Question 3 of 8: State space ↔ transfer function, Mason’s rule
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, December 2016, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard ζ–overshoot and ζ–resonant-peak design plots are supplied). Questions 1 and 2 are compulsory; a complete paper is five questions, so a candidate chooses three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — Routh–Hurwitz (Ch. 6), root locus (Ch. 8), steady-state error and static error constants (Ch. 7), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10), lead–lag and PID design (Ch. 9–11), state space and Mason’s rule (Ch. 3, 5); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and second-order correlations. All block diagrams, root loci, pole–zero maps, Bode sketches, Nyquist plots and the realization diagrams below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Questions 4, 6 and 7 are read from printed Bode/response plots, so those values carry normal chart-reading tolerance; the s-domain results (Q1, Q2, Q3, Q5, Q8) are exact.
Question 3 — State space ↔ transfer function, Mason’s rule [20]
Part A — realization of a proper (non-strictly-proper) $G(s)$
Given. numerator and denominator both cubic. Find. $A,B,C,D$. Approach. divide out the direct term to expose a strictly-proper remainder, then use the controllable-canonical (phase-variable) form.
Long division. $\dfrac{4s^3+25s^2+45s+34}{s^3+6s^2+10s+8}=4+\dfrac{s^2+5s+2}{s^3+6s^2+10s+8}$, so the direct-feedthrough term is $\boxed{D=4}$.
Controllable-canonical form for the strictly-proper part $\dfrac{s^2+5s+2}{s^3+6s^2+10s+8}$: place the denominator coefficients (with sign) in the bottom feedback row and the numerator coefficients on the output taps: $$A=\begin{bmatrix}0&1&0\\0&0&1\\-8&-10&-6\end{bmatrix},\ B=\begin{bmatrix}0\\0\\1\end{bmatrix},\ C=[\,2\ 5\ 1\,],\ D=4.$$
Check. $C(sI-A)^{-1}B=\dfrac{s^2+5s+2}{s^3+6s^2+10s+8}$, and adding $D=4$ reproduces $G(0)=4.25$ and the full $G(s)$ exactly.
Q3A phase-variable realization: three cascaded integrators, denominator gains $-8,-10,-6$ fed back to the input summer, numerator gains $2,5,1$ tapped to the output, and the direct term $D=4$.
Part B — $G(s)$ from $(A,B,C,D)$ by Mason’s rule
Given. upper-triangular $A$ with eigenvalues $-2,-3,-4$. Find. $G(s)=C(sI-A)^{-1}B$.
Solve $(sI-A)x=B$ by back-substitution (triangular): from the third row $x_3=\dfrac{1}{s+4}$; second row $x_2=\dfrac{2x_3}{s+3}=\dfrac{2}{(s+3)(s+4)}$; first row $x_1=\dfrac{3x_2+2x_3}{s+2}=\dfrac{2s+12}{(s+2)(s+3)(s+4)}$.
Output. $y=Cx=2x_1+3x_2+x_3$. Over the common denominator $(s+2)(s+3)(s+4)$: $2(2s+12)+6(s+2)+(s+2)(s+3)=s^2+15s+42$.
Transfer function. $$\boxed{G(s)=\frac{s^2+15s+42}{(s+2)(s+3)(s+4)}=\frac{s^2+15s+42}{s^3+9s^2+26s+24}.}$$ Mason’s rule on the state diagram (forward paths through each integrator, loop gains $-2/s,-3/s,-4/s$ and the coupling gains $3,2$) gives the identical result; $G(0)=42/24=1.75$.
Q3B controllable-canonical realization of the derived third-order $G(s)$: feedback gains $-24,-26,-9$, output taps $42,15,1$; equivalent to the given triangular $(A,B,C)$ after a similarity transform.
Item
Result
Part A
$D=4$; $A,B,C$ phase-variable with den. $s^3+6s^2+10s+8$, num. $s^2+5s+2$
