Process Control

4% of exam

Process dynamics (first- and second-order processes, gains and time constants, stability, damping, transfer functions), control strategies (feedback, feedforward, cascade, ratio, PID tuning), and control-loop design and hardware (sensors, control valves, interlocks, and conceptual DCS and PLC programming).

3 concepts

Exam areas

A. Dynamics

Process Dynamics: First- and Second-Order Response

Process gain and time constant, the 63.2% first-order step, second-order damping and overshoot, transfer functions, and what makes a loop stable.

Before you can control a process you must know how it ·moves· — how fast a tank temperature climbs when you open a steam valve, whether a level loop will overshoot and ring, how much dead time a long pipe adds. The FE handbook packs this into two compact models on the Instrumentation, Measurement, and Control pages: the first-order lag

K/(\tau s + 1)

and the standard second-order form. Almost every Dynamics question is a substitution into one of them, but only if you can read a transfer function, identify

K

\tau

\zeta

, and

\omega_n

from it, and translate those into a time response. Points are lost when candidates confuse the gain with the time constant, forget that the step response is measured in ·deviation· from the initial value, or mix up which

\zeta

band rings and which crawls. This concept builds the fluency the handbook assumes (see 'Control Systems', First- and Second-Order Control System Models).

Process gain and time constant

Two numbers characterize a self-regulating first-order process. The steady-state (process) gain

K

is how far the output ultimately moves per unit of sustained input change —

K = \Delta y_{\infty}/\Delta u

— and it carries the units of output over input (e.g.

^\circ\text{C}

per % valve opening). The time constant

\tau

is how ·fast· it gets there: the time to traverse

63.2\%

of the total change. A bigger

\tau

means a more sluggish process. Gain sets the destination; time constant sets the pace. Keep them straight —

K

has process units,

\tau

has units of time only.

K = \dfrac{\Delta y_{\infty}}{\Delta u}, \qquad \tau = \text{time to reach } 63.2\% \text{ of the final change}

The first-order step response

A first-order system

Y(s)/R(s) = K/(\tau s + 1)

driven by a step of magnitude

M

responds with a single rising exponential. Working in deviation variables (so the pre-step value

y_0 = 0

, the handbook's convention), the output approaches its new steady state

KM

asymptotically. The landmark numbers are worth memorizing: at

t=\tau

the response is

63.2\%

complete, at

2\tau

it is

86.5\%

, at

3\tau

it is

95.0\%

, at

4\tau

it is

98.2\%

, and at

5\tau

it is

99.3\%

— for engineering purposes a first-order process has effectively settled after about

5\tau

. The slope is steepest at

t=0

, and the initial tangent would reach the final value in exactly one

\tau

y(t) = y_0 e^{-t/\tau} + KM\left(1 - e^{-t/\tau}\right)

Transfer functions, poles, and the dc gain

A transfer function is the ratio of Laplace transforms,

G(s) = Y(s)/X(s)

, written as a ratio of polynomials whose denominator roots are the ·poles· and numerator roots the ·zeros·. The poles govern the character of the response: a first-order pole sits at

s = -1/\tau

, and the farther left (more negative) it lies, the faster the mode decays. The steady-state or dc gain falls straight out by the Final Value Theorem — set

s \to 0

— provided every pole has a negative real part. For the first-order form,

\lim_{s\to 0} K/(\tau s + 1) = K

, which is exactly the process gain.

\text{dc gain} = \lim_{s\to 0} G(s), \qquad \text{first-order pole at } s = -\dfrac{1}{\tau}

Second-order systems: natural frequency and damping

Add a second energy-storage element (a thermowell on a sensor, an underdamped flow loop, a controller closing a loop on a first-order plant) and you get the standard second-order model. Two parameters now set the shape: the undamped natural frequency

\omega_n

(how fast it ·would· oscillate with no damping) and the damping ratio

\zeta

(how much that oscillation is suppressed). The damping ratio sorts behavior into three regimes:

\zeta > 1

is overdamped (two real poles, sluggish, no overshoot),

\zeta = 1

is critically damped (fastest non-oscillatory), and

0 < \zeta < 1

is underdamped (complex poles, the response overshoots and rings). The chemical-process literature often rewrites the same model with

\tau = 1/\omega_n

K/(\tau^2 s^2 + 2\zeta\tau s + 1)

\dfrac{Y(s)}{R(s)} = \dfrac{K\,\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

Overshoot, oscillation, and settling

For an underdamped step response the damped frequency is

\omega_d = \omega_n\sqrt{1-\zeta^2}

— always lower than the natural frequency. The first peak arrives at

t_p = \pi/\omega_d

, and the fractional overshoot above the final value depends ·only· on

\zeta

, not on

\omega_n

: lower damping means taller overshoot. The handbook's

2\%

settling time,

T_s = 4/(\zeta\omega_n)

, comes from the decay envelope

e^{-\zeta\omega_n t}

. Note the split of roles:

\zeta

alone fixes the overshoot and the shape, while

\omega_n

scales the whole response in time.

\%\text{OS} = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}, \qquad t_p = \dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}, \qquad T_s = \dfrac{4}{\zeta\omega_n}

Dead time (transport lag)

Real processes rarely respond the instant you move a valve — fluid must travel down a pipe, a sample must reach an analyzer. That pure delay

\theta

is dead time, and it multiplies the transfer function by

e^{-\theta s}

. In the time domain the entire first-order curve simply starts

\theta

seconds later. Dead time is the controllability villain: it adds phase lag without adding gain, so a loop with a large

\theta/\tau

ratio is much harder to control and must be detuned. Recognizing

e^{-\theta s}

as a delay (not a pole) is a frequent exam discriminator.

\dfrac{Y(s)}{R(s)} = \dfrac{K e^{-\theta s}}{\tau s + 1}, \qquad y(t) = KM\left(1 - e^{-(t-\theta)/\tau}\right) \text{ for } t \ge \theta

Stability in one sentence

An open-loop process whose transfer function has all poles in the left half plane is stable: every mode

e^{-t/\tau}

e^{-\zeta\omega_n t}

decays. A pole on the imaginary axis (

\zeta = 0

) sustains oscillation; a pole in the right half plane (which never happens for these self-regulating models alone, but can appear once a controller is added) grows without bound. Stability of the ·closed· loop is the subject of the next concept, but the rule is the same currency — left-half-plane poles, bounded response.

\text{Stable} \iff \operatorname{Re}(p_i) < 0 \ \text{ for every pole } p_i

Exam strategy

Read the transfer function first and tag every parameter: the constant term in the denominator must be

1

to read

K

directly off the numerator (divide through if it is not), then

\tau

is the coefficient of

s

. For second-order, match against

s^2 + 2\zeta\omega_n s + \omega_n^2

—

\omega_n

is the square root of the constant term and

\zeta

comes from the

s

-coefficient

= 2\zeta\omega_n

. Memorize the percent-complete ladder (

63/86.5/95/98.2/99.3

1

–

5\,\tau

) so first-order time questions need no calculator. For overshoot, remember it is a function of

\zeta

only. Always work step responses in deviation variables, then add the initial value back at the end. Watch the units of

K

— they are never just a number unless input and output share units.

Key equations

First-order transfer function

\dfrac{Y(s)}{R(s)} = \dfrac{K}{\tau s + 1}

$K$ = steady-state gain (output units / input units); $\tau$ = time constant (s). Single pole at $s=-1/\tau$ .

First-order step response

y(t) = y_0 e^{-t/\tau} + KM\left(1 - e^{-t/\tau}\right)

Response to a step of magnitude $M$ . In deviation variables $y_0=0$ . Reaches $63.2\%$ of $KM$ at $t=\tau$ .

Percent-complete landmarks

63.2,\ 86.5,\ 95.0,\ 98.2,\ 99.3\,\% \text{ at } t = 1,2,3,4,5\,\tau

Fraction of the final change reached after each time constant; practically settled by $5\tau$ .

First-order plus dead time (FOPDT)

\dfrac{Y(s)}{R(s)} = \dfrac{K e^{-\theta s}}{\tau s + 1}

$\theta$ = dead time (transport lag, s). The step response is delayed by $\theta$ before the exponential begins.

Second-order standard form

\dfrac{Y(s)}{R(s)} = \dfrac{K\,\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

$\omega_n$ = undamped natural frequency (rad/s); $\zeta$ = damping ratio (dimensionless); $K$ = steady-state gain.

Second-order (process form)

\dfrac{Y(s)}{R(s)} = \dfrac{K}{\tau^2 s^2 + 2\zeta\tau s + 1}

Equivalent form with $\tau = 1/\omega_n$ . Common in the chemical-process literature.

Damping classification

\zeta > 1\ \text{(over)}, \quad \zeta = 1\ \text{(critical)}, \quad 0<\zeta<1\ \text{(under)}

Only the underdamped band overshoots and oscillates; critically damped is fastest with no overshoot.

Damped natural frequency

\omega_d = \omega_n\sqrt{1-\zeta^2}

Actual ringing frequency of an underdamped response (rad/s); always less than $\omega_n$ .

Percent overshoot

\%\text{OS} = 100\,\exp\!\left(\dfrac{-\zeta\pi}{\sqrt{1-\zeta^2}}\right)

Overshoot above the final value for a unit step; depends on $\zeta$ only, not $\omega_n$ .

Peak time

t_p = \dfrac{\pi}{\omega_d} = \dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}

Time to the first (largest) peak of an underdamped step response (s).

Two-percent settling time

T_s = \dfrac{4}{\zeta\omega_n}

Time for the envelope $e^{-\zeta\omega_n t}$ to decay within $2\%$ of final value (s).

DC gain (Final Value Theorem)

\text{dc gain} = \lim_{s\to 0} G(s)

Steady-state gain, valid only when all poles have negative real parts. Equals $K$ for the first-order form.

Worked examples

First-order sensor step response

Problem. A bare thermocouple behaves as a first-order system with time constant $\tau = 8\,\text{s}$ . It is suddenly plunged from a $25\,^\circ\text{C}$ bath into an $80\,^\circ\text{C}$ bath. (a) What does it read $12\,\text{s}$ after immersion? (b) How long until it reads $75\,^\circ\text{C}$ ?

Solution. Treat the bath change as a step. The total change is

\Delta y_\infty = KM = 80 - 25 = 55\,^\circ\text{C}

, with initial value

y_0 = 25\,^\circ\text{C}

. (a)

y(12) = 25 + 55\left(1 - e^{-12/8}\right) = 25 + 55\left(1 - e^{-1.5}\right) = 25 + 55(0.7769) = 67.7\,^\circ\text{C}

. (b) Solve

75 = 25 + 55\left(1 - e^{-t/8}\right)

\,\frac{50}{55} = 0.9091 = 1 - e^{-t/8}

, so

e^{-t/8} = 0.0909

, giving

t = 8\ln(11) = 8(2.398) = 19.2\,\text{s}

. **Answers: (a)

67.7\,^\circ\text{C}

, (b)

19.2\,\text{s}

.** Sanity: at one time constant (

t=8\,\text{s}

) the reading is

25 + 0.632(55) = 59.8\,^\circ\text{C}

; at

12\,\text{s}

(1.5

\tau

) it should be somewhat higher, and

67.7\,^\circ\text{C}

fits. Reaching

75\,^\circ\text{C}

(91% of the change, between

2\tau

and

3\tau

) at

19.2\,\text{s}

is consistent.

y(12) = 25 + 55\left(1 - e^{-12/8}\right) = 67.7\,^\circ\text{C}

Second-order overshoot and settling

Problem. A closed level loop is modeled as second order with $K = 2$ , $\omega_n = 4\,\text{rad/s}$ , and $\zeta = 0.3$ . Find the damped frequency, the percent overshoot, the peak time, and the $2\%$ settling time for a step change.

Solution. Damped frequency:

\omega_d = \omega_n\sqrt{1-\zeta^2} = 4\sqrt{1 - 0.09} = 4(0.9539) = 3.82\,\text{rad/s}

. Overshoot:

\%\text{OS} = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 100\,e^{-(0.3)(3.1416)/0.9539} = 100\,e^{-0.988} = 37.2\%

. Peak time:

t_p = \pi/\omega_d = 3.1416/3.8158 = 0.823\,\text{s}

. Settling time:

T_s = 4/(\zeta\omega_n) = 4/(0.3\times 4) = 3.33\,\text{s}

. **Answers:

\omega_d = 3.82\,\text{rad/s}

\%\text{OS} = 37.2\%

t_p = 0.823\,\text{s}

T_s = 3.33\,\text{s}

.** Sanity:

\zeta = 0.3

is lightly damped, so a large overshoot near

35

–

40\%

and several rings before settling (≈

T_s/t_p \approx 4

oscillation half-cycles, i.e. about

2

overshoot peaks) is exactly what we expect.

\%\text{OS} = 100\,e^{-(0.3)\pi/\sqrt{1-0.3^2}} = 37.2\%

First-order plus dead time

Problem. A temperature process is identified as FOPDT with gain $K = 2\,^\circ\text{C}/\%$ , time constant $\tau = 25\,\text{s}$ , and dead time $\theta = 5\,\text{s}$ . Starting from a steady output of $50\,^\circ\text{C}$ , the valve is stepped open by $\Delta u = 10\%$ . (a) What is the output $20\,\text{s}$ after the step? (b) When does the output first reach $64\,^\circ\text{C}$ ?

Solution. Final change

= KM = (2)(10) = 20\,^\circ\text{C}

, so the output ultimately rises from

50

70\,^\circ\text{C}

. Nothing happens during the dead time, so use

t-\theta

in the exponential. (a) At

t = 20\,\text{s}

t-\theta = 15\,\text{s}

\ y = 50 + 20\left(1 - e^{-15/25}\right) = 50 + 20\left(1 - e^{-0.6}\right) = 50 + 20(0.4512) = 59.0\,^\circ\text{C}

. (b) Set

64 = 50 + 20\left(1 - e^{-(t-\theta)/25}\right)

\frac{14}{20} = 0.70 = 1 - e^{-(t-5)/25}

, so

e^{-(t-5)/25} = 0.30

, giving

t - 5 = 25\ln(10/3) = 25(1.204) = 30.1\,\text{s}

, thus

t = 35.1\,\text{s}

. **Answers: (a)

59.0\,^\circ\text{C}

, (b)

35.1\,\text{s}

.** Sanity: at

20\,\text{s}

only

15\,\text{s}

of effective time has elapsed (<

1\tau

), so we expect just under half the change — about

45\%

(

9.0

20\,^\circ\text{C}

), confirmed. Ignoring the dead time would have under-predicted the time to

64\,^\circ\text{C}

by exactly

5\,\text{s}

y(20) = 50 + 20\left(1 - e^{-(20-5)/25}\right) = 59.0\,^\circ\text{C}

Common pitfalls

•Confusing gain and time constant: $K$ sets how far the output moves (with process units), $\tau$ sets how fast (units of time only). They are independent.
•Reading $K$ and $\tau$ off an un-normalized transfer function. The denominator constant term must be $1$ to read $\tau$ as the $s$ -coefficient and $K$ off the numerator — divide through first.
•Forgetting deviation variables: the step response formula gives the change from the initial value. Add the pre-step value back to report an absolute temperature/level.
•Treating dead time $e^{-\theta s}$ as an extra pole. It is a pure delay that shifts the whole curve by $\theta$ ; it adds phase lag but no extra lag dynamics.
•Thinking overshoot depends on $\omega_n$ . Percent overshoot is a function of $\zeta$ only; $\omega_n$ scales the time axis (peak and settling time), not the height.
•Using $\omega_n$ where $\omega_d$ is required. The actual ringing frequency and peak time use $\omega_d = \omega_n\sqrt{1-\zeta^2}$ , which is smaller.
•Assuming a process has settled at $t=\tau$ . One time constant is only $63.2\%$ complete; full practical settling needs about $5\tau$ (or $T_s = 4/\zeta\omega_n$ for second order).

References

NCEES FE Reference Handbook — Instrumentation, Measurement, and Control (Control Systems; First- and Second-Order Control System Models)
Seborg, Edgar, Mellichamp & Doyle — Process Dynamics and Control — first- and second-order response, FOPDT identification
Coughanowr & LeBlanc — Process Systems Analysis and Control — transient response of standard transfer functions

B. Control strategies

Control Strategies, PID and Stability

Feedback, feedforward, cascade and ratio control; P/I/D action and offset; block-diagram algebra; closed-loop stability and qualitative PID tuning.

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C. Control loop design and hardware

Control Loop Hardware: Sensors, Valves and DCS/PLC

Temperature/pressure/level/flow/pH sensors and transmitters, control-valve sizing and characteristics, final control elements, interlocks, and DCS/PLC roles.

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