Lecture 3: Common Simple Dynamic Systems

Common simple dynamic systems - First order -Second order Outline of the lesson. Common simple dynamic systems - First order -Second order - Dead time - (Non) Self-regulatory

When I complete this chapter, I want to be able to do the following. Predict output for typical inputs for common dynamic systems

3.1. First-Order Differential Equation Models The model can be rearranged as where τ is the time constant and K is the steady-state gain. The Laplace transform is

3.1.1. Step Response of a First-Order Model In the previous lecture, we have learned about the step response of first order systems.

3.1.2. Impulse Response of a First-Order Model Consider an impulse input, u(t) = δ (t), and U(s) = 1; the output is now The time-domain solution is which implies that the output rises instantaneously to some value at t = 0 and then decays exponentially to zero.

3.1.3. Integrating Process When the coefficient a0 = 0 in the first order differential equation , we get where K = (b / a1). Here, the pole of the transfer function G(s) is at the origin, s = 0.

3.1.3. Integrating Process The solution of the Equation , could be written immediately without any transform as This is called an integrating (also capacitive or non-self-regulating) process. We can associate the name with charging a capacitor or filling up a tank.

SIMPLE PROCESS SYSTEMS: INTEGRATOR pump valve Level sensor Liquid-filled tank Plants have many inventories whose flows in and out do not depend on the inventory (when we apply no control or manual correction). These systems are often termed “pure integrators” because they integrate the difference between in and out flows.

SIMPLE PROCESS SYSTEMS: INTEGRATOR Plot the level for this scenario pump valve Level sensor Liquid-filled tank Plot the level for this scenario Fin Fout time

SIMPLE PROCESS SYSTEMS: INTEGRATOR pump valve Level sensor Liquid-filled tank Level Fin Fout time

SIMPLE PROCESS SYSTEMS: INTEGRATOR pump valve Level sensor Liquid-filled tank Let’s look ahead to when we apply control. Non-self-regulatory variables tend to “drift” far from desired values. We must control these variables.

3.2. Second-Order Differential Equation Models We have not encountered examples with a second-order equation, especially one that exhibits oscillatory behavior. One reason is that processing equipment tends to be self-regulating. An oscillatory behavior is most often the result of implementing a controller. For now, this section provides several important definitions.

3.2. Second-Order Transfer Function Models where ωn is the natural (undamped) frequency, ζ is the damping ratio or coefficient, K is the steady-state gain, and τ is the natural period of oscillation, where τ = 1/ωn. The characteristic equation is Which provides the poles

SIMPLE PROCESS SYSTEMS: 2nd ORDER 10 20 30 40 50 60 70 80 0.2 0.4 0.6 0.8 1 Time Controlled Variable Manipulated Variable 100 120 140 160 180 200 0.5 1.5 overdamped underdamped

WORKSHOP Four systems experienced an impulse input at t=2. Explain what you can learn about each system (dynamic model) from the figures below. 5 10 15 20 25 30 1 2 3 output (a) -1 (b) time (c) 0.5 1.5 2.5 (d)

3.4 Processes with dead time Many chemical processes involve a time delay between the input and the output. This delay may be due to the time required for a slow chemical sensor to respond or for a fluid to travel down a pipe. A time delay is also called dead time or transport lag. In controller design, the measured output will not contain the most current information, and hence systems with dead time can be difficult to control.

Let’s learn a new dynamic response & its Laplace Transform Let’s consider plug flow through a pipe. Plug flow has no backmixing; we can think of this a a hockey puck traveling in a pipe. What is the dynamic response of the outlet fluid property (e.g., concentration) to a step change in the inlet fluid property?

THE FIRST STEP: LAPLACE TRANSFORM  = dead time  What is the value of dead time for plug flow? Xout Xin time

THE FIRST STEP: LAPLACE TRANSFORM Is this a dead time? What is the value? 1 2 3 4 5 6 7 8 9 10 -0.5 0.5 time Y, outlet from dead time X, inlet to dead time

THE FIRST STEP: LAPLACE TRANSFORM Our plants have pipes. We will use this a lot! The dynamic model for dead time is The Laplace transform for a variable after dead time is

Pade approximation of the time delay There several methods to approximate the dead time as a ratio of two polynomials in s. On such method is the first-order Pade approximation.

Example 3.2 Use the first-order Pade approximation to plot the unit-step response of the first order with a dead-time function: Making use of the dirst order Pade approximation, we can construct a plot with the approximation

Matlab code th = 3; P1 = tf([-th/2 1],[th/2 1]); % First-order Padé approximation t = 0:0.5:50; taup = 10; G1 = tf(1,[taup 1]); y1 = step(G1*P1,t); % y1 is first order with Padé approximation of % dead time y2 = step(G1,t); t2 = t+th; % Shift the time axis for the actual time-delay function plot(t,y1,t2,y2,’r’);

The approximation is very good except near t = 0, where the approximate response dips below. This behavior has to do with the first-order Pade approximation, and we can improve the result with a second-order Pade approximation.