#34 Control Theory #34 — Pole-Zero Analysis (Worked Example 11)

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Given poles, zero and coefficient

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   Both factored transfer functions have coefficient K=−2 and zero +0.2.

   G1 has pole −0.2; G2 also has poles −0.4±0.3j.

   Narration transcript

   Here is the problem. We are given pole-zero plots for two transfer functions G1 and G2, both with constant coefficient K equals negative 2. G1 has one pole at negative 0.2 and one zero at positive 0.2, both on the real axis. G2 has the same real pole and zero, plus a pair of complex conjugate poles at negative 0.4 plus or minus 0.3 j. We need to write the transfer functions, find relative degrees, check stability, and match G2 to a state space model.

2. 2. Preserve the negative coefficient

   $\displaystyle G1=-2\left(s-0.2\right)/\left(s+0.2\right).$

   $\displaystyle G2=-2\left(s-0.2\right)/\left(\left(s+0.2\right)\left(s²+0.8s+0.25\right)\right).$

   The complex-pole factor follows from (s+0.4)²+0.3². G2 has DC gain +8, not −8.

   Narration transcript

   Part a: write the transfer functions from the pole-zero plots. For G1, we have one zero at s equals 0.2 and one pole at s equals negative 0.2. With K equals negative 2, the transfer function is G1 of s equals negative 2 times s minus 0.2, divided by s plus 0.2. For G2, we have the same zero at 0.2, the same real pole at negative 0.2, and complex conjugate poles at negative 0.4 plus or minus 0.3 j. The complex pair gives the quadratic s squared plus 0.8 s plus 0.25. So G2 of s equals 2 times s minus 0.2, divided by the quantity s plus 0.2 times s squared plus 0.8 s plus 0.25.

3. 3. Relative degree

   G1 has relative degree zero and direct feedthrough; G2 has relative degree two.

   Both are proper rational functions. Properness is the standard causal finite-dimensional state-space condition for this class of model.

   Narration transcript

   Part b: find the relative degree. The relative degree is the number of poles minus the number of zeros. For G1: 1 pole minus 1 zero equals 0. A relative degree of zero means the system is not strictly proper. For G2: 3 poles minus 1 zero equals 2. G2 is strictly proper with relative degree 2. This matters because a system must be proper for physical realizability.

4. 4. Stability

   Both reduced proper transfer functions have all poles strictly in the left half-plane and are BIBO stable.

   Their right-half-plane zero is nonminimum phase. This conclusion concerns these transfer functions, not unspecified hidden states or a newly connected feedback loop.

   Narration transcript

   Part c: check stability. A transfer function is stable if and only if all poles are in the open left half plane, meaning all poles have strictly negative real parts. For G1: the only pole is at s equals negative 0.2. Its real part is negative, so G1 is stable. For G2: the poles are at negative 0.2, and negative 0.4 plus or minus 0.3 j. All three have negative real parts. So G2 is also stable. Note that the zero at positive 0.2 is in the right half plane. This is a non-minimum phase zero, but it does not affect stability. Stability depends only on poles.

5. 5. State-space candidate

   A1 has the required three eigenvalues. The stated A2 and A3 candidates do not supply the required reduced poles.

   A1 is a possible state matrix; a complete realization also needs input, output and feedthrough matrices plus controllability/observability checks.

   Narration transcript

   Part d: which system matrix matches G2? The poles of G2 must be the eigenvalues of the system matrix A. The poles are negative 0.2, negative 0.4 plus 0.3 j, and negative 0.4 minus 0.3 j. Matrix A1 is 3 by 3. Computing its eigenvalues from the characteristic polynomial: det of s I minus A1 gives s plus 0.2 times s squared plus 0.8 s plus 0.25. The eigenvalues are negative 0.2, and negative 0.4 plus or minus 0.3 j. These match the poles of G2 exactly. Matrix A2 is 2 by 2, so it can only have 2 eigenvalues, but G2 has 3 poles. Not suitable. Matrix A3 is 3 by 3 but its eigenvalues are negative 0.2, negative 0.4, and negative 0.3, which do not match. The answer is A1.

6. 6. Review

   Preserve K=−2 in both numerators. Pole locations alone cannot determine the numerator sign.

   Relative degrees are zero and two; both reduced transfer functions are stable; A1 is the compatible state-matrix candidate.

   Narration transcript

   Let us recap. From the pole-zero plots with K equals negative 2: G1 equals negative 2 times s minus 0.2 over s plus 0.2, with relative degree 0. G2 equals 2 times s minus 0.2 over s plus 0.2 times s squared plus 0.8 s plus 0.25, with relative degree 2. Both are stable because all poles have negative real parts. The zero at 0.2 is non-minimum phase but does not affect stability. For the state space match, only A1 has eigenvalues matching G2's poles. Four parts, pole-zero analysis complete, done.

Source video: Control Theory #34 — Pole-Zero Analysis (Worked Example 11) (5:23)