n]
Alternatively, let
x[n] = −A δ[n] + D u[n − 1]
The transform of this input signal is
The output transform is
µ
¶
A +D
−A z −
A
X (z) =
z −1
¶
A +D
Az z−
A
Y (z) = H (z) X (z) =
(z − 1) (z − 1 − c)
µ
which can be expressed in partial fraction form as
Y (z) =
The output signal is
µ
(D/c) z (A − D/c) z
+
z −1
z −1−c
¶
µ
¶
D
D
u[n] + A −
(1 + c)n u[n]
c
c
If A − D/c 6= 0, the output signal would grow unbounded even though the input signal is bounded.
315
c.
For the second input signal in part (b), let
A−
D
=0
c
D = Ac
⇒
This results in the output signal x[n] = A u[n]. From a practical perspective this corresponds to paying
exactly the amount of interest each month, and yields a constant monthly balance. The result found is
intuitively satisfying.
8.40.
Frequency response of the allpass filter is
¯
¯
H (Ω) = H (z)¯
z=e j Ω
=
e j Ω − r e j Ω0
e j Ω − (1/r ) e j Ω0
Multiply both the numerator and the denominator with the complex conjugate of the denominator:
¯
¯
H (z)¯
z=e j Ω
=
µ
e j Ω − r e j Ω0
e j Ω − (1/r ) e j Ω0
¶ µ −jΩ
¶
e
− (1/r ) e − j Ω0
e − j Ω − (1/r ) e − j Ω0
2 − (1/r ) cos (Ω − Ω0 ) − r cos (Ω − Ω0 ) − j (1/r ) sin (Ω − Ω0 ) + j r sin (Ω − Ω0 )
1 + 1/r 2 − (2/r ) cos (Ω − Ω0 )
¡ 2
¢
¡ 2
¢
r − 1 sin (Ω − Ω0 )
2r − r + 1 cos (Ω − Ω0 )
¢
¢
=¡ 2
+j ¡ 2
r + 1 /r − 2 cos (Ω − Ω0 )
r + 1 /r − 2 cos (Ω − Ω0 )
=
The phase characteristic is
∡ H (Ω) = tan−1
·
" ¡
#
¢
¸
2
r
−
1
sin
−
Ω
)
(Ω
Im {H (Ω)}
0
¡
¢
= tan−1
Re {H (Ω)}
2r − r 2 + 1 cos (Ω − Ω0 )
8.41.
a.
W (z) 1
X(z)
b
−1
1+2z
H (z) =
1 − (1/2) z −1
1/2
z −1
2
b
Y (z)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
316
b.
W (z)
X(z)
−0.8
H (z) =
−1
Y (z)
z −1
1
b
−3
z +z
1 + 0.8 z −1 − 2.2 z −2 + 0.6 z −3
z −1
2.2
b
−0.6
z −1
1
b
c.
W (z) 0.5
X(z)
H (z) =
−1
b
−0.4
−2
1+2z −3z +4
2 + 0.8 z −1 + 1.8 z −2 + 3.2 z −3
z −1
1
b
0.5 + z −1 − 1.5 z −2 + 2
=
1 + 0.4 z −1 + 0.9 z −2 + 1.6 z −3
−0.9
z −1
−1.5
b
−1.6
z −1
2
b
8.42.
a.
H (z) = H1 (z) H2 (z)
H1 (z) =
H2 (z) =
z +1
1 + z −1
=
z + 1/2 1 + (1/2) z −1
z −1
1
=
z + 2/3 1 + (2/3) z −1
1
X (z)
Y (z)
b
−1/2
z −1
−2/3
1
z −1
1
b
b
H1 (z)
H2 (z)
Y (z)
317
b.
H (z) = H1 (z) H2 (z)
H1 (z) =
H2 (z) =
z
1
=
z − 0.4 1 − 0.4 z −1
z +1
1 + z −1
=
z + 0.7 1 + 0.7 z −1
1
X (z)
1
Y (z)
b
b
z −1
−0.7
0.4
z −1
1
b
H1 (z)
H2 (z)
c.
H (z) = H1 (z) H2 (z)
H1 (z) =
z
1
=
z + 0.6 1 + 0.6 z −1
z −1 + z −2
z +1
=
z 2 − 1.4, z + 0.85 1 − 1.4, z −1 + 0.85 z −2
H2 (z) =
1
X (z)
Y (z)
b
−0.6
z −1
z −1
1.4
1
b
H1 (z)
−0.85
z −1
1
b
H2 (z)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
318
8.43.
a.
b
The system function can be written as
−1/2
z −1
3
b
H (z) = H̄1 (z) + H̄2 (z)
=
X(z)
H̄1 (z)
3
−2
+
z + 1/2 z + 2/3
Y (z)
b
3 z −1
−2 z −1
=
+
1 + (1/2) z −1 1 + (2/3) z −1
−2/3
z −1
−2
b
H̄2 (z)
b.
1.2727
b
z −1
The system function can be written as
0.4
X(z)
Y (z)
H (z) = H̄1 (z) + H̄2 (z)
H̄1 (z)
1.2727 z −0.2727 z
+
=
z − 0.4
z + 0.7
1.2727
−0.2727
=
+
−1
1 − 0.4 z
1 + 0.7 z −1
−0.2727
b
−0.7
z −1
H̄2 (z)
c. Write H (z) as
H (z) = H̄1 (z) + H̄2 (z)
¡
¢
¡
¢
−0.0976 − j 1.0447 z
−0.0976 + j 1.0447 z
0.1951 z
+
+
=
z + 0.6
z − 0.7 − j 0.6
z − 0.7 + j 0.6
=
=
0.1951 z −0.1951 z 2 + 1.3902 z
+
z + 0.6
z 2 − 1.4 z + 0.85
−0.1951 + 1.3902 z −1
0.1951
+
1 + 0.6 z −1 1 − 1.4 z −1 + 0.85 z −2
319
0.1951
b
−0.6
z −1
X(z)
H̄1 (z)
Y (z)
−0.1951
b
z −1
1.4
1.3902
b
−0.85
z −1
H̄2 (z)
8.44.
a. Using the definition of the unilateral z-transform
∞
©
ª X
Z u x[n + 1] =
x[n + 1] z −n
n=0
Let a new variable m be defined as m = n + 1:
∞
©
ª X
Z u x[n + 1] =
x[m] z −m+1
m=1
It is possible to start the summation at m = 0 by adding and subtracting the missing term:
∞
©
ª X
Z u x[n + 1] =
x[m] z −m+1 − z x[0]
m=0
=z
∞
X
m=0
x[m] z −m − z x[0]
=z X u (z) − z x[0]
b.
∞
©
ª X
Z u x[n + 2] =
x[n + 2] z −n
n=0
Let a new variable m be defined as m = n + 2:
∞
©
ª X
Z u x[n + 2] =
x[m] z −m+2
m=2
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
320
It is possible to start the summation at m = 0 by adding and subtracting the two missing terms:
∞
©
ª X
x[m] z −m+2 − z 2 x[0] − z x[1]
Z u x[n + 2] =
m=0
=z 2
∞
X
m=0
x[m] z −m − z 2 x[0] − z x[1]
=z X u (z) − z 2 x[0] − z x[1]
2
c.
∞
©
ª X
x[n + k] z −n
Z u x[n + k] =
n=0
Let a new variable m be defined as m = n + k:
∞
©
ª X
x[m] z −m+k
Z u x[n + k] =
m=k
Since
∞
X
m=0
x[m] z −m+k =
=
k−1
X
x[m] z −m+k +
k−1
X
x[m] z k−m + z k X u (z)
m=0
m=0
∞
X
x[m] z −m+k
m=k
∞
k−1
X
©
ª X
Z u x[n + k] =
x[m] z −m+k = z k X u (z) −
x[n] z k−n
m=k
8.45.
a.
n=0
¢
¢
¡
¡
Yu (z) − 1.4 z −1 Yu (z) + y[−1] + 0.85 z −2 Yu (z) + z −1 y[−1] + y[−2] = 0
Using initial values we obtain
Yu (z) =
1.05 − 4.25 z −1
z (1.05 z − 4.25)
= 2
−1
−2
1 − 1.4 z + 0.85 z
z − 1.4 z + 0.85
The partial fraction form of Yu (z) is
¡
¢
¡
¢
0.5250 − j 2.9292 z
0.5250 + j 2.9292 z
+
Yu (z) =
z − 0.7 − j 0.6
z − 0.7 + j 0.6
´n
´n
¡
¢³
¡
¢³
y[n] = 0.5250 + j 2.9292 0.9220 e j 0.7086 u[n] + 0.5250 − j 2.9292 0.9220 e − j 0.7086 u[n]
=1.05 (0.9220)n cos (0.7086n) u[n] + 5.8583 (0.9220) n sin (0.7086n) u[n]
b.
¢
¢
¡
¡
Yu (z) − 1.6 z −1 Yu (z) + y[−1] + 0.64 z −2 Yu (z) + z −1 y[−1] + y[−2] = 0
321
Using initial values we obtain
Yu (z) =
5.12 − 1.28 z −1
z (5.12 z − 1.28) z (5.12 z − 1.28)
= 2
=
−1
−2
1 − 1.6 z + 0.64 z
z − 1.6 z + 0.64
(z − 0.8)2
The partial fraction form of Yu (z) is
Yu (z) =
2.8160 z
(z − 0.8)
2
+
5.12 z
z − 0.8
µ
¶
2.8160
y[n] =
n (0.8)n u[n] + 5.12 (0.8)n u[n]
0.8
8.46.
a. The transform of the input signal is
X u (z) =
z (z − cos (0.2π))
z 2 − 2 cos (0.2π) z + 1
=
z (z − 0.8090)
z 2 − 1.6180 z + 1
Using the unilateral z-transform we have
¤
£
Yu (z) − 2 z −1 Yu (z) + 5 = X u (z)
Solving for Yu (z) yields
¡
¢
11 z z 2 − 1.5445 z + 0.9091
¡
¢
Yu (z) =
(z − 2) z 2 − 1.6180 z + 1
The partial fraction form of Yu (z) is
Yu (z) =
11.3503 z 0.3765 e − j 2.0548 0.3765 e j 2.0548
+
+
z −2
z − e j 0.6283
z − e − j 0.6283
and the output signal is
y[n] = 11.3503 (2)n u[n] + 0.3765 e − j 2.0548 e j 0.6283 n + 0.3765 e j 2.0548 e − j 0.6283 n
= 11.3503 (2)n u[n] + 0.7530 cos (0.6283 n − 2.0548) u[n]
b.
The transform of the input signal is
X u (z) =
z
z −1
Using the unilateral z-transform we have
Solving for Yu (z) yields
¤
£
Yu (z) + 0.6 z −1 Yu (z) − 3 = X u (z) + z −1 X u (z)
Yu (z) =
z +1
1.8 z
2.8 z (z − 0.2857)
X u (z) +
=
z + 0.6
z + 0.6
(z − 1) (z + 0.6)
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
322
The partial fraction form of Yu (z) is
Yu (z) =
and the output signal is
c.
1.25 z 1.55 z
+
z − 1 z + 0.6
¤
£
y[n] = 1.25 + 1.55 (−0.6)n u[n]
The transform of the input signal is
X u (z) =
z
z −1
Using the unilateral z-transform we have
¤
¤
£
£
Yu (z) − 0.2 z −1 Yu (z) − 1 − 0.48 z −2 Yu (z) + 2 − z −1 = X u (z)
Solving for Yu (z) yields
z2
0.76 z 2 − 0.48 z
+
X
(z)
u
z 2 − 0.2 z − 0.48
z 2 − 0.2 z − 0.48
¡
¢
z 1.76 z 2 − 1.24 z + 0.48
¡
¢
=
(z − 1) z 2 − 0.2 z − 0.48
Yu (z) =
The partial fraction form of Yu (z) is
Yu (z) =
3.125 z 2.1943 z 0.8293 z
−
+
z −1
z − 0.8
z + 0.6
and the output signal is
¤
£
y[n] = 3.125 − 2.1943 (0.8)n + 0.8293 (−0.6)n u[n]
8.47.
a.
y[n] = y[n − 1] + y[n − 2]
y[n] − y[n − 1] − y[n − 2] = 0 ,
⇒
y[1] = 1 ,
b. Using unilateral z-transform yields
and
¢
¢ ¡
¡
Yu (z) − z −1 Yu (z) − y[−1] − z −2 Yu (z) − z −1 y[−1] − y[−2] = 0
1 + z −1
z (z + 1)
=
1 − z −1 − z −2 z 2 − z − 1
The transform can be put into partial fraction form as follows:
Yu (z) =
Yu (z) =
0.17082 z
1.17802 z
−
z − 1.618034 z + 0.618034
The sequence can be found as the inverse transform.
y[n] = 1.17082 (1.618034) n − 0.17082 (−0.618034) n
y[−2] = 0
323
c. Using the result found in part (b), y[n + 1] is
y[n + 1] = 1.17082 (1.618034) n+1 − 0.17082 (−0.618034) n+1
and the ratio of two consecutive numbers in the sequence is
y[n + 1] 1.17082 (1.618034) n+1 − 0.17082 (−0.618034) n+1
=
y[n]
1.17082 (1.618034) n − 0.17082 (−0.618034) n
For large n, the terms (−0.618034) n+1 and (−0.618034) n approach zero. Therefore we have
·
¸
y[n + 1]
1.17082 (1.618034) n+1
ϕ = lim
= lim
= 1.618034
n→∞ y[n]
n→∞
1.17082 (1.618034) n
8.48.
a. Begin with
©
ª
Z u[n] − u[n − 8] =
z8 − 1
z 7 (z − 1)
Using the multiplication by an exponential signal property of the z-transform
©
ª
X (z) = Z (0.8)n (u[n] − u[n − 8]) = ³
³ z ´8
−1
0.8
´
z ´7 ³ z
−1
0.8
0.8
The zeros of the transform X (z) are found by solving
³ z ´8
= 1 e j 2πk ,
k = 0, . . . , 7
z k = 1 e j 2πk/8; ,
k = 1, . . . , 7
0.8
The zeros are at
The zero for k = 1 is canceled by the denominator factor (z − 1) so that there is neither a zero nor a pole
at z = 1. The transform has seven poles at z = 0.
b.
1
2
3
4
5
6
7
8
9
c.
% Anonymous function f o r the transform
X = @( z ) ( ( z / 0 . 8 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 8 ) . ^ 7 ) . * ( ( z / 0 . 8 ) − 1 ) ) ;
% Create a grid o f z values
[ zr , z i ] = meshgrid ( [ − 1 . 5 : 0 . 0 5 : 1 . 5 ] , [ − 1 . 5 : 0 . 0 5 : 1 . 5 ] ) ;
z = ( zr+eps)+ j * ( z i +eps ) ; % Avoid d i v i s i o n by 0
% Evaluate the magnitude | X( z ) |
Xmag = abs (X( z ) ) ;
% Clip the peak
Xmag = Xmag. * ( Xmag<=10)+10. * (Xmag> 10) ;
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
324
1
2
3
4
5
6
7
clf ;
shading interp ;
colormap copper ;
m1 = mesh( zr , zi ,Xmag ) ;
xlabel ( ’ Real ( z ) ’ ) ;
ylabel ( ’ Imag( z ) ’ ) ;
zlabel ( ’ | X( z ) | ’ ) ;
d.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
r = 1;
omg = [ 0 : 0 . 0 0 5 : 1 ] * 2 * pi+eps ;
t r = r * exp ( j *omg ) ; % Circular t r a j e c t o r y
shading interp ;
colormap copper ;
m1 = mesh( zr , zi ,Xmag ) ;
hold on ;
p2 = plot3 ( . . .
real ( t r ) , imag ( t r ) , zeros ( s i z e ( t r ) ) , . . . % Draw unit c i r c l e
real ( t r ) , imag ( t r ) , abs (X( t r ) ) , . . .
% Draw | X( z ) | evaluated on the unit c i r c l e
[ −1.5 ,1.5] ,[0 ,0] ,[0 ,0] ,...
[0 ,0] ,[ −1.5 ,1.5] ,[0 ,0]);
set (m1, ’ FaceAlpha ’ , 0 . 4 , ’ EdgeAlpha ’ , 0 . 6 ) ;
set ( p2 ( 1 ) , ’ Color ’ , [ 0 . 8 , 0 , 0 ] , ’ Linewidth ’ , 1 . 5 ) ;
set ( p2 ( 2 ) , ’ Color ’ , [ 0 , 0 , 1 ] , ’ Linewidth ’ , 1 . 5 ) ;
set ( p2 ( 3 ) , ’ Color ’ , [ 0 , 0 , 0 ] ) ;
set ( p2 ( 4 ) , ’ Color ’ , [ 0 , 0 , 0 ] ) ;
hold o f f ;
xlabel ( ’ Real ( z ) ’ ) ;
ylabel ( ’ Imag( z ) ’ ) ;
zlabel ( ’ | X( z ) | ’ ) ;
8.49.
a. In the script in part (b), modify line 2 as follows:
2
X = @( z ) ( ( z / 0 . 6 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 6 ) . ^ 7 ) . * ( ( z / 0 . 6 ) − 1 ) ) ;
b. In the script in part (b), modify line 2 as follows:
2
X = @( z ) ( ( z / 0 . 4 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 4 ) . ^ 7 ) . * ( ( z / 0 . 4 ) − 1 ) ) ;
8.50.
a. A system object can be obtained as follows:
>> s f = zpk ( [ 0 ] , [ 0 . 9 6 ] , 0 . 0 4 , 1 )
Zero / pole / gain :
325
0.04 z
−−−−−−−−
( z −0.96)
Sampling time : 1
b. The output signal is computed and graphed with the following statements:
>>
>>
>>
>>
n = [0:49];
x = sin ( 0 . 0 1 * n ) ;
y = lsim ( s f , x , n ) ;
stem (n , y )
8.51.
a. The script below computes and displays the impulse response by iterating through the difference
equation:
% S e t i n i t i a l conditions to zero
ynm1 = 0 ; % y [ −1]
ynm2 = 0 ; % y [ −2]
xn = 1 ;
% x [ 0 ] = 1 s i n c e x [ n] i s a unit impulse
xnm1 = 0 ; % x[ −1] = 0 s i n c e x [n ] i s a unit impulse
out = [ ] ; % Empty v e c t o r to s t a r t
for n=0:10 ,
yn = 1.2944 * ynm1−0.64 * ynm2+xn−0.6472 *xnm1 ;
out = [ out , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
xnm1 = xn ;
xn = 0 ; % x [ n ] = 0 f o r n > 0
end ;
% Display output s i g n a l
n = [0:10];
[ n ’ , out ’ ]
For comparison, the specified impulse response can be computed from its analytical expression and
displayed for n = 0, . . . , 10 using the following code:
% A n a ly t i c a l r e s u l t f o r the impulse response
y = ( 0 . 8 ) . ^ n . * cos ( 0 . 2 * pi * n ) ;
[n’ , y ’ ]
b.
The script below computes and displays the unit-step response by iterating through the
difference equation:
% S e t i n i t i a l conditions to zero
ynm1 = 0 ; % y [ −1]
ynm2 = 0 ; % y [ −2]
326
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
xn = 1 ;
% x [ 0 ] = 1 s i n c e x [ n] i s a unit s t e p
xnm1 = 0 ; % x[ −1] = 0 s i n c e x [n ] i s a unit s t e p
out = [ ] ; % Empty v e c t o r to s t a r t
for n=0:10 ,
yn = 1.2944 * ynm1−0.64 * ynm2+xn−0.6472 *xnm1 ;
out = [ out , yn ] ;
ynm2 = ynm1;
ynm1 = yn ;
xnm1 = xn ;
end ;
% Display output s i g n a l
n = [0:10];
[ n ’ , out ’ ]
For comparison, the unit-step response can be computed from the analytical result found in Problem
8.31 and displayed for n = 0, . . . , 10 using the following code:
% Compute and d i s p la y the a n a l y t i c a l s o lu t i o n
y = 1. 0209+ 1. 36 08 * ( 0 . 8 ) . ^ n . * cos ( 0. 6283 * n−1.5861);
[n’ , y ’ ]
c.
The script below creates a system object and uses it to compute and display the impulse response:
sys = t f ( [ 1 , − 0. 6472 , 0] , [ 1 , − 1. 2944 , 0. 64] , − 1)
n = [0:10];
h = impulse ( sys , n ) ;
% Display the impulse response
[n’ ,h]
For the unit-step response, use the following:
y = step ( sys , n ) ;
% Display the unit−s t e p response
[n’ , y ]
8.52.
a. The code listed below can be used for computing and graphing the frequency response H (Ω):
% Anonymous function f o r H( z )
H = @( z ) ( z . * z+3 * z ) . / ( z . * z −1.4 * z + 0 . 8 5 ) ;
% Evaluate H( z ) at z=exp ( j * Omega)
Omg = [ − 1 : 0 . 0 0 1 : 1 ] * pi ;
s f = H( exp ( j *Omg) ) ;
% Graph the magnitude
sfmag = abs ( s f ) ;
plot (Omg, sfmag ) ;
t i t l e ( ’ Magnitude of the system function ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
327
ylabel ( ’ Magnitude ’ ) ;
grid ;
% Graph the phase
sfmag = angle ( s f ) ;
plot (Omg, sfmag ) ;
t i t l e ( ’ Phase of the system function ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
The code below graphs the magnitude and the phase of H (z) evaluated for z = 0.8 e j Ω . Critical points
|H (z 0 )| and ∡ H (z 0 ) are marked on magnitude and phase graphs.
% Evaluate H( z ) at z =0.8 * exp ( j *Omega)
s f1 = H( 0 . 8 * exp ( j *Omg) ) ;
% Graph the magnitude and the phase
sf1mag = abs ( s f1 ) ;
sf1phs = angle ( s f1 ) ;
z0 = 0 . 8 * exp ( j * 0 . 4 * pi ) ;
clf ;
subplot ( 2 , 1 , 1 ) ;
plot (Omg, sf1mag , 0 . 4 * pi , abs (H( z0 ) ) , ’ r * ’ ) ;
t i t l e ( ’H( z ) evaluated at z =0.8 * exp ( j *Omega) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
subplot ( 2 , 1 , 2 ) ;
plot (Omg, sf1phs , 0 . 4 * pi , angle (H( z0 ) ) , ’ r * ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
The code¯ below graphs the magnitude and the phase of H (z) evaluated for z = 0.9 e j Ω . Critical points
¡ ¢
¡ ¢¯
|H (z 1 )|, ¯ H z 1∗ ¯, ∡ H (z 1 ) ∡ H z 1∗ are marked on magnitude and phase graphs.
% Evaluate H( z ) at z =0.9 * exp ( j *Omega)
s f1 = H( 0 . 9 * exp ( j *Omg) ) ;
% Graph the magnitude and the phase
sf1mag = abs ( s f1 ) ;
sf1phs = angle ( s f1 ) ;
z1 = 0 . 9 * exp ( j * 0 . 3 * pi ) ;
clf ;
subplot ( 2 , 1 , 1 ) ;
plot (Omg, sf1mag , [ − 0. 3 * pi , 0 . 3 * pi ] , [ abs (H( conj ( z1 ) ) ) , abs (H( z1 ) ) ] , ’ r * ’ ) ;
t i t l e ( ’H( z ) evaluated at z =0.9 * exp ( j *Omega) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
subplot ( 2 , 1 , 2 ) ;
plot (Omg, sf1phs , [ − 0. 3 * pi , 0 . 3 * pi ] , [ angle (H( conj ( z1 ) ) ) , angle (H( z1 ) ) ] , ’ r * ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
328
b.
The steady state response for the input signal x[n] = (0.8)n e j 0.4πn is computed and graphed as
follows:
% Steady−s t a t e response found in part ( a ) o f Problem 8 . 3 5 :
n = [0:19];
y = H( z0 ) * ( 0 . 8 ) . ^ n . * exp ( j * 0 . 4 * pi * n ) ;
clf ;
subplot ( 2 , 1 , 1 ) ;
stem (n , real ( y ) ) ;
t i t l e ( ’ y [n] ’ ) ;
ylabel ( ’ Real part ’ ) ;
grid ;
subplot ( 2 , 1 , 2 ) ;
stem (n , imag ( y ) ) ;
xlabel ( ’ Sample index n ’ ) ;
ylabel ( ’ Imag . part ’ ) ;
grid ;
The steady state response for the input signal x[n] = (0.9)n cos (0.3πn) is computed and graphed as
follows:
% Steady−s t a t e response found in part ( b ) o f Problem 8 . 3 5 :
n = [0:19];
y = 0 . 5 *H( z1 ) * ( 0 . 9 * exp ( j * 0 . 3 * pi ) ) . ^ n + 0 . 5 *H( conj ( z1 ) ) * ( 0 . 9 * exp(− j * 0 . 3 * pi ) ) . ^ n ;
clf ;
stem (n , real ( y ) ) ; % Take r e a l part as a precaution
t i t l e ( ’ y [n] ’ ) ;
xlabel ( ’ Sample index n ’ ) ;
grid ;
8.53.
% Anonymous function f o r a l l p a s s H( z ) with s p e c i f i e d r and theta
H = @( r , theta , z ) ( z−r * exp ( j * theta ) ) . / ( z−1/ r * exp ( j * theta ) ) ;
% Create v e c t o r z=exp ( j *Omega)
Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi ;
z = exp ( j *Omg) ;
% Evaluate system function f o r r = 0 . 2 , 0 . 4 , 0 . 6 , 0 . 8
H1 = H( 0 . 2 , 0 , z ) ;
H2 = H( 0 . 4 , 0 , z ) ;
H3 = H( 0 . 6 , 0 , z ) ;
H4 = H( 0 . 8 , 0 , z ) ;
plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3) ,Omg, angle (H4 ) ) ;
legend ( ’ r =0.2 ’ , ’ r =0.4 ’ , ’ r =0.6 ’ , ’ r =0.8 ’ ) ;
t i t l e ( ’ Phase of f i r s t −order a l l p a s s system ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
329
8.54.
a.
% Anonymous function f o r a l l p a s s H( z ) with s p e c i f i e d r and theta
H = @( r , theta , z ) ( z−r * exp ( j * theta ) ) . / ( z−1/ r * exp ( j * theta ) ) ;
% Create v e c t o r z=exp ( j *Omega)
Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi ;
z = exp ( j *Omg) ;
% Evaluate system function
H1 = H( 0 . 4 , pi /6 , z ) . * H(0.4 , − pi /6 , z ) ;
H2 = H( 0 . 6 , pi /6 , z ) . * H(0.6 , − pi /6 , z ) ;
H3 = H( 0 . 8 , pi /6 , z ) . * H(0.8 , − pi /6 , z ) ;
plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3 ) ) ;
legend ( ’ r =0.4 ’ , ’ r =0.6 ’ , ’ r =0.8 ’ ) ;
t i t l e ( ’ Phase of second−order a l l p a s s system , \ theta =\ pi /6 ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
b.
% Evaluate system function
H1 = H( 0 . 8 , pi /6 , z ) . * H(0.8 , − pi /6 , z ) ;
H2 = H( 0 . 8 , pi /4 , z ) . * H(0.8 , − pi /4 , z ) ;
H3 = H( 0 . 8 , pi /3 , z ) . * H(0.8 , − pi /3 , z ) ;
plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3 ) ) ;
legend ( ’ \ theta =\ pi /6 ’ , ’ \ theta =pi /4 ’ , ’ \ theta=pi /3 ’ ) ;
t i t l e ( ’ Phase of second−order a l l p a s s system , r =0.8 ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
8.55.
a. The first difference equation can be solved iteratively with the following:
ynm1 = 5 ; % Placeholder f o r y [ n−1] , i n i t i a l l y s e t to y [ −1]
ynm2 = 7 ; % Placeholder f o r y [ n−2] , i n i t i a l l y s e t to y [ −2]
y = [];
% Empty array ( output stream )
for n=0:10 ,
yn = 1 . 4 * ynm1−0.85 * ynm2; % y [ n ] = 1. 4 y [ n−1]−0.85 y [ n−2]
y = [ y , yn ] ;
% Append to output stream
ynm2 = ynm1;
% Update y [ n−2] f o r next index
ynm1 = yn ;
% Update y [ n−1] f o r next index
end ;
[[0:10] ’ , y ’ ]
% Tabulate s o lu t i o n
For the second difference equation, use
330
CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS
ynm1 = 2 ;
% Placeholder f o r y [ n−1] , i n i t i a l l y s e t to y [ −1]
ynm2 = −3; % Placeholder f o r y [ n−2] , i n i t i a l l y s e t to y [ −2]
y = [];
% Empty array ( output stream )
for n=0:10 ,
yn = 1 . 6 * ynm1−0.64 * ynm2; % y [ n ] = 1. 6 y [ n−1]−0.64 y [ n−2]
y = [ y , yn ] ;
% Append to output stream
ynm2 = ynm1;
% Update y [ n−2] f o r next index
ynm1 = yn ;
% Update y [ n−1] f o r next index
end ;
[[0:10] ’ , y ’ ]
% Tabulate s o lu t i o n
b.
The first difference equation can be solved symbolically and then evaluated numerically with the
following:
syms z n Yz
Y1 = 5+z ^( −1) * Yz ;
% z−transform o f y [ n−1]
Y2 = 7+5 * z^(−1)+z ^( −2) * Yz ; % z−transform o f y [ n−2]
Yz = solve ( Yz−1.4 * Y1+0.85 * Y2 , Yz )
% Solve f o r Y ( z )
yn = iztrans ( Yz )
% I n v e r s e z−transform o f Y ( z )
y = eval ( subs ( yn , ’n ’ , [ 0 : 1 0 ] ) )
% Evaluate y [ n ] f o r n = 0 , . . , 1 0
For the second difference equation, use
syms z n Yz
Y1 = 2+z ^( −1) * Yz ;
% z−transform o f y [ n−1]
Y2 = −3+2* z^(−1)+z ^( −2) * Yz ; % z−transform o f y [ n−2]
Yz = solve ( Yz−1.6 * Y1+0.64 * Y2 , Yz )
% Solve f o r Y ( z )
yn = iztrans ( Yz )
% I n v e r s e z−transform o f Y ( z )
y = eval ( subs ( yn , ’n ’ , [ 0 : 1 0 ] ) )
% Evaluate y [ n ] f o r n = 0 , . . , 1 0
Chapter 9
State-Space Analysis of Systems
9.1.

and


−1
0


2 x (t ) + 0
−2
1
2 2

ẋ (t ) = −3 0
1 1
y (t ) =
·
1 0
0 1
1
1
¸
x (t ) +
·

1
−1  r (t )
1
¸
1 0
1 0
r (t )
9.2.
a.
z (t ) = x (t ) − F r (t )
x (t ) = z (t ) + F r (t )
⇒
Differentiating both sides yields
ẋ (t ) = ż (t ) + F
r (t )
dt
Substituting this result into the state equation we obtain
ż (t ) + F
£
¤
r (t )
d r (t )
= A z (t ) + F r (t ) + B r (t ) + E
dt
dt
¡
¢
¡
¢ d r (t )
ż (t ) = A z (t ) + B + A F r (t ) + E − F
dt
Choose F = E to eliminate the term d r (t )/d t so that
b.
¡
¢
ż (t ) = A z (t ) + B + A E r (t )
B+AE =
·
0
1
¸ ·
−1
+
−2
3
−2
¸·
2
1
=
·
0
1
¸ ·
+
1
−6
¸
=
·
1
−5
¸
¸
x (t ) +
·
1
−5
¸
r (t )
The state-space model is
ẋ (t ) =
·
−1
3
−2 −2
331
¸
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
332
9.3.
R1 = 1 Ω
R2 = 1 Ω
N1
N2
b
−
+
r(t)
+
vC1 (t)
−
+
+
1
C1 = F
2
1
C2 = F
3
vC2 (t)
−
y(t)
−
b
Begin by writing KCL at node “N1”:
v C 1 (t ) − r (t ) v C 1 (t ) − v C 2 (t )
d v C 1 (t )
=0
+
+C 1
R1
R2
dt
Rearranging terms we obtain
d v C 1 (t )
R1 + R2
1
1
=−
v C 1 (t ) +
v C 2 (t ) +
r (t )
dt
R 1 R 2C 1
R 2C 1
R 2C 1
Similarly, writing the KCL at node “N2” yields
d v C 2 (t )
v C 2 (t ) − v C 1 (t )
+C 2
=0
R2
dt
which can be rearranged to become
1
1
d v C 2 (t )
=
v C 1 (t ) −
v C 2 (t )
dt
R 2C 2
R 2C 2
Defining the two state variables as x 1 (t ) = v C 1 (t ) and x 2 (t ) = v C 2 (t ) and substituting numerical values
of circuit components we get
d x 1 (t )
= − 4 x 1 (t ) + 2 x 2 (t ) + 2 r (t )
dt
d x 2 (t )
=3 x 1 (t ) − 3 x 2 (t )
dt
In matrix form the state equation is
ẋ (t ) =
·
−4
2
3 −3
¸
x (t ) +
·
2
0
¸
r (t )
Since y (t ) = x 2 (t ) = v C 2 (t ) the output equation is
y (t ) =
£
0 1
¤
x (t )
9.4.
The inductor and the capacitor have the same voltage, therefore
L
d i L (t )
= v C (t )
dt
⇒
d i L (t ) 1
= v C (t )
dt
L
333
Next step is to write the KCL to obtain
r (t ) − v C (t )
d v C (t )
= i L (t ) +C
R
dt
which can be rearranged to yield
1
1
1
d v C (t )
= − i L (t ) −
v C (t ) +
r (t )
dt
C
RC
RC
Defining the two state variables as x 1 (t ) = i L (t ) and x 2 (t ) = v C (t ) and substituting numerical values of
circuit components we get
d x 1 (t )
=1.2 x 2 (t )
dt
d x 2 (t )
= − 5 x 1 (t ) − 5 x 2 (t ) + 5 r (t )
dt
In matrix form the state equation is
ẋ (t ) =
·
0 1.2
−5 −5
¸
x (t ) +
·
0
5
¸
r (t )
Since y (t ) = x 2 (t ) = v C (t ) the output equation is
y (t ) =
£
0 1
¤
x (t )
9.5.
a. State variables can be defined as
x 1 (t ) =y (t )
x 2 (t ) =
d y (t )
dt
d x 1 (t )
=x 2 (t )
dt
=⇒
Recognizing that
d 2 y (t ) d x 2 (t )
=
dt2
dt
the differential equation can be written as
d x 2 (t )
= −3 x 2 (t ) − 2 x 1 (t ) + 2 r (t )
dt
In matrix form, the state-space model is
ẋ (t ) =
·
0
1
−2 −3
y (t ) =
£
¸
x (t ) +
1 0
¤
·
0
2
x (t )
¸
r (t )
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
334
b.
State variables can be defined as
x 1 (t ) =y (t )
x 2 (t ) =
d y (t )
dt
d x 1 (t )
=x 2 (t )
dt
=⇒
Recognizing that
d 2 y (t ) d x 2 (t )
=
dt2
dt
the differential equation can be written as
d x 2 (t )
= −4 x 2 (t ) − 3 x 1 (t ) + 3 r (t )
dt
In matrix form, the state-space model is
ẋ (t ) =
·
0
1
−3 −4
y (t ) =
c.
£
¸
x (t ) +
1 0
¤
·
0
3
¸
r (t )
x (t )
State variables can be defined as
x 1 (t ) =y (t )
x 2 (t ) =
d y (t )
dt
=⇒
d x 1 (t )
=x 2 (t )
dt
Recognizing that
d 2 y (t ) d x 2 (t )
=
dt2
dt
the differential equation can be written as
d x 2 (t )
= −x 1 (t ) + r (t )
dt
In matrix form, the state-space model is
ẋ (t ) =
·
0 1
−1 0
y (t ) =
d.
£
¸
x (t ) +
1 0
¤
·
0
1
¸
r (t )
x (t )
State variables can be defined as
x 1 (t ) =y (t )
x 2 (t ) =
d y (t )
dt
=⇒
d x 1 (t )
=x 2 (t )
dt
x 3 (t ) =
d 2 y (t )
dt2
=⇒
d x 2 (t )
=x 3 (t )
dt
335
Recognizing that
d 3 y (t ) d x 3 (t )
=
dt3
dt
the differential equation can be written as
d x 3 (t )
= −6 x 3 (t ) − 11 x 2 (t ) − 6 x 1 (t ) + r (t )
dt
In matrix form, the state-space model is

0

0
ẋ (t ) =
−6
y (t ) =
e.



0
0
1  x (t ) +  0  r (t )
−6
1
1
0
−11
£
1 0
0
¤
x (t )
Begin by writing the differential equation as
d 3 y (t )
d r (t )
d 2 y (t )
d y (t )
−
2
=
−6
− 11
− 6 y (t ) + r (t )
3
2
dt
dt
dt
dt
Let
x 1 (t ) =y (t )
x 2 (t ) =
d y (t )
dt
x 3 (t ) =
d 2 y (t )
− 2 r (t ) =⇒
dt2
=⇒
d x 1 (t )
=x 2 (t )
dt
d x 2 (t )
=x 3 (t ) + 2 r (t )
dt
It follows that
£
¤
d x 3 (t )
= −6 x 3 (t ) + 2 r (t ) − 11 x 2 (t ) − 6 x 1 (t ) + r (t )
dt
In matrix form, the state-space model is




0
1
0
0
0
1  x (t ) +  2  r (t )
ẋ(t ) =  0
−6 −11 −6
−11
y (t ) =
9.6.
a.
G (s) =
Using residue formulas
and
£
1 0
0
¤
x (t )
k2
k1
Y (s) X 1 (s) X 2 (s)
+
=
+
=
R (s)
R (s)
R (s)
s +1 s +2
¯
2 ¯¯
k1 =
=2
s + 2 ¯s=−1
¯
2 ¯¯
= −2
k2 =
s + 1 ¯s=−2
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
336
Therefore
and
In matrix form
X 1 (s)
2
=
R (s)
s +1
⇒
d x 1 (t )
= −x 1 (t ) + 2 r (t )
dt
−2
X 2 (s)
=
R (s)
s +2
⇒
d x 2 (t )
= −2 x 2 (t ) − 2 r (t )
dt
ẋ =
·
−1
0
0
−2
y (t ) =
b.
G (s) =
Using residue formulas
In matrix form
r (t )
x (t )
d x 1 (t )
= −2 x 1 (t ) − 9.5 r (t )
dt
X 2 (s) −19.5
=
R (s)
s +4
⇒
d x 2 (t )
= −4 x 2 (t ) − 19.5 r (t )
dt
·
−2
0
0 −4
c.
G (s) =
Therefore
¸
¯
10 s + 1 ¯¯
k2 =
= −19.5
s + 2 ¯s=−4
¸
y (t ) =
and
¤
2
−2
⇒
ẋ =
Using residue formulas
1 1
·
¯
10 s + 1 ¯¯
= −9.5
k1 =
s + 4 ¯s=−2
X 1 (s) −9.5
=
R (s)
s +2
and
£
x (t ) +
Y (s) X 1 (s) X 2 (s)
k2
k1
+
=
+
=
R (s)
R (s)
R (s)
s +2 s +4
and
Therefore
¸
£
x (t ) +
1 1
·
¤
−9.5
−19.5
¸
r (t )
x (t )
k2
k3
k1
Y (s) X 1 (s) X 2 (s) X 3 (s)
+
+
=
+
+
=
R (s)
R (s)
R (s)
R (s)
s +1 s +2 s +3
¯
(s − 1) (s + 4) ¯¯
k1 =
= −3
(s + 2) (s + 3) ¯s=−1
¯
(s − 1) (s + 4) ¯¯
k2 =
=6
(s + 1) (s + 3) ¯s=−2
¯
(s − 1) (s + 4) ¯¯
= −2
k3 =
(s + 1) (s + 2) ¯s=−3
X 1 (s)
−3
=
R (s)
s +1
⇒
d x 1 (t )
= −x 1 (t ) − 3 r (t )
dt
337
and
In matrix form
X 2 (s)
6
=
R (s)
s +2
⇒
d x 2 (t )
= −2 x 2 (t ) + 6 r (t )
dt
−2
X 3 (s)
=
R (s)
s +3
⇒
d x 3 (t )
= −3 x 3 (t ) − 2 r (t )
dt




−1
0
0
−3
0  x (t ) +  6  r (t )
ẋ =  0 −2
0
0 −3
−2
£
¤
y (t ) = 1 1 1 x (t )
9.7.
a. Taking the Laplace transform of both sides of the differential equation yields
¡ 3
¢
s + 5 s 2 + 11 s + 15 Y (s) = (3 + 7 s) X (s)
which leads to the system function
G (s) =
Y (s)
3+7s
= 3
R (s) s + 5 s 2 + 11 s + 15
Note: We assume zero initial conditions in taking the Laplace transform of the differential equation,
consistent with the facts that the system is CTLTI.
b.
A simulation diagram can be constructed for G (s) as shown.
7
r(t)
dx3 (t)
dt R
x3 (t)
dt
b
−5
dx2 (t)
dt
R
dt
x2 (t)
b
dx1 (t)
dt
R
dt
x1 (t)
−11
−15
d x 1 (t )
=x 2 (t )
dt
d x 2 (t )
=x 3 (t )
dt
d x 3 (t )
= − 15 x 1 (t ) − 11 x 2 (t ) − 5 x 3 (t ) + r (t )
dt
y (t ) =3 x 1 (t ) + 7 x 2 (t )
b
3
y(t)
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
338
In matrix form we have
and



0
1
0
0
0
0
1  x (t ) +  0  r (t )
ẋ(t) = 
1
−15 −11 −5

y (t ) =
£
3
¤
7 0
x (t )
9.8.
a. In order to find the transformation matrix P we need to determine the eigenvalues of the state
matrix A. Eigenvalues are the solutions of
¯
¯
¯λ I − A¯ = 0
·
¸ ·
¸ ·
1 0
−4 3
λ+4
λI−A = λ
−
=
0 1
−1 0
1
−3
λ
¸
¯
¯
¯λ I − A¯ =λ (λ + 4) − (−3) (1)
=λ2 + 4 λ + 3 = 0
Eigenvalues are λ1 = −1 and λ2 = −3. Let the corresponding eigenvectors be
·
and
v2 =
·
a
1
·
−4 3
−1 0
¸·
1
b
¸
− 4 + 3 b = −1
⇒
v1 =
A v1 = λ1 v1
1
b
¸
⇒
A v2 = λ2 v2
·
⇒
− 4 a + 3 = −3 a
=−
·
1
b
¸
·
a
1
¸
b =1
¸·
3
0
−4
−1
¸
a
1
¸
= −3
a=3
⇒
The eigenvectors of the state matrix A are
v1 =
·
1
1
¸
and
v2 =
P=
·
¸
and the transformation matrix is
b.
1 3
1 1
·
3
1
¸
The similarity transformation z (t ) = P x (t ) transforms the state-space model from
ẋ (t ) = A x (t ) + B r (t )
y (t ) = C x (t ) + d r (t )
to
ż (t ) = Ã z (t ) + B̃ r (t )
339
y (t ) = C̃ z (t ) + d˜ r (t )
The coefficient matrices are
−1
à = P
AP =
·
¸·
1.5
−0.5
−0.5
0.5
B̃ = P−1 B =
·
−0.5
0.5
C̃ = C P =
£
3
1.5
−0.5
¤
1
¸·
−4 3
−1 0
·
1 3
1 1
¸
=
·
0.5
0.5
¸
¸·
2
1
¸
¸
=
£
4 10
·
0.5
0.5
¸
1 3
1 1
=
·
−1
0
0 −3
¸
¤
d˜ = d = 2
The equivalent state-space model is
ż (t ) =
·
¸
−1
0
0 −3
y (t ) =
£
P=
·
0 1
1 0
¸
à = P−1 A P =
·
0
1
¸·
z (t ) +
¤
4 10
r (t )
z (t ) + (2) r (t )
9.9.
Let x (t ) = P z (t ).
1
0
B̃ = P−1 B =
C̃ = C P =
£
−5
⇒
·
−6
1
P
=
·
¸·
0
1
1
0
¸
=
·
0
1
£
−7
−1
−5
0
0 1
1 0
¸·
1
0
¸
·
0 1
1 0
¸
−7
¤
=
0 1
1 0
¸
0
1
−5 −6
=
·
¸
¸
−5
¤
d˜ = d = 1
9.10.
a. Equations in open form are
d x 1 (t )
= −2 x 1 (t ) − 2 x 2 (t ) + r (t )
dt
d x 2 (t )
= x 1 (t ) − 5 x 2 (t )
dt
y (t ) = 5 x 2 (t )
(P.9.10.1)
(P.9.10.2)
(P.9.10.3)
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
340
Differentiating Eqn. (P.9.10.3) and using it in conjunction with Eqn. (P.9.10.2) yields
d y (t )
d x2
=5
dt
dt
(P.9.10.4)
=5 x 1 (t ) − 25 x 2 (t )
Differentiating Eqn. (P.9.10.4) one more time we obtain
d x 1 (t )
d x 2 (t )
d 2 y (t )
=5
− 25
2
dt
dt
dt
¤
¤
£
£
=5 − 2 x 1 (t ) − 2 x 2 (t ) + r (t ) − 25 x 1 (t ) − 5 x 2 (t )
= − 35 x 1 (t ) + 115 x 2 (t ) + 5 r (t )
b.
Solving for x 2 (t ) from Eqn. (P.9.10.3) we get
µ ¶
1
y (t )
x 2 (t ) =
5
Substituting this result into Eqn. (P.9.10.4) leads to
µ ¶
1 d y (t )
x 1 (t ) =
+ y (t )
5
dt
These two results can be used in Eqn. (P.9.10.5) to produce the differential equation
d 2 y (t )
d y (t )
= −7
− 12 y (t ) + 5 r (t )
2
dt
dt
or in scaled form
c.
7 d y (t ) 12
1 d 2 y (t )
=−
−
y (t ) + r (t )
2
5 dt
5 dt
5
New state variables can be defined as
µ ¶
1
y (t )
z 1 (t ) =
5
µ ¶
1 d y (t )
z 2 (t ) =
5
dt
Recognizing that
=⇒ z 2 (t ) =
d z 1 (t )
dt
µ ¶ 2
1 d y (t ) d z 2 (t )
=
5
dt2
dt
the differential equation can be written as
d z 2 (t )
= −7 z 2 (t ) − 12 z 1 (t ) + r (t )
dt
In matrix form, the state-space model is
ż (t ) =
·
0
1
−12 −7
¸
z (t ) +
·
0
1
¸
r (t )
(P.9.10.5)
341
y (t ) =
£
·
−2
1
¸
· ¸
−2
1
x (t ) +
r (t )
−5
0
5 0
¤
x (t )
9.11.
The original state-space model is
ẋ (t ) =
£
¤
y (t ) = 0 5 x (t )
and the state-space model in phase-variable canonical form is
·
0
ż (t ) =
−12
¸
· ¸
1
0
z (t ) +
r (t )
−7
1
£
y (t ) = 5
¤
0 z (t )
Let the new state vector z be defined through the transformation
z (t ) = P x (t )
where P is a transformation matrix in the form
P=
·
p 11
p 21
p 12
p 22
¸
The first state variable z 1 (t ) can be written as
z 1 (t ) = p 11 x 1 (t ) + p 12 x 2 (t )
Let P1 and P2 be the row vectors constructed using the rows of matrix P, that is,
P1 =
so that
£
p 11
p 12
¤
and
P2 =
£
p 21
p 22
¤
z 1 (t ) = P1 x (t )
(P.9.11.1)
z 2 (t ) = P2 x (t )
(P.9.11.2)
and
Differentiating Eqn. (P.9.11.1) we get
d z 1 (t )
=P1 ẋ(t )
dt
£
¤
=P1 A x (t ) + B r (t )
=P1 A x (t ) + P1 B r (t )
In the phase-variable canonical form d z 1 (t )/d t should not have a r (t ) term. Therefore we require that
P1 B = 0, and
d z 1 (t )
(P.9.11.3)
= z 2 (t ) = P1 A x (t )
dt
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
342
Similarly, differentiating Eqn. (P.9.11.2) yields
d z 2 (t )
=P2 ẋ(t )
dt
£
¤
=P2 A x (t ) + B r (t )
=P2 A x (t ) + P2 B r (t )
(P.9.11.4)
which leads to the requirement P2 B = 1. It should be noted that d z 2 (t ) /d t can also be obtained by
differentiating Eqn. (P.9.11.3) which results in
d z 2 (t )
=P1 A ẋ(t )
dt
£
¤
=P1 A A x (t ) + B r (t )
=P1 A2 x (t ) + P1 A B r (t )
(P.9.11.5)
Comparing Eqns. (P.9.11.4) and (P.9.11.5) we conclude that P2 = P1 A and P1 A B = 0. Therefore, the
transformation matrix P is in the form
"
#
P1
P=
P1 A
Furthermore, we have the condition
P1
which leads to
£
P1 =
¤
=
0 1
¤
¤£
B AB
¤−1
·
−2 −2
1 −5
¸·
1
0
¸
1 −2
0
1
¸
=
£
0
1
¤
−2 −2
1 −5
¸
=
£
1
−5
·
¸
B AB
£
0
1
Using the original state-space model, we have
B=
·
1
0
¸
and
AB =
The vector P1 is found as
P1 =
and the vector P2 is
P2 =
The transformation matrix is
£
£
0
0
1
1
¤
¤
·
·
P=
£
0
1
1 −5
=
·
−2
1
¸
¤
It can easily be verified that the matrix P converts the original state-space model to phase-variable
canonical form:
·
¸
· ¸
£
¤
0
1
0
à = P A P−1 =
B̃ = P B =
C̃ = C P−1 = 5 0
−12 −7
1
343
9.12.
a. State equations in open form are
d x 1 (t )
= −2 x 1 (t ) − 2 x 2 (t ) + R (s)
dt
d x 2 (t )
= x 1 (t ) − 5 x 2 (t )
dt
and the output equation is
y (t ) = 5 x 2 (t )
Taking the Laplace transform of each state equation yields
s X 1 (s) = −2 X 1 (s) − 2 X 2 (s) + R (s)
(P.9.12.1)
s X 2 (s) = X 1 (s) − 5 X 2 (s)
(P.9.12.2)
The Laplace transform of the output equation is
Y (s) = 5 X 2 (s)
b.
(P.9.12.3)
Solving for X 1 (s) from Eqn. (P.9.12.2) we obtain
X 1 (s) = (s + 5) X 2 (s)
Substituting this result into Eqn. (P.9.12.1) leads to
s (s + 5) X 2 (s) = −2 (s + 5) X 2 − 2 X 2 + R (s)
and therefore
R (s)
s 2 + 7 s + 12
Substituting this result into the s-domain output equation given by Eqn. (P.9.12.3) yields the system
function
5
Y (s)
= 2
G (s) =
R (s) s + 7 s + 12
X 2 (s) =
c.
From the system function we get
¡ 2
¢
s + 7 s + 12 Y (s) = 5 R (s)
and the corresponding differential equation is
d y (t )
d 2 y (t )
+7
+ 12 y (t ) = 5 r (t )
dt2
dt
9.13.
Choosing the state variables as
x 1 (t ) = y 1 (t )
and
x 2 (t ) = y 2 (t ) − r 1 (t )
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
344
the state equations become
¤
£
d x 1 (t )
= − 2 x 1 (t ) − x 2 (t ) + r 1 (t ) + r 1 (t ) + r 2 (t )
dt
= − 2 x 1 (t ) − x 2 (t ) + r 2 (t )
and
d x 2 (t )
= x 1 (t ) − 3 x 2 (t ) − 3 r 1 (t )
dt
Output equations are
y 1 (t ) = x 1 (t )
y 2 (t ) = x 2 (t ) + r 1 (t )
In matrix form we have
·
−2
1
y (t ) =
·
ẋ (t) =
and
−1
−3
¸
x (t ) +
·
0
−3
1
0
¸
1 0
0 1
¸
·
0
1
¸
r (t )
x (t ) +
0
0
r (t )
9.14.
a. In order to find the transformation matrix P we need to determine the eigenvalues of the state
matrix A. Eigenvalues are the solutions of
¯
¯
¯λ I − A¯ = 0
·
¸ ·
¸ ·
1 0
−1
2
λ+1
λI−A = λ
−
=
0 1
−3 −6
3
−2
λ+6
¸
¯
¯
¯λ I − A¯ = (λ + 1) (λ + 6) − (3) (−2)
=λ2 + 7 λ + 12 = 0
Eigenvalues are λ1 = −3 and λ2 = −4. Let the corresponding eigenvectors be
·
¸
·
¸
v 11
v 21
v1 =
and
v2 =
v 12
v 22
A v1 = λ1 v1
⇒
·
−1
−3
− v 11 + 2 v 12 = −3 v 11
A v2 = λ2 v2
⇒
·
− v 21 + 2 v 12 = −4 v 21
−1
−3
¸·
v 11
v 12
⇒
v 11 = −v 12
¸·
v 21
v 22
2
−6
2
−6
⇒
¸
¸
= −3
= −4
·
·
v 11
v 12
¸
v 21
v 22
¸
v 21 = − (2/3) v 22
Arbitrarily choosing v 12 = 1 and v 22 = 3 the other elements of vectors v1 and v2 can be determined from
these relationships, and lead to eigenvectors
·
¸
·
¸
−1
−2
v1 =
and
v2 =
1
3
345
and the transformation matrix is
P=
·
−1 −2
1
3
¸
The similarity transformation x (t ) = P z (t ) transforms the state-space model from
ẋ(t ) = A x (t ) + B r (t )
y (t ) = C x (t ) + D r (t )
to
ż (t ) = Ã z (t ) + B̃ r (t )
y (t ) = C̃ z (t ) + D̃ r (t )
The coefficient matrices are
−1
à = P
AP =
·
−3 −2
1
1
B̃ = P−1 B =
C̃ = C P =
·
·
¸·
¸·
−3 −2
1
1
−1 −3
2
0
2
−6
−1
−3
¸
−1 −2
1
3
1
2
0 −1
¸
−3 −2
1
1
¸
·
¸
¸·
D̃ = D =
¸·
=
·
−3
0
=
·
−3 −4
1
1
¸
=
·
−2
−2
¸
0
4
1
−1
−3 −4
1
1
¸
0
1
4 −1
¸
−7
−4
0
−4
¸
The equivalent state-space model is
ż (t ) =
·
−3
0
0
−4
¸
z (t ) +
·
·
−2
−2
−7
−4
¸
z (t ) +
·
⇒
s Z1 (s) = − 3 Z1 (s) − 3 R 1 (s) − 4 R 2 (s)
y (t ) =
b.
r (t )
r (t )
Using the state equation
d z 1 (t )
= −3 z 1 (t ) − 3 r 1 (t ) − 4 r 2 (t )
dt
Z1 (s) = −
d z 2 (t )
= −4 z 2 (t ) + r 1 (t ) + r 2 (t )
dt
⇒
3
4
R 1 (s) −
R 2 (s)
s +3
s +3
s Z2 (s) = − 4 Z2 (s) + R 1 (s) + R 2 (s)
Z2 (s) =
1
1
R 1 (s) +
R 2 (s)
s +4
s +4
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
346
Using the output equation
y 1 (t ) = − 2 z 1 (t ) − 7 z 2 (t ) + r 2 (t )
Y1 (s) = − 2 Z1 (s) − 7 Z2 (s) + R 2 (s)
¸
·
¸
·
4
1
1
3
R 1 (s) −
R 2 (s) − 7
R 1 (s) +
R 2 (s) + R 2 (s)
=−2 −
s +3
s +3
s +4
s +4
·
· 2
¸
¸
s + 11
s + 6 s + 15
= 2
R 1 (s) + 2
R 2 (s)
s + 7 s + 12
s + 7 s + 12
y 2 (t ) = − 2 z 1 (t ) − 4 z 2 (t ) + 4 r 1 (t ) − r 2 (t )
Y2 (s) = − 2 Z1 (s) − 4 Z2 (s) + 4 R 1 (s) − R 2 (s)
·
¸
·
¸
4
1
3
1
R 1 (s) −
R 2 (s) − 4
R 1 (s) +
R 2 (s) + 4 R 1 (s) − R 2 (s)
=−2 −
s +3
s +3
s +4
s +4
· 2
¸
¸
· 2
−s − 3 s + 8
4 s + 30 s + 60
+
R
R 2 (s)
=
(s)
1
s 2 + 7 s + 12
s 2 + 7 s + 12
c.

s 2 + 6 s + 15
s 2 + 7 s + 12


Y (s) = 
 4 s 2 + 30 s + 60
s 2 + 7 s + 12

s + 11
s 2 + 7 s + 12 

 R (s)
−s 2 − 3 s + 8 
s 2 + 7 s + 12
9.15.
The state variables were found as
µ
¶
7 5 −t 11 −2t 11 −3t
x 1 (t ) = + e −
e
+
e
u (t )
6 2
2
6
µ
¶
7
11 −2t
−t
x 2 (t ) = + 5 e −
e
u (t )
2
2
¢
¡
x 3 (t ) = 1 + e −t u (t )
Evaluating each state variable at t = 0 we get
7 5 11 11
+
=0
x 1 (0) = + −
6 2 2
6
7
11
x 2 (0) = + 5 −
=3
2
2
x 3 (0) =1 + 1 = 2
347
matching the desired initial state vector. Derivatives of state variables are
¶
µ
d x 1 (t )
5 −t
11 −3t
−2t
u (t )
= − e + 11 e
−
e
dt
2
2
¢
d x 2 (t ) ¡
= −5 e −t + 11 e −2t u (t )
dt
d x 3 (t )
= − e −t u (t )
dt
It can be shown that
d x 1 (t )
= − 3 x 1 (t ) + x 2 (t )
dt
d x 2 (t )
= − 2 x 1 (t ) + 5 x 2 (t ) + 2 u (t )
dt
d x 3 (t )
= − x 3 (t ) + u (t )
dt
9.16.
The state equations in open form are
d x 1 (t )
= − 3 x 1 (t ) + 2 x 2 (t )
dt
d x 2 (t )
= − x 2 (t ) + r (t )
dt
The second equation can be solved with r (t ) = u (t ) and x 2 (0) = 3 to yield
x 2 (t ) =e
−t
=3 e
ˆ t
e −(t −τ) u (τ) d τ
ˆ t
eτ dτ
x 2 (0) +
−t
+e
−t
0
0
¢
¡
= 1 + 2 e −t u (t )
Now the first state equation can be written as
¡
¢
d x 1 (t )
= − 3 x 1 (t ) + 2 1 + 2 e −t u (t )
dt
¢
¡
= − 3 x 1 (t ) + 2 + 4 e −t u (t )
It can be solved with the initial condition x 1 (0) = −2 to yield
x 1 (t ) =e
−3t
x 1 (0) +
= −2e
−3t
+e
ˆ t
0
−3t
¡
¢
e −3 (t −τ) 2 + 4 e −τ u (τ) d τ
ˆ t
0
¡
¢
2 e 3τ + 4 e 2τ d τ
¶
14 −3t
2
−t
e
u (t )
= +2e −
3
3
µ
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
348
The output signal is
y (t ) =x 1 (t ) + x 2 (t ) + r (t )
¶
µ
14 −3t
8
−t
e
u (t )
= +4e −
3
3
9.17.
The eigenvalues of the state matrix A are found as follows:
λI−A = λ
¯
¯
¯λ I − A¯ = 0
·
¸ ·
¸ ·
1 0
−3
2
λ+3
−
=
0 1
0 −1
0
¯
¯
¯λ I − A¯ = (λ + 3) (λ + 1) = 0
−2
λ+1
¸
Eigenvalues are λ1 = −3 and λ2 = −1. Let the corresponding eigenvectors be
A v2 = λ2 v2
1
b
¸
= −3
·
1
a
¸
⇒
a=0
·
1
b
¸
⇒
b=1
1
a
¸
and
⇒
·
−3
2
0 −1
¸·
1
a
¸
⇒
·
−3
0
2
−1
¸·
1
b
¸
·
1 1
0 1
¸
v1 =
A v1 = λ1 v1
·
·
Thus the transformation matrix is
P=
v2 =
=−
The similarity transformation x (t ) = P z (t ) transforms the state-space model from
ẋ(t ) = A x (t ) + B r (t )
y (t ) = C x (t ) + d r (t )
to
ż (t ) = Ã z (t ) + B̃ r (t )
y (t ) = C̃ z (t ) + d˜ r (t )
The coefficient matrices are
à = P−1 A P =
·
1
0
−1
B̃ = P
−1
1
¸·
B=
·
C̃ = C P =
£
1
−3
0
2
−1
¸·
1
0
1
1
¸
=
·
·
−1
1
¸
1 2
¤
1
0
−1
1
¸·
0
1
¸
1
¤
·
1 1
0 1
¸
=
=
£
−3
0
0 −1
¸
349
d˜ = d = 1
The equivalent state-space model is
ż (t ) =
·
−3
0
0
−1
¸
y (t ) =
£
1
2
·
1 −1
0
1
z (t ) +
¤
·
−−1
1
¸
r (t )
z (t ) + r (t )
The initial conditions must also be translated to the new state-space model.
−1
z (0) = P
x (0) =
¸·
−2
3
¸
=
·
−5
3
¸
The two state equations are now decoupled, and can be solved independently of each other. With
r (t ) = u (t ) the first state equation is
d z 1 (t )
= −3 z 1 (t ) − u (t ) ,
dt
z 1 (0) = −5
It can be solved for z 1 (t ) as
z 1 (t ) =e −3t z 1 (0) −
= −5e
−3t
−e
ˆ t
e −3 (t −τ) u (τ) d τ
0
−3t
ˆ t
e 3τ d τ
0
µ
¶
1 14 −3t
= − −
u (t )
e
3 3
The second state equation is
d z 2 (t )
= −z 2 (t ) + u (t ) ,
dt
z 2 (0) = 3
and can be solved to yield
z 2 (0) +
ˆ t
e −(t −τ) u (τ) d τ
=3 e −t + e −t
ˆ t
eτ dτ
z 2 (t ) =e
The output signal is
−t
0
0
¡
¢
= 1 + 2 e −t u (t )
y (t ) =z 1 (t ) + 2 z 2 (t ) + u (t )
¶
µ
14 −3t
8
−t
u (t )
e
= +4e −
3
3
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
350
9.18.
a.
s I−A = s
·
1 0
0 1
¸ ·
−3
−
0
Resolvent matrix:
State transition matrix:
£
¤−1
Φ (s) = s I − A
=
¸
=
·
1
(s + 1) (s + 3)
·
s +1
0
2
−1
s +3
0
−2
s +1
2
s +3
¸
¸
φ (t ) = L −1 {Φ (s)}
The elements of the state transition matrix are:
φ11 (t ) = L
φ12 (t ) = L −1
½
−1
½
¾
1
= e −3t u (t )
s +3
½
¾
¾
¢
¡
1
2
1
= L −1
= e −t − e −3t u (t )
−
s +1 s +3
(s + 1) (s + 3)
φ21 (t ) = L −1 {0} = 0
φ11 (t ) = L
−1
½
Thus, the state transition matrix is
"
e −3t u (t )
"
e −6
"
e −15
φ (t ) =
b.
x (2) = φ (2) x (0) =
c.
x (5) = φ (5) x (0) =
¾
1
= e −t u (t )
s +1
#
¢
¡ −t
e − e −3t u (t )
e −t u (t )
0
0
0
¡ −2
¢ #·
¸
e − e −6
−2
=
·
0.3936
0.4060
¸
¡ −5
¢ #·
¸
e − e −15
−2
·
0.0202
0.0202
¸
3
e −2
3
e −5
=
9.19.
a.
Let the vector X (s) be defined as
X (s) =
·
X 1 (s)
X 2 (s)
¸
Taking unilateral Laplace transform of both sides of the state equation yields
s X (s) − x (0) = A X (s) + B R (s)
(P.9.19.1)
351
b.
Rearranging the terms in Eqn. (P.9.19.1) we get
£
¤
s I − A X (s) = x (0) + B R (s)
Multiplying both sides of this result by [s I − A]−1 leads to
£
¤−1
£
¤−1
X (s) = s I − A
x (0) + s I − A
B R (s)
Recall that
L −1
Therefore
L −1
and
L −1
n£
s I−A
¤−1
n£
n£
s I−A
s I−A
¤−1 o
= φ (t ) = e A t
o
x (0) = e A t x (0)
¤−1
(P.9.19.2)
n
o
n£
o
¤−1 o
∗ L −1 B R (s)
B R (s) = L −1 s I − A
=e A t ∗ B r (t )
=
ˆ t
e A (t −τ) B r (τ) d τ
0
Combining Eqns. (P.9.19.2) and (P.9.19.3) we get
x (t ) = e A t x (0) +
ˆ t
e A (t −τ) B r (τ) d τ
0
c.
Y (s) =C X (s) + d R (s)
£
¤−1
£
¤−1
=C s I − A
x (0) + C s I − A
B R (s)
d.
y (t ) = C e
At
x (0) +
ˆ t
C e A (t −τ) B r (τ) d τ
0
9.20.
The state variables were found as
¶
5 28 −3t 27 −4t
u (t )
+
e
−
e
x 1 (t ) =
12 3
4
¶
µ
1 14 −3t 27 −4t
u (t )
+
e
−
e
x 2 (t ) =
12 3
4
µ
Evaluating each state variable at t = 0 we get
5 28 27
+
−
=3
12 3
4
1 14 27
−
= −2
x 2 (0) = +
12 3
4
x 1 (0) =
(P.9.19.3)
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
352
matching the desired initial state vector. Derivatives of state variables are
¢
d x 1 (t ) ¡
= −28 e −3t + 27 e −4t u (t )
dt
¢
d x 2 (t ) ¡
= −14 e −3t + 27 e −4t u (t )
dt
It can be shown that
d x 1 (t )
= − 2 x 1 (t ) − 2 x 2 (t ) + u (t )
dt
d x 2 (t )
= x 1 (t ) − 5 x 2 (t )
dt
9.21.
G (s) =
·
1 0
0 1
¸ ·
−2
−
26
s I−A
¤−1
s I−A = s
£
The system function is
£
¤−1
Y (s)
= C s I−A
B+d
R (s)
=
−1
0
¸
1
2
s + 2 s + 26
·
s
26
·
s
26
£
¤
1
−2 0
G (s) = 2
s + 2 s + 26
=
·
s +2 1
−26 s
−1
s +2
−1
s +2
9.22.
a. Expressing y[n] as a function of the other terms leads to
y[n] = 0.9 y[n − 1] + r [n]
Define the state variable x 1 [n] as
x 1 [n] = y[n − 1]
so that the difference equation can be written as
x 1 [n + 1] = 0.9 x[n] + r [n]
In matrix form, the state-space model is
x[n + 1] =
£
y[n] =
0.9
£
¤
0.9
x[n] +
¤
£
1
¤
x[n] + r [n]
¸
¸·
s 2 + 26
= 2
s + 2 s + 26
r [n]
¸
1
0
¸
+1
353
b.
Expressing y[n] as a function of the other terms leads to
y[n] = 1.7 y[n − 1] − 0.72 y[n − 2] + 3 r [n]
State variables can be defined as
x 1 [n] =y[n − 2]
x 2 [n] =y[n − 1] =⇒ x 1 [n + 1] =x 2 [n]
Recognizing that
y[n] = x 2 [n + 1]
the difference equation can be written as
x 2 [n + 1] = 1.7 x 2 [n] − 0.72 x 1 [n] + 3 r [n]
In matrix form, the state-space model is
x[n + 1] =
·
0
−0.72
1
1.7
¸
£
−0.72
1.7
¤
y[n] =
c.
x[n] +
·
0
3
¸
r [n]
x[n] + 3 r [n]
Expressing y[n] as a function of the other terms leads to
y[n] = y[n − 1] − 0.11 y[n − 2] − 0.07 y[n − 3] + r [n]
State variables can be defined as
x 1 [n] =y[n − 3]
x 2 [n] =y[n − 2] =⇒ x 1 [n + 1] =x 2 [n]
x 3 [n] =y[n − 1] =⇒ x 2 [n + 1] =x 3 [n]
Recognizing that
y[n] = x 3 [n + 1]
the difference equation can be written as
x 3 [n + 1] = x 3 [n] − 0.11 x 2 [n] − 0.07 x 1 [n] + r [n]
In matrix form, the state-space model is

0

0
x[n + 1] =
−0.07
y[n] =
£
−0.07
1
0
−0.11
−0.11

 
0
0


1 x[n] + 0  r [n]
1
1
1
¤
x[n] + r [n]
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
354
9.23.
a. Partial fraction expansion of the system function is
H (z) =
Y (z) X 1 (z) X 2 (z)
k1
k2
=
+
=
+
R (z)
R (z)
R (z)
z + 1/2 z + 1/3
The residues are
¯
z + 1 ¯¯
=3
k1 =
z + 2/3 ¯z=−1/2
and
Therefore
and
X 1 (z)
3
=
R (z)
z + 1/2
¯
z + 1 ¯¯
k2 =
−2
z + 1/2 ¯z=−2/3
2
X 2 (z)
=−
R (z)
z + 2/3
In matrix form
x[n + 1] =
·
⇒
1
x 1 [n + 1] = − x 1 [n] + 3 r [n]
2
⇒
2
x 2 [n + 1] = − x 2 [n] − 2 r [n]
3
¸
·
¸
0
3
x[n] +
r [n]
−2/3
−2
£
¤
y[n] = 1 1 x[n]
−1/2
0
b.
In this case the numerator order is the same as the denominator order and we will therefore use a
slightly different form of partial fraction expansion compared to part (a):
H (z) =
Y (z) X 1 (z) X 2 (z)
k1 z
k2 z
=
+
=
+
R (z)
R (z)
R (z)
z + 1/2 z + 1/3
The residues are
¯
z + 1 ¯¯
= 1.2727
k1 =
z + 0.7 ¯z=0.4
and
k2 =
Therefore
and
¯
z + 1 ¯¯
= −0.2727
z − 0.4 ¯z=−0.7
W1 (z) 1.2727 z
=
R (z)
z − 0.4
⇒
w 1 [n + 1] = 0.4 w 1 [n] + 1.2727 r [n + 1]
0.2727 z
W2 (z)
=−
R (z)
z + 2/3
⇒
w 2 [n + 1] = −0.7 w 2 [n] − 0.2727 r [n + 1]
The terms w 1 [n] and w 2 [n] can not be selected as the state variables of the system since we do not want
the r [n + 1] terms to appear in state equations. Let us substitute n → n − 1 and write the two equations
as
w 1 [n] =0.4 w 1 [n − 1] + 1.2727 r [n]
w 2 [n] = − 0.7 w 2 [n − 1] − 0.2727 r [n]
355
and define x 1 [n] and x 2 [n] as follows:
x 1 [n + 1] = w 1[n]
x 2 [n + 1] = w 2 [n]
It follows that
x 1 [n + 1] =0.4 x 1 [n] + 1.2727 r [n]
x 2 [n + 1] = − 0.7 x 2 [n] − 0.2727 r [n]
The output signal is
y[n] =w 1 [n] + w 2[n]
=x 1 [n + 1] + x 2 [n + 1]
=0.4 x 1 [n] + 1.2727 r [n] − 0.7 x 2 [n] − 0.2727 r [n]
=0.4 x 1 [n] − 0.7 x 2 [n] + r [n]
In matrix form
x[n + 1] =
·
0.4
0
y[n] =
c.
£
0
−0.7
¸
0.4
−0.7
x[n] +
¤
·
1.2727
−0.2727
¸
r [n]
x[n] + r [n]
Partial fraction expansion of the system function is
H (z) =
The residues are
and
Y (z) X 1 (z) X 2 (z) X 3 (z)
k2
k3
k1
+
+
=
+
+
=
R (z)
R (z)
R (z)
R (z)
z + 3/4 z − 1/2 z − 3/2
¯
¯
z (z + 1)
¯
= −1/15
k1 =
(z − 1/2) (z − 3/2) ¯z=−3/4
¯
¯
z (z + 1)
¯
k2 =
= −3/5
(z + 3/4) (z − 3/2) ¯z=1/2
k3 =
Therefore
In matrix form
¯
¯
z (z + 1)
¯
= 5/3
(z + 3/4) (z − 1/2) ¯z=3/2
⇒
3
1
x 1 [n + 1] = − x 1 [n] −
r [n]
4
15
X 2 (z)
−3/5
=
R (z)
z − 1/2
⇒
x 2 [n + 1] =
1
3
x 2 [n] − r [n]
2
5
5/3
X 3 (z)
=
R (z)
z − 3/2
⇒
x 3 [n + 1] =
3
5
x 3 [n] + r [n]
2
3
X 1 (z) −1/15
=
R (z)
z + 3/4

−3/4
x[n + 1] =  0
0
0
1/2
0



0
−1/15
0  x[n] +  −3/5  r [n]
3/2
1/3
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
356
y[n] =
£
1 1
1
¤
x[n]
9.24.
a.
The system function is
r[n]
z +1
H (z) = 2
z + (7/6) z + 1/3
y[n]
x2 [n + 1]
D
x2 [n]
−7/6
1
b
x 1 [n + 1] =x 2 [n]
x 2 [n + 1] = −
x1 [n + 1]
1
7
x 1 [n] − x 2 [n] + r [n]
3
6
D
x1 [n]
−1/3
1
b
y[n] =x 1 [n] + x 2 [n]
In matrix form
x[n + 1] =
·
b.
¸
· ¸
1
0
x[n] +
r [n]
−7/6
1
£
¤
y[n] = 1 1 x[n]
0
−1/3
The system function is
H (z) =
z2 + z
z 2 + 0.3 z − 0.28
1
r[n]
b
x2 [n + 1]
D
x2 [n]
x 1 [n + 1] =x 2 [n]
−0.3
x 2 [n + 1] =0.28 x 1 [n] − 0.3 x 2 [n] + r [n]
x1 [n + 1]
b
D
y[n] =x 2 [n + 1] + x 2 [n]
=0.28 x 1 [n] − 0.3 x 2 [n] + r [n] + x 2[n]
0.28
=0.28 x 1 [n] + 0.7 x 2 [n] + r [n]
In matrix form
·
¸
· ¸
0
1
0
x[n + 1] =
x[n] +
r [n]
0.28 −0.3
1
£
¤
y[n] = 0.28 0.7 x[n] + r [n]
x1 [n]
1
y[n]
357
c.
r[n]
The system function is
H (z) =
y[n]
x3 [n + 1]
z2 + z
z 3 − (5/4) z 2 − (3/4) z + 9/16
D
x3 [n]
5/4
1
b
x2 [n + 1]
x 1 [n + 1] =x 2 [n]
D
x 3 [n + 1] = −
x2 [n]
3/4
x 2 [n + 1] =x 3 [n]
1
b
3
5
9
x 1 [n] + x 2 [n] + x 3 [n] + r [n]
16
4
4
x1 [n + 1]
D
x1 [n]
−9/16
y[n] =x 2 [n] + x 3[n]
In matrix form




1
0
0
0
1  x[n] +  0  r [n]
1
3/4 5/4
£
¤
y[n] = 0 1 1 x[n]
0
0
x[n + 1] = 
−9/16
9.25.
a.
The system function is
H (z) =
z +1
1 + z −1
=
−1
1 − 0.9 z
z − 0.9
1
r[n]
b
x1 [n + 1]
x 1 [n + 1] =0.9 x 1 [n] + r [n]
D
b
=x 1 [n] + 0.9 x 1 [n] + r [n]
=1.9 x 1 [n] + r [n]
In matrix form
x1 [n]
0.9
y[n] =x 1 [n] + x 1 [n + 1]
¤
£ ¤
0.9 x[n] + 1 r [n]
£
¤
y[n] = 1.9 x[n] + r [n]
x[n + 1] =
£
1
y[n]
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
358
b.
The system function is
H (z) =
3 + 2 z −2
3 z2 + 2
=
1 − 1.7 z −1 + 0.72 z −2 z 2 − 1.7 z + 0.72
3
r[n]
b
y[n]
x2 [n + 1]
D
x2 [n]
x 1 [n + 1] =x 2 [n]
1.7
x 2 [n + 1] = − 0.72 x 1 [n] + 1.7 x 2 [n] + r [n]
x1 [n + 1]
b
D
y[n] =2 x 1 [n] + 3 x 2[n + 1]
¡
¢
=2 x 1 [n] + 3 − 0.72 x 1 [n] + 1.7 x 2 [n] + r [n]
x1 [n]
−0.72
2
b
= − 0.16 x 1 [n] + 5.1 x 2 [n] + 3 r [n]
In matrix form
x[n + 1] =
·
y[n] =
c.
£
0
−0.72
1
1.7
¸
−0.16
5.1
¤
x[n] +
·
0
1
¸
r [n]
x[n] + 3 r [n]
The system function is
H (z) =
−1
−2
1−z +z
−1
1 − z + 0.11 z −2 + 0.07 z −3
z3 − z2 + z
= 3
z − z 2 + 0.11 z + 0.07
1
r[n]
b
x3 [n + 1]
D
b
x2 [n + 1]
D
x 1 [n + 1] =x 2 [n]
x2 [n]
−0.11
x 2 [n + 1] =x 3 [n]
b
x 3 [n + 1] = − 0.07 x 1 [n] − 0.11 x 2 [n] + x 3 [n] + r [n]
y[n] =x 2 [n] − x 3 [n] + x 3[n + 1]
x1 [n + 1]
D
−0.07
= − 0.07 x 1 [n] + 0.89 x 2 [n] + r [n]
In matrix form
x3 [n] −1
1


 
0
1
0
0



0
0
1 x[n] + 0  r [n]
x[n + 1] =
−0.07 −0.11 1
1
£
¤
y[n] = −0.07 0.89 0 x[n] + r [n]
x1 [n]
1
y[n]
359
9.26.
x[1] = A x[0] + B r [0] =
·
0
0.25
y[0] = C x[0] =
·
−0.5
0.75
¸·
£
1
3
¤
2
0
·
¸ ·
·
¸
¸
2
2
+
(1) =
1
1.5
¸
2
=6
0
¸·
¸ ·
·
¸
¸
2
2
1.25
x[2] = A x[1] + B r [1] =
+
(1) =
1.5
1
2.625
·
¸
£
¤
2
y[1] = C x[1] = 3 1
= 7.5
1.5
0
0.25
·
0
0.25
·
0
0.25
−0.5
0.75
¸·
¸ ·
·
¸
¸
1.25
2
0.6875
x[3] = A x[2] + B r [2] =
+
(1) =
2.625
1
3.2813
¸
·
£
¤
1.25
= 6.375
y[2] = C x[2] = 3 1
2.625
−0.5
0.75
¸·
¸ ·
·
¸
¸
2
0.3594
+
(1) =
1
3.6328
·
¸
£
¤ 0.6875
y[3] = C x[3] = 3 1
= 5.3438
3.2813
x[4] = A x[3] + B r [3] =
9.27.
a. Resolvent matrix is found as
£
¤
The first step is to find the matrix z I − A .
z I−A = z
The resolvent matrix is
£
·
z I−A
0.6875
3.2813
£
¤−1
Φ (z) = z z I − A
¸ ·
−0.1
−
−0.8
1
0
0
1
¤−1
1
=
(z − 0.7) (z + 0.8)
£
¤−1
Φ (z) = z z I − A
=
b.
−0.5
0.75
−0.7
0
·
1
(z − 0.7) (z + 0.8)
¸
·
z + 0.1
0.8
0.7
z
z
−0.8
−0.7
z + 0.1
¸
·
=
z2
−0.8 z
−0.7 z
z (z + 0.1)
The state transition matrix is
φ[n] = [Φ (z)]
Φ11 (z) =
−1
=
·
φ11 [n] φ12 [n]
φ21 [n] φ22 [n]
¸
¸
z2
(7/15) z (8/15) z
=
+
z − 0.7
z + 0.8
(z − 0.7) (z + 0.8)
¸
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
360
φ11 [n] =
µ
µ ¶
¶
7
8
(0.7)n u[n] +
(−0.8)n u[n]
15
15
−0.7 z
(7/15) z (7/15) z
+
=−
z − 0.7
z + 0.8
(z − 0.7) (z + 0.8)
Φ12 (z) =
µ
µ ¶
¶
7
7
n
φ12 [n] = −
(0.7) u[n] +
(−0.8)n u[n]
15
15
−0.8 z
(8/15) z (8/15) z
+
=−
z − 0.7
z + 0.8
(z − 0.7) (z + 0.8)
Φ21 (z) =
µ ¶
¶
8
8
n
φ21 [n] = −
(0.7) u[n] +
(−0.8)n u[n]
15
15
µ
Φ22 (z) =
z (z + 1)
(8/15) z (7/15) z
+
=
z − 0.7
z + 0.8
(z − 0.7) (z + 0.8)
µ
µ ¶
¶
8
7
n
φ22 [n] =
(0.7) u[n] +
(−0.8)n u[n]
15
15
c.
Using Eqn. (9.177) we have
X (z) = Φ (z) x (0) + z −1 Φ (z) B R (z)
The resolvent matrix was found in part (a).
Φ (z) x (0) =
1
(z − 0.7) (z + 0.8)
·
z2
−0.8 z
−0.7 z
z (z + 0.1)
¸·
¸
2
0
=
1
(z − 0.7) (z + 0.8)
·
2 z2
−1.6 z
The second term in the expression for X (z) is
z
−1
Φ (z) B R (z) = z
=
−1
1
−
0.7)
(z
(z + 0.8)
·
1
(z − 1) (z − 0.7) (z + 0.8)
z2
−0.8 z
·
Thus X (z) is found as
1
X (z) =
−
1)
−
0.7)
(z
(z
(z + 0.8)
and the output transform is
Y (z) =
−0.7 z
z (z + 0.1)
3 z 2 − 0.7 z
z 2 − 2.3 z
·
¸·
3
1
¸
2 z 3 + z 2 − 0.7 z
−0.6 z 2 − 0.7 z
¸
¡
¢
z 4 z 2 + 2.6 z − 0.7
(z − 1) (z − 0.7) (z + 0.8)
The output signal can be found using partial fraction expansion:
y[n] = 10.9259 − 0.0815 (−0.8)n − 6.8444 (0.7)n ,
n≥0
¸³
z ´
z −1
¸
361
9.28.
The system function is found as
£
¤−1
H (z) = C z I − A
B
£
¤
The first step is to find the matrix z I − A and its inverse:
z I−A = z
£
H (z) =
=
£
·
z I−A
2
¸ ·
−0.1
−
−0.8
1
0
0
1
¤−1
1
=
(z − 0.7) (z + 0.8)

z
¤  (z − 0.7) (z + 0.8)
−1 

−0.8
(z − 0.7) (z + 0.8)
5 z + 0.9
(z − 0.7) (z + 0.8)
¸
−0.7
0
·
·
z + 0.1
0.8
0.7
z
z
−0.8
−0.7
z + 0.1
¸
=
¸

−0.7
· ¸
(z − 0.7) (z + 0.8) 
 3
 1
z + 0.1
(z − 0.7) (z + 0.8)
9.29.
The system function must first be written using powers of z −1 :
H (z) =
0 + z −1 + 0 z −2 − 7 z −3 + 6 z −4
1 − 0.2 z −1 − 0.93 z −2 + 0.198 z −3 − 0.1296 z −4
It is important to ensure that numerator and denominator polynomials have the same length for the
use of the function tf2ss (..) . We also account for missing terms using zero-valued coefficients. The
state-space model for the system is found with the statements
>>
>>
>>
num = [ 0 , 1 , 0 , − 7 , 6] ;
den = [ 1 , − 0. 2 , − 0. 93 , 0. 198 , 0. 1296] ;
[ A , B , C, d ] = t f 2 s s (num, den )
A =
0.2000
1.0000
0
0
B =
1
0
0
0
C =
0.9300
0
1.0000
0
−0.1980
0
0
1.0000
−0.1296
0
0
0
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
362
1
0
−7
6
d =
0
The transformation matrix P to convert this state-space model to an alternative one with a diagonal
state matrix is found using the following:
>>
[ P , lambda ] = eig (A)
P =
0.4211
−0.4678
0.5198
−0.5776
0.3367
0.4209
0.5262
0.6577
0.1743
0.2904
0.4841
0.8068
−0.0258
0.0859
−0.2862
0.9540
0
0.8000
0
0
0
0
0.6000
0
0
0
0
−0.3000
lambda =
−0.9000
0
0
0
Coefficient matrices Ã, B̃ and C̃ for the alternative state-space model are determined using the
following:
>>
A _ t i l d e = inv (P ) * A * P
A_tilde =
−0.9000
0.0000
−0.0000
0.0000
>>
−0.0000
0.8000
0.0000
−0.0000
−0.0000
−0.0000
0.6000
−0.0000
−0.0000
−0.0000
−0.0000
−0.3000
1.6265
7.7014
B _ t i l d e = inv (P ) * B
B_tilde =
1.1316
4.0654
−4.5906
1.7647
>>
C _ t i l d e = C* P
C_tilde =
−6.6833
0.5998
363
9.30.
Code for the script is listed below:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
% S e t up the s t a t e space model
A = [0 , −0.5;0.25 ,0.75];
B = [2;1];
C = [3 ,1];
xn = [ 2 ; 0 ] ; % "xn " r e p r e s e n t s the current s t a t e v e c t o r x [ n ]
% Solve i t e r a t i v e l y
n = [0:99];
out = [ ] ;
% Placeholder f o r the output s i g n a l f o r n = 0 , . . . , 9 9
rn = 1 ;
% Unit−s t e p input ; always 1 f o r n>=0
for nn=0:99 ,
xnp1 = A * xn+B * rn ; % "xnp1" r e p r e s e n t s v e c t o r x [n+1]
yn = C* xn ;
out = [ out , yn ] ;
% Append y [ n ] to the output stream
xn = xnp1 ;
% Get ready f o r next i t e r a t i o n
end ;
% Graph the output s i g n a l
stem (n , out ) ;
9.31.
Set up the state space model:
1
2
3
A = [ − 0. 1 , − 0. 7; − 0. 8 , 0] ;
B = [3;1];
C = [ 2 , − 1] ;
Compute the resolvent matrix and the state transition matrix.
1
2
3
4
z = sym( ’ z ’ ) ;
tmp = z * eye(2) −A ;
rsm = z * inv (tmp)
stm = iztrans (rsm)
% Resolvent matrix
% S t a t e t r a n s i t i o n matrix
Find X (z), the z-transform of the state vector.
1
2
Xz = rsm * [ 2 ; 0 ] + 1 / ( z −1) * rsm * B
xn = iztrans ( Xz )
Find Y (z), the output transform. Afterwards compute and graph the output signal y[n].
1
2
3
4
5
6
7
8
Yz = C* Xz
yn = iztrans ( Yz )
n = [0:60];
y = subs ( yn , n ) ;
stem (n , y ) ;
axis ( [ − 0 . 5 , 6 0 . 5 , 0 , 1 2 ] ) ;
xlabel ( ’n ’ ) ;
ylabel ( ’ y [ n ] ’ ) ;
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
364
9.32.
Code for the script is listed below:
1
2
3
4
5
6
7
8
9
%% S e t up the s t a t e −space model
A = [ − 0. 1 , − 0. 7; − 0. 8 , 0] ;
B = [3;1];
C = [ 2 , − 1] ;
% Find H( z ) using symbolic math functions
z = sym( ’ z ’ ) ;
tmp = z * eye(2) −A ;
H = C * inv (tmp) * B ;
s i mpl i fy (H)
The answer produced by the script is
31/ ( 3 * ( 5 * z + 4 ) ) + 88/ ( 3 * ( 10 * z − 7 ) )
and is equivalent to
H (z) =
31/3
88/3
5 z + 0.9
+
=
5 z + 4 10 z − 7 (z − 0.7) (z + 0.8)
9.33. Set up the continuous-time state space model and convert to a discrete-time model using
Euler’s method with T = 0.02 seconds:
1
2
3
4
5
6
7
8
9
A = [ −2 , −2;1 , −5];
B = [1;0];
C = [0 ,5];
d = 0;
Ts = 0 . 0 2 ;
A_bar = eye ( 2) +A * Ts
B_bar = B * Ts
C_bar = C
d_bar = d
Solve the approximate discrete-time model iteratively:
1
2
3
4
5
6
7
8
xn = [ 3 ; − 2 ] ;
% I n i t i a l value o f s t a t e v e c t o r
n = [0:150];
% Vector o f i n d i c e s
yn = [ ] ;
% Empty v e c t o r to s t a r t
for nn=0:150 ,
xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1]
yn = [ yn , C* xn ] ;
% Append to v e c t o r ’ yn ’
xn = xnp1 ;
% New becomes old f o r next i t e r a t i o n
end ;
Graph actual and approximate solutions together:
365
1
2
3
4
5
t = [0:0.01:3];
y = 5/12+70/3 * exp(−3 * t ) −135/4 * exp(−4 * t ) ;
plot (n * Ts , yn , ’ r . ’ , t , y , ’b− ’ ) ; grid ;
ht = t i t l e ( ’ y_ { a } ( t ) and y [ n ] ’ ) ;
hx = xlabel ( ’ t ( sec ) ’ ) ;
% From Example 9.13
9.34.
a.
Set up the continuous-time state space model and convert to a discrete-time model with T = 0.1
seconds:
1
2
3
4
5
6
7
8
9
b.
1
2
3
4
5
6
7
8
A = [0 ,1 ,0;0 ,0 ,1; −15 , −11 , −5]
B = [0;0;3]
C = [1 ,0 ,0]
d = 0
Ts = 0 . 1 ;
A_bar = expm(A * Ts )
B_bar = inv (A ) * ( A_bar−eye ( 3 ) ) * B
C_bar = C
d_bar = d
Solve the approximate discrete-time model iteratively:
xn = [ 0 ; 0 ; 0 ] ;
% I n i t i a l value o f s t a t e v e c t o r
n = [0:60];
% Vector o f i n d i c e s
yn = [ ] ;
% Empty v e c t o r to s t a r t
for nn=0:60 ,
xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1]
yn = [ yn , C* xn ] ;
% Append to v e c t o r ’ yn ’
xn = xnp1 ;
% New becomes old f o r next i t e r a t i o n
end ;
The system function for the continuous-time system is
H (s) =
3
s 3 + 5 s 2 + 11 s + 15
Create a system object “sys” and use it to compute the unit-step response of the system through the use
of the function lsim(..) Afterwards graph the output signal along with that obtained through the iterative
solution of the discretized state space model:
1
2
3
4
5
6
7
sys = t f ( [ 3 ] , [ 1 , 5 , 1 1 , 1 5 ] ) ;
t = [0:0.01:6];
xa = ones ( s i z e ( t ) ) ;
ya = lsim ( sys , xa , t ) ;
plot (n * Ts , yn , ’ r . ’ , t , ya , ’b− ’ ) ; grid ;
ht = t i t l e ( ’ Actual and approximate s o l u t i o n s ’ ) ;
hx = xlabel ( ’ t ( sec ) ’ ) ;
CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS
366
9.35.
a. Set up the continuous-time state space model and convert to a discrete-time model using Euler’s
method with T = 0.1 seconds:
1
2
3
4
5
6
7
8
9
b.
1
2
3
4
5
6
7
8
A = [0 ,1 ,0;0 ,0 ,1; −15 , −11 , −5];
B = [0;7; −32];
C = [1 ,0 ,0];
d = 0;
Ts = 0 . 1 ;
A_bar = eye ( 3) +A * Ts
B_bar = B * Ts
C_bar = C
d_bar = d
Solve the approximate discrete-time model iteratively:
xn = [ 0 ; 0 ; 0 ] ;
% I n i t i a l value o f s t a t e v e c t o r
n = [0:60];
% Vector o f i n d i c e s
yn = [ ] ;
% Empty v e c t o r to s t a r t
for nn=0:60 ,
xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1]
yn = [ yn , C* xn ] ;
% Append to v e c t o r ’ yn ’
xn = xnp1 ;
% New becomes old f o r next i t e r a t i o n
end ;
Create a system object “sys” and use it to compute the unit-step response of the system through the use
of the function lsim(..) Afterwards graph the output signal along with that obtained through the iterative
solution of the discretized state space model:
1
2
3
4
5
6
7
sys = t f ( [ 7 , 3 ] , [ 1 , 5 , 1 1 , 1 5 ] ) ;
t = [0:0.01:6];
xa = ones ( s i z e ( t ) ) ;
ya = lsim ( sys , xa , t ) ;
plot (n * Ts , yn , ’ r . ’ , t , ya , ’b− ’ ) ; grid ;
ht = t i t l e ( ’ Actual and approximate s o l u t i o n s ’ ) ;
hx = xlabel ( ’ t ( sec ) ’ ) ;
Chapter 10
Analysis and Design of Filters
10.1.
a.
ωc =
1
1
=
= 1000 rad/s
RC 1000 × 10−6
The peak magnitude occurs at ω = 0 and its value is |H (0)| = 1. For at most 1 percent deviation from the
peak magnitude
s
¶
µ
100 2
− 1 = 142.5 rad/s
ω0 = 1000
100 − 1
Therefore we require |ω| ≤ 142.5 rad/s.
b. The slope of the phase characteristic at ω = 0 is
¯
d Θ ¯¯
1
1
=−
=−
¯
d ω ω=0
ωc
1000
If the phase characteristic were perfectly linear, the phase at ω0 = 142.5 rad/s would be
Θ̂ (142.5) = −
142.5
= −0.1425 radians
1000
The actual phase at ω0 = 142.5 rad/s is
Θ (142.5) = − tan−1
µ
¶
142.5
== −0.1415 radians
1000
and the percent deviation from linear phase is
µ
¶
−0.1415 + 0.1425
100
= −0.67 percent
−0.1425
c. If the phase characteristic were linear, the time delay for the frequency ω0 would be
t̂ d (ω0 ) = −
Θ̂ (ω0 ) 0.1425
=
= 0.001 sec
ω0
142.5
The actual time delay for the frequency ω0 is
t d (ω0 ) = −
Θ (ω0 ) 0.1415
= 0.000993 sec
=
ω0
142.5
367
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
368
d. For at most 2 percent deviation from the peak magnitude
ω0 = 1000
s
µ
¶
100 2
− 1 = 203.1 rad/s
100 − 2
Therefore we require |ω| ≤ 203.1 rad/s. The slope of the phase characteristic at ω = 0 is
¯
d Θ ¯¯
1
1
=−
=−
¯
d ω ω=0
ωc
1000
If the phase characteristic were perfectly linear, the phase at ω0 = 203.1 rad/s would be
Θ̂ (203.1) = −
203.1
= −0.2031 radians
1000
The actual phase at ω0 = 203.1 rad/s is
−1
Θ (203.1) = − tan
µ
¶
203.1
== −0.2003 radians
1000
and the percent deviation from linear phase is
µ
¶
−0.2003 + 0.2031
100
= −1.34 percent
−0.2031
If the phase characteristic were linear, the time delay for the frequency ω0 would be
t̂ d (ω0 ) = −
Θ̂ (ω0 ) 0.2031
= 0.001 sec
=
ω0
203.1
The actual time delay for the frequency ω0 is
t d (ω0 ) = −
Θ (ω0 ) 0.2003
=
= 0.000987 sec
ω0
203.1
10.2.
a.
H (ω) =
¯
¯
1
1
¯
=¡
p
p
¢
¯
s 2 + 2 s + 1 s= j ω
1 − ω2 + j 2ω
¯
¯
¯ H (ω)¯ = q
¡
1
1
=p
¢2
1 + ω4
1 − ω2 + 2 ω2
For the magnitude response to stay within p percent of the of the peak we need
1
100 − p
≤
p
p
1 + ω4
For p = 1
ω0 =
½µ
⇒
ω0 =
½µ
¾1/4
¶
100 2
−1
100 − p
¶
¾1/4
100 2
−1
= 0.3775 rad/s
99
369
b.
Θ (ω) = ∡ H (ω) = − tan
It can be shown that
−1
Ãp
2ω
1 − ω2
!
p ¡
¢
d Θ (ω) − 2 1 + ω2
=
dω
1 + ω4
At ω = 0 we get
¯
p
d Θ (ω) ¯¯
=− 2
¯
d ω ω=0
If the phase characteristic were perfectly linear, the phase at ω0 = 0.3775 rad/s would be
p
Θ̂ (0.3775) = − 2 (0.3775) = −0.5338 radians
The actual phase at ω0 = 0.3775 rad/s is
Θ (0.3775) = − tan−1
Ãp
!
2 (0.3775)
= −0.5568 radians
1 − 0.37752
and the percent deviation from linear phase is
µ
¶
−0.5568 + 0.5338
100
= 4.31 percent
−0.5338
c. If the phase characteristic were linear, the time delay for the frequency ω0 would be
Θ̂ (ω0 ) 0.5338
= 1.4142 sec
=
ω0
0.3775
t̂ d (ω0 ) = −
The actual time delay for the frequency ω0 is
Θ (ω0 ) 0.5568
= 1.4751 sec
=
ω0
0.3775
t d (ω0 ) = −
d. For at most 2 percent deviation from the peak magnitude
ω0 =
½µ
¾1/4
¶
100 2
−1
= 0.4506 rad/s
98
Therefore we require |ω| ≤ 0.4506 rad/s. The slope of the phase characteristic at ω = 0 is
¯
p
d Θ (ω) ¯¯
=− 2
¯
d ω ω=0
If the phase characteristic were perfectly linear, the phase at ω0 = 0.4506 rad/s would be
p
Θ̂ (0.4506) = − 2 (0.4506) = −0.6373 radians
The actual phase at ω0 = 0.4506 rad/s is
−1
Θ (0.4506) = − tan
Ãp
!
2 (0.4506)
= −0.6745 radians
1 − 0.45062
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
370
and the percent deviation from linear phase is
¶
µ
−0.6745 + 0.6373
= 5.85 percent
100
−0.6373
If the phase characteristic were linear, the time delay for the frequency ω0 would be
t̂ d (ω0 ) = −
Θ̂ (ω0 ) 0.6373
=
= 1.4142 sec
ω0
0.4506
The actual time delay for the frequency ω0 is
t d (ω0 ) = −
Θ (ω0 ) 0.6745
= 1.4969 sec
=
ω0
0.4506
10.3.
a. a. Using Eqn. (10.29)
HLP (ω) = Π
or in terms of f
¡ ¢
HLP f = Π
³ ω ´
e − j 0.1ω
400 π
µ
¶
f
e − j 0.2πω
200
b. Using Eqn. (10.31)
µ
¶
¡
¢
200 π
200 π
h LP (t ) =
sinc
(t − 0.1) = 200 sinc 200 (t − 0.1)
π
π
c.
hLP (t)
250
Amplitude
200
150
100
50
0
−50
−0.1
−0.05
0
0.05
0.1
0.15
t (sec)
10.4.
a. Using Eqn. (10.33) with ωb = 450 π rad/s and 2 ω0 = 500 pi rad/s and incorporating the time delay of
t d = 0.3 seconds we obtain
· µ
¶
µ
¶¸
ω − 450 π
ω + 450 π
HB P (ω) = Π
+Π
e − j 0.3ω
500 π
500 π
371
or in terms of f
· µ
¶
µ
¶¸
¡ ¢
f − 225
f + 225
HB P f = Π
+Π
e − j 0.6π f
250
250
b. Using Eqn. (10.37) with f b = 225 Hz and 2 f 0 = 250 Hz we obtain
¡
¢
¡
¢
h B P (t ) = 500 sinc 250 (t − 0.3) cos 450 π (t − 0.3)
c.
hBP (t)
Amplitude
400
200
0
−200
0.28
0.29
0.3
0.31
0.32
t (sec)
10.5.
Using parameters ωc = 2 rad/s and N = 4, the Butterworth squared magnitude function is
¯
¯
¯ H (ω)¯2 =
Using Eqn. (10.49)
¯
¯2 ¯¯
H (s) H (−s) = ¯ H (ω)¯ ¯
1
³ ω ´8
1+
2
ω2 =−s 2
1
=
1+
µ
−s
4
256
=
¶
4
2
256 + s 8
The poles of H (s) H (−s) are the solutions of
s 8 = −256
which can be written in the equivalent form
s 8 = 256 e j π e j 2πk = 256 e j (2k+1)π
The poles of H (s) H (−s) are at
p k = 4 e j (2k+1)π/8 ,
k = 0, . . . , 7
The poles in the left half s-plane are associated with H (s).
H (s) = ¡
s − p2
¢¡
s − p3
K
¢¡
s − p4
¢¡
s − p5
¢
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
372
Poles p 2 and p 5 form a conjugate pair:
p 2 =e j 5π/8
p 5 =e j 11π/8 = e − j 5π/8
¡
s − p2
¢¡
´
´³
¢ ³
s − p 5 = s − 2 e j 5π/8 s − 2 e − j 5π/8
=s 2 + 1.5307 s + 4
Similarly, poles p 3 and p 4 form a conjugate pair:
p 3 =e j 7π/8
p 4 =e j 9π/8 = e − j 7π/8
¡
s − p3
¢¡
´
´³
¢ ³
s − p 4 = s − 2 e j 7π/8 s − 2 e − j 7π/8
=s 2 + 3.6955 s + 4
The system function is
K
¢¡
¢
H (s) = ¡ 2
s + 1.5307 s + 4 s 2 + 3.6955 s + 4
16
= 4
s + 5.2263 s 3 + 13.6569 s 2 + 20.9050 s + 16.0000
where the gain factor K is adjusted to achieve |H (0)| = 1.
10.6.
Use Eqn. (10.65) to determine N :
N≥
log10
q¡
¢ ¡
¢
103 − 1 / 100.1 − 1
log10 (4.5/3)
= 10.18
Filter order must be chosen as N = 11. If the excess tolerance that results from rounding up N to 11 is to
be used for improving the stopband response, we need to obtain ωc from Eqn. (10.62):
µ
¶
3 22
= 101/10 − 1
ωc
⇒
ωc = 3.19 rad/s
Alternatively, the excess tolerance can be used for improving the response in the passband by solving
for ωc from Eqn. (10.63):
µ
¶
4.5 22
= 1030/10 − 1
ωc
⇒
ωc = 3.2875 rad/s
373
10.7.
¡
¢
In order to evenly distribute the excess tolerance, we need a passband ripple of R p − ∆ dB and a
stopband attenuation of (A s + ∆) dB where ∆ is a positive quantity. Using Eqns. (10.62) and (10.63) we
get
µ
¶
ω1 2N
= 10(R p −∆)/10 − 1
(P.10.7.1)
ωc
and
µ
¶
ω2 2N
= 10(A s +∆)/10 − 1
ωc
(P.10.7.2)
Combining the two requirements yields
µ
Let
K=
µ
¶
ω1 2N
,
ω2
¶
ω1 2N 10R p /10 10−∆/10 − 1
=
ω2
10 A s /10 10∆/10 − 1
A 1 = 10R p /10 ,
A 2 = 10 A s /10
and
D = 10∆/10
The condition for equal distribution of the excess tolerance between the passband and the stopband is
K=
A 1 /D − 1
A2 D − 1
or, equivalently
A 2 K D 2 + (1 − K ) D − A 1 = 0
Solving for D we obtain
D=
(K − 1) ±
p
(1 − K )2 + 4A 1 A 2 K
2A 2 K
Use the positive solution since ∆ = 10 log10 (D). Once ∆ is found, solve for ωc from either Eqn. (P.10.7.1)
or Eqn. (P.10.7.2).
10.8.
Parameter αk is found as
αk =
(2k + 1) π
,
8
k = 0, . . . , 7
and the parameter βk is
sinh−1 (1/0.3)
= 0.4797
4
Poles of H (s) H (−s) are found through the use of Eqn. (10.86) as
·
µ
¶
µ
¶
¸
(2k + 1) π
(2k + 1) π
p k = 2 cos
cosh (0.4797) − j sin
sinh (0.4797) ,
8
8
βk =
k = 0, . . . , 7
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
374
Locations of the poles of H (s) H (−s) are as follows:
p 0 =0.3814 + j 2.0645
p 1 =0.9208 + j 0.8551
p 2 =0.9208 − j 0.8551
p 3 =0.3814 − j 2.0645
p 4 = − 0.3814 − j 2.0645
p 5 = − 0.9208 − j 0.8551
p 6 = − 0.9208 + j 0.8551
p 7 = − 0.3814 + j 2.0645
Poles p 4 , p 5 , p 6 and p 7 are in the left half s-plane, and are therefore associated with H (s). System
function is constructed as
H (s) = ¡
A
¢¡
¢¡
¢¡
¢
s + 0.3814 + j 2.0645 s + 0.9208 + j 0.8551 s + 0.9208 − j 0.8551 s + 0.3814 − j 2.0645
A
= 4
3
s + 2.6044 s + 7.3915 s 2 + 9.3217 s + 6.9602
The gain factor A is adjusted to achieve |H (ω)|max = 1. With N = 4, this requires that (see Fig. 10.21)
¯
¯
1
¯ H (0)¯ = p 1
= 0.9578
=p
1 + ε2
1 + (0.3)2
and therefore
A
= 0.9578
6.9602
A = 6.6667
⇒
The system function is
H (s) =
6.6667
s 4 + 2.6044 s 3 + 7.3915 s 2 + 9.3217 s + 6.9602
10.9.
Parameter αk is found as
αk =
(2k + 1) π
,
8
k = 0, . . . , 7
and the parameter βk is
sinh−1 (1/0.3)
= 0.4797
4
Poles of H (s) H (−s) are found through the use of Eqn. (10.109) as
βk =
pk =
j2
µ
¶
,
(2k + 1) π
(2k + 1) π
cosh (0.4797) − j sin
sinh (0.4797)
cos
8
8
µ
¶
k = 0, . . . , 7
375
Locations of the poles of H (s) H (−s) are as follows:
p 0 = − 0.3461 − j 1.8736
p 1 = − 2.3324 − j 2.1661
p 2 = − 2.3324 + j 2.1661
p 3 = − 0.3461 + j 1.8736
p 4 =0.3461 + j 1.8736
p 5 =2.3324 + j 2.1661
p 6 =2.3324 − j 2.1661
p 7 =0.3461 − j 1.8736
Poles p 0 , p 1 , p 2 and p 3 are in the left half s-plane, and are therefore associated with H (s). Zeros of the
system function H (s) are found using Eqns. (10.111) and (10.112):
zk =
± j ω2
¶ ,
(2k − 1) π
cos
2N
µ
k = 1, 2
Locations of the zeros of H (s) are as follows:
z 1 =0 ± j 2.1648
z 2 =0 ± j 5.2263
System function is constructed as
¡
¢¡
¢¡
¢¡
¢
A s + j 2.1648 s − j 2.1648 s + j 5.2263 s − j 5.2263
¢¡
¢¡
¢¡
¢
H (s) = ¡
s + 0.3461 + j 1.8736 s + 2.3324 + j 2.1661 s + 2.3324 − j 2.1661 s + 0.3461 − j 1.8736
0.2873 s 4 + 9.1951 s 2 + 36.7805
= 4
s + 5.3571 s 3 + 16.9916 s 2 + 23.9481 s + 36.7805
The peak of the system function occurs at ω = 0. The gain factor A is adjusted to achieve |H (0)| = 1.
10.10.
a. The minimum filter order is found using Eqn. (10.65):
N≥
log10
q¡
¢ ¡
¢
1020/10 − 1 / 101/10 − 1
log10 (3.5/2)
= 5.3129
Filter order must be chosen as N = 6. If the excess tolerance that results from rounding up N to 6 is to be
used for improving the passband response, we need to obtain ωc from Eqn. (10.63):
¶
µ
3.5 12
= 1020/10 − 1
⇒
ωc = 2.3865 rad/s
ωc
The poles of H (s) H (s) are at
p k = 2.3865 e j (2k+1)π/12 ,
k = 0, . . . , 11
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
376
Using the poles in the left half s-plane, H (s) is constructed as
H (s) =
184.75
s 6 + 9.2208 s 5 + 42.5116 s 4 + 124.256 s 3 + 242.124 s 2 + 299.109 s + 184.75
b. Minimum filter order is found using Eqn. (10.94) and Eqn. (10.101):
F=
N≥
s
1020/10 − 1
101/10 − 1
= 19.5538
cosh−1 (19.5538)
cosh−1 (3.5/2)
= 3.1633
which must be rounded up to N = 4. The parameter ε is found using Eqn. (10.89):
ε=
p
101/10 − 1 = 0.5088
The poles of the product H (s) H (−s) can now be found using Eqns. (10.82), (10.84) and (10.85).
Parameter αk is found as
(2k + 1) π
,
k = 0, . . . , 7
αk =
8
and the parameter βk is
sinh−1 (1/0.5088)
= 0.3570
βk =
4
Poles of H (s) H (−s) are found through the use of Eqn. (10.86) as
·
µ
¶
µ
¶
¸
(2k + 1) π
(2k + 1) π
p k = 2 cos
cosh (0.3570) − j sin
sinh (0.3570) ,
k = 0, . . . , 7
8
8
Locations of the poles of H (s) H (−s) are as follows:
p 0 =0.2791 − j 1.9668
p 1 =0.6737 − j 0.8147
p 2 =0.6737 + j 0.8147
p 3 =0.2791 + j 1.9668
p 4 = − 0.2791 + j 1.9668
p 5 = − 0.6737 + j 0.8147
p 6 = − 0.6737 − j 0.8147
p 7 = − 0.2791 − j 1.9668
Poles p 4 , p 5 , p 6 and p 7 are in the left half s-plane, and are therefore associated with H (s). System
function is constructed as
H (s) = ¡
A
¢¡
¢¡
¢¡
¢
s + 0.2791 − j 1.9668 s + 0.6737 − j 0.8147 s + 0.6737 + j 0.8147 s + 0.2791 + j 1.9668
A
= 4
3
s + 1.9056 s + 5.8157 s 2 + 5.9410 s + 4.4100
377
The gain factor A is adjusted to achieve |H (ω)|max = 1. With N = 4, this requires that (see Fig. 10.21)
¯
¯
1
¯ H (0)¯ = p 1
= 0.8913
=p
1 + ε2
1 + (0.5088)2
and therefore
A
= 0.8913
4.4100
A = 3.9305
⇒
The system function is
H (s) =
3.9305
s 4 + 1.9056 s 3 + 5.8157 s 2 + 5.9410 s + 4.4100
c. Minimum filter order is found using Eqn. (10.117) and Eqn. (10.118):
F=
N≥
s
1020/10 − 1
101/10 − 1
= 19.5538
cosh−1 (19.5538)
cosh−1 (3.5/2)
= 3.1633
which must be rounded up to N = 4. The parameter ε is found using Eqn. (10.114):
ε= p
Parameter αk is found as
αk =
1
1020/10 − 1
(2k + 1) π
,
8
= 0.1005
k = 0, . . . , 7
and the parameter βk is
sinh−1 (1/0.1005)
= 0.7483
4
Poles of H (s) H (−s) are found through the use of Eqn. (10.109) as
βk =
pk =
j2
µ
¶
,
(2k + 1) π
(2k + 1) π
cosh (0.7483) − j sin
sinh (0.0.7483)
cos
8
8
µ
¶
Locations of the poles of H (s) H (−s) are as follows:
p 0 = − 0.7198 + j 2.7402
p 1 = − 3.2378 + j 2.1149
p 2 = − 3.2378 − j 2.1149
p 3 = − 0.7198 − j 2.7402
p 4 =0.7198 − j 2.7402
p 5 =3.2378 − j 2.1149
p 6 =3.2378 + j 2.1149
p 7 =0.7198 + j 2.7402
k = 0, . . . , 7
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
378
Poles p 0 , p 1 , p 2 and p 3 are in the left half s-plane, and are therefore associated with H (s). Zeros of the
system function H (s) are found using Eqns. (10.111) and (10.112):
zk =
± j ω2
¶ ,
(2k − 1) π
cos
2N
k = 1, 2
µ
Locations of the zeros of H (s) are as follows:
z 1 =0 ± j 3.7884
z 2 =0 ± j 9.1459
System function is constructed as
¡
¢¡
¢¡
¢¡
¢
A s + j 3.7884 s − j 3.7884 s + j 9.1459 s − j 9.1459
¢¡
¢¡
¢¡
¢
H (s) = ¡
s + 0.7198 − j 2.7402 s + 3.2378 − j 2.1149 s + 3.2378 + j 2.1149 s + 0.7198 + j 2.7402
0.1 s 4 + 9.8 s 2 + 120.05
= 4
s + 7.9152 s 3 + 32.3049 s 2 + 73.5079 s + 120.05
The peak of the system function occurs at ω = 0. The gain factor A is adjusted to achieve |H (0)| = 1.
10.11.
Let the system functions for lowpass and bandpass filters be HLP (s) and HB P (λ) respectively.
Furthermore, let ω and p be the radian frequency variables in s and λ domains so that s = j ω and
λ = j p. The lowpass to bandpass frequency transformation is
s=
λ2 + ω20
Bλ
jω=
⇒
−p 2 + ω20
⇒
jBp
−ω =
−p 2 + ω20
Bp
p (rad/s)
Lowpass to bandpass frequency transformation
ωL1
−ωL1
−ωB3
−ωB2
ωB2
ωB3
ω (rad/s)
The goal is to map the band of frequencies −ωL1 < ω < ωL1 for the lowpass filter to the band of
frequencies ωB 2 < p < ωB 3 for the bandpass filter. At the two critical frequencies ω = ±ωL1 we have
− ωL1 =
−ω2B 3 + ω20
B ωB 3
and
ωL1 =
−ω2B 2 + ω20
B ωB 2
379
which can be solved together to yield
ω2B 3 − ω20
B ωB 3
=
−ω2B 2 + ω20
B ωB 2
and
B=
ω0 =
and
p
ωB 2 ωB 3
ωB 3 − ωB 2
ωL1
10.12.
Let the system functions for lowpass and band-reject filters be HLP (s) and HB R (λ) respectively.
Furthermore, let ω and p be the radian frequency variables in s and λ domains so that s = j ω and
λ = j p. The lowpass to bandpass frequency transformation is
s=
Bλ
⇒
λ2 + ω20
jω=
jBp
⇒
−p 2 + ω20
ω=
−B p
p 2 − ω20
p (rad/s)
Lowpass to band-reject frequency transformation
ωL1
−ωL1
−ωS4 −ω0 −ωS1
ωS1 ω0
ωS4
ω (rad/s)
The goal is to map the band of frequencies −ωL1 < ω < ωL1 for the lowpass filter to the three bands of
frequencies, namely
Band 1:
− ∞ < p < ωS4
Band 2:
ωS1 < p < ωS1
Band 3:
ωS4 < p < ∞
for the band-reject filter. At the two critical frequencies ω = ±ωL1 we have
− ωL1 =
−B ωS4
ω2S4 − ω20
and
ωL1 =
−B ωS1
ω2S1 − ω20
which can be solved together to yield
B ωS4
ω2S4 − ω20
=
−B ωS1
ω2S1 − ω20
and
ω0 =
and
B = (ωS4 − ωS1 ) ωL1
p
ωS1 ωS4
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
380
10.13.
a.
ω0 = ωL1 ωH2 = (2) (5) = 10
The transformation is
s=
b. Let s = j ω and λ = j p so that
jω =
10
jp
10
λ
⇒
ω=−
10
p
If ω1 = 6 rad/s, the corresponding frequency of the highpass filter is
p1 = −
10
= −1.6667 rad/s
6
c.
H (λ) = µ
At λ = j 5
Similarly, for G (s) at s = j 2 we have
λ
2
¶
=
10
λ+5
+2
λ
¯
¯
¯ j5 ¯
¯ = p1
= ¯¯
j5+5¯
λ= j 5
2
¯
¯¯
¯ H (λ)¯¯¯
¯
¯¯
¯G (s)¯¯¯
¯
¯
¯ j2 ¯
¯
¯ = p1
=¯
j2+2¯
s= j 2
2
10.14.
a. Let the passband edge frequency for the lowpass filter be ωL1 = 1 rad/s. For the lowpass to highpass
transformation we have
ω0 = ωL1 ωH2 = (1) (5) = 5
Therefore, the transformation is
5
λ
The frequency ωL1 for the lowpass filter corresponds to the frequency −ωH2 for the highpass filter.
Similarly, the frequency ωL2 for the lowpass filter should correspond to the frequency −ωH1 for the
highpass filter, therefore
5
= 2.5 rad/s
ωL2 =
ωH1
s=
The specifications for the lowpass prototype are
ωL1 = 2 rad/s
ωL2 = 2.5 rad/s
R p = 1 dB
A s = 30 dB
381
b. The minimum filter order is found using Eqn. (10.65):
N≥
log10
q¡
¢ ¡
¢
1030/10 − 1 / 101/10 − 1
log10 (2.5/1)
= 4.5062
Filter order must be chosen as N = 5. If the excess tolerance that results from rounding up N to 5 is to be
used for improving the passband response, we need to obtain ωc from Eqn. (10.63):
µ
¶
2.5 10
= 1030/10 − 1
ωc
⇒
ωc = 1.2531 rad/s
c. The poles of G (s) G (s) are at
p k = 1.2531 e j kπ/5 ,
k = 0, . . . , 9
Using the poles in the left half s-plane, G (s) is constructed as
G (s) =
3.0897
s 5 + 4.0551 s 4 + 8.2219 s 3 + 10.3028 s 2 + 7.9791 s + 3.0897
d.
¯
¯
H (λ) = G (s)¯
3.0897
= µ ¶5
µ ¶4
µ ¶3
µ ¶2
µ ¶
s=5/λ
5
5
5
5
5
+ 4.0551
+ 8.2219
+ 10.3028
+ 7.9791
+ 3.0897
λ
λ
λ
λ
λ
=
λ5
λ5 + 12.91 λ4 + 83.36 λ3 + 332.63 λ2 + 820.28 λ + 1011.4
e.
¯
¯
20 log10 ¯H (2)¯ = − 30 dB
¯
¯
20 log10 ¯H (5)¯ = − 0.43 dB
10.15.
a. The partial fraction form of G (s) is
G (s) =
2
2
−
s +1 s +2
and the corresponding impulse response is
g (t ) = 2 e −t u (t ) − 2 e −2t u (t )
b.
h[n] = 0.5 g (0.5 n) = e −0.5 n u[n] − e −n u[n]
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
382
c.
z
z
0.2387 z
−
= 2
−0.5
−1
z −e
z −e
z − 0.9744 z + 0.2231
H (z) =
10.16.
s=
The system function is
H (z) = µ
=
=
2 1 − z −1 4 (z − 1)
=
T 1 + z −1
z +1
2
¶µ
¶
4 (z − 1)
4 (z − 1)
+1
+2
z +1
z +1
2 (z + 1)2
(5 z − 3) (6 z − 2)
2 z2 + 4 z + 2
30 z 2 − 28 z + 6
10.17.
a. The system function is
H (Ω) =
NX
−1
h[n] e − j Ωn
n=0
Since N is even, the summation can be split into two halves:
H (Ω) =
NX
/2−1
n=0
h[n] e − j Ωn +
NX
−1
h[n] e − j Ωn
n=N /2
Using the variable change n = N − 1 − m, the second summation can be written as
NX
−1
n=N /2
h[n] e − j Ωn =
0
X
m=N /2−1
h[N − 1 − m] e − j Ω (N −1−m)
Recognizing that h[N − 1 − m] = h[m] and that the summation limits can be swapped, we have
NX
−1
n=N /2
h[n] e − j Ωn = e − j (N −1)
NX
/2−1
h[m] e j Ω m
m=0
and H (Ω) is
H (Ω) =
=
NX
/2−1
n=0
h[n] e − j Ωn + e − j (N −1)
NX
/2−1 n
n=0
NX
/2−1
h[n] e j Ωn
n=0
h[n] e − j Ωn + e − j (N −1) h[n] e j Ωn
o
383
Factoring out e − j (N −1)/2 the result can be written as
H (Ω) =e − j (N −1)/2
=e
− j (N −1)/2
NX
/2−1 n
n=0
h[n] e − j Ωn e j Ω (N −1)/2 + h[n] e j Ωn e − j Ω (N −1)/2
o
¶¸
· µ
N −1
2 h[n] cos Ω n −
2
n=0
|
{z
}
NX
/2−1
Purely real
=e − j (N −1)/2 A (Ω)
b. The system function is
H (Ω) =
NX
−1
h[n] e − j Ωn
n=0
Since N is even, the summation can be split into two halves:
H (Ω) =
NX
/2−1
n=0
h[n] e − j Ωn +
NX
−1
h[n] e − j Ωn
n=N /2
Using the variable change n = N − 1 − m, the second summation can be written as
NX
−1
n=N /2
h[n] e − j Ωn =
0
X
m=N /2−1
h[N − 1 − m] e − j Ω (N −1−m)
Recognizing that h[N − 1 − m] = −h[m] and that the summation limits can be swapped, we have
NX
−1
n=N /2
h[n] e − j Ωn = −e − j (N −1)
NX
/2−1
h[m] e j Ω m
m=0
and H (Ω) is
H (Ω) =
=
NX
/2−1
n=0
h[n] e − j Ωn − e − j (N −1)
NX
/2−1 n
n=0
NX
/2−1
h[n] e j Ωn
n=0
h[n] e − j Ωn − e − j (N −1) h[n] e j Ωn
o
Factoring out e − j (N −1)/2 the result can be written as
H (Ω) =e − j (N −1)/2
NX
/2−1 n
n=0
h[n] e − j Ωn e j Ω (N −1)/2 − h[n] e j Ωn e − j Ω (N −1)/2
¶¸
· µ
N −1
− j 2 h[n] sin Ω n −
2
n=0
· µ
¶¸
NX
/2−1
N −1
−2 h[n] sin Ω n −
=e − j (N −1)/2 e j π/2
2
n=0
|
{z
}
=e − j (N −1)/2
NX
/2−1
Purely real
=e − j (N −1)/2 e j π/2 B (Ω)
o
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
384
10.18.
a. The system function is
H (Ω) =
NX
−1
h[n] e − j Ωn
n=0
Since N is odd, the summation must be split as follows:
H (Ω) =
−3)/2
(N X
n=0
h[n] e − j Ωn + h[(N − 1) /2] e − j ω(N −1)/2 +
NX
−1
h[n] e − j Ωn
n=(N +1)/2
Using the variable change n = N − 1 − m, the second summation can be written as
NX
−1
n=(N +1)/2
h[n] e − j Ωn =
0
X
m=(N −3)/2
h[N − 1 − m] e − j Ω (N −1−m)
Recognizing that h[N − 1 − m] = h[m] and that the summation limits can be swapped, we have
NX
−1
n=(N +1)/2
h[n] e − j Ωn = e − j (N −1)
−3)/2
(N X
h[m] e j Ω m
m=0
and H (Ω) is
H (Ω) =
=
−3)/2
(N X
n=0
h[n] e − j Ωn + h[(N − 1) /2] e − j ω(N −1)/2 + e − j (N −1)
−3)/2 n
(N X
−3)/2
(N X
h[n] e j Ωn
n=0
o
h[n] e − j Ωn + e − j (N −1) h[n] e j Ωn + h[(N − 1) /2] e − j ω(N −1)/2
n=0
Factoring out e − j (N −1)/2 the result can be written as
#
"
o
−3)/2 n
(N X
− j Ωn j Ω (N −1)/2
j Ωn − j Ω (N −1)/2
− j (N −1)/2
+ h[(N − 1) /2]
h[n] e
e
+ h[n] e
e
H (Ω) =e
n=0
=e
− j (N −1)/2
"
|
−3)/2
(N X
n=0
#
¶¸
· µ
N −1
+ h[(N − 1) /2]
2 h[n] cos Ω n −
2
{z
}
Purely real
=e − j (N −1)/2 A (Ω)
b. The system function is
H (Ω) =
NX
−1
h[n] e − j Ωn
n=0
It should be noted that, for the odd symmetry to work in this case, the center sample must be zero, that
is, h[(N − 1) /2] = 0 as no other value would be equal to its own negative. The summation can be split
into two halves as follows:
H (Ω) =
−3)/2
(N X
n=0
h[n] e − j Ωn +
NX
−1
n=(N +1)/2
h[n] e − j Ωn
385
Using the variable change n = N − 1 − m, the second summation can be written as
NX
−1
n=(N +1)/2
h[n] e − j Ωn =
0
X
m=(N −3)/2
h[N − 1 − m] e − j Ω (N −1−m)
Recognizing that h[N − 1 − m] = −h[m] and that the summation limits can be swapped, we have
NX
−1
n=(N +1)/2
h[n] e − j Ωn = −e − j (N −1)
−3)/2
(N X
h[m] e j Ω m
m=0
and H (Ω) is
H (Ω) =
=
−3)/2
(N X
n=0
h[n] e − j Ωn − e − j (N −1)
−3)/2 n
(N X
n=0
−3)/2
(N X
h[n] e j Ωn
n=0
h[n] e − j Ωn − e − j (N −1) h[n] e j Ωn
o
Factoring out e − j (N −1)/2 the result can be written as
H (Ω) =e − j (N −1)/2
−3)/2 n
(N X
n=0
h[n] e − j Ωn e j Ω (N −1)/2 − h[n] e j Ωn e − j Ω (N −1)/2
o
¶¸
· µ
N −1
− j 2 h[n] sin Ω n −
=e
2
n=0
·
µ
¶¸
−3)/2
(N X
N −1
− j (N −1)/2 j π/2
−2 h[n] sin Ω n −
=e
e
2
n=0
|
{z
}
− j (N −1)/2
−3)/2
(N X
Purely real
=e − j (N −1)/2 e j π/2 B (Ω)
10.19.
Spectrum of an ideal bandpass filter can be expressed as the difference of the spectra of two lowpass
filters:
¶
¶
µ
µ
Ω2
Ω2
Ω2 n
Ω2 n
HB P (Ω) =
−
sinc
sinc
π
π
π
π
=0.7 sinc (0.7 n) − 0.3 sinc (0.3 n)
Since N = 2M + 1 = 19 we have M = 9. Hamming window with M = 9 is
¶
µ
π (n + M )
w [n] = 0.54 − 0.46 cos
M
Coefficients are listed in the table on the left. The impulse response of the designed filter is
h[n] = h T [n − 9] w [n − 9]
and is listed in the table on the right.
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
386
n
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
7
8
9
h T [n]
0.0000
0.0757
0.0000
−0.0624
0.0000
−0.0935
0.0000
0.3027
0.0000
−0.4000
0.0000
0.3027
0.0000
−0.0935
0.0000
−0.0624
0.0000
0.0757
0.0000
w [n]
0.0800
0.1077
0.1876
0.3100
0.4601
0.6199
0.7700
0.8924
0.9723
1.0000
0.9723
0.8924
0.7700
0.6199
0.4601
0.3100
0.1876
0.1077
0.0800
h T [n] w [n]
0.0000
0.0082
0.0000
−0.0193
0.0000
−0.0580
0.0000
0.2702
0.0000
−0.4000
0.0000
0.2702
0.0000
−0.0580
0.0000
−0.0193
0.0000
0.0082
0.0000
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
h[n]
0.0000
0.0082
0.0000
−0.0193
0.0000
−0.0580
0.0000
0.2702
0.0000
−0.4000
0.0000
0.2702
0.0000
−0.0580
0.0000
−0.0193
0.0000
0.0082
0.0000
10.20.
Using the inverse DTFT with the desired spectrum Hd (Ω) yields
1
h d [n] =
2π
=
ˆ π
ˆ π
Hd (Ω) e j Ωn d Ω
−π
(1) e − j Ω (N −1)/2 e j Ωn d Ω
−π
µ
¶¸
·
N −1
Ωc
Ωc
n−
sinc
=
π
π
2
If normalized frequency F c is used, then
µ
·
¶¸
N −1
h d [n] = 2F c sinc 2F c n −
2
Truncating the ideal impulse response h d [n] to keep only the samples for n = 0, . . . , N − 1 yields
h T [n] =
½
h d [n] ,
0,
n = 0, . . . , N − 1
otherwise
387
Check for linear phase:
¶¸
µ
·
N −1
h d [N − 1 − n] =2F c sinc 2F c N − 1 − n −
2
µ
·
¶¸
N −1
=2F c sinc 2F c −n +
2
=h d [n] ,
n = 0, . . . , N − 1
10.21.
a. Triangular window:
¯
¯
w [n] = 1 − α ¯n + β¯
The center of the window function needs to be at n (N − 1) /2. Therefore
β=−
N −1
2
We also need w [0] = w [N − 1] = 0. Using the value at either end results in
α=
2
N −1
Using the values of α and β found, the triangular window function is
¯
¶¯
µ
¯
N − 1 ¯¯
2
¯
n−
w [n] = 1 −
N −1 ¯
2 ¯
b. Hanning window
¡
¢
w [n] = 0.5 − 0.5 cos α n + β
The center of the window function needs to be at n (N − 1) /2. Therefore
¶
µ
N −1
N −1
+ β = −1
⇒
α
+β = π
cos α
2
2
We also need w [0] = w [N − 1] = 0. Using the value at n = 0 results in
¡ ¢
cos β = 1
⇒
β=0
⇒
α=
2π
N −1
Using the values of α and β found, the Hamming window function is
µ
¶
2πn
w [n] = 0.5 − 0.5 cos
N −1
c. Hamming window
¡
¢
w [n] = 0.54 − 0.46 cos α n + β
The center of the window function needs to be at n (N − 1) /2. Therefore
µ
¶
N −1
N −1
cos α
+ β = −1
⇒
α
+β = π
2
2
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
388
We also need w [0] and w [N − 1] to be minimum, that is, w [0] = w [N − 1] = 0.08. Using the value at n = 0
results in
¡ ¢
2π
cos β = 1
⇒
β=0
⇒
α=
N −1
Using the values of α and β found, the Hamming window function is
¶
µ
2πn
w [n] = 0.54 − 0.46 cos
N −1
d. Blackman window
¡
¢
¡
¢
w [n] = 0.42 − 0.5 cos α n + β + 0.08 cos α n + β
Parameters α and β are the same as in parts (b) and (c):
α=
2π
,
N −1
0.42 − 0.5 cos
µ
β=0
¶
µ
¶
4πn
2πn
+ 0.08 cos
N −1
N −1
10.22. a.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
p = 1;
% Determine the frequency omg0 f o r p percent magnitude v a r i a t i o n
omg0 = 1000 * sqrt ((100/(100 − p))^2 −1);
% Define anonymous function f o r H(omg)
H = @(omg) 1. / ( 1+ j *omg/ 1000) ;
% Graph the magnitude in the range −omg0, +omg0
omg = [−omg0 : 0 . 1 : omg0 ] ;
subplot ( 3 , 1 , 1 ) ;
plot (omg, abs (H(omg ) ) ) ;
axis ([ −omg0, omg0, 0 . 9 9 , 1 ] ) ;
t i t l e ( ’ |H( \omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
% Graph the phase in the range −omg0, +omg0
subplot ( 3 , 1 , 2 ) ;
plot (omg, angle (H(omg ) ) ) ;
axis ([ −omg0, omg0,− pi /18 , pi / 1 8 ] ) ;
t i t l e ( ’ Phase of H( \omega) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
% Graph the time delay in the range −omg0, omg0
subplot ( 3 , 1 , 3 ) ;
omg = omg+eps ; % Avoid d i v i s i o n by 0
tdelay = −angle (H(omg ) ) . / omg;
plot (omg, tdelay ) ;
axis ([ −omg0, omg0, 0 . 0 0 0 9 , 0 . 0 0 1 1 ] ) ;
389
29
30
31
32
t i t l e ( ’Time delay fo r H( \omega) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Delay ( sec ) ’ ) ;
grid ;
b. Modify line 1 as follows and repeat.
1
p = 2;
10.23.
a.
p = 1;
% Determine the frequency omg0 f o r 1 percent magnitude v a r i a t i o n
tmp = (100/(100 −p))^2 −1;
omg0 = sqrt ( sqrt (tmp ) ) ;
% Define anonymous function f o r H(omg)
H = @(omg) 1./((1 −omg. * omg)+ j * sqrt ( 2 ) * omg ) ;
% Graph the magnitude in the range −omg0, +omg0
omg = [−omg0 : 0 . 0 1 :omg0 ] ;
subplot ( 3 , 1 , 1 ) ;
plot (omg, abs (H(omg ) ) ) ;
axis ([ −omg0, omg0, 0 . 9 9 , 1 ] ) ;
t i t l e ( ’ |H( \omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
% Graph the phase in the range −omg0, +omg0
subplot ( 3 , 1 , 2 ) ;
plot (omg, angle (H(omg ) ) ) ;
axis ([ −omg0, omg0, − 0 . 7 5 , 0 . 7 5 ] ) ;
t i t l e ( ’ Phase of H( \omega) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
% Graph the time delay in the range −omg0, omg0
subplot ( 3 , 1 , 3 ) ;
omg = omg+eps ; % Avoid d i v i s i o n by 0
tdelay = −angle (H(omg ) ) . / omg;
plot (omg, tdelay ) ;
axis ([ −omg0, omg0, 1 . 2 , 1 . 5 ] ) ;
t i t l e ( ’Time delay fo r H( \omega) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Delay ( sec ) ’ ) ;
grid ;
b. Modify line 1 as follows and repeat.
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
390
1
p = 2;
10.24.
a.
function h = s s _ i l p (omg0, td , t )
f0 = omg0/ ( 2 * pi ) ;
h = 2 * f0 * sinc ( 2 * f0 * ( t−td ) ) ;
end
% Eqn . ( 1 0 . 3 2 )
b. Duplicate the result of Problem 10.3.
t = [ −0.1:0.0001:0.2];
hLP = s s _ i l p (200 * pi , 0 . 1 , t ) ;
plot ( t , hLP ) ;
grid ;
10.25.
a.
function h = ss_ibp (omg1, omg2, td , t )
h = s s _ i l p (omg2, td , t )− s s _ i l p (omg1, td , t ) ;
end
b. Duplicate the result of Problem 10.4.
t = [0.275:0.00005:0.325];
hBP = ss_ibp (200 * pi , 700 * pi , 0 . 3 , t ) ;
plot ( t , hBP ) ;
axis ( [ 0. 275 , 0. 325 , − 350 , 550] ) ;
xlabel ( ’ t ( sec ) ’ ) ;
ylabel ( ’ Amplitude ’ ) ;
t i t l e ( ’ h_ {BP } ( t ) ’ ) ;
grid ;
10.26.
The filter can be designed using the statement
[num, den ] = butter ( 4 , 2 , ’ s ’ )
which results in numerator and denominator polynomials
391
num =
0
0
0
0
16.0000
den =
1.0000
5.2263
13.6569
20.9050
16.0000
The system function can be evaluated at a set of radian frequencies using the following code:
omg = [ 0 : 0 . 0 1 : 5 ] ;
H = freqs (num, den ,omg ) ;
Graph the magnitude response:
plot (omg, abs (H) ) ;
axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
t i t l e ( ’ |H( \omega ) | ’ ) ;
grid ;
Graph the phase response:
plot (omg, angle (H) ) ;
axis ([0 ,5 , − pi , pi ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
t i t l e ( ’ Phase of H( \omega) ’ ) ;
grid ;
10.27.
The filter is designed using the following statements:
N = 11;
omgc = 3. 2875;
[num, den ] = butter (N, omgc, ’ s ’ )
Evaluate the system function for ω = 0, . . . , 6 rad/s.
omg = [ 0 : 0 . 0 1 : 6 ] ;
H = freqs (num, den ,omg ) ;
Graph the dB magnitude of the system function.
plot (omg, 2 0 * log10 ( abs (H) ) ) ;
axis ( [ 0 , 6 , − 6 0 , 1 0 ] ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude (dB) ’ ) ;
t i t l e ( ’ |H( \omega ) | _ {dB} ’ ) ;
grid ;
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
392
10.28.
a. The system function for the filter is found using the following statements:
omg1 = 2 ;
epsilon = 0 . 3 ;
% Maximum passband r i p p l e
Rp = 10 * log10 (1+ epsilon ^2)
[num, den ] = cheby1 ( 4 ,Rp, omg1, ’ s ’ )
b. Evaluate the system function in the interval ω = 0, . . . , 5 rad/s.
omg = [ 0 : 0 . 0 1 : 5 ] ;
H = freqs (num, den ,omg ) ;
Graph |H (ω)|:
plot (omg, abs (H) ) ;
axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( s ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
Graph ∡ H (ω):
plot (omg, angle (H) ) ;
axis ([0 ,5 , − pi , pi ] ) ;
t i t l e ( ’ Phase of H( s ) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
c. Pole-zero plot is obtained through the following:
p = roots ( den )
z = roots (num)
plot ( real (p ) , imag (p ) , ’ rx ’ , real ( z ) , imag ( z ) , ’ ro ’ ) ;
axis ( [ − 2 , 2 , − 3 , 3] ) ;
xlabel ( ’ Real ’ ) ;
ylabel ( ’ Imag ’ ) ;
grid ;
393
10.29.
a. The system function for the filter is found using the following statements:
omg2 = 2 ;
epsilon = 0 . 3 ;
% Minimum stopband attenuation
As = 10 * log10 ( ( 1+ epsilon ^2)/ epsilon ^ 2 ) ;
% Compute system function using MATLAB function cheby2 ( )
[num, den ] = cheby2 ( 4 , As , omg2, ’ s ’ )
b. Evaluate the system function in the interval ω = 0, . . . , 5 rad/s.
omg = [ 0 : 0 . 0 1 : 5 ] ;
H = freqs (num, den ,omg ) ;
Graph |H (ω)|:
plot (omg, abs (H) ) ;
axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( s ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
Graph ∡ H (ω):
plot (omg, angle (H) ) ;
axis ([0 ,5 , − pi , pi ] ) ;
t i t l e ( ’ Phase of H( s ) ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Phase ( rad ) ’ ) ;
grid ;
c. Pole-zero plot is obtained through the following:
p = roots ( den )
z = roots (num)
plot ( real (p ) , imag (p ) , ’ rx ’ , real ( z ) , imag ( z ) , ’ ro ’ ) ;
axis ( [ − 3 , 3 , − 6 , 6] ) ;
xlabel ( ’ Real ’ ) ;
ylabel ( ’ Imag ’ ) ;
grid ;
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
394
10.30.
a.
omg1 = 2 ;
omg2 = 3 . 5 ;
Rp = 1 ;
As = 20;
% Butterworth design
[N1, omgc] = buttord (omg1, omg2, Rp , As , ’ s ’ )
[num1, den1 ] = butter (N1, omgc, ’ s ’ )
% Chebyshev type−I design
N2 = cheb1ord (omg1, omg2, Rp , As , ’ s ’ )
[num2, den2 ] = cheby1 (N2, Rp , omg1, ’ s ’ )
% Chebyshev type−I I design
N3 = cheb2ord (omg1, omg2, Rp , As , ’ s ’ )
[num3, den3 ] = cheby2 (N3, As , omg2, ’ s ’ )
b. The following code evaluates the system function for each design.
omg = [ 0 : 0 . 0 1 : 6 ] ;
H1 = freqs (num1, den1 ,omg ) ;
H2 = freqs (num2, den2 ,omg ) ;
H3 = freqs (num3, den3 ,omg ) ;
Graph |H (s)| for the Butterworth design.
plot (omg, abs (H1 ) ) ;
axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( s ) | fo r the Butterworth design ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
Graph |H (s)| for the Chebyshev type-I design.
plot (omg, abs (H2 ) ) ;
axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( s ) | fo r the Chebyshev type−I design ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
Graph |H (s)| for the Chebyshev type-II design.
plot (omg, abs (H3 ) ) ;
axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( s ) | fo r the Chebyshev type−I I design ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
395
c. Plot the dB magnitude of all three designs simultaneously.
plot (omg, 2 0 * log10 ( abs (H1) ) ,omg, 2 0 * log10 ( abs (H2) ) , omg, 2 0 * log10 ( abs (H3 ) ) ) ;
axis ( [ 0 , 6 , − 5 0 , 1 0 ] ) ;
t i t l e ( ’ |H( s ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’dB Magnitude ’ ) ;
legend ( ’ Butterworth ’ , ’ Chebyshev type−I ’ , ’ Chebyshev type−I I ’ ) ;
grid ;
10.31.
Design the filter:
% S e t up f i l t e r s p e c i f i c a t i o n s
omg1 = 1 ;
omg2 = 2 . 5 ;
Rp = 1 ;
As = 30;
% Design the lowpass prototype
[N, omgc] = buttord (omg1, omg2, Rp , As , ’ s ’ )
[numL, denL ] = butter (N, omgc, ’ s ’ )
% Lowpass to highpass transformation
[numH, denH] = lp2hp (num, den , 5 )
Evaluate and graph the dB magnitude:
omg = [ 1 : 0 . 0 1 : 1 0 ] ;
H = freqs (numH, denH,omg ) ;
plot (omg, 2 0 * log10 ( abs (H) ) ) ;
axis ( [ 0 , 10 , − 70 , 10] ) ;
t i t l e ( ’ |H( \omega ) | _ {dB} ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude (dB) ’ ) ;
grid ;
10.32.
a. The discrete-time filter can be designed using the following statements:
num = [ 2 ] ;
den = [ 1 , 3 , 2 ] ;
[numz, denz ] = impinvar (num, den , 2 )
MATLAB response is
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
396
numz =
0
denz =
1.0000
0.2387
−0.9744
0.2231
and corresponds to
H (z) =
0.2387 z
0 + 0.2387 z −1
=
1 − 0.9744 z −1 + 0.2231 z −2 z 2 − 0.9744 z + 0.2231
b.
omg = [ 0 : 0 . 0 1 : 2 * pi ] ;
G = freqs (num, den ,omg ) ;
plot (omg, abs (G ) ) ;
axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
c.
Omg = [ 0 : 0 . 0 1 : 1 ] * pi ;
H = freqz (numz, denz ,Omg) ;
plot (Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( \Omega ) | ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
d. Use ω = Ω/T s = 2 Ω. Graph |G (ω)| and |Ya (ω) /X a (ω)| up to f max = f s /2 = 1 Hz, or equivalently, up
to ωmax = ωs /2 = 2π rad/s.
plot (omg, abs (G) , ’b ’ , 2 *Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
grid ;
10.33.
a. The discrete-time filter can be designed using the following statements:
397
num = [ 2 ] ;
den = [ 1 , 3 , 2 ] ;
[numz, denz ] = impinvar (num, den , 4 )
MATLAB response is
numz =
0
denz =
1.0000
0.0861
−1.3853
0.4724
and corresponds to
H (z) =
0 + 0.0861 z −1
0.0861 z
= 2
−1
−2
1 − 1.3853 z + 0.4724 z
z − 1.3853 z + 0.4724
b.
omg = [ 0 : 0 . 0 1 : 4 * pi ] ;
G = freqs (num, den ,omg ) ;
plot (omg, abs (G ) ) ;
axis ( [ 0 , 4 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
c.
Omg = [ 0 : 0 . 0 1 : 1 ] * pi ;
H = freqz (numz, denz ,Omg) ;
plot (Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( \Omega ) | ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
d. Use ω = Ω/T s = 4 Ω. Graph |G (ω)| and |Ya (ω) /X a (ω)| up to f max = f s /2 = 2 Hz, or equivalently, up
to ωmax = ωs /2 = 4π rad/s.
plot (omg, abs (G) , ’b ’ , 4 *Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , 4 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
grid ;
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
398
This design has a lower degree of aliasing compared to the one in Problem 10.32 due to the use of a
higher sampling rate. It should also be noted, however, that the bandwidth of the equivalent analog
system is approximately halved in the process.
10.34.
a. The discrete-time filter can be designed using the following statements:
num = [ 2 ] ;
den = [ 1 , 3 , 2 ] ;
[numz, denz ] = b i l i n e a r (num, den , 2 )
MATLAB response is
numz =
0.0667
denz =
1.0000
0.1333
0.0667
−0.9333
0.2000
and corresponds to
H (z) =
0.0667 + 0.1333 z −1 + 0.0667 0.0667 z 2 + 0.1333 z + 0.0667
=
1 − 0.9333 z −1 + 0.2 z −2
z 2 − 0.9333 z + 0.2
b.
omg = [ 0 : 0 . 0 1 : 2 * pi ] ;
G = freqs (num, den ,omg ) ;
plot (omg, abs (G ) ) ;
axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
c.
Omg = [ 0 : 0 . 0 1 : 1 ] * pi ;
H = freqz (numz, denz ,Omg) ;
plot (Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( \Omega ) | ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
d. Use ω = Ω/T s = 2 Ω. Graph |G (ω)| and |Ya (ω) /X a (ω)| up to f max = f s /2 = 1 Hz, or equivalently, up
to ωmax = ωs /2 = 2π rad/s.
399
plot (omg, abs (G) , ’b ’ , 2 *Omg, abs (H) , ’ r ’ ) ;
axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
xlabel ( ’ \omega ( rad / s ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ;
grid ;
10.35.
a.
1
2
3
4
5
6
7
8
9
10
11
h = [0.0000 ,0.0082 ,0.0000 , −0.0193 ,0.0000 , −0.0580 ,0.0000 ,0.2702 , . . .
0.0000 , −0.4000 ,0.0000 ,0.2702 ,0.0000 , −0.0580 ,0.0000 , −0.0193 ,...
0. 0000 , 0. 0082 , 0. 0000] ;
Omg = [ −256:255]/256 * pi ;
% Compute DFT f r e q u e n c i e s
Hmag = abs ( f f t s h i f t ( f f t ( h , 5 1 2 ) ) ) ;
% 512−point DFT
plot (Omg,Hmag) ;
axis ([ − pi , pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( \Omega ) | ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
b.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
c.
% I d e a l lowpass f i l t e r impulse response
hLP = @(n ,Omgc) Omgc/ pi * sinc (Omgc* n/ pi ) ;
% Eqn . ( 10. 172)
% I d e a l bandpass f i l t e r impulse response
hBP = @(n ,Omg1,Omg2) hLP (n ,Omg2)−hdLP(n ,Omg1) ; % Bandpass
% Anonymous function f o r Hamming window
wHamm = @(n ,M) 0.54 −0.46 * cos ( pi * ( n+M) /M) ;
% Eqn . ( 10. 181)
% Truncated impulse response f o r bandpass f i l t e r
n = [ − 22: 22] ’ ;
% N=45=2M+1 t h e r e f o r e M=22
hT = hBP(n , 0 . 7 * pi , 0 . 3 * pi ) ;
% Let h1 be the impulse response o f the bandpas f i l t e r designed using
% the Hamming window
h1 = hT . *wHamm(n , 2 2 ) ;
% Compute and graph the magnitude c h a r a c t e r i s t i c
H1mag = abs ( f f t s h i f t ( f f t ( h1 , 5 1 2 ) ) ) ;
plot (Omg, 1 0 * log10 (Hmag) ,Omg, 1 0 * log10 (H1mag ) ) ;
axis ([ − pi , pi , − 50 , 20] ) ;
t i t l e ( ’ |H( \Omega ) | _ {dB} and | H_ { 1 } ( \Omega ) | _ {dB} ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
legend ( ’ |H( \Omega ) | _ {dB} ’ , ’ | H_{ 1 } ( \Omega ) | _ {dB} ’ , ’ Location ’ , ’ North ’ ) ;
grid ;
CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS
400
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
% Anonymous function f o r Blackman window
wBlk = @(n ,M) 0.42 −0.5 * cos ( pi * ( n+M) /M) + . . .
0. 08 * cos ( 2 * pi * ( n+M) /M) ;
% Eqn . ( 10. 183)
% Let h2 be the impulse response o f the bandpas f i l t e r designed using
% the Blackman window
h2 = hT . * wBlk (n , 2 2 ) ;
% Compute and graph the magnitude c h a r a c t e r i s t i c
H2mag = abs ( f f t s h i f t ( f f t ( h2 , 5 1 2 ) ) ) ;
plot (Omg, 1 0 * log10 (H1mag) ,Omg, 1 0 * log10 (H2mag ) ) ;
axis ([ − pi , pi , − 50 , 20] ) ;
t i t l e ( ’ | H_{ 1 } ( \Omega ) | _ {dB} and | H_{ 2 } ( \Omega ) | _ {dB} ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
legend ( ’ | H_{ 1 } ( \Omega ) | _ {dB} ’ , ’ | H_ { 2 } ( \Omega ) | _ {dB} ’ , ’ Location ’ , ’ North ’ ) ;
grid ;
10.36.
a.
function hd = s s _ f i r 1 (Omgc,N)
n = [ 0 : N−1];
tdelay = (N−1)/2;
hd = Omgc/ pi * sinc (Omgc * ( n−tdelay ) / pi ) ;
end
b. Compute and graph the impulse response hd [n]:
n = [0:35];
hd = s s _ f i r 1 ( 0 . 4 * pi , 3 6 ) ;
stem (n , hd ) ;
t i t l e ( ’ Impulse response ’ ) ;
xlabel ( ’n ’ ) ;
ylabel ( ’h[ n ] ’ ) ;
Compute and graph the magnitude |Hd ()|:
Omg = [ −256:255]/256 * pi ;
Hmag = abs ( f f t s h i f t ( f f t (hd , 5 1 2 ) ) ) ;
plot (Omg,Hmag) ;
axis ([ − pi , pi , 0 , 1 . 2 ] ) ;
t i t l e ( ’ |H( \Omega ) | ’ ) ;
xlabel ( ’ \Omega ( rad ) ’ ) ;
ylabel ( ’ Magnitude ’ ) ;
grid ;
Chapter 11
Amplitude Modulation
11.1.
a.
µ=
Am
2
=
= 0.7
Ac
Ac
⇒
Ac =
b.
µ=
Am
8
=
= 0.85
Ac
Ac
⇒
Ac =
c.
µ=
Am Am
= 0.9
=
Ac
2
⇒
A m = 1.8
d.
µ=
A m 2.5
=
= 0.83
Ac
3
2
= 2.86
0.7
3
= 3.53
0.85
11.2.
a.
5
0
−5
0
0.5
1
1.5
2
1.5
2
t (ms)
b.
5
0
−5
0
0.5
1
t (ms)
401
CHAPTER 11. AMPLITUDE MODULATION
402
c.
10
5
0
−5
−10
0
0.2
0.4
0.6
0.8
1
t (ms)
11.3.
a.
m (t)
5
m min = −2.5746
0
Ac =
|m min | 2.5746
=
= 3.68
µ
0.7
−5
0
0.5
1
1.5
2
2.5
3
3.5
4
2.5
3
3.5
4
t (sec)
b.
m (t)
5
m min = −4
0
Ac =
|m min |
4
=
= 5.71
µ
0.7
−5
0
0.5
1
1.5
2
t (sec)
403
c.
m (t)
5
m min = −4.971
0
Ac =
|m min | 4.971
=
= 7.10
µ
0.7
−5
0
0.5
1
1.5
2
2.5
3
3.5
4
t (sec)
11.4.
a.
The signal x AM (t ) is
£
¡
¢¤
¡
¢
x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t
£
¤
= 3 1 + 0.6 cos (4πt ) cos (20πt )
= 3 cos (20πt ) + 0.9 cos (16πt ) + 0.9 cos (24πt )
The carrier signal can be written as
n
o
n
o
x car (t ) = Re X car e j 2π f c t = Re 3 e j 20πt
Similarly, lower and upper sideband signals are
n
o
n
o
x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 0.9 e j 16πt
and
n
o
n
o
x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 0.9 e j 24πt
At the time instant t = 0.35 seconds we get
o
n
©
ª
x car (0.35) = Re 3 e j 21.9911 = Re −3 + j 0
n
o
©
ª
x lsb (0.35) = Re 0.9 e j 17.5929 = Re 0.2781 − j 0.8560
n
o
©
ª
x usb (0.35) = Re 0.9 e j 26.3894 = Re 0.2781 + j 0.8560
im
X̃ usb
fm
X car
−fm
fc
X̃ lsb
re
CHAPTER 11. AMPLITUDE MODULATION
404
b.
The signal x AM (t ) is
£
¡
¢¤
¡
¢
x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t
£
¤
= 3 1 + 1.3 cos (4πt ) cos (20πt )
= 3 cos (20πt ) + 1.95 cos (16πt ) + 1.95 cos (24πt )
The carrier signal can be written as
o
o
n
n
x car (t ) = Re X car e j 2π f c t = Re 3 e j 20πt
Similarly, lower and upper sideband signals are
o
o
n
n
x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 1.95 e j 16πt
o
o
n
n
x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 1.95 e j 24πt
and
At the time instant t = 0.6 seconds we get
n
o
©
ª
x car (0.6) = Re 3 e j 37.6991 = Re 3 + j 0
n
o
©
ª
x lsb (0.6) = Re 1.95 e j 30.1593 = Re 0.6026 − j 1.8546
n
o
©
ª
x usb (0.35) = Re 1.95 e j 45.2389 = Re 0.6026 + j 1.8546
im
X̃ usb
fm
fc
X car
re
−fm
X̃ lsb
c.
The modulation index is
µ=
Am 4
= = 0.8
Ac
5
and the signal x AM (t ) is
£
¡
¢¤
¡
¢
x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t
£
¤
= 5 1 + 0.8 cos (16πt ) cos (50πt )
= 5 cos (50πt ) + 2 cos (34πt ) + 2 cos (66πt )
The carrier signal can be written as
n
o
n
o
x car (t ) = Re X car e j 2π f c t = Re 5 e j 50πt
405
Similarly, lower and upper sideband signals are
o
o
n
n
x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 2 e j 34πt
o
o
n
n
x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 2 e j 66πt
and
im
At the time instant t = 150 milliseconds we get
o
n
©
ª
x car (0.6) = Re 5 e j 23.5619 = Re − j 5
o
n
©
ª
x lsb (0.6) = Re 2 e j 16.0221 = Re −1.9021 − j 0.6180
o
n
©
ª
x usb (0.35) = Re 2 e j 31.1018 = Re 1.9021 − j 0.6180
re
X car
fm
fc
−fm
X̃ usb
X̃ lsb
11.5.
The AM signal is in the form
¤ ¢ µA c
¤ ¢
¡ £
¡ £
¡
¢ µA c
cos 2π f c − f m t +
cos 2π f c + f m t
x AM (t ) = A c cos 2π f c t +
2
2
and the frequency spectrum is in the form
¡ ¢ Ac ¡
¢ Ac ¡
¢ µA c ¡
¢ µA c ¡
¢
X AM f =
δ f − fc +
δ f + fc +
δ f − fc − fm +
δ f − fc + fm
2
2
4
4
+
a.
¢ µA c ¡
¢
µA c ¡
δ f + fc + fm +
δ f + fc − fm
4
4
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 1.4286 δ f − f c + 1.4286 δ f + f c + 0.5 δ f − f c − f m + 0.5 δ f − f c + f m
¢
¢
¡
¡
+ 0.5 δ f + f c + f m + 0.5 δ f + f c − f m
XAM (f )
1.4286
1.4286
−fc−fm −fc −fc+fm
fc−fm fc fc+fm
f
CHAPTER 11. AMPLITUDE MODULATION
406
b.
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 1.7647 δ f − f c + 1.7647 δ f + f c + 0.75 δ f − f c − f m + 0.75 δ f − f c + f m
¢
¢
¡
¡
+ 0.75 δ f + f c + f m + 0.75 δ f + f c − f m
XAM (f )
1.7647
1.7647
−fc−fm −fc −fc+fm
fc−fm fc fc+fm
f
c.
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = δ f − f c + δ f + f c + 0.45 δ f − f c − f m + 0.45 δ f − f c + f m
¢
¢
¡
¡
+ 0.45 δ f + f c + f m + 0.45 δ f + f c − f m
XAM (f )
1
1
−fc−fm −fc −fc+fm
fc−fm fc fc+fm
f
d.
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 1.5 δ f − f c + 1.5 δ f + f c + 0.625 δ f − f c − f m + 0.625 δ f − f c + f m
¢
¢
¡
¡
+ 0.625 δ f + f c + f m + 0.625 δ f + f c − f m
XAM (f )
1.5
1.5
−fc−fm −fc −fc+fm
fc−fm fc fc+fm
f
407
11.6.
a. Using A c = 3.68 we obtain
x AM (t ) = 3.68 cos (20πt ) + cos (4πt ) cos (20πt ) + 2 cos (6πt ) cos (20πt )
= 3.68 cos (20πt ) +
1
1
cos (16πt ) + cos (24πt ) + cos (14πt ) + cos (26πt )
2
2
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 1.84 δ f − 10 + 1.84 δ f + 10 + 0.25 δ f − 8 + 0.25 δ f + 8 + 0.25 δ f − 12 + 0.25 δ f + 12
¡
¢
¡
¢
¡
¢
¡
¢
+ 0.5 δ f − 7 + 0.5 δ f + 7 + 0.5 δ f − 13 + 0.5 δ f + 13
|XAM (f )|
1.84
−20
−10
1.84
0
10
20
f (Hz)
b.
Let us write m (t ) as
m (t ) = cos (4πt ) + 3 cos (6πt − π/2)
Using A c = 5.71 we obtain
x AM (t ) = 5.71 cos (20πt ) + cos (4πt ) cos (20πt ) + 3 cos (6πt − π/2) cos (20πt )
= 5.71 cos (20πt ) +
1
3
3
1
cos (16πt ) + cos (24πt ) + cos (14πt + π/2) + cos (26πt − π/2)
2
2
2
2
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 2.855 δ f − 10 + 2.855 δ f + 10 + 0.25 δ f − 8 + 0.25 δ f + 8 + 0.25 δ f − 12 + 0.25 δ f + 12
¡
¢
¡
¢
¡
¢
¡
¢
+ j 0.75 δ f − 7 − j 0.75 δ f + 7 − j 0.75 δ f − 13 + j 0.75 δ f + 13
|XAM (f )|
2.885
−20
−10
2.885
0
f (Hz)
10
20
CHAPTER 11. AMPLITUDE MODULATION
408
6
XAM (f )
3
Phase (rad)
2
1
0
−1
−2
−3
−20
−15
−10
−5
0
5
10
15
20
f (Hz)
c.
Let us write m (t ) as
m (t ) = 3 cos (4πt − π/2) + 2 cos (6πt − π/3)
Using A c = 7.10 we obtain
x AM (t ) = 7.10 cos (20πt ) + 3 cos (4πt − π/2) cos (20πt ) + 2 cos (6πt − π/3) cos (20πt )
= 7.10 cos (20πt ) +
3
3
cos (16πt + π/2) + cos (24πt − π/2) + cos (14πt + π/3) + cos (26πt − π/3)
2
2
¡ ¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
¡
¢
X AM f = 3.55 δ f − 10 + 3.55 δ f + 10 + j 0.75 δ f − 8 − j 0.75 δ f + 8 − j 0.75 δ f − 12 + j 0.75 δ f + 12
¡
¢
¡
¢
¡
¢
¡
¢
+ 0.5 e j π/3 δ f − 7 + 0.5 e − j π/3 δ f + 7 + 0.5 e − j π/3 δ f − 13 + 0.5 e j π/3 δ f + 13
|XAM (f )|
3.55
−20
3.55
−10
0
10
20
f (Hz)
6
XAM (f )
3
Phase (rad)
2
1
0
−1
−2
−3
−20
−15
−10
−5
0
f (Hz)
5
10
15
20
409
11.7.
¡
¢
¡
¢
x AM (t ) = A c cos 2π f c t + m (t ) cos 2π f c t
¢ Ac ¡
¢ 1 ¡
¢ 1 ¡
¢
¡ ¢ Ac ¡
δ f − fc +
δ f + fc + M f − fc + M f + fc
X AM f =
2
2
2
2
XAM (f )
0.75
0.75
0.5
−13
−10
−7
7
10
13
f (kHz)
11.8.
a.
The terms in the spectrum of the AM signal are listed in the table below:
Index
1
2
3
4
Term
¤
3 1 + 0.8 m (t ) cos (20πt )
2 m (t )
1.8 cos2 (20πt )
0.8 m 2 (t )
£
Frequencies
7 ≤ f ≤ 13 kHz
−3 ≤ f ≤ 3 kHz
f = 0, and f = 20 kHz
−6 ≤ f ≤ 6 kHz
M (f )
(a)
−3
f (kHz)
3
Vout (f )
(b)
1
3
3
2
1
3
4
−10
f (kHz)
10
BT = 6 kHz
CHAPTER 11. AMPLITUDE MODULATION
410
b.
To select the desirable terms (row 1 of the table), the passband of the filter should span the
frequency range
7 ≤ f ≤ 13 kHz
11.9.
a.
The carrier completes 15 full cycles in a duration of 1 ms. Therefore the carrier frequency is
f c = 15 kHz.
b.
Observing the envelope, we see that it completes one full cycle in about 0.84 ms. The message
frequency is approximately f m = 1.2 kHz.
c.
For a single-tone modulated AM carrier the maximum value of the positive envelope is
¯
¯
¯x AM (t )¯
max ≈ A c + A m
and the minimum value of the positive envelope is
¯
¯
¯x AM (t )¯
From the graph we observe
Ac + Am ≈ 7
min ≈ |A c − A m |
and
Ac − Am ≈ 7
and
Am ≈ 3
yielding carrier and message amplitudes
Ac ≈ 4
The modulation index is
µ=
d.
Am
≈ 0.75
Ac
Approximate efficiency is
η≈
(0.75)2
2 + (0.75)2
= 0.2195
11.10.
a.
A 2c = (3.68)2 = 13.5424
® 1 4
m 2 (t ) = + = 2.5
2 2
­ 2 ®
m (t )
2.5
η= 2 ­ 2 ® =
= 0.1558
13.5424 + 2.5
A c + m (t )
­
b.
A 2c = (5.71)2 = 32.6041
­
® 1 9
m 2 (t ) = + = 5
2 2
411
­ 2 ®
m (t )
5
= 0.1330
η= 2 ­ 2 ® =
32.6041 + 5
A c + m (t )
c.
A 2c = (7.10)2 = 50.41
® 9 4
m 2 (t ) = + = 6.5
2 2
­ 2 ®
m (t )
6.5
= 0.1142
η= 2 ­ 2 ® =
50.41 + 6.5
A c + m (t )
­
11.11.
a.
vin (t)
10
5
0
−5
−10
0
5
10
15
20
25
30
20
25
30
t (ms)
b.
vout (t)
10
5
0
−5
−10
0
5
10
15
t (ms)
c.
The message bandwidth is W = 100 Hz. The carrier frequency is f c = 800 Hz. The significant
frequency components of the modulated carrier are at 700, 800 and 900 Hz. A reasonable choice for a
bandpass filter to isolate the AM signal may be one with a passband for
600 < f < 1000 Hz
CHAPTER 11. AMPLITUDE MODULATION
412
11.12.
The output of the first AM modulator:
£
¤
¡
¢
w 1 (t ) = A c + m (t ) cos 2π f c t
The output of the second AM modulator:
£
¤
¡
¢
w 2 (t ) = A c − m (t ) cos 2π f c t
Adding the two signals yields
¡
¢
x DSB (t ) = w 1 (t ) + w 2 (t ) = 2 A c cos 2π f c t
11.13.
a.
¡
¢
x DSB (t ) = A c m (t ) cos 2π f c t
¢ Ac ¡
¢
¡ ¢ Ac ¡
M f − fc +
M f + fc
X DSB f =
2
2
XDSB (f )
0.75
−13
−10
−7
7
10
13
10
13
f (kHz)
b.
XSSB,U (f )
0.75
−13
−10
f (kHz)
c.
XSSB,L (f )
0.75
−10
−7
7
10
f (kHz)
413
11.14.
The modulation index is
3.2
= 0.8
4
Using Eqn. (11.64) with µ = 0.8 and f m = 100 Hz, we obtain
µ=
1
τ≤
200π
Ãp
!
1 − (0.8)2
= 0.0012
0.8
Let C = 1 µF. Required resistor value is R = 1200 Ω.
11.15.
The signal r (t ) is
£
¤
¡
¢
¡
¢
r (t ) = 2 A c + m (t ) cos 2π f c t cos 2π f c t + θ
Using the appropriate trigonometric identity
¡
¢
¡
¢
r (t ) = A c cos 4π f c t + θ + A c cos (θ) + m (t ) cos 4π f c t + θ + m (t ) cos (θ)
The bandpass filter removes the first three terms, resulting in
m̂ (t ) = m (t ) cos (θ)
The effect of the phase error is multiplication of the message signal by the factor cos (θ) which is a
constant as long as the phase error θ is constant. For values of θ close to π/2 radians, the factor cos (θ) is
small, potentially degrading the performance of the demodulator in the presence of random noise.
11.16.
The signal r (t ) is
£
¤
¡
¢
¡ £
¤ ¢
r (t ) = 2 A c + m (t ) cos 2π f c t cos 2π f c + ∆ f t
Using the appropriate trigonometric identity
¡ £
¤ ¢
¡
¢
¡ £
¤ ¢
¡
¢
r (t ) = A c cos 2π 2 f c + ∆ f t + A c cos 2π ∆ f t + m (t ) cos 2π 2 f c + ∆ f t + m (t ) cos 2π ∆ f t
The bandpass filter removes the first three terms, resulting in
¡
¢
m̂ (t ) = m (t ) cos 2π ∆ f t
¡
¢
The effect of the frequency error is multiplication of the message signal by the factor cos 2π ∆ f t . The
output signal m̂ (t ) is still a modulated term, with the carrier frequency ∆ f .
11.17.
Let us begin by defining an anonymous function for the AM signal:
1
xAM = @( t , Ac ,mu, fc , fm) Ac * (1+mu* cos ( 2 * pi * fm * t ) ) . * cos ( 2 * pi * fc * t ) ;
CHAPTER 11. AMPLITUDE MODULATION
414
Afterwards, each AM signal in question can be produced using this anonymous function.
a.
1
2
3
4
t = [ 0 : 2 e −6:2e −3];
plot (1000 * t ,xAM( t , 3 , 0 . 6 , 1 0 0 0 0 , 2 0 0 0 ) ) ;
axis ( [ 0 , 2 , − 8 , 8 ] ) ;
xlabel ( ’ t (ms) ’ ) ;
b.
1
2
3
4
t = [ 0 : 2 e −6:2e −3];
plot (1000 * t ,xAM( t , 3 , 1 . 3 , 1 0 0 0 0 , 2 0 0 0 ) ) ;
axis ( [ 0 , 2 , − 8 , 8 ] ) ;
xlabel ( ’ t (ms) ’ ) ;
c.
1
2
3
4
5
mu = 4/ 5;
t = [ 0 : 1 e −6:1e −3];
plot (1000 * t ,xAM( t , 3 ,mu, 25000 , 8000) ) ;
axis ( [ 0 , 1 , − 1 0 , 1 0 ] ) ;
xlabel ( ’ t (ms) ’ ) ;
11.18.
a.
1
2
3
4
5
6
7
8
t = [0:0.005:4];
Ac = 3 . 6 8 ;
m = cos ( 4 * pi * t )+2 * cos ( 6 * pi * t ) ;
x_am = ( Ac+m) . * cos (20 * pi * t ) ;
plot ( t , x_am ) ;
xlabel ( ’ t ( sec ) ’ ) ;
t i t l e ( ’ x_ {AM} ( t ) ’ ) ;
grid ;
b.
1
2
3
4
5
6
7
Ac = 5 . 7 1 ;
m = cos ( 4 * pi * t )+3 * sin ( 6 * pi * t ) ;
x_am = ( Ac+m) . * cos (20 * pi * t ) ;
plot ( t , x_am ) ;
xlabel ( ’ t ( sec ) ’ ) ;
t i t l e ( ’ x_ {AM} ( t ) ’ ) ;
grid ;
415
c.
1
2
3
4
5
6
7
Ac = 7 . 1 0 ;
m = 3 * sin ( 4 * pi * t )+2 * cos ( 6 * pi * t−pi / 3 ) ;
x_am = ( Ac+m) . * cos (20 * pi * t ) ;
plot ( t , x_am ) ;
xlabel ( ’ t ( sec ) ’ ) ;
t i t l e ( ’ x_ {AM} ( t ) ’ ) ;
grid ;
11.19.
a.
Using A c = 3.68 we obtain
x AM (t ) = 3.68 cos (20πt ) + cos (4πt ) cos (20πt ) + 2 cos (6πt ) cos (20πt )
= 3.68 cos (20πt ) +
1
1
cos (16πt ) + cos (24πt ) + cos (14πt ) + cos (26πt )
2
2
The frequency components in x AM (t ) are
f 1 = 10 Hz ,
f 2 = 8 Hz ,
f 3 = 12 Hz ,
f 4 = 7 Hz ,
f 5 = 13 Hz
Fundamental frequency f 0 is found through (see Section 1.3.4)
m1 m2 m3 m4 m5
1
=
=
=
=
=
f0
10
8
12
7
13
Using the integers
m 1 = 10 ,
m2 = 8 ,
m 3 = 12 ,
m4 = 7 ,
m 5 = 13
the fundamental frequency is found as f 0 = 1 Hz corresponding to a fundamental period of T0 = 1 s. In
order to generate 1024 samples in one period, samples need to be taken 1/1024 seconds apart.
1
2
3
4
5
6
7
8
b.
t = [ 0: 1023] / 102 4 ;
x_am = 3. 68 * cos (20 * pi * t )+ cos ( 4 * pi * t ) . * cos (20 * pi * t )+2 * cos ( 6 * pi * t ) . * cos (20 * pi * t ) ;
k =[ −20:20];
c = ss_efsapprox (x_am , k ) ;
stem ( k , c )
axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 3 ] ) ;
xlabel ( ’ k ’ ) ;
t i t l e ( ’ | c_ { k } | ’ ) ;
Writing m (t ) as
m (t ) = cos (4πt ) + 3 cos (6πt − π/2)
and using A c = 5.71 we obtain
x AM (t ) = 5.71 cos (20πt ) + cos (4πt ) cos (20πt ) + 3 cos (6πt − π/2) cos (20πt )
= 5.71 cos (20πt ) +
1
3
3
1
cos (16πt ) + cos (24πt ) + cos (14πt + π/2) + cos (26πt − π/2)
2
2
2
2
As in part (a) of the problem, the fundamental frequency and the fundamental period are f 0 = 1 Hz and
T0 = 1 s respectively.
CHAPTER 11. AMPLITUDE MODULATION
416
1
2
3
4
5
6
7
8
c.
t = [ 0: 1023] / 102 4 ;
x_am = 5. 71 * cos (20 * pi * t )+ cos ( 4 * pi * t ) . * cos (20 * pi * t )+3 * cos ( 6 * pi * t−pi / 2 ) . * cos (20 * pi * t ) ;
k =[ −20:20];
c = ss_efsapprox (x_am , k ) ;
stem ( k , abs ( c ) ) ;
axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 3 ] ) ;
xlabel ( ’ k ’ ) ;
t i t l e ( ’ | c_ { k } | ’ ) ;
Writing m (t ) as
m (t ) = 3 cos (4πt − π/2) + 2 cos (6πt − π/3)
and using A c = 7.10 we obtain
x AM (t ) = 7.10 cos (20πt ) + 3 cos (4πt − π/2) cos (20πt ) + 2 cos (6πt − π/3) cos (20πt )
3
3
3
3
cos (16πt + π/2) + cos (24πt − π/2) + cos (14πt + π/3) + cos (26πt − π/3)
2
2
2
2
As in parts (a) and (b) of the problem, the fundamental frequency and the fundamental period are f 0 = 1
Hz and T0 = 1 s respectively.
= 7.10 cos (20πt ) +
1
2
3
4
5
6
7
8
t = [ 0: 1023] / 102 4 ;
x_am = 7. 10 * cos (20 * pi * t )+3 * cos ( 4 * pi * t−pi / 2 ) . * cos (20 * pi * t )+2 * cos ( 6 * pi * t−pi / 3 ) . * cos (20 * pi * t ) ;
k =[ −20:20];
c = ss_efsapprox (x_am , k )
stem ( k , abs ( c ) ) ;
axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 4 ] ) ;
xlabel ( ’ k ’ ) ;
t i t l e ( ’ | c_ { k } | ’ ) ;
11.20. Compute and graph the signal v i n (t ).
1
2
3
4
5
6
7
8
9
10
11
t = [0:0.05:30]/1000;
Bc = 5 ;
fc = 800;
c a r r i e r = Bc * cos ( 2 * pi * fc * t ) ;
message = 3 * cos (200 * pi * t ) ;
vin = c a r r i e r +message ;
plot (1000 * t , vin ) ;
axis ( [ 0 , 30 , − 10 , 10] ) ;
xlabel ( ’ t (ms) ’ ) ;
t i t l e ( ’ v_ { in } ( t ) ’ ) ;
grid ;
Compute and graph the signal v out (t ).
1
2
3
4
5
6
vout = vin . * ( vin >=0);
plot (1000 * t , vout ) ;
axis ( [ 0 , 30 , − 10 , 10] ) ;
xlabel ( ’ t (ms) ’ ) ;
t i t l e ( ’ v_ { out } ( t ) ’ ) ;
grid ;
417
Compute and graph the signal x AM (t ) using the function ss_switchmod(..)
1
2
3
4
5
6
x_am = ss_switchmod ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000);
plot (1000 * t , x_am ) ;
axis ( [ 0 , 3 0 , − 5 , 5 ] ) ;
xlabel ( ’ t (ms) ’ ) ;
t i t l e ( ’ x_ {AM} ( t ) ’ ) ;
grid ;
11.21. Function ss_switchmod2(..).
1
2
3
4
5
6
7
8
9
10
11
12
function x_am = ss_switchmod2 (msg, Bc , fc , Ts , f1 , f2 )
nSamp = length (msg ) ;
% Number o f samples in "msg " .
t = [ 0 : nSamp−1] * Ts ;
% Vector o f time i n s t a n t s .
c a r r i e r = Bc * cos ( 2 * pi * fc * t ) ;
% Compute input to the diode switch .
v_in = c a r r i e r +msg ;
% Eqn . ( 1 1 . 3 6 )
% Simulate the diode switch .
v_out = ( v_in − 0 . 6 ) . * ( v_in > = 0 . 6 ) ;
% Design the bandpass f i l t e r .
[numz, denz ] = butter ( 5 , [ 2 * f1 * Ts , 2 * f2 * Ts ] , ’ bandpass ’ ) ;
% P r o c e s s switch output through bandpass f i l t e r .
x_am = f i l t e r (numz, denz , v_out ) ;
Compute and graph the signal x AM (t ) using the functions ss_switchmod(..) and ss_switchmod2(..)
1
2
3
4
5
6
x_am1 = ss_switchmod ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000);
x_am2 = ss_switchmod2 ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000);
plot (1000 * t , x_am1 , ’b ’ ,1000 * t , x_am2 , ’ r ’ ) ;
axis ( [ 0 , 3 0 , − 5 , 5 ] ) ;
xlabel ( ’ t (ms) ’ ) ;
grid ;
11.22.
Script to simulate envelope detector using the value τ = 0.0012 seconds found in Problem 11.14:
1
2
3
4
5
6
7
t = [ 0 : 2 0 e−6:20e −3];
x_am = ( 4+ 3. 2 * cos (200 * pi * t ) ) . * cos (2000 * pi * t ) ;
Ts = 20e−6;
tau = 0. 0012;
x_env = ss_envdet (x_am , Ts , tau ) ;
plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ;
grid ;
Repeat using half the time constant value found:
CHAPTER 11. AMPLITUDE MODULATION
418
1
2
3
x_env = ss_envdet (x_am , Ts , 0 . 5 * tau ) ;
plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ;
grid ;
Repeat using twice the time constant value found:
1
2
3
x_env = ss_envdet (x_am , Ts , 2 * tau ) ;
plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ;
grid ;
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