The Stability Of Linear Feedback Systems
In the S-plane, if a pole is located at the left side of the plane then the system is stable.
S=-α±jωd If -α is negative then the system is stable.
i.e. G(s)=S+2S3+2S2+50S+1
If one pole is at the unstable region, then the system is unstable.
S3+2S2+50S+1=0 is called the characteristic equation.
Note: For system stability the following conditions must be satisfied (but not sufficient):
1) All coefficients of the characteristic equation must have the same sign.
2) None of the polynomial coefficients be zero.
Not sufficient means we immediately know that the system is unstable if any of these conditions is not satisfied.
Consider:
T(s)=N(s)Q(s)=N(s)anSn+an-1Sn-1+an-2Sn-2+...... +a1S+a0
Notes:
1) The # of roots of Q(s) which are positive (unstable) real roots = the # of sign change in the first column.
2) A stable system should have no sign change.
Case (1): No zeros in the first column.
Ex: T(s)=S+0.532S3+S2+2S+24
S3S2 / 1 2
1 24
S
1 / b1 b2
C1
Two sign change in the first column ---> there are two roots in the positive s-plane (unstable)
Case (2) : some zeros in the first column with at least one non-zero element in the same row containing the zero.
EX: T(s)=S+0.356S5+2S4+2S3+4S2+11S+10
Since b1is zero, we assume it a small number (ϵ) which has the sign of the element located directly above it.
The first column has no sign change with a zero, then the system is marginally stable.
Case (3): All the elements in a row are zero.
EX: Q(s)=S3+2S2+4S+K
A- For a stable system, we require that:
1) K>0
2) 8-K2>0
⇒0<K<8
B- When K=8 we obtain a row with all zeros, so we define the auxiliary polynomial U(s) which is the equation of the row preceding the row of zeros.
U(s)=2S2+8S0
dU(s)ds=4S+0 , This equation replaces the all zero row.
This occurs when factors are symmetrically located about the origin.
Notice that the auxiliary polynomial has no roots on the right half plane (which means stable!!). However, for the auxiliary, it is necessary to factor its equation to determine its poles.
2S2+8=0
S=±j2
The general form
Y(s)R(s)=KG(s)1+KG(s)
Δ(s)=1+KG(s)=0
KG(s)=-1±j0
|KG(s)|=1
angle(KG(s))=±180+n 360o, n=0,±1,±2,......
For the closed loop,
Δ(s)=1+KG(s)=1+K1S(S+2)=0
Δ(s)=S2+2S+K=0
S1,2=-1±jK-1
Notes:
If K=0 ⇒S1,2=0,-2 open loop poles
If K=5 ⇒S1,2=-1±j2
If K=1 ⇒S1,2=-1
If K=0.5 ⇒S1,2=-1.7,-0.3
If we are looking for the K value (knowing the root value and S is given),
K=|S||S+2|
@S=-1+j2 ⇒K=|-1+j2||-1+j2+2|=5
Procedure for the root locus:
1) Write the characteristic equation as Δ(s)=1+KP(s)
EX:
For this example K=10,
Δ(s)=1+101S(s+a)=0
S(S+a)+10=0
S2+aS+10=0
aS=-(10+S2)
aSS2+10=-1
⇒1+aSS2+10=1+KP(s)
2) Factor P(s) for poles and zeros
Δ(s)=1+Ki=1nz (S+Zi)j=1np (S+Pj)
3) Locate on the S-plane the poles and zeros.
Important Note:
Locus of the roots (poles) of (1+KP(s)=0) begins at the poles of P(s) and ends at the zero of P(s) as K increases (zero at ∞)
4) The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros.
EX: A single loop feedback control system processes the characteristic equation
1+K2(S+2)S(S+4)=0
Here we have,
Z1=-2
P1=0
P2=-4
5) Determine the number of separate loci( SL=# of poles).
6) Root locus must be symmetrical around the real ( X-axis).
7) The loci proceeds to the zero at ∞ along asymptotes centered at (σA) with angle (φA) if N=np-nz, then an N sections of locies proceed to the zero at ∞ along an asymptote as K→∞
Where σA=i=1np (pi)-j=1nz (Zj)np-nz
np: # of poles
nz: # of zeros
φA=2q+1np-nz180o , q=0,1,2,.....,(np-nz-1)
EX: Δ(s)=1+K1S(S+2)=0
nz=0
np=2 , P=0,-2
σA=(0+-2)-02-0=-1
φ0=2(0)+12180=90o
φ1=2(1)+12180=270o
* Determine the Breakaway point(B.P) where the root locus leaves the X-axis.
For example, consider a unity feedback closed loop system with an open loop transfer function G(s)=K(S+2)(S+4)
Δ(s)=1+K1(S+2)(S+4)=0
P(s)=K=-(S+2)(S+4)
B.P=roots(dP(s)dS=0)=roots(2S+6=0)⇒S=-3
* Determine the angle of departure from the complex poles:
For a point in the S-plane to be on the root locus:
(all the angles from finite zeros)- (all the angles from finite poles)
=±180+r360
EX: Δ(s)=1+KS4+12S3+64S2+128S=0
nz=0
np=4
P1=0
P2=-4
P3,4=-4±j4
σA=0-4-4+j4-j4-44-0=-3
q=0,1,2,3
φ0=14180=45o
φ1=135o
φ2=225o
φ3=315o
B.P⇒P(s)=-(S4+12S3+64S2+128S)
dP(s)ds=0=-(4S3+36S2+128S+128)
S1,2=-3.71±j2.5
S3=-1.576 we take this value because it’s located on the X-axis as the definition.
Angle of departure for complex poles:
0-(θ+90+90+135)=180
θ=-135o
Parameter Design by the Root Locus Method (Two Parameters)
EX:
Find K1, K2 such that the following are satisfied:
1) eSS≤35% of a ramp input slope.
2) ζ≥12
3) ts≤3 sec.
Sol:
Let G1(s)=G(s)1+G(s)H1(s)=K1S2+2S+K1K2S
eSS=limS→0SE(s)
eSS=limS→0S[(1-T(s))R(s)]=limS→0[S11+G1(s)AS2]=limS→0[AS+SG1(s)]
eSS=limS→0AS+SK1S2+2S+K1K2S
eSSA=2+K1K2K1≤0.35
For ts≤3 sec,
TS=4τ=4ζωn≤3
؞ ζωn≥43=1.33
Keep in mind: S 1,2=-α±jωd=-ζωn±jωn1-ζ2
Given that ζ=0.707
ζωn≥43
⇒ωn=1.88
⇒S1,2=-1.33±j1.881-(12)2
Now,
T(s)=G1(s)1+G1(s)
1+K1S2 +2S+K1K2S=0
Δ(s)=S2+2S+K1K2S+K1=0
Δ(s)=S2+2S+βS+α=0
Since we have two parameters that we have to find, we can do:
1) β=0,
⇒Δ(s)=S2+2S+α=0
1+α1S(S+2)=0
2) Choose any value of α on the
root locus (such that =20 )
S1,2=-1±j4.36
3) Back to β
1+βSS2+2S+20
β=K1K2
4.3=20K2
K2=0.215
PID Controller
P: Proportional (Kp)
I : Integrator (KI1S)
D: Differentiator (KDS)
P controller:
VoutVin=-RfRin=Kp=Gc(s)
Standard closed second order system:
YR=ωn2S2+2ζωnS+ωn2
Standard open second order system:
YR=ωn2S(S +2ζωn)=G(s)
PD controller:
Gc(s)=Kp+KDS
Gc(s)=VoVin=-[R2R1+R2CS]
⇒Kp=-R2R1 , KD=-R2C
The open second order system transfer function will be:
YR=ωn2(Kp+KDS)S(S+2ζωn)
⇒Here a new zero has been added to the system.
PI controller:
Gc(s)=Kp+KIS
Gc(s)=VoVin=-[R2R1+1R1CS]
⇒Kp=-R2R1 , KI=-1R1C
The open second order system transfer function will be:
YR=SKp+KISωn2S(S+2ζωn)
⇒Here one zero and one pole have been added.
PID controller:
Gc(s)=Kp+KIS+KDS=KDS2+KpS+KIS
⇒Here two zeros and one pole have been added.
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