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Home Work 4 Solution
- We start by noting that 2 3
G s s^ s s s s
^
has a triple pole at s = 0, and single poles at -1 and -2. Therefore, the partial fraction expansion will have the form
( ) (^2 3) ( 1) ( 2) G s A^ B^ C^ D^ E s s s s s
Now we have many procedures to calculate A, B, C, D, and E.
(i) We can bring the partial fraction expansion to a common denominator as 2 3 3 3 4 3 2 3 2 2 4 3 4 3 3 4 3 2 3
( ) (^ 1)(^ 2)^ (^ 1)(^ 2)^ (^ 1)(^ 2)^ (^ 2)^ (^ 1)
G s As^ s^ s^ Bs s^ s^ C s^ s^ Ds^ s^ Es^ s s s s A s s s B s s s C s s D s s E s s s s s A D E s A B D E s A B C s B C s C s s s
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
Now equating coefficients of the numerator gives us the following simultaneous equations:
0 3 2 0 2 3 2 2 3 3 2 5
A D E
A B D E
A B C
B C
C
We can now solve these equations to get: C=5/2, B=-9/4, A=25/8, D=-4, E=7/
(ii) We can use the reside formulae to first calculate D and E:
D=G(s)(s+1)|s=-1 = -4, E = G(s)(s+2)|s=-2 = 7/8, C=G(s)s^3 |s=0 = 2.5.
Now we cannot calculate A and B so easily unless we use the more complicate formula for residues for repeated poles involving derivatives. A simple solution at this point is to note that
( ) (^3) ( 1) ( 2) 2 G s C^ D^ E^ A^ B s s s s s
We can calculate the left hand side since we know C, D, E and G(s). After doing this algebra, we get
4 3 2 3 3
2 2
s s s s s s s s s s s s s s s s s
Therefore, A = 25/8 and B =-9/4. [Note a few things about the second method. If you notice there are cancellations of factors s(s+1)(s+2) in the above calculation. This had to happen! The reason is that since C, D, E had been calculate to make the partial fraction expansion true. Therefore, once we took terms containing C, D, and E on the left, the net could only have poles at s=0 and that too a double pole. This is exactly what happened!
(iii) We can evaluate both sides of the partial fraction expansion at 5 randomly chosen values of s avoiding the poles of G(s). For example, we can pick s=1, s=2, s=3, s=4, s=5. (These are not so random as a matter of fact.) Now, we can get 5 equations in 5 unknowns A, B, C, D and E. I suggest you get these 5 equations and verify that the solution is consistent with answers in (i) and (ii) above.
- In this case, the numerator degree exceeds the denominator degree. So, first we use long division to convert the problem to the case where the denominator degree exceeds the numerator degree. Therefore,
4 2 4 2 3 2 2
G s s^ s^ s^ s^ s^ s s s s s s s s s^ s s s s
^ ^ ^ ^ ^
^
We can now compute partial fraction expansion of the last term
22 2 40 2 ( ) ( 2)( 3) ( 2) ( 3)
R s s^ s^ A^ B^ C s s s s s s
^
Again, there are several ways to solve for A, B, and C. A simple way, since there are no repeated roots, is to use the residue formula: A =R(s)s|s=0 = 1/3, B=R(s)(s+2)|s=-2 = -5/3, and C=R(s)(s+3)|s=-3 = 40/ Thus, the partial fraction expansion of G(s) is
( ) 5 1 5 40 3 3( 2) 3( 3)
G s s s s s
Verify for your own understanding that the right hand side equals the left hand side.
[You are free to skip past this. Note for those who are interested in a deeper understanding of the underlying math in this case, first note that G(s) has a pole at infinity. Then there should be a term in
