Modeling — Question 8

Question 8 – Solution

Strategy. The accumulated error supplies an additional state; its equilibrium condition eliminates the temperature offset of proportional feedback.

Step 1: Introduce the controller state. Differentiation gives $$\boxed{T'=-T+P,\quad P'=k(1-T),\quad (T(0),P(0))=(0,0).}$$ At equilibrium, $T=1$ and $P=1$. Heater output depends on accumulated past error, so two states with the same temperature can have different future motion if their current $P$ values differ.

Step 2: Check stability and the critical gain. For $x=T-1$, $u=P-1$, the matrix is $\left(\begin{smallmatrix}-1&1\\-k&0\end{smallmatrix}\right)$. Its roots are $(-1\pm\sqrt{1-4k})/2$. For $0<k<1/4$ both are real and negative; at $k=1/4$ there is a repeated root $-1/2$; for $k>1/4$ they are complex with real part $-1/2$. Every homogeneous mode decays, including the polynomial times exponential at the repeated root. Thus the equilibrium attracts every initial state within this ideal linear model.

Step 3: Solve the critical-gain response. Eliminating $P$ gives $T''+T'+T/4=1/4$, with $T(0)=T'(0)=0$. The repeated-root solution and $P=T'+T$ yield $$\boxed{T=1-(1+t/2)e^{-t/2},\qquad P=1-(1+t/4)e^{-t/2}.}$$ The initial values vanish and $T'=te^{-t/2}/4\ge 0$, $P'=(1/4+t/8)e^{-t/2}>0$. The latter equals $(1-T)/4$, checking the controller equation. The positive subtracted terms and monotonicity imply $0\le T,P<1$ at finite $t$, and both tend to $1$. Here the heater output stays nonnegative without needing a saturation rule.

Step 4: Compare steady errors. Proportional feedback instead gives $T'=k-(1+k)T$, so $T\to k/(1+k)$ and the steady error is $\boxed{1/(1+k)>0}$. The memory controller can retain $P=1$ even when the current error is zero; a proportional controller would then set $P=0$. Real actuator limits and delays can change stability or feasibility, so the ideal conclusions apply under the stated assumptions, not to every physical controller.