11
Problem Set 2.1
In Problems 1–4,
(a) Perform 12
/zz
and express the result in rectangular form,
(b) Verify that
12 1 2
//zz z z
,
(c) Repeat Part (a)in MATLAB.
1.3
2
j
j
Solution
(a)
31313
2222
jj j j
jj
22
1 3 1 3 10 10
§·§·
310
j
2.
2
12
j
j
Solution
(a)
2
12
jj
j
3.
3
23
j
j
Solution
(a)
396
2 3 13 13
jj
j
4.
4
43j
12
Solution
(a)44316121612
4 3 4 3 25 25 25
jj j
jj
In Problems 5–8 express each complex number in its polar form.
5.
33j
Solution
To calculate phase, we first find
11
3
tan 3
S
. Since
33j
is located in the 3rd quadrant, the phase is taken as
6.
3
2
1j
Solution
7.
33j
Solution
8.
1
2
1j
Solution
In Problems 9 – 16 perform using polar form and express the result in rectangular form.
9.
32
13
j
j
Solution
10.33
33
j
j
13
Solution
11.
35
2
j
j
Solution
12.
3
1
j
j
Solution
13.
3
(4 3 )j
Solution
14.
10
(0.9511 0.3090 )j
Solution
15.
3
2
(1 3 )
(1 2)
j
j
Solution
16.
3
5
(1 4 )
j
j
Solution
In Problems 17–20, find all possible values for each expression.
17.
1/6
(1)
Solution
The goal is to find 6
wz where
1z
. Noting that
1z
is located on the negative real axis, one unit from the
14
origin, we have
1r
and
TS
, hence
1
j
ze
S
. Then,
18.
1/3
(1 )j
Solution
33
1/3
1/3 44
22
( 1 ) 2 cos sin , 0,1, 2
33
kk
jjk
SS SS
ªº
«»
«»
¬¼
19.
1/ 2
(3 3)j
Solution
20.
13i
Solution
The goal is to find
wz
where
13zi
. Since /3
132
j
zi e
S
, we have
Problem Set 2.2
In Problems 1–10 solve the initial-value problem.
1.
sin , (0) 1xx t x
Solution
Since
() 1gt
, we have ()ht t and
2.11
33
0 , (0)xx x
Solution
Writing the ODE in standard form yields
() 3gt
so that
() 3ht t
and
3.2 , (0) 2yty t y
Solution
Writing the ODE in standard form yields
1
2
() ()gt t f t
so that
2
1
4
()ht t
and
4.
1
2
(1 )sin , 2uutu
S
Solution
Writing the ODE in standard form yields
() sin ()gt t f t
so that () cosht t and
5.
( 1) 2 , (0) 1tytyty
Solution
Writing the ODE in standard form yields
() 1
t
gt t
and
2
() 1
t
ft t
so that
() ln( 1)
1
t
ht dt t t
t
³
and
6.
2
2 , (0) 1 , (0) 1
t
xxxe x x
Solution
Case (2)
7.
4 17 cos , (0) 1 , (0) 0xx tx x
Solution
Case (1) 4
8.
sin 2 , (0) 1 , (0) 0uu t u u
Solution
9.
4 3 4 , (0) 0 , (0) 1
t
uuu e u u
Solution
Characteristic values are
1, 3
O
so that
3
12
tt
h
uce ce
. Because t
ecoincides with an independent
10.
1
2
2 3 0 , (0) 0 , (0)yyy y y
Solution
Characteristic values are
1
1,
O
so that
/2
12
tt
h
yce ce
. Since the ODE is homogeneous, () ()
h
yt y t . By
In Problems 11–14 write the expression in the form
sin( )Dt
ZI
.
11.
cos 3sintt
Solution
Write
cos 3sin sin( ) sin cos cos sinttDtDt Dt
III
and compare the two sides to find
1st quadrant
10 1
D
12.
cos 2 sin 2tt
Solution
Write cos 2 sin 2 sin(2 ) sin 2 cos cos 2 sinttDt Dt Dt
III
. Comparing the two sides,
2nd quadrant
2
sin 1 sin 0
D
D
II
!
13.
1
2
sin 2 cos 2tt
Solution
Write
1
2
sin 2 cos 2 sin(2 ) sin 2 cos cos 2 sinttDtDtDt
III
. Comparing the two sides,
17
14.
3sin costt
ZZ
Solution
Expand
3sin cos sin( ) sin cos cos sinttDtDtDt
ZZ ZI ZI ZI
. Comparison gives
4th quadrant
10 1
sin 1 sin 0
D
D
II
15.2
3cos sintt
Solution
Write
2
3
cos sin cos( ) cos cos sin sinttDt Dt Dt
III
and compare the two sides to find
16.
4cos 3sintt
Solution
Write 4 cos 3sin cos( ) cos cos sin sinttDtDt Dt
III
and compare the two sides to find
Problem Set 2.3
In Problems 1–8,
(a) Find the Laplace transform of the given function. Use Table 2.2 when applicable.
(b) Confirm the result in MATLAB.
1.
at b
e
,
,a b const
Solution
(a)
^`^ ` ^`
linearity b
at b at b at b
e
eee ee
sa
LL L
18
2.
2
2
31t
Solution
(a)
^`
2
2
333
22 1 4 1
133
tss
L
3.
sin( )t
ZI
,
,const
ZI
Solution
(a) Using trigonometric expansion, we find
4.cos( )t
ZI
,
,const
ZI
Solution
(a) Using trigonometric expansion, we find
{cos( )} {cos }cos {sin }sin
ttt
ZI Z I Z I
LLL
5.
2
cos t
Solution
(a) Noting that
21
2
cos (1 cos 2 )tt
, we find
^`
2
Simplify
2
22
111 2
cos {1 cos 2 }
22
4(4)
ss
tt
ssss
ªº
«»
¬¼
LL
6.
costt
Solution
(a) Following Eq. (2.16),
^`
2
222
1
cos 1( 1)
ds s
tt ds ss
§·
¨¸
©¹
L
7.2sintt
Z
Solution
(a) Following the general form of Eq. (2.16) with () singt t
Z
and 2n ,
^`
222
2
22 2 2 23
2(3 )
sin ()
ds
tt
ds s s
ZZZ
ZZZ
§·
¨¸
©¹
L
8.
sinhtt
Solution
(a) Comparing with Eq. (2.16), we have
() sinhgt t
so that
^`
^`
Simplify
2
11111
() sinh 2211
1
tt
Gs t e e ss s
ªº
«»
¬¼
LL
Then,
9.
1 if 0 1
( ) 1 if 1 2
0 otherwise
t
gt t
°
®
°
¯
20
Solution
(a)
() () 2 ( 1) ( 2)gt ut ut ut
10.
if 0 1
() 0 otherwise
tt
gt
®
¯
Solution
(a) We construct
()gt
using the strategy outlined in Figure PS2-3 No10, resulting in
11.
0 if 0
( ) if 0 1
1 if 1
t
gt t t
t
°
®
°!
¯
Solution
12.
0 if 0
( ) 1 if 0 1
0 if 1
t
gt t t
t
°
®
°!
¯
Solution
(a) Construct
()gt
using the strategy shown in Figure PS2-3 No12, leading to () (1 ) () (1 ) ( 1)gt tut tut .
21
In Problems 13–16 find the Laplace transform of each periodic function whose definition in one period is given.
13.
1 if 0 1
() 1 if 1 2
t
ft t
®
¯
Solution
14.
() 2(1 ), 0 1ft t t
Solution
The period is
1P
. Using the description of ()ft, we have
15.
if 0 1
() 1 if 1 2
tt
ft tt
®
¯
Solution
The period is
2P
. Using the description of ()ft, we have
22
16.
1 if 0 1
() 2 if 1 2
t
ft tt
®
¯
Solution
The period is
2P
. Using the description of ()ft, we have
17. Find the Laplace transform of the periodic function
()ft
in Figure 2.14.
Solution
The period is
1P
. Using the description of ()ft, we have
18. Find the Laplace transform of the periodic function ()ft in Figure 2.15.
Solution
The period is
2Pb
. Using the description of ()ft, we have