Solutions for Signals and Systems, A Primer with MATLAB®, 2nd Edition by Sadiku (All Chapters included)

Signals and Systems, A Primer with
MATLAB®, 2nd Edition by
Matthew N. O. Sadiku



Complete Chapter Solutions Manual
are included (Ch 1 to 7)




** Immediate Download
** Swift Response
** All Chapters included

,Table of Contents are given below



1.Basic Concepts

2.Convolution

3.The Laplace Transform

4.Fourier Series

5.The Fourier Transform

6.Discrete Fourier Transform

7.z-Transform

, Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch7-1)

CHAPTER 7

P. P. 7.1

−11 −42 4 + 2 z 32
X ( z) = 2z + 4z =
z 42

P. P. 7.2

(
2 1 + z + z 2 + z3 + z 4 )
( −2 −3
X ( z )= 2 z + z + z + z + z = −4 −5 −6
) z 6




P.P. 7.3

X (=
z ) 1=
zo 1


P.P. 7.4

1 1 1 1 z
Z {u[n]} =1 + + 2
+ 3 + = =
z z z 1 − 1/ z z − 1

Alternatively, we may regard this as a special case of Example 7.3 when a =1.
z
Z {u[n]}
= , | z |> 1
z −1

P.P. 7.5
Using the result from Example 7.3,

4z 5z 12 z 15 z 2
Y ( z) = − = − , | z |>
z − 1/ 3 z − 3 z − 1 3 z − 2 3

P. P. 7.6
z (1 − z − m )
p[n] ⇔
z −1
The frequency scaling property states that
z
a n x[n] ⇔ X 
a
Applying this gives

z / a 1 − ( z / a ) − m  z (1 − a m z − m )
=Q( z ) =
z / a −1 z−a

, 241



P.P. 7.7
z
Let x[n] a nu[n]
= → X ( z)

=
z−a
d a
X ( z) = −
dz ( z − a)2
d2 2a
2
X ( z) =
dz ( z − a )3
We use the multiplication-by-n2 property, i.e.
2
2 d 2 d
n x[n] ⇔ z X ( z ) + z X ( z)
dz dz 2
−az 2az 2
Y ( z ) Z  n 2 a=
= n
u[n] +
( z − a ) 2 ( z − a )3
az ( z + a )
=
( z − a )3


P.P. 7.8
1 z −1 + z −2 + z −3
X ( z ) =+
H ( z ) =0 + z −1 + 0.5 z −2 + z −3 + 1.5 z −4
Y ( z ) =H ( z ) X ( z ) =0 + z −1 + 1.5 z −2 + 2.5 z −3 + 4 z −4 + 3 z −5 + 2.5 z −6 + 1.5 z −7
Thus,
y[n] = [ 0, 1, 1.5, 2.5, 4, 3, 2.5, 1.5 ]


P.P. 7.9
1+ 0
x[0] lim
= = X ( z) = 1
z →∞ 1− 0 − 0
−1 (1 − z −1 )(1 + 2 z −1 )
x[=
∞] lim(1 − z ) X (= z ) lim
z →1 z →1 1 − 1.6 z −1 + 0.6 z −2

(1 − z −1 )(1 + 2 z −1 ) (1 + 2 z −1 ) 3
= lim 1
=1
lim = 1
= 7.5
z →1 (1 − z )(1 − 0.6 z )
− − z →1 (1 − 0.6 z )
−
0.4

P.P. 7.10

Applying linearity and time-shifting properties, we get
z 3Y ( z ) + 2 zY ( z ) + Y ( z )= z 2 X ( z ) − X ( z )
( z 3 + 2 z + 1)Y ( z ) = ( z 2 − 1) X ( z )
Y ( z) z2 −1
H
= ( z) =
X ( z ) ( z 3 + 2 z + 1)