Problem Sheet 2

This document provides the solutions to Problem Sheet 2. It includes step-by-step explanations and answers to all the questions. The topics covered in this problem sheet include partial fraction decomposition, solving equations, and finding the Fourier transform.

Running head: PROBLEM SHEET 2 1
Problem Sheet 2
Name
Institution

PROBLEM SHEET 2 2
Problem sheet 2
Part a
5
(s+5)( s+1)= A
s+5 + B
s +1
5= A (s +5)(s +1)
s+5 + B(s+ 5)( s+1)
s +1
5=A ( s+1)+ B (s +5)
When s=−5 ,
5=A (−5+1)+B (0)
5=−4 A
A=−5
4
When s=−1 ,
5=B (−1+5 ) +0
5=4 B
B= 5
4
5
( s+5)( s+1) =
−5
4
s+5 +
5
4
s +1 = 5
4 (s +1)− 5
4 (s+5)
Part b
sym s

PROBLEM SHEET 2 3
ans=partfrac(5/((s+5)*(s+1))) % partial fraction decomposition
The answer obtained in part b is similar to the one obtained in part a.
Part c
3 s+1
( s2 +s+1)(s +4 )= Bs+ A
s2 + s+1 + C
s+ 4
3 s +1= Bs+ A
s2 +s +1 (s2+ s +1)(s+ 4)+ C
s +4 ( s2+s+ 1)(s+ 4)
3 s +1=(Bs + A )(s +4 )+C (s2 +s+1)
When s=0
3( 0)+1=(B(0)+ A)(0+4)+C (02 +0+1)
1=4 A+C

PROBLEM SHEET 2 4
When s=−4
3(−4)+1=(−4 B+ A )(−4 +4 )+C ( ( −4 ) 2 −4+ 1)
−11=0+13C
13 C=−11
C=−11
13
1=4 A−11
13
4 A=1+ 11
13 = 24
13
A=24
13 × 1
4 = 6
13
3 s +1= ( Bs+ 6
13 ) ( s+ 4 ) − 11
13 (s2+ s+1)
3 s +1= ( Bs+ 6
13 ) ( s+ 4 ) − 11
13 (s2+ s+1)
3 s +1=B s2 +4 Bs+ 6
13 s+ 24
13 − 11
13 s2−11
13 s− 11
13
3 s +1=B s2− 11
13 s2 +4 Bs+ 6
13 s− 11
13 s+ 24
13 − 11
13
We can equate similar coefficients to get:
3 s=4 Bs+ 6
13 s− 11
13 s

PROBLEM SHEET 2 5
3=4 B+ 6
13 − 11
13
4 B=3+ 11
13 − 6
13 =3+ 5
13 = 44
13
B= 44
13 × 1
4 = 11
13
A= 6
13 , B=11
13 ,∧C=−11
13
3 s+ 1
( s2 +s+1)(s +4 ) = Bs+ A
s2 + s+1 + C
s+ 4 =
11
13 s+ 6
13
s2+ s +1 +
−11
13
s +4
¿ 11 s+6
13(s ¿¿ 2+ s+1)− 11
13(s +4 ) ¿
Part d
F ( s ) = ωk
(s2 +ω2 )(sT +1)= Bs+ A
(s2 +ω2)+ C
(sT +1)
ωk = Bs + A
( s2 +ω2) ( s2 +ω2)(sT +1)+ C
( sT + 1) (s2 +ω2 )(sT +1)
ωk =(Bs+ A)(sT +1)+C (s2 +ω2)
ωk =BT s2+Bs+ ATs+ A+C s2+ C ω2
0 s2 +0 s+ωk=(BT + C) s2 +(B+ AT ) s+( A+C ω2 )
( BT +C ) s2=0 s2
BT +C=0 ,

PROBLEM SHEET 2 6
C=−BT … equation 1
( B+ AT ) s=0 s
B+ AT =0 … equation 2
( A+C ω2 ) =ωk … equation 3
Substituting equation 1 into equation 3 we get:
A−BT ω2=ωk … equation 4
We Solve equations 2 and 4 simultaneously as follows.
First, multiply equation 4 by T then subtract equation 2.
( AT −BT 2 ω2 ) − ( B+ AT ) =Tωk −0
−B T2 ω2−B=Tωk
B(−T 2 ω2−1)=Tωk
B= Tωk
−T 2 ω2−1
Then, we Substitute the value of B obtained into equation 2
B+ AT =0 … equation 2
AT =−B , A=−B
T
A=−B
T =
− Tωk
−T2 ω2 −1
T = ωk
T2 ω2+1

PROBLEM SHEET 2 7
C=−BT= −T 2 ωk
−T 2 ω2−1 = T2 ωk
T2 ω2+1
A= ωk
T 2 ω2 +1 , B= Tωk
−T2 ω2−1 , C= T 2 ωk
T 2 ω2 +1
F ( s ) = ωk
( s2+ ω2 ) ( sT +1 ) = Bs+ A
( s2 +ω2 ) + C
( sT +1 )
¿
Tωks
−T 2 ω2−1 + ωk
T 2 ω2 +1
( s2 + ω2) +
T 2 ωk
T 2 ω2 +1
(sT +1)
¿
ωk
T 2 ω2 +1 − Tωks
T 2 ω2 +1
(s2 +ω2) +
T 2 ωk
T 2 ω2+ 1
( sT + 1)
¿
ωk
T 2 ω2 +1
(s2+ ω2 ) −
Tωks
T 2 ω2 +1
(s2 +ω2) +
T 2 ωk
T 2 ω2 +1
( sT +1)
F (s)= k
T 2 ω2 +1 × ω
(s2+ ω2 )− Tωk
T 2 ω2 +1 × s
(s2+ ω2) + T 2 ωk
T2 ω2 +1 × 1
(sT +1)
F ( t )=L−1 {F (s )}
F ( t )=L−1 { k
T2 ω2+1 × ω
( s2+ ω2 ) − Tωk
T2 ω2+1 × s
( s2+ ω2 ) + T 2 ωk
T 2 ω2 +1 × 1
( sT +1 ) }
¿ k
T2 ω2 +1 L−1
{ ω
( s2 +ω2 ) }− Tωk
T 2 ω2 +1 L−1
{ s
( s2 +ω2 ) }+ T 2 ωk
T 2 ω2 +1 L−1
{ 1
( sT +1 ) }
¿ k
T2 ω2 +1 sin (ωt )− Tωk
T2 ω2 +1 cos (ωt)+ T 2 ωk
T2 ω2 +1 × 1
T e
−t
T

PROBLEM SHEET 2 8
F (t)= k
T 2 ω2+1 sin (ωt)− Tωk
T 2 ω2+ 1 cos (ωt)+ Tωk
T 2 ω2+1 e
−t
T
F (t)= 1
T 2 ω2+1 (k sin ( ωt )−Tωk cos ( ωt )+ Tωk e
−t
T )