Routh stability
Diperkenalkan| English | Makna |
|---|---|
| Routh array | Array Routh |
| auxiliary equation/ɔːkˈsɪlɪəri ɪˈkweɪʒn/ | Persamaan pembantu |
| marginal stability | Stabilitas marginal |
Stable first, then good
- Stability comes before every other specification. A loop that rings itself into saturation is not badly tuned; it is broken.
- The Routh test tells you whether a loop is stable without solving for a single root.
The stability rule
- A closed loop is stable when every root of its characteristic polynomial has a negative real part.
- The Routh array 劳斯阵列 counts the right-half-plane (RHP) roots: the number of sign changes in its first column.
Which set of closed-loop poles is stable?
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Every pole needs a negative real part. A pole at 0 or on the imaginary axis is only marginal, not stable.
Building the array
- Write the coefficients in two rows, alternating: $s^{n}$ takes the 1st, 3rd, 5th…; $s^{n-1}$ takes the 2nd, 4th….
- Each new entry is a cross-multiply-and-divide of the two rows above it. A missing power gets a 0.
The polynomial s³ + 4s + 10 (no s² term) can still be stable.
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
A stable polynomial needs every coefficient present and of the same sign. A missing power puts a 0 in the first column at once.
Worked example: count the RHP roots
- $s^{4} + 2s^{3} + 3s^{2} + 4s + 5$: rows 1, 3, 5 and 2, 4. Then $s^{2}$: 1, 5; $s^{1}$: $-6$; $s^{0}$: 5.
- First column 1, 2, 1, −6, 5: two sign changes, so two RHP roots. The exact roots, $0.29 \pm j1.42$, agree.
How many RHP roots does s³ + s² + 2s + 8 have?
Kesulitan 2/5 · 140 koin + bonus kecil pada penyelesaian pertama hari ini
Rows 1, 2 and 1, 8; then $s^{1}$: $(1 \times 2 - 1 \times 8)/1 = -6$; $s^{0}$: 8. The column 1, 1, −6, 8 changes sign twice. The roots are $-2$ dan $0.5 \pm j1.94$.
Worked example: the gain range
- $s^{3} + 5s^{2} + 6s + K$: the $s^{1}$ entry is $\dfrac{30 - K}{5}$, so stability needs $0 < K < 30$.
- At $K = 30$ the row vanishes. The auxiliary equation 辅助方程 $5s^{2} + 30 = 0$ gives $s = \pm j2.45$: marginal stability 临界稳定, a steady oscillation.
s³ + 4s² + 5s + K is stable for 0 < K < K_max. What is K_max?
Kesulitan 2/5 · 140 koin + bonus kecil pada penyelesaian pertama hari ini
The $s^{1}$ entry is $(4 \times 5 - K)/4$, positive while $K < 20$.
At K = 20 that loop oscillates. At what frequency, in rad/s? Give 2 decimal places.
Kesulitan 2/5 · 140 koin + bonus kecil pada penyelesaian pertama hari ini
The auxiliary equation $4s^{2} + 20 = 0$ gives $s = \pm j\sqrt{5} = \pm j2.24$.
A design picture
- For $s^{3} + as^{2} + 8s + K$, the $s^{1}$ entry $\dfrac{8a - K}{a}$ gives $0 < K < 8a$: a straight boundary.
- The course's running plant $s(s+2)(s+4)$ has $a = 6$, so its limit is $K = 48$. Root locus and Nyquist will find the same 48.
In the family s³ + as² + 8s + K, take a = 3. What is the largest stable K?
Kesulitan 2/5 · 140 koin + bonus kecil pada penyelesaian pertama hari ini
The boundary is $K = 8a = 24$. A smaller $a$ leaves less room for gain.
Three common slips
- A missing power. Every power from $s^{n}$ down needs a coefficient, even 0.
- Dividing a row by a negative number. That flips the very signs you are counting.
- "Stable" at the boundary. Poles on the imaginary axis mean marginal: the oscillation never dies.
Put the Routh test in order.
Kesulitan 3/5 · 200 koin + bonus kecil pada penyelesaian pertama hari ini
The count of sign changes is the number of RHP roots; none means stable.
Recap
- Build the array; count sign changes in the first column.
- One row gives the stable range of a gain.
- A zero row means poles on the axis: use the auxiliary equation.