The initial value theorem is given by: (a)The final value theorem states that Ma

Question

The initial value theorem is given by: (a)The final value theorem states that Matlab returns the transforms by The partial fraction expansion can be accomplished using Matlab using the function R,P,K]- residue(num,den), returns the residues, poles and direct term of a partial fraction expansion of the ratio of two polynomials mun(s)l/den(s). Similarly the function [num,den]- residue(R,P.K), with 3 input arguments and 2 output arguments, converts the partial fraction expansion back to the polynomials mun(s)/den(s), were R represent residue values, P represent the system poles, and K represent the stand alone term 8.4. EXPERIMENT Part (A) Verify by hand and by using Matlab the following partial fraction expansion: 3S+7 4 s2-2s - 3 2s2-4 (s +1D(s-2s-3 s+3 s-1 s-3 552-155-11=-1/3 +--7 1/3 (c) S+0(S Part(B) A series RLC circuit with R-3 Ohms, L-2 Henries, and C=136 Faradays, with the following initial conditions: Vc(0)-90 volts, and IL(0) 0 Amps. 32

Explanation / Answer

matlab verification :

nums1=[3 7]
dens1=[1 -2 -3]
nums2=[2 0 -4]
dens2=[1 -4 1 6]
nums3=[5 -15 -11]
dens3=[1 -5 6 4 -8]
[r1 p1 k1]=residue(nums1,dens1)
[r2 p2 k2]=residue(nums2,dens2)
[r3 p3 k3]=residue(nums3,dens3)

r1 =

4.0000
-1.0000

p1 =

3.0000
-1.0000

k1 =

[]

(i) 4 / (s - 3) - 1 / (s + 1)

r2 =

3.5000
-1.3333
-0.1667

p2 =

3.0000
2.0000
-1.0000

k2 =

[]

(ii) 3.5 / (s - 3) - 1.33 / (s - 1) - 0.1667 / (s + 1)

r3 =

0.3333
4.0000
-7.0000
-0.3333

p3 =

2.0000
2.0000
2.0000
-1.0000

k3 =

[]

(ii) 0.3333 / (s - 2) - 4 / (s - 2)^2 - 7 / (s - 1)^3 - 0.333 / (s + 1)

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