Preview of the most important points required for dynamic process modelling
LaPlace transforms, Transfer Function Diagrams, Response to changes in input variable
7. Degrees of Freedom Analysis
Structured Approach
1. List all quantities in the model that are known constants (or
parameters that can be specified) on the basis of equipment
dimensions, known physical properties, and so on.
2. Determine the number of equations NE and the number of
process variables, NV. Note that time t is not considered to be
a process variable, because it is neither an input nor an
output.
3. Calculate the number of degrees of freedom, š š¹ (Eq. 2-27)
4. Identify the NE output variables (include dependent variables
in the ODEs) obtained by solving the process model.
5. Identify the š š¹ input variables that must be specified as
either disturbance variables (DVs) or manipulated variables
(MVs). 7Rami Bechara
8. Chapter 3
Definition of LaPlace Transform
ā¢ Definition of Laplace Transform
ā¢ F(s): symbol for the Laplace transform
ā¢ s : complex independent variable
ā¢ f(t) : function of time to be transformed
ā¢ operator, defined by the integral.
ā¢ f(t) must satisfy mild conditions that include mainly
being piecewise continuous for 0 < š” < ā
ā Requirement almost always holds for functions that are
useful in process modeling and control
8Rami Bechara
9. Properties of LaPlace Tranform
ā¢ Inverse Laplace transform: operates on the
function F(s) and converts it to f(t).
ā¢ F(s) contains no information about f(t) for t <
0.
ā¢ ļ not defined for t < 0
ā¢ Linearity of LaPlace Transform
9Rami Bechara
10. ā¢ The asymptotic value of y(t) for large values of
time y(ā) can be found
ā¢ if lim
š ā0
š š š exists for all š š(š ) ā„ 0 and has
a limit for a real value for every s ā„ 0
ā¢ Proved using derivative
Final Value Theorem
Sļ 0
10Rami Bechara
11. Initial Value Theorem
ā¢ Similar to Final Value Theorem
ā¢ Conditions and development similar to that of
the Final Value Theorem
ā¢ Both theorems are useful for checking
mathematical errors that may occur in
obtaining Laplace transform solutions.
11Rami Bechara
15. Laplace Transforms
Rectangular Pulse Function - Definition
ā¢ The pulse has height ā and width
š” š¤
ā¢ This type of signal can depict the
opening and closing of a valve
regulating flow into a tank.
ā¢ The flow rate would be held at h
for a duration of š” š¤ units of time.
ā¢ The area under the curve could
be interpreted as the amount of
material delivered to the tank (=
āš” š¤)
ā¢ For a unit rectangular pulse,
ā =
1
š” š¤
; area under pulse = unity.
15Rami Bechara
17. Laplace Transforms
Impulse or Dirac delta Function
ā¢ Limiting case of the unit rectangular.
symbol Ī“(t)
ā¢ Obtained when š” š¤ ā 0 with area under
pulse =1
ā¢ ļ Pulse of infinite height and infinitesimal
width
ā¢ Mathematically accomplished by
substituting ā =
1
š” š¤
in Eq. 3-15
ā¢ ļ Laplace transform of Ī“(t)
Evaluated by applying
LāHospitalās rule
ā¢ š” š¤ā = š ļ
š ā 0
17Rami Bechara
18. Partial Fraction Expansion
Heaviside Method - General Form
ā¢ Conditions: factors are real and distinct (no complex
or repeated factors appear), the following expansion
formula applies:
ā¢ Alternative Expression
ā¢ The denominator D(s), an nth-order polynomial, is
denoted as the characteristic polynomial.
ā¢ The numerator N(s) has a maximum order of n ā 1. 18Rami Bechara
19. General Procedure for Solving
Differential Equations
ā¢ Laplace transforms are
used as an intermediate
step.
ā¢ Step 3 can be bypassed
if the transform found in
Step 2 matches an entry
in Table 3.1.
ā¢ In order to factor D(s) in
Step 3, software such as
MATLAB, Mathematica,
or Mathcad can be used
19Rami Bechara
20. Expression of Time Delay
ā¢ Liquid velocity v = 1 m/s
ā¢ Time delay (š = šæ/š£) =10 s.
ā¢ f(t) = 1st sensor transient
temperature response
ā¢ š š(t) = 2nd sensor temperature
response
ā¢ š š = 0 for t < š.
ā¢ Therefore, š š and f are related by
=0, t < š
=1, t > š 20Rami Bechara
21. Time Delay
La Place Transform
ā¢ La Place expansion
ā¢ (t ā Īø) is now the artificial variable of
integration, it can be replaced by tā.
ā¢ Real Translation Theorem
21Rami Bechara
23. Chapter 4 Transfer Functions
ā¢ The time-domain model that relates u and y is
an ODE.
ā¢ For a linear ODE, there is an equivalent model
in the Laplace domain, the TF model
Input/ Output
Model
23Rami Bechara
27. Linearization method
ā¢ Suppose a nonlinear dynamic model derived
from first principles y output
u input
ā¢ Linear approximation of this equation can be
obtained by using a Taylor series expansion
and truncating after the first-order terms.
ā¢ Final expression
Deviation Variables
27Rami Bechara
28. TF reminder Blending Process
ā¢ Blending case
ā¢ 1st order TF:
ā¢ TF advantages:
ā¢ Makes it easy to compare effects of different inputs
ā¢ The dynamic behavior of a given process can be
generalized easily.
ā Once the process response to an input change is analyzed,
the response of any other process described by the same
generic transfer function is then known.
28Rami Bechara
29. Chapter 5
1st order TF
ā¢ General Expression
ā¢ An analytical time-domain solution can be
found once the nature of the input change is
specified
ā¢ Solution can be applied to multiple cases:
blending tanks and liquid surge tanks
ā¢ Another benefit: not necessary to re-solve the
ODE when K, Ļ, or U(s) change.
29Rami Bechara
30. 1- Step Input
ā¢ Sudden and sustained input changes
ā¢ Reactor feedstock may be changed quickly
from one supply to another
ā¢ Causing a corresponding change in important
input variables such as feed concentration and
feed temperature
ā¢ Best approximated by the step change
t=0 zero time / time of step change
When M Step Magnitude occurs
us: deviation variable 30Rami Bechara
31. Step Input Example
ā¢ Heat input to a stirred-tank heating unit
suddenly changes from 8000 to 10,000 kcal/h
ā¢ S(t) Unit step function, Qā deviation variable
ā¢ La Place Transform
31Rami Bechara
32. 2- Ramp Input
Functions
ā¢ Time-Domain function
ā¢ La Place Transform : (From the famous Table
3.1)
uR: deviation variable
a slope
32Rami Bechara
33. 3- Rectangular Pulse Input
ā¢ Processes sometimes are subjected to a
sudden step change that then returns to its
original value.
ā¢ Example : a feed to a reactor is shut off for a
certain period of time or a natural-gas-fired
furnace experiences a brief interruption in fuel
gas
ā¢ Approximate Equation
33Rami Bechara
34. 3- Rectangular Pulse Input
Equations
ā¢ š” š¤ : pulse width - can range from very short
(approximation to an impulse) to very long.
ā¢ Alternative Equation
ā¢ š(š”) = 0 (š” < 0) ššš = 1 (š” ā„ 0)
ā¢ š š” ā š” š¤ , shifted unit step input, equal to 1
for š” ā„ š” š¤ and equal to zero for š” < š” š¤.
ā¢ Equation
ā¢ La Place Transform
34Rami Bechara
36. 4-Sinusoidal Input
ā¢ Inputs that vary periodically.
ā¢ Example: the drift in cooling water temperature
discussed earlier often tied to diurnal (day-to-
night-to-day) fluctuations in ambient conditions.
ā¢ Cyclic process changes within a 24-h period often
caused by variations in cooling water T
approximated as a sinusoidal function:
A amplitude of the sinusoidal function
Ļ angular frequency (Ļ in radians/time).
Related to the period P by P = 2Ļ/Ļ
36Rami Bechara
37. 4-Sinusoidal Input
Equations
ā¢ On a shorter time scale, high-frequency
disturbances are associated with mixing and
pumping operations, and with 60-Hz electrical
noise arising from AC electrical equipment and
instrumentation.
ā¢ Sinusoidal inputs are particularly important,
because they play a central role in frequency
response analysis ( Chapter 14).
ā¢ The Laplace transform obtained by multiplying
entry 14 in Table 3.1 by the amplitude A to obtain
37Rami Bechara
38. 5- Impulse Input
ā¢ (Chap.3) Simplest Laplace transform:
ā¢
Exact impulse functions are not encountered in normal
plant operations.
ā¢ To obtain an impulse input, it is necessary to inject a
finite amount of energy or material into a process in an
infinitesimal length of time, which is not possible.
ā¢ However, this type of input can be approximated
through the injection of a concentrated dye or other
tracer into the process
ā¢ We have an example though (current cuts)
38Rami Bechara
39. 6- Random Inputs
ā¢ Many process inputs change in such a complex manner that it is not
possible to describe them as deterministic functions of time.
ā¢ If an input u exhibits apparently random fluctuation, it is convenient
to characterize it in statistical termsāthat is, to specify its mean
value Ī¼ and standard deviation Ļ.
ā¢ The mathematical analysis of such random or stochastic processes
is beyond the scope of this book. GOOD NEWS
ā¢ See Maybeck (1997) and Box et al. (1994) for more details.
ā¢ Control systems designed assuming deterministic inputs usually
perform satisfactorily for random inputs; hence that approach
ā¢ is taken in this book
ā¢ Monitoring techniques based on statistical analysis are discussed in
Chap 21.
39Rami Bechara
40. Response Of First-order
Processes
ā¢ General 1st-order TF
ā¢ K = steady-state gain and Ļ =time constant.
ā¢ Useful in describing the dynamics of the
blending system(Cha p4 ā Sec 4.4)
ā¢ Investigate responses to process inputs
40Rami Bechara
41. 1st order TF
Response to Step Input Functions
ā¢ u(s) expressed:
ā¢
ā¢ Y(s) obtained:
ā¢ Transform into time-domain:
ā¢ No instantaneous response to a sudden change in its input.
ā¢ At t = Ļ, the process response is still only 63.2% complete
ā¢ Theoretically, process output never reaches the new
steady-state value except as š” ā ā
ā¢ It does however approximate the final steady-state value
when t ā 5Ļ
41Rami Bechara
42. 1st order TF
Response to Step Input Illustrated
ā¢ Dimensionless or normalized
Figure
ā¢ Time divided by the time
constant
ā¢ Output change divided by
the product of the process
gain and magnitude of input
change.
42Rami Bechara
43. 1st order TF
Response to Ramp Input
ā¢ u(s) expressed:
ā¢ Y(s) obtained:
ā¢ Partial Fraction Expansion / Heaviside Expansion
ā¢ Transform into time-domain
43Rami Bechara
44. 1st order TF
Response to Ramp Input
ā¢ For large values of time (š” >> š)
ā¢ Equation implies : that after an initial transient period, the
ramp input yields a ramp output with slope equal to Ka, but
shifted in time by the process time constant Ļ
ā¢ An unbounded ramp input will ultimately cause some process
component to saturate, so ramp duration is limited.
ā¢ A process input frequently will be ramped from one value to
another in a fixed amount of time so as to avoid the sudden
change associated with a step change.
ā¢ Ramp inputs of this type are particularly useful during the
start-up of a continuous process or in operating a batch
process.
44Rami Bechara
45. 1st order TF
Response to Sinusoidal Input
ā¢ u(s) expressed:
ā¢ Y(s) obtained:
ā¢ Inverse LaPlace Transform
ā¢ Transform into time-domain
45Rami Bechara
46. 1st order TF
Response to Sinusoidal Input
ā¢ Exponential term (šā
š”
š) goes to zero as š” ā ā, leaving
a pure sinusoidal response. (sin š¤š” + š )
ā¢ This property is exploited in Chapter 14 for frequency
response analysis.
ā¢ Students often have difficulty imagining how a real
ā¢ process variable might change sinusoidally.
ā¢ How can the flow rate into a reactor be negative as
well as positive?
ā¢ Remember that we have defined the input u and
output y in these relations to be deviation variables
46Rami Bechara
47. Response Of Integrating
Processes
ā¢ Chap 2 liquid-level system with a pump
attached to the outflow line
ā¢ Assume that outflow rate q can be set at any
time by adjusting the speed of the pump
ā¢ Equation (deviation var) becomes
ā¢ LaPlace
47Rami Bechara
48. Second order processes Expression
ā¢ Alternatively, a second-order process transfer
function will arise upon transforming either a
second-order differential equation process
model such as the two coupled first-order
differential equations, for the CSTR
ā¢ Global Expression:
48Rami Bechara
49. 2nd order processes
Parameters
ā¢ K and Ļ have the same importance as for a first-order
transfer function.
ā¢ K is the steady-state gain, and Ļ determines the
speed of response (or, equivalently, the response
time) of the system.
ā¢ The damping coefficient Ī¶ (zeta) is dimensionless.
ā¢ Ī¶ measures the amount of damping in the systemā
that is, the degree of oscillation in a process
response after an input change.
49Rami Bechara
50. 2nd order processes
Parameters: Ī¶
ā¢ Small values of Ī¶ imply little damping and a large
amount of oscillation, as, for example, in an
automobile suspension system with ineffective shock
absorbers.
ā¢ Hitting a bump causes a vehicle to bounce up and
down dangerously.
ā¢ In some textbooks, the G(s) equation is written in
terms of Ļn = 1/Ļ, the undamped natural frequency of
the system.
ā¢ This name arises because it represents the frequency
of oscillation of the system when there is no damping
(Ī¶=0).
50Rami Bechara
51. Classes of 2nd order processes
ā¢ Three important classes of 2nd-order systems
ā¢ The case Ī¶ < 0 omitted here because it corresponds to an
unstable second-order system that has an unbounded response
to any input (effects of instability are covered in Chapter 11).
ā¢ The overdamped (Ī¶ >1) and critically damped (Ī¶ = 1) forms of the
second-order transfer function most often appear when two
first-order systems occur in series 51Rami Bechara
52. Identities for systems in series
ā¢ Equating the denominators (slides 43-44) ļ
52Rami Bechara
53. Underdamped Systems
0<š<1
ā¢ The underdamped form can arise from some
mechanical systems
ā¢ From flow or other processes such as a pneumatic (air)
instrument line with too little line capacity, or from a
mercury manometer, where inertial effects are
important.
ā¢ For process control problems the underdamped form is
frequently encountered in investigating the properties
of processes under feedback control.
ā¢ Next we develop the relations for the step responses of
all three classes of second-order processes.
53Rami Bechara
54. 2nd order process
Step Response
54
ā¢ u(s) expressed:
ā¢
ā¢ Y(s) obtained:
Rami Bechara
55. 2nd order process
Step Response Time Domain
ā¢ Overdamped (Identity Slide 48)
ā¢ Critically Damped
ā¢ Underdamped
55
š1= š2
Rami Bechara
56. 2nd order process
Step Response Plot Underdamped
ā¢ Normalized plots (
š”
š
ššš
š¾
š
)
56Rami Bechara
57. 2nd order process Step Response Plots
Over and Critically Damped
ā¢ Normalized plots (
š”
š
ššš
š¾
š
)
57Rami Bechara
58. 2nd order process Step Response
Plot Analysis
ā¢ Responses exhibit a higher degree of
oscillation and overshoot (y/KM > 1) as
š approaches zero.
ā¢ Large values of š yield a sluggish (slow)
response.
ā¢ The fastest response without overshoot is
obtained for the critically damped case (Ī¶ = 1).
58Rami Bechara
59. Step Response characteristics of a
2nd -order underdamped process
ā¢ Important terms
ā¢ X-axis: time relate terms
ā¢ Rise Time (tr) is the time the process output
takes to first reach the new steady-state
value.
ā¢ Time to First Peak. (tp) is the time required
for the
ā¢ output to reach its first maximum value.
ā¢ Settling Time. ts is the time required for the
process output to reach and remain inside a
band whose width is equal to Ā±5% of the
total change in y (Ā±1% sometimes used).
59
ā¢ Period of Oscillation. P time between 2 successive peaks or valleys of the response.
ā¢ Y-axis terms
ā¢ Overshoot. OS = a/b (% overshoot is 100 a/b).
ā¢ Decay Ratio.DR = c/a (where c is the height of the second peak).
ā¢ True for the step response of any underdamped process.
ā¢ If no overshoot: rise time definition: time to go from 10% to 90% of steady-state
response Rami Bechara
60. Expressions of terms for
2nd order underdamped processes
60
ā¢ Expressions for terms in case of 2nd order underdamped
process:
ā¢ For an underdamped second-order transfer function,
Equations and figures can be used t obtain estimates of
Ī¶ and Ļ based on step response characteristicsRami Bechara
61. Relation between parameters and š
ā¢ OS and DR are
functions of Ī¶ only.
ā¢ For a 2nd-order
system, DR constant
for each successive
pair of peaks.
ā¢ Figure illustrates the
dependence of
overshoot and
decay ratio on
damping coefficient.
61Rami Bechara
62. Chapter 6
Roots and Poles
ā¢ Roots of example TF
ā¢ For control engineers roots of the
denominator polynomial as poles oftransfer
function G(s).
62Rami Bechara
63. ā¢ Useful to plot the roots (poles) and to discuss process
response characteristic in terms of root locations in the
complex s plane.
ā¢ Figure , ordinate = imaginary part of
each root; abscissa = real part.
ā¢ Figure indicates the presence of
four poles: an integrating element
(pole at the origin), one real pole
(at ā1/Ļ1), and a pair of complex
conjugate poles, s3 and s4.
63Rami Bechara
64. Pole Location and speed of response
ā¢ The real pole is closer to the
imaginary axis than the
complex pair, indicating a
slower response mode
(eātāĻ1 decays slower than
eāĪ¶tāĻ2 ).
ā¢ In general, the speed of
response for a given mode
increases as the pole
location moves farther away
from the imaginary axis.
64Rami Bechara
65. Importance of previous analysis
ā¢ Previous have played an important role in the design of mechanical
and electrical control systems,
ā¢ Rarely used in designing process control systems.
ā¢ However, it is helpful to develop some intuitive feeling for the
influence of pole locations.
ā¢ A pole to the right of the imaginary axis (called a right-half plane
pole), for example, s = +1/Ļ, indicates that one of the system
response modes is et/Ļ.
ā¢ This mode grows without bound as t becomes large, a characteristic
of unstable systems.
ā¢ As a second example, a complex pole always appears as part of a
conjugate pair, (s3 and s4 in example Equation).
ā¢ The complex conjugate poles indicate that the response will contain
sine and cosine terms; that is, it will exhibit oscillatory modes.
65Rami Bechara
66. TF extension: lead-lag
ā¢ All of the transfer functions discussed so far
can be extended to represent more complex
process dynamics simply by adding numerator
terms.
ā¢ For example, some control systems contain a
leadālag element, with differential equation
ā¢ TF function
Term added to standard
equation
Processes with
numerator dynamics
66Rami Bechara
67. Effects of Integration
ā¢ Integral of u included in differential equation
ā¢ The transfer function becomes, assuming zero
initial conditions
ā¢ Values of s that cause the numerator of G(s) to
become zero are called the zeros of G(s). They
have an important role in process dynamics
67Rami Bechara
68. Different expressions of G(s)
ā¢ Standard TF form
ā¢ Alternatively
zi and pi are zeros and poles
poles of G(s) are also the roots of the characteristic
equation
This equation is obtained by setting the denominator
of G(s), the characteristic polynomial, equal to zero.
68Rami Bechara
69. Different expressions of G(s)
ā¢ It is convenient to express transfer functions in
gain/time constant form; that is, b0 is factored out of
the numerator and a0 out of the denominator to
show the steady-state gain explicitly (K = b0/a0 =
G(0)).
ā¢ Then resulting expressions are factored to give
ā¢ Relation between zeroes and poles
69Rami Bechara
70. Effect of zeroes
ā¢ The presence or absence of system zeros has no effect on
the number and location of the poles and their associated
response modes
ā¢ Exception : an exact cancellation of a pole by a zero with
the same numerical value.
ā¢ However, the zeros exert a profound effect on the
coefficients of the response modes (i.e., how they are
weighted) in the system response.
ā¢ Such coefficients are found by partial fraction expansion.
ā¢ For practical control systems the number of zeros is less
than or equal to the number of poles (m ā¤ n).
ā¢ When m = n, the output response is discontinuous after a
step input change.
70Rami Bechara
71. 2nd-Order Processes with
Numerator Dynamics
ā¢ The presence of a zero in the first-order system causes a
jump discontinuity in y(t) at t = when a step input is
applied.
ā¢ Such an instantaneous step response is possible only when
the numerator and denominator polynomials have the
same order, which includes the case, G(s) = K.
ā¢ Industrial processes have higher-order dynamics in the
denominator, causing them to exhibit some degree of
inertia.
ā¢ This feature prevents them from responding
instantaneously to any input, including an impulse input.
Thus, we say that m ā¤ n for a system to be physically
realizable.
71Rami Bechara
72. Example Resolved
ā¢ Response to step change (in time domain)
ā¢ Note that š¦(š” ā ā) = š¾š as expected
ā¢ Thus, the effect of including the single zero does
not change the final value, nor does it change the
number or locations of the poles.
ā¢ But the zero does affect how the response modes
(exponential terms) are weighted in the solution
72Rami Bechara
73. Example resolved
Response Types
ā¢ Mathematical analysis (see Exercise 6.3)
shows that three response types can occur
ā¢ Ļ1 > Ļ2 is arbitrarily chosen
73Rami Bechara
74. Example resolved
Response Types Illustrated
74
Case 1 (Ļa = 8, 16)
Case 2 (Ļa = 0.5, 1, 2, 4)
Case 3 (Ļa = ā1,ā4)
Rami Bechara
75. Approximation of time delays
ā¢ For small values of s, truncating the expansion
after the first-order term provides a suitable
approximation
ā Note that this time-delay approximation is a right-
half plane (RHP) zero at s = +Īø.
ā¢ An alternative 1st -order approximation
consists of the transfer function
75ApproximationRami Bechara
76. Skogestadās āHalf Ruleā
Approximation
ā¢ Approximation method for higher-order models that
contain multiple time constants.
ā¢ He approximates the largest neglected time constant in the
denominator in the following manner.
ā¢ One-half of its value is added to the existing time delay (if
any), and the other half is added to the smallest retained
time constant.
ā¢ Time constants smaller than the largest neglected time
constant are approximated as time delays, along with plane
zero according to previous equations.
ā¢ The motivation for this āhalf ruleā is to derive approximate
low-order models more appropriate for control system
design.
76Rami Bechara
77. Skogestadās āHalf Ruleā
Approximation
ā¢ Largest neglected time constant = 3
ā¢ According to his āhalf rule,ā half of this value is
added to the next largest time constant to
generate a new time constant,
Ļ = 5 + 0.5(3) = 6.5
ā¢ The other half provides a new time delay of
0.5(3) = 1.5.
ā¢ Total time delay: Īø = 1.5 + 0.1 + 0.5 = 2.1
ā¢ Final Transfer function
77Rami Bechara
78. Left-plane Zeroes
ā¢ Skogestad (2003) has also proposed
approximations for left-half plane zeros of the
form, Ļas + 1, where Ļa > 0.
ā¢ However, these approximations are more
complicated and beyond the scope of this
book.
ā¢ In these situations, a simpler model can be
obtained by empirical fitting of the step
response using the techniques in Chapter 7.
78Rami Bechara
79. Linear State Space Model
ā¢ Matrix presentation
ā¢ x is the state vector; u is the input vector of manipulated
variables (also called control variables); d is the disturbance
vector; and y is the output vector of measured variables.
ā¢ The elements of x are referred to as state variables.
ā¢ The elements of y are typically a subset of x, namely, the
state variables that are measured.
ā¢ In general, x, u, d, and y are functions of time.
ā¢ Matrices A, B, C, and E are constant matrices.
ā¢ Vectors have different dimensions (or ālengthsā) and are
usually written as deviation variables.
79Rami Bechara
80. Chapter 7
TF estimation
ā¢ Consider the problem of estimating the time
constants for first-order and overdamped
second-order dynamic models based on the
measured output response to a step input
change of magnitude M. (Chap 5)
80Rami Bechara
81. Variable Transformation
ā¢ Sometimes a variable transformation can be employed to
transform a nonlinear model so that linear regressioncan
be used (Montgomery and Runger, 2013).
ā¢ For example, if K is assumed to be known, the first-order
step response can be rearranged:
ā¢ Because ln(1 ā y/KM) can be evaluated at each time ti, this
model is linear in the parameter 1/Ļ.
ā¢ Standard linear form, where the left-hand side is Yi,
Ī²1 = 0, and ui = ti.
ā¢ Fraction incomplete response method of determining first-
order models
81Rami Bechara
82. Linking the equation and the figure
ā¢ The normalized
step response is
shown in the figure
ā¢ The intercept of the
tangent at t = 0 with
the horizontal line,
y/KM = 1, occurs at
t = Ļ.
ā¢ As an alternative, Ļ can be estimated from a step
response plot using the value of t at which the
response is 63.2% complete; following example
82Rami Bechara
83. Steps to determine FOPTD
ā¢ The process gain K is found by calculating the ratio of
the steady-state change in y to the size of the input
step change, M.
ā¢ 2. A tangent is drawn at the point of inflection of the
step response; the intersection of the tangent line and
the time axis (where y = 0) is the time delay.
ā¢ 3. If the tangent is extended to intersect the steady-
state response line (where y = KM), the point of
intersection corresponds to time t = Īø + Ļ.
ā¢ Therefore, Ļ can be found by subtracting Īø from the
point of intersection.
83Rami Bechara
85. Sundaresan and Krishnaswamy
Method
ā¢ Avoids use of the point of inflection construction entirely to
estimate the time delay.
ā¢ They proposed that two times, t1 and t2, be estimated from
a step response curve.
ā¢ These times correspond to the 35.3% and 85.3% response
times, respectively.
ā¢ The time delay and time constant are then calculated from
the following equations:
ā¢ These values of Īø and Ļ approximately minimize the
difference between the measured response and the model
response, based on a correlation for many data sets.
85Rami Bechara
86. Second order functions
Approximating š and š
ā¢ Smithās method requires the
times (with apparent time
delay removed) at which the
normalized response reaches
20% and 60%, respectively
ā¢ Using the figure, the ratio of
t20/t60 gives the value of Ī¶.
ā¢ An estimate of Ļ can be
obtained from the plot of t60/Ļ
vs. t20/t60.
86Rami Bechara