Routh-Hurwitz Criterion: Special Cases
Zero only in the first column
When forming the Routh table, replace the
zero with a small number ε and evaluate thefist column for positive or negative values of ε.
Problem: Determine the stability of the closed-
loop transfer function:
T (s) =10
s5+2s4+3s3+6s2+5s+3(1)
s5 1 3 5
s4 2 6 3
s3−
1 32 6
2=6 0→ ε
−1 52 3
2= 7
2
−1 02 0
2= 0
s2
−2 6ε 7
2ε
= 6ε−7ε
−2 3ε 0
ε= 3
−2 0ε 0
ε= 0
s1
−ε 7
26ε−7
ε3
6ε−7ε
= 42ε−49−6ε2
12ε−14
−ε 0
6ε−7ε
06ε−7
ε
= 0
−ε 0
6ε−7ε
06ε−7
ε
= 0
s0 3 0 0
1
Label First Column ε = + ε = −
s5 1 + +
s4 2 + +
s3 ε + −
s2 6ε−7ε − +
s1 42ε−49−6ε2
12ε−14 + +
s0 3 + +
The changes in sign occur in both instances so
it doesn’t matter which we choose. they both
indicate the system is unstable with two poles
in the rhp.
2
Another, less computationally expensive method
to use when a zero occurs in the first column
is to create the Routh table using the polyno-
mial that has the reciprocal roots of the origi-
nal polynomial. i.e.
T (s) =10
s5+2s4+3s3+6s2+5s+3(2)
by replacing s with 1/s.
D(s) = 3s5+5s4+6s3+3s2+2s+1 (3)
s5 3 6 2
s4 5 3 1
s3 4.2 1.4 0
s2 1.33 1 0
s1 −1.75 0 0
s0 1 0 0
since there are two sign changes, the system is unstable and has
two right half plane (rhp) poles. This is the same as the result
obtained by using ε for the original polynomial.
3
Entire row is zero
Problem: Determine the number of right hand
plane poles in the closed-loop transfer func-
tion:
T (s) =10
s5+7s4+6s3+42s2+8s+56(4)
s5 1 6 8
s4 6 7→ 1 6 42→ 6 6 56→ 8
s3 0→ 0→ 0
s2
s1
s0
Now we are faced with the problem of zeros in
the third row.
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1. Form a new polynomial using the entries inthe row above zeros. The polynomial will startwith power of s in that row, and continue byskipping every other power of s, i.e.
P (s) = s4+6s2+8 (5)
2. Next we differentiate the polynomial withrespect to s and obtain
dP (s)
ds= 4s3+12s+0 (6)
3. Finally the row with all zeros in the Routhtable is replaced with the coefficients in Eq.(6),and continue the table.
s5 1 6 8
s4 6 7→ 1 6 42→ 6 6 56→ 8
s3 6 0→6 4 → 1 6 0→6 12→ 3 6 0→ 0
s2 3 8 0
s1 13
0 0
s0 8 0 0
We see no sign changes hence no rhp poles.
5
• Why does an entire row of zeros occur?When a purely odd or even polynomial is afactor of the original polynomial. (s4+6s2+8is an even polynomial as it only has even powerof s.)
σ
j ω
j ωωj
σ
quadrantal and symmetrical
imaginarysymmetrical and symmetrical and real
σ
• Some polynomial only have roots symmetri-cal about the origin.
• Routh table from the even polynomial (s4 →
s0) is a test of the even polynomial.
• The rows of zeros indicates the possilbilityof jω roots.
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Look at a larger example:
T (s) =20
s8 + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s + 20(7)
s8 1 12 39 48 20
s7 1 22 59 38 0
s6 − 6 10→ −1 − 6 20→ −2 6 10→ 1 6 20→ 2 0
s5 6 20→ 1 6 60→ 3 6 40→ 2 0 0
s4 1 3 2 0 0
s3 6 0→6 4 → 2 6 0→6 6→ 3 6 0→ 0 0 0
s2 32
2 0 0 0
s1 13
0 0 0 0
s0 4 0 0 0 0
1. Rows of zeros of s3, Form a new polynomial
using the entries in the row above zeros, i.e.
s4.
P (s) = s4+3s2+2 (8)
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2. Next we differentiate the polynomial with
respect to s and obtain
dP (s)
ds= 4s3+6s+0 (9)
3. Finally the row with all zeros in the Routh
table is replaced with the coefficients in Eq.(9),
and continue the table.
• As the entries from s4 to s0 are looking at
the even polynomials and there are no sign
changes. So all four poles must exits on the
jω axis.
• There are two sign changes from s8 to s4, so
there are 2 right hand plane poles and 2 left
hand plane poles.
• Final results: 2rhp, 2lhp, 4 jω poles.
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