Nate Olson
ECET 307
Homework 5
Professor Lin
October 22, 2006
Grade 99/100, Excellent!
In problems 1-6, determine the Laplace transforms of the functions given using Table 5-1
on page 170 and Table 5-2 on page 173.
1) v(t) = 8
(Problem 5-1)
1
8
Lv(t) =8( ) = √, 5/5
s
s
2) i(t) = 4 ε −2t (Problem 5-2)
1
4
)=
√, 5/5
Li(t) = 4(
s+2
s+2
3) e(t) = 5t
(Problem 5-3)
1
5
Le(t) = 5( 2 ) = 2 √, 5/5
s
s
4) i(t) = 4 ε − t sin(3t − 30 o )
(Problem 5-12)
sin(A-B) = sinAcosB-cosAsinB
Li(t) = 4 ε − t (sin 3t cos 30 o − cos 3t sin 30 o )
Li(t) = 4 ε −t (.866 sin 3t − .5 cos 3t )
Li(t) = 3.464 ε −t sin 3t − 2ε − t cos 3t
3
s +1
Li(t) = 3.464
−2
2
( s + 1) 2 + 9
( s + 1) + 9
10.392
2s + 2
Li(t) = 2
− 2
s + 2s + 10 s + 2s + 10
8.392 − 2s
Li(t) = 2
√, 5/5
s + 2s + 10
5) Write an equation for the figure below and take the Laplace transform of it.
(Problem 5-17)
v(t)
10 V
Time (sec)
2
4
6
f(t-a)u(t-a) = ε − as F (s)
1
F(s) = 2
s
10
=5
Slope =
2
f(t) = 5tu(t) – 5(t-2)u(t-2) – 5(t-4)u(t-4) + 5(t-6)u(t-6)
5 5ε −2 s 5ε −4 s 5ε −6 s
Lf(t) = 2 − 2 − 2 + 2
s
s
s
s
5(1 − ε −2 s − ε −4 s + ε −6 s )
Lf(t) =
√, 10/10
s2
6) Write an equation for the figure below and take the Laplace transform of it.
(Problem 5-18)
i(t)
4A
3
6
9
12 15 18
Time (sec)
u(t-a) = ε − as
f(t) = 4u(t) – 4u(t-3) + 4u(t-6) – 4u(t-9) + 4u(t-12) – 4u(t-15) + 4u(t-18)
4 4ε −3 s 4ε −6 s 4ε −9 s 4ε −12 s 4ε −15 s 4ε −18 s
Lf(t) = −
+
−
+
−
+
s
s
s
s
s
s
s
4
−3s
−6s
−9 s
−12 s
−15 s
−18 s
Lf(t) = (1 − ε + ε − ε + ε
−ε
+ ε ) √, 10/10
s
7) Determine the roots of the following polynomials and classify them according to real,
complex, or imaginary. Write each polynomial in factored form. Also use MATLAB
function to verify answer. (Problem 5-19a & 5-19e). 10/10
(a) 2s+8 √
2s = -8 ⇒ s = -4; 2(s+4) , real
>> a=[2 8]
a=
2
8
>> roots(a)
ans =
-4
(e) s 2 +25 √
s 2 = -25 ⇒ s = ± j 5; ( s − j 5)( s + j 5), imaginary
>> e=[1 0 25]
e=
1
0
25
>> roots(e)
ans =
0 + 5.0000i
0 - 5.0000i
8) Determine the roots of the following polynomials and classify them according to real,
complex, or imaginary. Write each polynomial in factored form. Also use MATLAB
function to verify answer. (Problem 5-20). 10/10
(a) s 2 + 200s + 50,000
2
2
−b
− 200
⎛b⎞
⎛ 200 ⎞
± ⎜ ⎟ −c =
± ⎜
s=
⎟ − 50,000
2
2
⎝2⎠
⎝ 2 ⎠
s = -100 ± j 40,000 = −100 ± j 200
s = (-100 ± j 200) , complex, (-100s+j200)(-100s-j200)√
>> a=[1 200 50000]
a=
1
200
50000
>> roots(a)
ans =
1.0e+002 *
-1.0000 + 2.0000i
-1.0000 - 2.0000i
(b) s 2 + 10,000√
s 2 = −10,000 → s = j 10,000 → s = ± j100
s = ± j100 , (s+j100)(s-j100), imaginary
>> b=[1 0 10000]
b=
1
0
>> roots(b)
ans =
1.0e+002 *
0 + 1.0000i
0 - 1.0000i
10000
(c) s 3 +6 s 2 + 33s + 50 (s=-2 is a root)√
(s+2)(s 2 +4s + 25)
2
2
−b
−4
⎛b⎞
⎛4⎞
s=
± ⎜ ⎟ −c →
± ⎜ ⎟ − 25 ⇒ s = −2 ± j 21
2
2
⎝2⎠
⎝2⎠
s = -2, -2+ j 21,−2 − j 21 , (s+2)(s 2 +4s + 25) , complex
>> c=[1 6 33 50]
c=
1
6
33
50
>> roots(c)
ans =
-2.0000 + 4.5826i
-2.0000 - 4.5826i
-2.0000
(d) s 4 + 4s 3 + 14 s 2 + 20 s + 25 (s=-1+j2 is a root)√
(s+1-j2)(s+1+j2) → s 2 + 2s + 5 ,complex conjugate
By factoring s 2 + 2s + 5 from the original s 4 + 4s 3 + 14s 2 + 20s + 25
We find that the other factor is s 2 + 2 s + 5 and therefore
( s 2 + 2 s + 5 )( s 2 + 2 s + 5 ) = s 4 + 4s 3 + 14s 2 + 20s + 25
∴ s = -1+j2, -1+j2, -1-j2, -1-j2, ( s 2 + 2 s + 5) 2 , complex
>> d=[1 4 14 20 25]
d=
1
4
14
20
>> roots(d)
ans =
-1.0000 + 2.0000i
-1.0000 - 2.0000i
-1.0000 + 2.0000i
-1.0000 - 2.0000i
25
(e) s 4 + 13s 2 + 36 , Since the multiplication of 4 and 6 and the summation of 4
and 6 fit the above equation. The factors are as follows:√
( s 2 + 4)( s 2 + 9) , s = ± j 2, s = ± j 3 , Imaginary
>> e=[1 0 13 0 36]
e=
1
0
13
0
36
>> roots(e)
ans =
0 + 3.0000i
0 - 3.0000i
0 + 2.0000i
0 - 2.0000i
In problems 9-12, determine the inverse Laplace transforms of the functions given using
Table 5-1 on page 170 and Table 5-2 on page 173.
9) V(s) =
8
; L −1 V(s) = 8 ε −3t
s+3
(Problem 5-21)√ 10/10
3 4
(Problem 5-22), 9/10
+
; L −1 I(s) = 2δ(t)+3+4t
s s2
4(8)
32
11) F(s) = 2
; L −1 F(s) = 2
= 4 sin(8t ) (Problem 5-23)√, 10/10
s + 64
s + 82
10s
10( s)
12) I(s) = 2
; L −1 I(s) = 2
= 10 cos(5t ) (Problem 5-24)√, 10/10
s + 25
s + 52
10) I(s) = 2 +
Deleted:
dt +