ELEN90055 · Control Systems
Pole-placement synthesis
This chapter of ELEN90055 Control Systems at the University of Melbourne is the most direct design method in the course: instead of nudging a lead or lag compensator until a Bode plot looks right, you name the closed-loop poles you want and solve for the controller that delivers them. The device is the Diophantine equation Acl = A·L + B·P, whose solvability is governed by the Sylvester resultant matrix.
It sits in Part IV (design of feedback control systems, loosely following Goodwin, Graebe & Salgado), and connects pole placement to the Internal Model Principle for zero steady-state error and to the familiar PID controller for a second-order plant.
What this chapter covers
- 01Write the Diophantine (Bezout) equation A_cl = A L + B P and read the closed-loop poles as its roots
- 02Choose the controller order deg L = deg P = n - 1 from the plant order n so the solution is unique
- 03Build the 2n x 2n Sylvester (resultant) matrix from the coefficients of A and B
- 04Apply the theorem: the Sylvester matrix is invertible if and only if A and B are coprime
- 05Solve the square linear system S theta = c for the controller polynomials L(s) and P(s)
- 06See why a stable target A_cl gives internal stability with no forbidden unstable pole-zero cancellation
- 07Combine pole placement with the Internal Model Principle by augmenting the plant with the generating polynomial Gamma(s)
- 08Recover robust zero steady-state error to steps (Gamma = s) or sinusoids (Gamma = s^2 + omega^2)
- 09Realise a PID-form controller C = (p2 s^2 + p1 s + p0)/(s(l1 s + l0)) as pole placement for a second-order plant
- 10Check that the uncancelled controller zeros carried into T0 = B P / A_cl do not spoil the transient
Place the closed-loop poles of a second-order plant
- +2Coprimeness and order rule. B = 1 shares no root with A, so the pair (A, B) is coprime and the Sylvester matrix is invertible: a solution exists and is unique. With deg A = n = 2, take L = s + l0 and P = p1 s + p0 (each degree n - 1 = 1). Then deg A_cl = 2n - 1 = 3, placing three poles.
- +1Target polynomial. A_cl = (s^2 + 4s + 8)(s + 5) = s^3 + 9s^2 + 28s + 40. The quadratic s^2 + 4s + 8 has the dominant pair -2 +/- j2 rad/s (2*zeta*omega_n = 4, omega_n^2 = 8, so omega_n = 2.83 rad/s, zeta = 0.707); the factor (s + 5) is the fast real pole.
- +3Match coefficients of A L + B P = A_cl. Expand (s^2 + 6s + 8)(s + l0) + (p1 s + p0) and match: s^3 gives 1 (ok); s^2 gives l0 + 6 = 9, so l0 = 3; s^1 gives 6*l0 + 8 + p1 = 28, so 18 + 8 + p1 = 28 and p1 = 2; s^0 gives 8*l0 + p0 = 40, so 24 + p0 = 40 and p0 = 16.
- +1Controller. C(s) = P/L = (2s + 16)/(s + 3). It is proper, with a zero at s = -8 rad/s (left-half plane, so minimum-phase) and a pole at s = -3 rad/s. No plant pole is cancelled (plant poles -2 and -4; controller pole -3), so the placement is legal.
- +1Verify. Back-substitute: (s^2 + 6s + 8)(s + 3) + (2s + 16) = s^3 + 9s^2 + 26s + 24 + 2s + 16 = s^3 + 9s^2 + 28s + 40 = A_cl, whose roots are -2 +/- j2 and -5 (all in the left-half plane, so stable). Note the DC closed-loop gain T0(0) = B(0)P(0)/A_cl(0) = 16/40 = 0.4, so this controller (no integrator) still leaves a 60% steady-state step error; add Gamma = s (IMP) to remove it.
Key terms
- Diophantine (Bezout) equation
- The pole-placement design equation A(s) L(s) + B(s) P(s) = A_cl(s), where G = B/A is the plant and C = P/L the controller. Its roots are the closed-loop poles; solving it for L and P places those poles at the chosen target A_cl.
- Characteristic polynomial A_cl(s)
- The denominator of the closed-loop transfer function, A_cl = A L + B P. The closed-loop poles are its roots; a design is internally stable exactly when every root of A_cl has Re < 0.
- Sylvester (resultant) matrix
- The 2n x 2n coefficient matrix S that turns coefficient-matching of A L + B P = A_cl into the linear system S theta = c, with theta the controller coefficients and c the target coefficients. Its columns are the coefficients of A (for L) and of B (for P), shifted one row per lower power of s. Its determinant is the resultant Res(A, B).
- Coprime polynomials
- Two polynomials A and B that share no common factor (equivalently, no common root, i.e. no plant pole-zero cancellation). The Sylvester matrix is invertible if and only if A and B are coprime, which is the precise condition for arbitrary pole placement to be possible.
- Controller order rule
- For a strictly proper plant with deg A = n, choose deg L = deg P = n - 1. This makes the Sylvester system square (2n equations in 2n unknowns) and gives a proper controller placing 2n - 1 closed-loop poles.
- Internal Model Principle (IMP)
- To achieve zero steady-state error to a persistent input, the controller must contain a model of that input's generating dynamics. In pole placement, the plant is augmented as B/(A*Gamma) and the final controller is C = P/(Gamma L), so the error is rejected robustly.
- Generating polynomial Gamma(s)
- The polynomial whose imaginary-axis roots generate a persistent signal: Gamma = s for a step (one integrator), Gamma = s^2 for a ramp, and Gamma = s^2 + omega^2 for a sinusoid at omega rad/s. Including Gamma as a controller-denominator factor forces S0(gamma_p) = 0 at each root.
- PID-form pole placement
- For a second-order plant with no finite zeros, the controller C = (p2 s^2 + p1 s + p0)/(s(l1 s + l0)) performs pole placement: the denominator factor s is the integrator (internal model for a step) and the extra pole l1 s + l0 rolls off the derivative term to keep C proper. In the ideal limit it is the parallel PID C = Kp + Ki/s + Kd s.
Pole-placement synthesis FAQ
Why must the plant polynomials A(s) and B(s) be coprime for pole placement to work?
Coprimeness means A and B share no common factor, which is exactly the condition for the Sylvester (resultant) matrix to be invertible. When it is invertible, the Diophantine equation A L + B P = A_cl has a unique solution for the controller coefficients, so any target characteristic polynomial of the right degree is achievable and the closed-loop poles can be placed anywhere. If A and B did share a factor, that factor would divide A L + B P as well, so it would be forced to appear in A_cl: you cannot move a plant pole that is cancelled by a plant zero. In state-space language, coprimeness of A and B is the transfer-function version of the plant being controllable and observable.
How is pole placement different from tuning a PID controller with a rule like Ziegler-Nichols?
Ziegler-Nichols and similar rules are experimental recipes: you measure a reaction curve or the ultimate gain and read the PID gains off a table, aiming for a roughly acceptable response. Pole placement is analytic and exact: you write the closed-loop characteristic polynomial A L + B P, set it equal to a polynomial whose roots are the poles you actually want, and solve for the controller. The two connect neatly for a second-order plant with no zeros, where the pole-placement controller has PID form, so you can compute Kp, Ki and Kd directly by matching coefficients rather than by tuning. The trade-off is that pole placement needs a model, and it fixes the poles but not the closed-loop zeros, so you should still check the time-domain response.
Can AI help me with pole-placement synthesis in ELEN90055?
Yes, as a study aid. An AI tutor like Sia can explain the Diophantine equation A_cl = A L + B P step by step, show why the Sylvester matrix is invertible exactly when A and B are coprime, and check your coefficient-matching and units on a small placement or IMP problem. Use it to understand the method, the order rule and the internal-model augmentation, and to practise; it does not sit your open-book exam for you or guarantee a mark, and you should confirm every formula and the exam details against the unit materials on Canvas.
Studying with AI? Sia — free AI electrical engineering tutor works through ELEN90055 step by step.
Exam move
Anchor the whole chapter to one equation: A_cl = A L + B P. Be able to state, in order, the recipe you would write in the exam: model the plant as G = B/A and the controller as C = P/L; fix the orders (deg L = deg P = n - 1, so deg A_cl = 2n - 1); write the target A_cl from the poles you want (all with Re < 0); confirm A and B are coprime so the Sylvester matrix inverts; solve the square linear system for L and P; and always back-substitute to verify A L + B P equals your target. Practise the two extensions that carry marks: fold in the Internal Model Principle by augmenting the plant with the generating polynomial Gamma (Gamma = s for a step, s^2 + omega^2 for a sinusoid) so the final controller is C = P/(Gamma L) and the steady-state error goes to zero; and specialise to a second-order plant with no zeros, where the placement controller has PID form and you can read off Kp, Ki and Kd by matching coefficients. Watch the closed-loop zeros: the controller numerator P appears in T0 = B P / A_cl and can cause overshoot or a wrong-way start, so push the poles only as fast as the transient can tolerate. The exam is 3 hours, open book, worth 70% with a hurdle, and you answer four questions summing to 50 marks, so budget about 3.6 minutes per mark. Because it is open book, marks reward method shown with correct orders, coefficients, coprimeness and stability checks rather than a memorised controller, so state each check in one clear line. Confirm the exact date of the next (Semester 1, ~June 2027) sitting on Canvas.
