Process control (IN) formulas

Holds at every $\omega$ ($\omega$ in rad/s, $T$ in s). The magnitude plot is untouched. The phase lag grows linearly with $\omega$, so on a logarithmic frequency axis it falls away more and more steeply.

$P{M}_0$ is the margin of the delay-free loop and ${\omega }_{gc}$ is its gain-crossover frequency (where $|GH|=1$). This is valid because the delay leaves ${\omega }_{gc}$ where it was. Read stability from $P{M}_T>0$ only when the magnitude crosses 0 dB once and the open loop has no right-half-plane poles.

$P{M}_0$ is the phase margin of the delay-free loop, so $T_{\max }$ is the TOTAL loop delay at which the phase margin reaches zero and the closed loop becomes marginally stable. Convert the margin to radians first. If the loop already contains a delay $T_0$, the extra delay it can tolerate is $T_{\max }-T_0$, which equals (present phase margin in rad)$/{\omega }_{gc}$. Assumes the delay-free closed loop is stable ($P{M}_0>0$) and the magnitude crosses 0 dB only once.

Negative feedback, $K>0$, $T>0$. The gain crossover is at ${\omega }_{gc}=K$, where the phase margin is $\pi /2-KT$ rad. The phase crossover is where $\pi /2+\omega T=\pi$, i.e. ${\omega }_{pc}=\pi /(2T)$, and there $|GH|=2KT/\pi$. Equivalently, the delay margin is $T_{\max }=\pi /(2K)$.

For the loop $K{G}_0(s)e^{-sT}$ under negative feedback, $K>0$. Solve the phase condition (usually by trial) for its lowest root ${\omega }_u$ in rad/s, then read off $K_u$: the gain at which the closed loop is marginally stable and oscillates at ${\omega }_u$. Valid when $G_0$ has no right-half-plane poles and $|G_0(j\omega )|$ decreases with frequency. The Pade example in the next section works one out exactly.

$K$ = steady-state gain, $\tau$ = time constant, $\theta$ = dead time. The ratio $\theta /\tau$ measures how hard the process is to control: the larger it is, the smaller the loop gain $K{K}_c$ at which a proportional loop becomes unstable.

Nothing happens for $\theta$ seconds. After that the response is an ordinary first-order rise, reaching 63.2% of $KA$ at $t=\theta +\tau$.

An all-pass: magnitude exactly 1, phase $-2{\tan }^{-1}(\omega T/2)$. It puts a right-half-plane zero at $s=+2/T$ (non-minimum phase) and a pole at $s=-2/T$. Accurate while $\omega T$ is well below 1.

Also all-pass, with two right-half-plane zeros (a complex pair). It follows the true phase to higher frequencies than the first-order form.

$R$ is the rangeability. The ideal equal-percentage law gives $f=1/R$ at $x=0$, so it never reaches zero; the real valve shuts by seating. The square-root law is one common model of quick opening.

Equal steps of travel change the current flow by equal percentages. The gain is proportional to flow: small near closed, large near fully open.

Ratio of the largest to the smallest flow the valve can control. For the ideal equal-percentage valve, $R=f(1)/f(0)$.

$q$ in US gal/min, $\Delta P$ across the valve in psi, $G$ = specific gravity relative to water. $C_v$ is the water flow in gal/min through the fully open valve ($f=1$) at a 1 psi drop. The metric coefficient $K_v$ uses m³/h and bar, with $C_v\approx 1.156K_v$.

Piston area $A$. Neglects the load-pressure term $K_c{p}_L$ (no load), oil compressibility and leakage. The actuator is then an integrator, so a unity-feedback position loop with a proportional controller is type 1 and has zero steady-state error to a step command.

$K_t$ = torque constant (N·m/A), $K_b$ = back-emf constant (V·s/rad); the two are numerically equal in SI units. $J$ and $B$ are the total inertia and viscous friction referred to the motor shaft. Speed is $\omega (s)=s\theta (s)$.

$K_m$ in (rad/s)/V is the steady-state speed per volt. Back emf shortens the time constant from $J/B$ to ${\tau }_m$.

Armature current is constant and the torque is $T=K_{tf}{i}_f$. With no back-emf feedback there is no electrical damping, and the field time constant $L_f/R_f$ is significant. Used for small motors.

The reference winding has a fixed voltage, and the control-winding voltage $e_c$ is 90° out of phase with it. $K_1$ = stall torque per volt of $e_c$. $K_2$ = magnitude of the slope of the torque–speed line (stall torque ÷ no-load speed), which acts like extra viscous friction. The rotor has a high resistance-to-reactance ratio so that this line slopes downward over the whole speed range.

$m$ = number of stator phases, $N_r$ = rotor teeth, $N_s$ = stator teeth (poles). The two forms agree for the usual designs, where $N_s-N_r=N_s/m$.

Two phases and 50 rotor teeth give $1.8^{\circ }$, i.e. 200 steps per revolution.

$f$ = stepping rate in steps (pulses) per second, $\beta$ in degrees. Half-step drive halves $\beta$ and doubles $S$.

Subscript 1 = outer (primary), 2 = inner (secondary). $G_v$ = valve, $H$ = measuring elements. Convention: negative feedback in both loops, $Y_2=G_{p2}{G}_{v}U$ and $Y_1=G_{p1}{Y}_2$ (the inner variable drives the outer process).

Obtained by clearing the fractions above. Apply Routh to this equation for cascade stability limits.

Obtained by setting the disturbance's total effect $\left(G_d+G_t{G}_{ff}{G}_v{G}_p\right)D$ to zero. The minus sign makes the correction oppose the disturbance.

For $G_p=K_p{e}^{-{\theta }_{p}s}/({\tau }_{p}s+1)$ and $G_d=K_d{e}^{-{\theta }_{d}s}/({\tau }_{d}s+1)$ with $G_t=G_v=1$, the result is a lead–lag plus delay. It is exactly realisable only if ${\theta }_d\ge {\theta }_p$. If ${\theta }_d<{\theta }_p$ it would need a prediction $e^{+sT}$, so the delay term is dropped. A right-half-plane zero in $G_p$ would make the ideal controller unstable, and an approximation is used instead.

$K_R$ is the setting of a ratio station working on normalised (0–1) signals from linear transmitters with spans 0 to $F_{w,\max }$ and 0 to $F_{c,\max }$. With unlinearised orifice (square-law) signals, the setting becomes $K_R^2$.

An input that changes just after the input scan is read only on the next scan and written out at the end of that scan, which takes almost two scans. A pulse shorter than one scan can fall between two input scans and be missed entirely. Input-filter and output-module delays add to both figures.