ELECTRICAL ENGINEERING 2012 MNS UET MULTAN Reg. No: 2012-EE-727 Page 13

Lab.03

Study of polynomial and Partial fractions

Introduction

Improper fraction:

When the degree of denominator is greater or equal to degree of

numerator is called improper fraction.

Partial fraction:

Partial fraction is used to solve the improper fractions.

Order:

Highest degree of variable in a polynomial equation.

Operations:

1. 2nd Order polynomial equation: A=[1 5 6];

B=roots(A) Ans=

B = -3.0000 -2.0000

2. 3rd order polynomial equation: c= [1 2 -1 -2] d=roots(c)

Ans= d =

1.0000 -2.0000 -1.0000

ELECTRICAL ENGINEERING 2012 MNS UET MULTAN Reg. No: 2012-EE-727 Page 14

3. Matrix to polynomial: E=[1 2 3; 4 5 6;7 8 9] F=poly(E) G=roots(F) Ans= E = 1 2 3 4 5 6 7 8 9 F = 1.0000 -15.0000 -18.0000 -0.0000 G =

16.1168 -1.1168 -0.0000

4. Find partial fraction& Roots of 2nd order b(s)/a(s)=[3x2+x-3]/[2x2+x-4] b=[3 1 -3] a=[2 1 -4] [r, p, k]=residue(b,a) [b,a]=residue(r,p,k) Ans= b = 3 1 -3 a = 2 1 -4 r = -0.6690 0.4190 p = -1.6861 1.1861 k = 1.5000 b = 1.5000 0.5000 -1.5000 a = 1.0000 0.5000 -2.0000

ELECTRICAL ENGINEERING 2012 MNS UET MULTAN Reg. No: 2012-EE-727 Page 15

Example 2: find partial fraction& Roots of following equation: b(s)/a(s)=[2x3-5x2-x+5]/[x2-1]

b=[2 -5 -1 5]

c=[1 0 -1]

[r, p, k]=residue(b,c) [b,c]=residue(r,p,k) Ans= B =

2 -5 -1 5

C =

1 0 -1

r =

0.5000

0.5000

p =

-1

1

k =

2 -5

B =

2 -5 -1 5

C =

1 0 -1