Process Dynamics and Control
1. Mathematical Preliminaries and Dynamic Response
Linearization of nonlinear models
\[ A\frac{dh}{dt}=q_i-\beta\sqrt h,\qquad h(0)=h_s \]
Steady state: \(0=q_{i,s}-\beta\sqrt{h_s}\). Taylor-expand \(\sqrt h\) about \(h_s\): \[ A\frac{dh}{dt}\simeq q_i-\beta\left[\sqrt{h_s}+\frac{1}{2\sqrt{h_s}}(h-h_s)\right] \]
Subtract the steady state and use \(\bar q_i=q_i-q_{i,s}\), \(\bar h=h-h_s\) FIX: \[ A\frac{d\bar h}{dt}+\frac{\beta}{2\sqrt{h_s}}\bar h=\bar q_i,\qquad \bar h(0)=0 \]
General case (one state \(y\), inputs \(\mathbf u\)): with \(y'=y-y_s\), \(\mathbf u'=\mathbf u-\mathbf u_s\), \[ \frac{dy'}{dt}=\left.\frac{\partial f}{\partial y}\right|_s y'+\nabla_{\mathbf u}^Tf\big|_s\,\mathbf u' \] The model is accurate only near the operating point.
Inputs common to control problems
- Pulse: \(A[u(t)-u(t-T)]\ \Rightarrow\ \dfrac As(1-e^{-sT})\)
- Impulse: \(\delta(t)=\displaystyle\lim_{T\to0}\frac{u(t)-u(t-T)}{T}\), so \(\mathcal L[\delta]=\displaystyle\lim_{T\to0}\frac{1-e^{-sT}}{Ts}=1\)
- \(\displaystyle\int_{-\infty}^\infty\delta\,dt=1\), \(\quad\displaystyle\int_{-\infty}^\infty f(t)\delta(t-t_0)\,dt=f(t_0)\)
- Dead time (time shift) multiplies by \(e^{-st_0}\): an exponential, not a polynomial, in \(s\) (§1.7).
Example 1.5 Case 1: distinct real roots
\(s^3+s^2-4s-4=(s+1)(s+2)(s-2)\). Two numerators, same denominator:
| \(\dfrac{6s^2-12}{(s+1)(s+2)(s-2)}\) | \(\dfrac{6s}{(s+1)(s+2)(s-2)}\) | |
|---|---|---|
| \(\alpha_1=\left.\dfrac{q}{(s+2)(s-2)}\right\vert_{-1}\) | \(\dfrac{-6}{-3}=2\) | \(\dfrac{-6}{-3}=2\) |
| \(\alpha_2=\left.\dfrac{q}{(s+1)(s-2)}\right\vert_{-2}\) | \(\dfrac{12}{4}=3\) | \(\dfrac{-12}{4}=-3\) |
| \(\alpha_3=\left.\dfrac{q}{(s+1)(s+2)}\right\vert_{2}\) | \(\dfrac{12}{12}=1\) | \(\dfrac{12}{12}=1\) |
| \(f(t)\) | \(2e^{-t}+3e^{-2t}+e^{2t}\) | \(2e^{-t}-3e^{-2t}+e^{2t}\) |
The same procedure gives \(\dfrac{6}{(s+1)(s+2)(s+3)}=\dfrac3{s+1}-\dfrac6{s+2}+\dfrac3{s+3}\).
Poles tell the shape of the response
- real pole \(p\) → \(e^{pt}\); repeated \(m\) times → \((\alpha_1+\cdots+\alpha_mt^{m-1})e^{pt}\)
- complex pair \(-\zeta/\tau\pm j\sqrt{1-\zeta^2}/\tau\) → \(e^{-\zeta t/\tau}\sin(\sqrt{1-\zeta^2}\,t/\tau+\phi)\)
- the discussion of poles is independent of the input; the actual \(y(t)\) also depends on \(x(t)\)
- zeros change only the coefficients
- Stability: all poles must have negative real parts (Ch 4).
Effect of \(\tau\) and \(K\)
The larger the time constant, the slower the response. The larger the gain, the more sensitive the output to the input.
Ramp input and the integrating process
Ramp \(x=at\), \(X=a/s^2\): \[ Y=\frac{Ka}{s^2(\tau s+1)}=\frac{Ka\tau^2}{\tau s+1}-\frac{Ka\tau}{s}+\frac{Ka}{s^2} \] \[ y(t)=Ka\tau\left(e^{-t/\tau}-1\right)+Kat \]
\(\to Ka(t-\tau)\) for \(t\gg\tau\): the output lags the input by \(\tau\).
Integrating process (\(a_0=0\)): \[ \frac{dy}{dt}=K_Ix(t)\ \Rightarrow\ G(s)=\frac{K_I}{s} \] a pole at the origin.
\(A\dfrac{d\bar h}{dt}=\bar q_{in}-\bar q\ \Rightarrow\ H(s)=\dfrac1A\dfrac{Q_{in}(s)-Q(s)}{s}\)
A step input gives a ramp; an impulse is not forgotten, the level keeps whatever was added. Not self-regulating: levels need control.
Example 1.8 Two stirred tanks in series
\(\tau_1\dfrac{d\bar c_1}{dt}+\bar c_1=\bar c_i,\quad \tau_2\dfrac{d\bar c_2}{dt}+\bar c_2=\bar c_1\) \[ \frac{C_2(s)}{C_i(s)}=\frac{1}{(\tau_1s+1)(\tau_2s+1)} \]
\(\tau_1=V_1/q=2\) min, \(\ \tau_2=V_2/q=1.5\) min
Feed: \(c_i=6\) for \(0<t<0.25\) min, otherwise 1, so \(\bar c_i\) is a pulse of height 5: \[ C_i(s)=\frac5s\left(1-e^{-s/4}\right)\quad\text{or, as an impulse,}\quad C_i(s)=\frac54 \] \[ C_2(s)=\frac{5(1-e^{-s/4})}{s(2s+1)(1.5s+1)}\quad\text{or}\quad\frac{5/4}{(2s+1)(1.5s+1)} \]
The original MATLAB script used pulselength = 1.25, a pulse 5× too long. FIX
Step response of a second-order model
\(\zeta>1\): \(\ y=KM\left[1-e^{-\zeta t/\tau}\left(\cosh\dfrac{\sqrt{\zeta^2-1}}{\tau}t+\dfrac{\zeta}{\sqrt{\zeta^2-1}}\sinh\dfrac{\sqrt{\zeta^2-1}}{\tau}t\right)\right]\) (simpler with \(e^{-t/\tau_1}\), \(e^{-t/\tau_2}\))
\(\zeta=1\): \(\ y=KM\left[1-\left(1+\dfrac t\tau\right)e^{-t/\tau}\right]\)
\(0\le\zeta<1\): \(\ y=KM\left[1-\dfrac{e^{-\zeta t/\tau}}{\sqrt{1-\zeta^2}}\sin\left(\dfrac{\sqrt{1-\zeta^2}}{\tau}t+\phi\right)\right]\), \(\ \phi=\tan^{-1}\dfrac{\sqrt{1-\zeta^2}}{\zeta}\)
Rarely used directly; the underdamped one yields the characteristics on the next slide.
Features of the underdamped step response
| overshoot | peak time | rise time | period | settling time | decay ratio |
|---|---|---|---|---|---|
| \(\dfrac ab=\exp\left(\dfrac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right)\) | \(t_p=\dfrac{\pi\tau}{\sqrt{1-\zeta^2}}\) | \(t_r=\dfrac{\tau(\pi-\cos^{-1}\zeta)}{\sqrt{1-\zeta^2}}\) | \(P=\dfrac{2\pi\tau}{\sqrt{1-\zeta^2}}\) | \(t_s\approx\) \(\dfrac{3\tau}{\zeta}\) (5%) \(\dfrac{4\tau}{\zeta}\) (2%) |
\(\dfrac ca=\text{OS}^2\) |
Settling: the envelope \(e^{-\zeta t/\tau}\) behaves like a first-order term with time constant \(\tau/\zeta\). Smaller \(\zeta\): faster rise, larger overshoot, longer settling.
Example 1.10 Padé approximation of \(e^{-3s}/(10s+1)\)
\[ \frac{Y}{X}\approx\frac{-1.5s+1}{(10s+1)(1.5s+1)} \]
Far from \(t=0\) both approximations are excellent; at early times 1/1 dips to \(-0.04\), and 2/2 wiggles around zero.
Non-interacting tanks in series
\(A_i\dfrac{dh_i}{dt}=q_{i-1}-q_i,\ q_i=\dfrac{h_i}{R_i}\) (linear resistance): \[ \frac{H_i(s)}{H_{i-1}(s)}=\frac{K_i}{\tau_is+1},\quad \tau_i=A_iR_i,\ K_i=\frac{R_i}{R_{i-1}} \] \[ \frac{H_n(s)}{Q_1(s)}=\frac{R_n}{(\tau_1s+1)\cdots(\tau_ns+1)} \]
The more tanks, the longer the \(n\)-th tank waits to “see” a change: the response looks like a delay followed by a lag.
Example 1.11 Low-order approximation with dead time
Keep the dominant (large) time constants and lump the small ones into a dead time: \[ G(s)\approx\frac{Ke^{-t_ds}}{(\tau_1s+1)(\tau_2s+1)},\quad t_d\approx\sum_{i\ne1,2}\tau_i \]
\[ \frac YX=\frac{3}{(0.1s+1)(0.5s+1)(s+1)(3s+1)} \]
- FOPDT: \(\dfrac{3e^{-1.6s}}{3s+1}\)
- SOPDT: \(\dfrac{3e^{-0.6s}}{(s+1)(3s+1)}\)
Half rule (Skogestad) NEW: split the largest neglected time constant, half to the lag, half to the delay, e.g. \(\dfrac{3e^{-0.35s}}{(1.25s+1)(3s+1)}\).
The second-order approximation is clearly better (IAE 0.14 vs 0.70). FOPDT models are the basis of the empirical tuning rules of Ch 3.
Interacting tanks
\(A_1\dfrac{dh_1}{dt}=q_i-\dfrac{h_1-h_2}{R_1},\quad A_2\dfrac{dh_2}{dt}=\dfrac{h_1-h_2}{R_1}-\dfrac{h_2}{R_2}\) \[ \frac{H_1}{Q_i}=\frac{R_1\tau_2s+R_1+R_2}{p(s)},\qquad \frac{H_2}{Q_i}=\frac{R_2}{p(s)} \] \[ p(s)=\tau_1\tau_2s^2+\left(\tau_1+\tau_2+\tau_1\frac{R_2}{R_1}\right)s+1 \]
- Even the first tank is second order (with a zero).
- Interaction adds \(\tau_1R_2/R_1\) to the middle coefficient; the poles stay real and negative.
Which damps inlet-flow disturbances better? The interacting pair: the extra term pushes one pole closer to the origin, so the response is slower and smoother. NEW
Lead-lag element
Poles govern the inherent dynamics; zeros add finer “fingerprints” by changing the coefficients. \[ \frac{Y}{X}=\frac{\tau_zs+1}{\tau s+1}\ \xrightarrow{X=1/s}\ Y=\frac1s+\frac{\tau_z-\tau}{\tau s+1} \] \[ y(t)=1-\left(1-\frac{\tau_z}{\tau}\right)e^{-t/\tau},\qquad y(0^+)=\frac{\tau_z}{\tau} \]
- the exponential still comes from the pole \(-1/\tau\)
- \(\tau_z>\tau\): jump above 1, overshoot
- \(0<\tau_z<\tau\): jump, then rise
- \(\tau_z<0\) (RHP zero): inverse response
- \(\tau_z=\tau\): pole-zero cancellation, \(y=1\)
Transfer functions in parallel
Two competing effects with different time constants: \[ \frac{Y}{X}=\frac{K_1}{\tau_1s+1}+\frac{K_2}{\tau_2s+1}=\frac{K(\tau_zs+1)}{(\tau_1s+1)(\tau_2s+1)} \] \(K=K_1+K_2,\quad \tau_z=\dfrac{K_1\tau_2+K_2\tau_1}{K_1+K_2}\)
With opposing effects (\(K_2<0\)), superposition gives a rough picture right away.
Interactive: poles, zeros, dead time NEW
Interactive version: seoultechpse.github.io/ProCon/ch1/ch1_dynamics.html — first or second order, \(K\), \(\tau\), \(\zeta\), zero \(\tau_z\), dead time \(\theta\).
