EE571 Cheat Sheet II
Root Locus: Let KGH K sbs ...
sas ...
mm1
nn1
=++
++
−
−be the open-loop transfer function.
Angle Condition: A point s1 is on the R.L. if ∠GH(s1) = 180° x (odd number)
Magnitude Condition: If a point s1 lies on the R.L., then the value of K can be found from |KGH(s1)| = 1
10 rules:1. Starts @ open loop poles @ K=0
2. Ends @ open loop zeroes @ K=∞
3. # of branches = n
4. R.L. is symmetric WRT σ-axis and any other axis about which the open-loop pole zero configuration is symmetric
5. ∠ of asymptotes = 180° (odd #)/(n-m)
6. asymptotes are centered at σ = (b-a)/(n-m)
7. R.L. exists at a pt. on σ-axis if # of poles and zeroes of GH to the right is odd
8. To find angle of departure - draw a pt. very close to pole or zero and use ∠ condition
9. Use Routh array to find pt. where R.L. crosses jω-axis
10. Break-away and break-in pts. @ ∂K/∂s = 0
Root Locus Compensation: ∠Gc(s1) = angle of deficiency = 180° x (odd #) - ∠GH|desired poles; GPID = Ki/s + Kp + Kds
Error Analysis: (Valid for unity-feedback with closed loop system stable):
KlimG(s)
ps0
=→ K limsG(s)
vs0
=→ K lims G(s)
as0
2
=→ ess| 1/ (1 K )
step p
=+ ess| 1/K
ramp v
= ess| 1 / K
parabola a
=
Routh Array: Check the RHP roots of b s b s b s ... b s b 0
nnn1n1 n2n2 10
+++++=
−−−−
Frequency Response Methods:
Mapping Theorem: Let F(s) be a ratio of two polynomials in s. Choose a
closed contour and map it into the F(s) - plane. Then, the number of
encirclements the mapped contour makes about the origin in the F(s)-plane
(in the same direction as the original contour), in N=Z-P where Z= # of
zeroes of F(s) enclosed by the original contour and P = # of poles of F(s)
enclosed by the original contour
Decibel: 20log(|⋅|)
gain margin = 1/|GH(jωcp)| where ∠GH(jωcp) = -180° phase margin = 180°+∠GH(jωcg) where |GH(jωcg)| = 1 or 0 dB
Conversion Factors: 2 + j= 5 26.6 3 + j = 2 30 1+ j= 2 45 1+ j 3 = 2 60 1+ j2 = 5 63.4
ooo o o
∠∠∠∠∠;;; ;
1 = 0 dB 2 = 6 dB 3 = 9.5 dB 5 = 14 dB 8 = 18 dB
Lead Compensation Lag Compensation
Gs K Ts
Ts
cc
()=+
+
1
1
α
0 < α < 1 GsK Ts
Ts
lag lag
()=+
+
1
1
β
1 <
β
< ∞
Maximum phase: Find ωcgnew such that
sin m
φ
α
α
=−
+
1
1 occurs at
ωα
m1/( T)= ∠GH(jωcg) = -180°+ desired pm + fudge factor
Extra gain at ωm= 1/
α
Attenuation at high frequencies = 1/
β
Sensitivity: Sx
y
Y
X
y
xXnom Ynom
=
∂
∂
|,
Observer Equation: xAxBwKyCx
o
^()
^^
⋅
=++ −
s
n
b
n b
n-2 b
n-4 ...
sn-1 b
n-1 b
n-3 b
n-5 ...
sn-2 c
1 c
2 ...
.
: c1
bn1
bn2 bnbn3
bn1
=−−
−−
−
s1 c2
bn1
bn4 bnbn5
bn1
=−−
−−
−
s0