undefined | unit 4 stability analysis

Unit-4Stability Analysis Q1) For the given polynomials below determine the stability of the system S4+2S3+3S2+4S+5=0A1)

Arranging Coefficient in Rows.

S4	1	3	5
S3	2	4	0
S2	1	5	0
S1	-6	0	0
S0	5	0	0

For row S2 first term

S2 = = 1

For row S2 Second term

S2 = = 5

For row S1:

S1 = = -6

For row S0

S0 = = 5

As there are two sign change in the first column, So there are two roots or right half of S-plane making system unstable.

Q2. Using Routh criterion determine the stability of the system with characteristic equation S4+8S3+18S2+16S+S = 0 A2)

Arrange in rows.

S4	1	18	5
S3	8	16	0
S2	16	5
S1	13.5
S0	5

For row S2 first element

S1 = = 16

Second terms = = 5

For S1

First element = = 13.5

For S0

First element = = 5

As there is no sign change for first column so all roots are is left half of S-plane and hence system is stable.

Q3) Determine the stability of the system represent by following characteristic equations using Routh criterionS4 + 3s3 + 8s2 + 4s +3 = 0S4 + 9s3 + 4S2 – 36s -32 = 0 A3) 1. S4+3s3+8s2+4s+3=0

S4	1	8	3
S3	3	4	0
S2	6.66	3
S1	2.650	0
S0	3

No sign change in first column to no rows on right half of S-plane system stable.S4+9S3+4S2-36S-32 = 0

S4	1	4	-32
S3	9	-36	0
S2	8	-32
S1	0	0
S0

Special case II of Routh Hurwitz criterion forming auxiliary equation

A1 (s) = 8S2 – 32 = 0

= 16S – 0 =0

S4	1	4	-32
S3	9	-36	0
S2	8	-32
S1	16	0
S0	-32

One sign change so, one root lies on right half S-plane hence system is unstable. Q4) For using feedback open loop transfer function G(s) =

find range of k for stabilityA4) Findlay characteristics equation.CE = 1+G (s) H(s) = 0H(s) =1 using feedbackCE = 1+ G(s)1+

= 0S(S+1)(S+3)(S+4)+k = 0(S2+5)(S2+7Sα12)αK = 0S4α7S3α1252+S3α7S2α125αK = 0S4+8S3α19S2+125+k = 0

By Routh Hurwitz Criterion

S4	1	19	K
S3	8	12	0
S2	17.5	K
S1		0
S0	k

For system to be stable the range of K is 0< K < .

Q5) The characteristic equation for certain feedback control system is given. Find range of K for system to be stable.

S4	1	12	K
S3	4	36

S2	3	K
S1
S0	K

A5) S4+4S3α12S2+36SαK = 0For stability K>0

> 0 K < 27Range of K will be 0 < K < 27 Q6) Check if all roots of equation S3+6S2+25S+38 = 0, have real poll more negative than -1.A6)

S3	1	25
S2	6	38
S1	18.67
S0	38

No sign change in first column, hence all roots are in left half of S-plane.Replacing S = Z-1. In above equation (Z-1)3+6(Z-1)2+25(Z-1)+38 = 0Z3+ Z23+16Z+18=0

Z3	1	16
Z2	3	18
Z1	10
Z0	18

No sign change in first column roots lie on left half of Z-plane hence all roots of original equation in S-domain lie to left half 0f S = -1 Q7) Explain relative stability?A7) Routh stability criterion deals about absolute stability of any closed loop system. For relative stability we need to shift the S-plane and the apply the Routh criterion.

Fig 1 Location of Pole for relative stabilityThe above fig 1 shows the characteristic equation is modified by shifting the origin of S-plane to S1= -

.S = Z-S1After substituting new valve of S =(Z-S1) applying Routh stability criterion, the number of sign changes in first column is the number of roots on right half of S-plane Q8) In detail explain the concept of stability?A8) A stable system always gives bounded output for bounded input and the system is known as BIBO stableA linear time invariant (LTI) system is stable if, The system is BIBO stableIn absence of the input the output tends towards zero

For system

1) R(S)

Fig 2 Control system with G(s) = 1/s2 So, system unstable.

Eg.2 R(s)

[R(s) = 1]

C(t) = e-10t C(t)

t= 0

e-10t

e-10t C(t) 1

1 0

tFig 5 BIBO stable figure

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