1-7
Solutions to Problems
14.
Desired
HT's
+
Amplifier
Actual
Gyroscopic
15.
16.
17.
di
a. L
dt
+ Ri = u(t)
b. Assume a steady-state solution i ss = B. Substituting this into the differential equation yields RB =
1,
1
from which B =
R
R
. The characteristic equation is LM + R = 0, from which M = -
Copyright ©
2011 by John Wiley & Sons, Inc.
L
. Thus, the total
1-8
Chapter 1:
Introduction
1
solution is i(t) = Ae-(R/L)t +
R
1
1
-
R
1
. The final solution is i(t) =
R
. Solving for the arbitrary constants, i(0) = A +
1
--
R
e-(R/L)t =
1
R
(1 e
 ( R/ L) t 
).
c.
18.
1
idt v (0) v(t)

C
dt  C
2
ddt 2
i 2 di  25i 0 
b. Differentiating and substituting values,
dt
a. Writing the loop equation, Ri L
di
Writing the characteristic equation and factoring,
M 2  2M 25 (M1
24i) . 
24i)(M1
The general form of the solution and its derivative is
i Ae
di
t 
cos(
24t) Be
di
Using i(0) 0;
i
dt
dt
(0)
24t)
24t) (
24A B)et  sin(
v L (0) 1
1 
L
L
= A =0
0
di
sin(
24B)e t  cos(
 (A
dt
t 
(0)A
24B =1
Thus, A 0 and B
1
24
.
The solution is
i
1
24 e
t 
sin(
24t)

Copyright ©
2011 by John Wiley & Sons, Inc.
24t) 
R
= 0. Thus, A =
1-9
Solutions to Problems
c.
19.
a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
From which, C =
35
53
and D =
10
.
53
The characteristic polynomial is
Thus, the total solution is
Solving for the arbitrary constants, x(0) = A +
35
= 0. Therefore, A = 53
b. Assume a particular solution of
Copyright ©
2011 by John Wiley & Sons, Inc.
35
. The final solution is
53
1-10
Chapter 1:
Introduction
xp = Asin3t + Bcos3t
Substitute into the differential equation and obtain
(18A B)cos(3t) (A 18B)sin(3t) 5sin(3t)
Therefore, 18A - B = 0 and -(A + 18B) = 5. Solving for A and B we obtain
xp = (-1/65)sin3t + (-18/65)cos3t
The characteristic polynomial is
M2 + 6 M + 8 =
M+4
M+2
Thus, the total solution is
18
1
cos 3 t sin 3 t
65
65
18
Solving for the arbitrary constants, x(0) C D
0. 
65
x=Ce
-4t
+De
-2t
+ -
Also, the derivative of the solution is
3
54
dx
=cos 3 t +
sin 3 t - 4 C e
dt
65
65
.
Solving for the arbitrary constants, x(0)
3
65
-4t
-2De
 4C 2D 0 , or C =
-2t
3
and D = 
10
15
26
The final solution is
18
1
3 - 4 t 15 - 2 t
cos 3 t sin 3 t e +
e
65
65
10
26
x=-
c. Assume a particular solution of
xp = A
Substitute into the differential equation and obtain 25A = 10, or A = 2/5.
The characteristic polynomial is
2
M + 8 M + 25 =
M+4+3i
M+4-3i
Thus, the total solution is
x=
2
-4t
+e
B sin 3 t + C cos 3 t
5
Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the
solution is
Copyright ©
2011 by John Wiley & Sons, Inc.
.