Spring ECE202: 2012
HW 2 Solutions
page 1
Problem 5:
a)
= 15
= 15
=
+
=
+
|
+
|
= 15 +
+
+
+
=
+
= 15
+
+
= 15
=
+
b)
=
=
+
,
,
,
+
+
,
=
+
+1
=
,
=
+
,
=1
+
=
|
2
+
=
+
+
,
=
=
=
+
=
+1
=
|
|
+
=
.
.
+
+
+
+1
= ,
=
=
+
+
+
=
=0
+
=
+
.
+
=
+
= , ,
.
ECE202: Spring 2012
=
=
2
= lim
=
=
+
|
HW 2 Solutions
= 1
+2 |
12
=2 + +
=2
=2
2 |
=
2 +4 8 2
=
4
=2
+2
page 2
2 +4 8
=
2
+2
+
+
+
2
+
+2
=1
= 1
1
1
2
+2
+
MATLAB Code:
1
B=[2 -1 -2 4 -8]; % coefficients of the polynomial (numerator)
A=[1 0 -4 0 0];
% coefficients of the polynomial (denominator)
[R,P,K] = RESIDUE(B,A) % function which calculates the partial fraction expansion
parameters
R=
1.0000
% corresponds to (s-2)
-1.0000
% corresponds to (s+2)
-1.0000
% corresponds to (s)
2.0000
% corresponds to (s^2)
P=
2.0000
-2.0000
0
0
%
%
%
%
corresponds
corresponds
corresponds
corresponds
to
to
to
to
(s-2)
(s+2)
(s)
(s^2)
K=
2
% corresponds to Lim F(s), as s->inf
Therefore, the MATLAB code verifies our answers.
ECE202: Spring 2012
HW 2 Solutions
page 3
e
6 + 17 + 115
=
+ 5 + 33 + 29
=
=
=
+1
+1 |
6 + 17 + 115
=
+
+1
+ 4 + 29
+1
+2
+
+ 2 + 25
+
+2
+2 +
+ 2 + 25
+ 25
=4
To find A and B, we should generate two equations to get A and B (evaluate F(s=-2) and F(s=0) then you
can calculate A,B).
4
2 5
4
+
=
+
+1
+ +2 2 + 25
+1
=
=4
= 2,
A=2, B=-5
+2
= 1,
= 5,
cos 5
= 2,
15
= 5,
1
2 +2
+ 2 + 25
sin 5
5
+ 2 + 25
= 4.
Problem 6:
a)
= 120 ,
=
=
1
2
=
140
+
1
4
= 240,
+
||
+ 10
= 140
:
||
= 240
+
+
||
||
+ 12 + 8
24 4
+
+2
+4
4
+2
= 24 2 sin 2
8
+2
:
2
+ + 1 + 24
+ cos 2
+ 140 8
+4
+4
4
+4
+2
+ 140
ECE202: Spring 2012
HW 2 Solutions
page 4
b)
= 90 ,
=
=
3
5
=
12
1
5
= 30,
+
+
12
,
+
= 12
:
+
+
+1
18
1 18
:
+
+2 1
+4
+1
18
=
= 30
1
cos
3
+
12
=
+4
18
1
sin
3
1
3
1
+ sin 2
6
Problem 7:
a)
12
,
+ 16
=
=
b)
:
12 + 2
+ 2 + 16
:
=
=
=
12
+
+ 16 + 2 12
+ 16
=
12
192
+ 16
1
1 2 +1
+
3
+1
+4
2
+ cos 2
3
ECE202: Spring 2012
:
=
:
=
=
Now put
Rule used
= 0,
=
=
+
page 5
=
+
12
192
+ 16
=
=
=
=
:
0
HW 2 Solutions
=
12
+ 2 192
+ 2 + 16
+
.This shows that M(s)=H(s+a)
=
=
0 =
+
12
+2
+ 2 + 16
e)
=
First of all we find Inverse Laplace of
=
=
+2
+
=
=
=
,
+
(by the frequency shift property)
:
0
ECE202: Spring 2012
HW 2 Solutions
:
a)
i)
=
=
1 +
1
+
=
=
+
=
4 +2
=
+
=
2
0
+
1
+0=
+
b)
i)
=
=
+1 +
1
1
++
+2
2 +2
+1
+
2
page 6
ECE202: Spring 2012
=
=
HW 2 Solutions
1
+
1
=
1
+
1
1
1
1= +
=
1
+
+
1
= +1+
1
+
=
+
0
page 7
+
+
1
3
2
1
0