1. ETEC 4501 Chapter 4 Laplace Transforms and Transfer Functions

2. 1.Establish a methodology to mathematically describe and analyze control systems. Chapter 4 Objectives 2.Equations establish a time relationship between input and output. 3.Often these equations contain integral and derivative terms. 4. Laplace transforms provide a convenient way to handle these equations. Introduction 1.The defining equations will be introduced for a number of systems. 2.These equations will contain: a.Derivative terms - mathematically describe the rate of change of a variable with respect to time. b.Integral terms – mathematical expression for an accumulation of an amount of a variable.

3. 3.Transfer functions describe the relationship between the input and output. 4.Frequency response is a description of how a component responds to a sinusoidal input. 5.To evaluate the response for a specific frequency, the complex frequency parameter (S) can be replaced with jw, then rectangular and polar forms can be utilized.

4. 4.2 Input / Output Relationships In this section the equations that describe the input / output relationship of five simple control systems are introduced.  Self-regulating liquid tank.  Non-regulating liquid tank.  Electrical RC circuit.  Liquid-filled thermometer.  Process control valve.

5. Self-Regulating Liquid Tank The self-regulating liquid tank is shown in Figure 4.1 below. Tank level remains constant when the inflow rate (q in ) equals the outflow rate (q out ). If the flows are different the liquid in the tank will change by an amount The change occurs over a time interval

6. The change in liquid level in the tank is equal to the change in volume divided by the cross-sectional area of the tank (A). The average rate of change of the level in the tank is equal to the change in level divided by the time interval. As the time interval diminishes to a instant in time, the average rate of change becomes the instantaneous rate of change. In mathematics, the instantaneous rate of change of liquid level is called the derivative of level (h) with respect to time (t) and is designated dh/dt.

7. If the equation for flow out of the tank is linear, the outflow rate (q out ) is given by Equation 4.5 ( see Chapter 3). Substituting Equation 4.5 into Equation 4.4 gives us

8. The term is the capacitance (C L ) of the liquid tank (see Chapter 3) The entire term is called the time constant of the liquid tank. The term is the steady-state gain (G) of the system. Substituting time constant and gain into the preceding equation the final form of the differential equation for the liquid tank.

9. Non-Regulating Liquid Tank The non-regulating liquid tank is shown in Figure 4.2 below. A positive displacement pump provides a constant flow rate, q out. Equation 4.2 developed for the self-regulating liquid tank also applies to the non-regulating tank.

10. For convenience, q(t) is defined as: q(t) = q in - q out In other words, q(t) is the difference between the input flow rate and the output flow rate. If the time interval, delta t, begins at t 0 and ends at time t 1, then Using calculus, the accumulation of liquid is given by

11. Electrical Circuit Consider the RC circuit of Figure 4.3. If the output terminals are connected to a high impedance load, i R is approximately equal to i C. For the resistor. For the capacitor.

12. Equating the right-hand side of Equations 4.8 and 4.9 and setting RC = tau results in the following differential equation for the circuit.

13. Liquid-Filled Thermometer A liquid-filled thermometer is illustrated in Figure 4.4. The amount of heat transferred from the fluid surrounding the bulb to the liquid inside the bulb depends on the thermal resistance (R T ) between the two fluids, the difference in temperature (T a – T m ), and the time interval.

14. The change in temperature of the liquid in the bulb is equal to the amount of heat added divided by the thermal capacity (C T ) of the liquid inside the bulb.

15. Process Control Valve A control valve is illustrated in Figure 4.5. Valve positioning is achieved through a balance of applied air pressure at the top of the diaphragm and opposing forces.

16. The opposing forces consist of: the inertial force of the moving mass (F I ), the resistive force of the seal (F R ), and the compressive force of the spring (F C ). The inertial force is equal to the mass of the moving parts (valve stem and diaphragm plate) multiplied by the acceleration of the moving mass. If x is the position of the moving mass, the acceleration is expressed by the second derivative of x with respect to time.

17. The resistive force of the seal (F R ), is equal to the resistance multiplied by the velocity of the moving mass. The spring force is equal to the compression of the spring divided by the mechanical capacitance value of the spring (C m ). If the position of the moving valve stem at x = 0 results in a spring force of 0, the equation for the spring force is: Substituting Equations 4.12, 4.14, 4.15, and 4.16 into Equation 4.13 yields the following equation for the control valve:

18. Equation 4.17 is a second-order differential equation. Equation 4.5, 4.10, and 4.11 are first-order differential equations. Since many components, systems, and processes are modeled by some type of differential equation, a brief discussion on differential equation terminology is in order. Differential Equations – an equation that has one or more derivatives of the dependent variable with respect to the independent variable (ex: dh/dt in Equation 4.6) A solution of a differential equation is an expression for the dependent variable in terms of the independent variable that satisfies the differential equation.

19. Example, consider the following differential equation. Assumptions; GK is a constant and h = 0 when t = 0 (initial conditions). Solution: The order of the differential equation is the order of the highest derivative that appears in the equation. Note the similarities between Equations 4.6, 4.10, and 4.11, even though they describe three very different systems. A linear differential equation has only first-degree terms in the dependent variable and its derivatives.

20. If any term is raised to a power other than 1, the equation is non-linear. A linear differential equation of the second order can be written in the following standard form: If F(t) = 0, the equation is called homogeneous, otherwise it is called nonhomogeneous. If A(t) and B(t) are constants, the equation is said to have constant coefficients. Many components of a control system are modeled by linear differential equation with constant coefficients.

21. The Laplace transform is a tool both for solving these differential equations and for converting the equations into transfer functions. Once the conversion to transfer function is made, the operations of calculus are replaced by algebraic operations. Often the equation that models a component includes both derivative and integral terms. Equations that have at least one integral term and one derivative term are called integro-differential equations. The Laplace transform method also applies to integro-differential equations.

24. 4.3 Laplace Transforms The Laplace transform provides a transform from integral/differential equations into simple algebraic equations. By solving the algebraic equation for the ratio of the output over the input, the transfer function of the component can be obtained. With the transfer function, the frequency response of the component can be computed.

25. The solution outlined in Figure 4.7 is for a step change in the inlet flow rate of the self-regulating liquid tank of Figure 4.1.

26. Equation 4.6 is the time-domain equation for this component. Before t = 0 s, the tank is empty and the input flow is zero. After t = 0 s, the input valve is opened and the input flow changes to K m 3 /s. This type of change is called a step change in the input signal (q in ). A graph of the step change is shown at the top of Figure 4.7. Now the question is: what is the level of the tank after the step change in input?

27. Step 1 – Transform the input q in and Equation 4.6 into the frequency domain. Step 2 – Solve the two frequency-domain equations for H. Step 3 – Use the table of Laplace transforms (Table 4.1) to transform Equation 4.18 into the time domain.

28. The time-domain graph of h(t) is shown at the bottom of Figure 4.7.

29. Functional Laplace Transforms Laplace transforms can be divided into two types: functional and operational. The first type is simply the Laplace transform of a particular function such as: The second involves the transform of the result of some operation, such as the sum of two functions, the derivative of a function, or the integral of a function. The Laplace transform F(s) of the function f(t) is defined by the following relationship:

30. As an illustration, the defining equation above is used to determine the Laplace transform of f(t) = K, where K is a constant. Table 4.1 contains commonly used transform pairs. It is used to determine the frequency-domain mate to a time-domain function. It also is used to perform the inverse Laplace transform, i.e., convert from frequency to time-domain.

33. Operational Laplace Transforms The operational Laplace transform involves the transform of the result of some operation, such as the sum of two functions, the derivative of a function, or the integral of a function. Table 4.2 provides a number of commonly used operational transforms where; f(t) can be any function of t and F(s) is the Laplace transform of f(t).

34. For example, if f(t) = t, then F(s) = 1/s 2. Entry 12 states that the Laplace transform of the sum of two or more functions is the sum of the transforms of each term taken alone. The term f(0), which appears in entries 13, 14, and 15, is the value of f(t) when t = 0. In a mechanical system where f(t) is the position of the object, f(0) is the position at t = 0 (i.e. initial position). The term df(0)/dt, which appears in entries 14 and 15, is the value of the derivative of f(t) when t = 0. In a mechanical system this would be the initial velocity of the object. The term d 2 f(0)/dt 2, which appears in entry 15, is the initial acceleration of the object. In control systems, the usual practice is to assume that all initial conditions are zero.

35. The transfer function is defined as the s-domain ratio of the output over the input with all initial conditions set to zero. When using Laplace transforms to get the transfer function of a component, we use only the first term in entries 13, 14, and 15.

40. 4.4 Inverse Laplace Transforms The inverse Laplace transform converts a frequency-domain function of (s) into a time-domain function of (t). If the frequency-domain function is in the table of Laplace transform pairs, the inverse transform is the mating time-domain function in the table. For the example in Figure 4.7, the frequency-domain equation is given by: Since G and K both represent numeric values, the value of K in table entry 9 can be replaced with GK. The inverse transform is then given by:

41. A common example of a control system function is given in Equation 4.19. To determine the inverse transform, convert the right-hand side of the equation to a sum of terms that are given in a table of Laplace transform pairs. Partial fraction expansion is used to break the ratio of polynomials into a sum of terms. K 1, K 2, and K 3 are undetermined numerical constants, so all that remains is to determine their values.

42. In general, there will be a term on the right-hand side for each root of the polynomial in the denominator of the left-hand side. Multiple roots for factors such as (s+2) n will have a term for each power of the factor from 1 to n as demonstrated in Equation 4.22. Complex roots are common, and they always appear in conjugate pairs. The two constants in the numerator are also complex conjugates. Expansion of a polynomial with complex conjugate roots is illustrated in Equation 4.23 below. Recall that j is the square root of -1. K is a complex constant and K* is the complex conjugate of K.

43. A two-step procedure is used to find each distinct root, real or complex. 1.For evaluation of the constant in a term (K n ), multiply both sides of the equation by the factor in the denominator of that term. 2.Replace s on both sides of the equation by the root of the (K n ) term in step 1. All terms on the right-hand side will be zero, leaving just the single constant for term (K n ). This process is repeated for each term on the right-hand side.

47. 4.5 Transfer Function The transfer function introduced in Chapter 1 can be used to compute the frequency response of the component. If the input is a sinusoidal with frequency of w rad/s and the frequency parameter, s is replaced with jw, the result is a complex number The magnitude is the gain and the angle is the phase difference of the component. Therefore, the frequency response of a component can be computed from its transfer function. A major use of the Laplace transform in control systems is to obtain the transfer function of a component. The transfer function is obtained by solving the frequency-domain algebraic equation for the ratio of the output signal over the input signal with all initial conditions set to zero.

48. As an example, consider the liquid system described by equation 4.6. The frequency-domain algebraic equation is obtained by applying the Laplace transformation to each term in Equation 4.6 and solving for the ration H/Q in. It is understood that h is a function of t and H is a function of s even though the (t) and (s) are not used. (t) and (s) are often dropped for convenience in writing the long expression.

53. The valve has a very underdamped response. To show the underdamped response, the time-domain response of the output, x(t) will be determined for a step change in input, i(t).

56. 4.6 Initial and Final Value Theorems The initial value theorem states that if X(s) is multiplied by s and take the limit as s approaches infinity, the result will be the initial value of x(t); that is, x(0). The final value theorem states that if X(s) is multiplied by s and take the limit as s approaches zero, the result will be the final value of x(t); that is, x(00). These theorems allow for determination of the initial and final values without performing the inverse Laplace transformations.

58. 4.7 Frequency Response: Bode Plots The response of a component to a sinusoidal input signal is called the frequency response of the component. A graph of the frequency response is called a Bode diagram. A sinusoidal input to a linear component will produce a sinusoidal output of the same frequency. Only the amplitude and the phase angle may change. Gain and Phase Angle Figure 4.9 shows typical input and output signals for a linear component.

59. Gain is often expressed in decibel (dB) units. Phase angle is usually measure in degrees. It is positive when the output leads the input and negative when the output lags the input. Bode Diagram The frequency response of a component is the set of values of gain and phase angle that occur when a sinusoidal input signal is varied over a range of frequencies. For each frequency, there is a gain and phase angle that give the response of the component at that frequency.

60. Figure 4.10 shows a typical Bode diagram.

61. Determination of the Frequency Response The frequency response of a component can be obtained from the transfer function by replacing s with jw. The transfer function reduces to a single complex number for each value of w with the magnitude representing the gain and the angle representing the phase angle at that frequency.

62. A straight-line graphical approximation of the Bode diagram is shown in Figure 4.12.

65. Program BODE Obtain the program BODE and execute Example 4.19 (reference pages 145 – 147 in your text. Compare your results with those on page 147.