System Dynamics Dr. Mohammad Kilani Class 4 Dynamic Systems and Block Diagrams

System Dynamics Dr. Mohammad Kilani Midterm Exam: Sunday, 2/August 12:30 – 2:00

Transfer Functions

Transfer Functions The complete response of a linear ODE is the sum of the free and the forced responses. For zero initial conditions the free response is zero, and the complete response is the same as the forced response. Thus we can focus our analysis on the effects of the input only by taking the initial conditions to be zero temporarily. When we have finished analyzing the effects of the input, we can add to the result the free response due to any nonzero initial conditions

Transfer Functions The concept of the transfer function is useful for analyzing the effects of the input. Consider the model: The function T(s) is called the transfer function. The transfer function is the transform of the forced response divided by the transform of the input. It can be used as a multiplier to obtain the forced response transform from the input transform; that is, X(s)=T(s)F(s). The transfer function is a property of the system model only. Its independent of the input function and the initial conditions.

Transfer Functions Zero Order First Order Second Order Third Order

Steam Turbine Speed Control System (Watt’s Governor)

Watt’s Governor Block Diagram Sleeve’s Position Disk’s Position Steam Flow Rate + Sleeve -Spring System Lever Main Valve Turbine Desired Speed (Initial Spring Compression) - Centrifugal Force Fly Ball Axis Speed Turbine Speed Fly Balls Bevel Gears

Steam Turbine Speed Control System (Watt’s Governor)

(Initial Spring Compression) Load on Turbine Nut Angle Nut Position Spring Force Net force on Sleeve Turbine Speed Displacement Disk’s Position Steam Flow Rate Driving Torque - + Sleeve -Spring System Screw and Nut Compression Spring Σ Lever Main Valve Turbine Blades Shaft Σ - + Net Torque Desired Speed (Initial Spring Compression) Centrifugal Force Fly Ball Axis Speed Turbine Speed Fly Balls Bevel Gears

(Initial Spring Compression) Desired Speed (Initial Spring Compression) Load on Turbine Nut Angle Nut Position Spring Force Net force on Sleeve - Turbine Speed Disk’s Position Steam Flow Rate Driving Torque Displacement + 1 ------------ ms2+bs+ksp + 1 --------- τ1s+1 Σ kvalve kblades Σ knut ksp klever - Net Torque Centrifugal Force Fly Ball Axis Speed kgear Turbine Speed kballs

Transfer Functions It is important to realize that the transfer function is equivalent to the ODE. If we are given the transfer function we can reconstruct the corresponding ODE. For example, the transfer function: corresponds to the equation Note that the denominator of the transfer function is the characteristic polynomial, and thus the transfer function tells us something about the intrinsic behavior of the model, apart from the effects of the input and specific values of the initial conditions. In the previous equation, the characteristic polynomial is s2 + 7s + 10 and the roots are −2 and −5. The roots are real, and this tells us that the free response does not oscillate and that the forced response does not oscillate unless the input is oscillatory. Because the roots are negative, the model is stable and its free response disappears with time

Transfer Functions Multiple Inputs and Outputs Obtaining a transfer function from a single ODE is straightforward, as we have seen. Sometimes, however, models have more than one input or occur as sets of equations with more than one dependent variable. It is important to realize that there is one transfer function for each input- output pair. If a model has more than one input, a particular transfer function is the ratio of the output transform over the input transform, with all the remaining inputs ignored (set to zero temporarily).

Transfer Functions Multiple Inputs and Outputs When there is more than one input, the transfer function for a specific input can be obtained by temporarily setting the other inputs equal to zero (this is another aspect of the superposition property of linear equations). Thus, we obtain: Note that the denominators of both transfer functions have the same roots. Example: Obtain the transfer functions X(s)/F(s) and X(s)/G(s) for the following equation: Solution: Using the derivative property of the Laplace transform, we have Solving for X(s), we obtain

Transfer Functions The transfer function of a linear, time-invariant differential- equation system is defined as the ratio of the Laplace transform of the output (response function) to the Laplace transform of the input (driving function) under the assumption that all initial conditions are zero. If the highest power of s in the denominator of the transfer function is equal to n, the system is called an nth-order system.

Properties and Applications of the Transfer Functions The applicability of the concept of the transfer function is limited to linear, time-invariant differential-equation systems. Still, the transfer- function approach is used extensively in the analysis and design of linear systems. The transfer function of a system is a mathematical model of that system, in that it is an operational method of expressing the differential equation that relates the output variable to the input variable. The transfer function is a property of a system itself, unrelated to the magnitude and nature of the input or driving function.

Properties and Applications of the Transfer Functions The transfer function includes the units necessary to relate the input to the output; however, it does not provide any information concerning the physical structure of the system. (The transfer functions of many physically different systems can be identical.) If the transfer function of a system is known, the output or response can be studied for various forms of inputs with a view toward understanding the nature of the system. If the transfer function of a system is unknown, it may be established experimentally by introducing known inputs and studying the output of the system. Once established, a transfer function gives a full description of the dynamic characteristics of the system, as distinct from its physical description.

Example 1 y x Consider the spring-mass- dashpot system mounted on a cart as shown. Derive the transfer function G(s) relating mass displacement (x) to cart displacement (y). (bs+k)/(ms2+bs+k) Y(s) X(s)

Example 1 Find the steady response in x for a unit step input in y. (bs+k)/(ms2+bs+k) Y(s) X(s)

Example 1 >> num=[2 10]; >> den=[1 2 10]; y x Use MATLAB’s step function to obtain the response x(t) of the system for a unit step input y(t) when m = 10 kg, b = 20 N.s/m and k = 100 N/m. >> num=[2 10]; >> den=[1 2 10]; >> step(num,den) (bs+k)/(ms2+bs+k) Y(s) X(s)

Example 1 Obtain an analytical expression for x(t) for a unit step input on the system shown when m = 10 kg, b = 20 N.s/m and k = 100 N/m. Use Matlab’s residue function to obtain the partial fraction expansion of the expression for X(s). >> num=[2 10]; >> den=[1 2 10 0]; >> [r,p,k]=residue(num,den) r = -0.5000 - 0.1667i -0.5000 + 0.1667i 1.0000 p = -1.0000 + 3.0000i -1.0000 - 3.0000i k = []

Example 1 What conditions need to be satisfied in order to justify considering the system to be a) A zero-order system, b) A first order system. When m/k << 1 s2, and b/k << 1 s, the system may be modeled as a zero- order system. This happens when the spring stiffness k is very large. When m/k << 1 s2, but b/k is not, the system may be modeled as a first order system. This happens when the mass m is very small. y x (bs+k)/(ms2+bs+k) Y(s) X(s)

Example 2 – Automobile Accelerometer Consider the accelerometer used in seismic and vibration engineering to determine the motion of large bodies to which the accelerometer is attached. The acceleration of the large body places the piezoelectric crystal into compression or tension, causing a surface charge to develop on the crystal. The charge is proportional to the compression or tension. As the large body accelerates, the mass of the accelerometer will move with an inertial response. The stiffness of the spring, k, provides a restoring force to move the accelerometer mass back to equilibrium while internal frictional damping, c, opposes any displacement away from equilibrium. Piezoelectric crystal m k c xi xo

Example 2 – Automobile Accelerometer k c xi xo

Numerator Dynamics The following model contains a derivative of the input y(t): Its transfer function is: Note that the input derivative y(t) results in an s term in the numerator of the transfer function, and such a model is said to have numerator dynamics. So a model with input derivatives has numerator dynamics, and vice versa, and thus the two terms describe the same condition.

Numerator Dynamics With such models we must proceed carefully if the input is discontinuous, as is the case with the step function, because the derivative produces an impulse when acting on a discontinuous function. It can be shown that unit impulse δ(t) is the time- derivative of the unit-step function us(t); Taking the Laplace transform of the unit step us(t), we have: This result does not contradict common sense, because the step function changes from 0 at t = 0 to 1 at t = 0+ in an infinitesimal amount of time.

Numerator Dynamics Thus an input derivative will create an impulse in response to a step input. For example, consider the model If the input y(t) = us(t), the model is equivalent to: which has an impulse input. Numerator dynamics can significantly alter the response, and the Laplace transform is a convenient and powerful tool for analyzing models having numerator dynamics

Numerator Dynamics Example: A 1st order model with numerator dynamics: Obtain the transfer function and investigate the response of the following model in terms of the parameter a. The input y(t) is a unit-step function.

Numerator Dynamics Example: A 1st order model with numerator dynamics: Obtain the transfer function and investigate the response of the following model in terms of the parameter a. The input g(t) is a unit-step function. Solution: Transforming the equation with x(0) = 0 and solving for the ratio X(s)/Y(s) gives the transfer function: For a unit step, Y(s) = 1/s, and Thus the response is From this solution or the initial- value theorem we find that x(0+) = a/5, which is not equal to x(0) unless a = 0 (which corresponds to the absence of numerator dynamics).

Numerator Dynamics y x The plot of the response is given for several values of a. The initial condition is different for each case, but for all cases the response is essentially constant for t >2 because of the term e−2t . For the system below, when the mass is negligible, the transfer function is as shown. The pot describe the response of this system to a step input (bs+k)/(bs+k)

Impulse Response Function The impulse response of a mechanical system can be observed when the system is subjected to a very large force for a very short time, for instance, when the mass of a spring-mass-dashpot system is hit by a hammer or a bullet. Mathematically, such an impulse input can be expressed by an impulse function

Impulse Response Function The impulse function represents signals that last for a short time (Δt) but has a large amplitude (h), so that the area (A = hΔt) in a time plot is not negligible. The impulse input is usually denoted by a vertical arrow to indicate that it has a very short duration and a very large height.

Impulse Response Function The transfer function of a linear, time- invariant system is as shown, where X(s) is the Laplace transform of the input and Y(s) is the Laplace transform of the output and where we assume that all initial conditions involved are zero. Consider the output (response) of the system to a unit-impulse input when the initial conditions are zero. Since the Laplace transform of the unit-impulse function is unity, or Y(s) = 1, the Laplace transform of the output of the system is the same as T(s). The inverse Laplace transform of T(s) is called the impulse-response function, or the weighting function, of the system.

Impulse Response Function The impulse-response function is thus the response of a linear system to a unit-impulse input when the initial conditions are zero. The Laplace transform of g(t) gives the transfer function. Therefore, the transfer function and impulse-response function of a linear, time- invariant system contain the same information about the system dynamics. It is hence possible to obtain complete information about the dynamic characteristics of a system by exciting it with an impulse input and measuring the response. In practice, a large pulse input with a very short duration compared with the significant time constants of the system may be considered an impulse.

Example A body is stationary at x(0) = 2. Find the position of a body knowing that its speed v(t) is described by a unit impulse function applied at time t=0.

Example 3 mB mM A bullet of mass mB is shot into a block of mass m. Assume that when the bullet hits the block, it becomes embedded there. Determine the speed of the block immediately after it is hit by the bullet.

Example 3 Determine the displacement (x) of the block after it is hit by the bullet. The displacement x is measured from the equilibrium position before the bullet hits it. Draw a curve x(t) versus t. Assume the following numerical values mM = 50 kg, mB = 0.01 kg, b = 100 N.s/m, k = 2500 N/m, and v(0-) = 800 m/s, mB mM

Block Diagrams A block diagram of a dynamic system is a pictorial representation of the functions performed by each component of the system and of the flow of signals within the system. Such a diagram depicts the interrelationships that exist among the various components. Differing from a purely abstract mathematical representation, a block diagram has the advantage of indicating the signal flows of the actual system more realistically. Transfer Function G(s)

Block Diagrams In a block diagram, all system variables are linked to each other through functional blocks. The functional block, or simply block, is a symbol for the mathematical operation on the input signal to the block that produces the output. The transfer functions of the components are usually entered in the corresponding blocks, which are connected by arrows to indicate the direction of the flow of signals. Note that a signal can pass only in the direction of the arrows. Transfer Function G(s)

Block Diagrams The arrowhead pointing toward the block indicates the input to the block, and the arrowhead leading away from the block represents the output of the block. Note that the dimension of the output signal from a block is the dimension of the input signal multiplied by the dimension of the transfer function in the block. Transfer Function G(s)

Block Diagrams The advantages of the block diagram representation of a system lie in the fact that it is easy to form the overall block diagram for the entire system merely by connecting the blocks of the components according to the signal flow and that it is possible to evaluate the contribution of each component to the overall performance of the system. In general, the functional operation of a system can be visualized more readily by examining a block diagram of the system than by examining the physical system itself. Transfer Function G(s)

Block Diagrams A block diagram contains information concerning dynamic behavior, but it does not include any information about the physical construction of the system. Consequently, many dissimilar and unrelated systems can be represented by the same block diagram. Note that in a block diagram the main source of energy is not explicitly shown and that the block diagram of a given system is not unique. A number of different block diagrams can be drawn for a system, depending on the point of view of the analysis. Transfer Function G(s)

Block Diagrams: Summing Points A summing point is shown on the block diagram as a circle with a cross, the symbol that stands for a summing operation. The plus or minus sign at each arrowhead indicates whether the associated signal is to be added or subtracted. It is important that the quantities being added or subtracted have the same dimensions and the same units. R(s) Transfer Function G(s) C(s) C(s) = G(s)R(s) a b + - a - b

Block Diagrams Reduction Rules

Closed Loop Transfer Function For the system shown, the output C(s) and input R(s) are related as follows: R(s) B(s) + - E(s) G(s) H(s) C(s) G/(1+GH)

Example 3 The figure shows a proposed block diagram for a spring-mass-damper system. Use the block diagram reduction rules to obtain the overall transfer function X(s)/F(s)

Example 4 Find the overall transfer function for the block diagram below by utilizing the block diagram reduction rules

Example 4 Find the overall transfer function for the block diagram below by utilizing the block diagram reduction rules

Steam Turbine Speed Control System (Watt’s Governor)

(Initial Spring Compression) Load on Turbine Nut Angle Nut Position Spring Force Net force on Sleeve Turbine Speed Displacement Disk’s Position Steam Flow Rate Driving Torque - + Sleeve -Spring System Screw and Nut Compression Spring Σ Lever Main Valve Turbine Blades Shaft Σ - + Net Torque Desired Speed (Initial Spring Compression) Centrifugal Force Fly Ball Axis Speed Turbine Speed Fly Balls Bevel Gears

(Initial Spring Compression) Desired Speed (Initial Spring Compression) Load on Turbine Nut Angle Nut Position Spring Force Net force on Sleeve - Turbine Speed Disk’s Position Steam Flow Rate Driving Torque Displacement + 1 ------------ ms2+bs+ksp + 1 --------- τ1s+1 Σ kvalve kblades Σ knut ksp klever - Net Torque Centrifugal Force Fly Ball Axis Speed kgear Turbine Speed kballs

(Initial Spring Compression) Desired Speed (Initial Spring Compression) Load on Turbine Spring Force Net force on Sleeve Driving Torque Net Torque Turbine Speed + Σ Nut Angle ka kb ---------------- ms2+bs+ka Σ 1 ---------- τ1s+1 - Centrifugal Force kc

Σ - + Spring Force Net force on Sleeve Net Torque Turbine Speed Nut Angle ka kb ---------------- ms2+bs+ks 1 ---------------- τ1s+1 - Centrifugal Force kc

----------------------------- kb=0.1; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Turbine Speed kb ----------------------------- s3+ 2s2+2s+(kb + 1) Nut Angle

----------------------------- >> p = [1 2 2 1.1]; >> roots(p) ans = -1.0910 + 0.0000i -0.4545 + 0.8954i -0.4545 - 0.8954i >> 4./min(abs(real(roots(p)))) 8.8006 kb=0.1; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Turbine Speed 0.1 ----------------------------- s3+ 2s2+2s+1.1 Nut Angle

----------------------------- >> p = [1 2 2 2]; >> roots(p) ans = -1.5437 + 0.0000i -0.2282 + 1.1151i -0.2282 - 1.1151i >>4./min(abs(real(roots(p)))) 17.5319 kb=1; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Turbine Speed 1 ----------------------------- s3+ 2s2+2s+2 Nut Angle

----------------------------- >> p = [1 2 2 3]; >> roots(p) ans = -1.8105 + 0.0000i -0.0947 + 1.2837i -0.0947 - 1.2837i >> 4*max(-1 ./ real(roots(p))) 42.2243 kb=2; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Turbine Speed 2 ----------------------------- s3+ 2s2+2s+3 Nut Angle

----------------------------- >> p=[1 2 2 4]; >> roots(p) ans = -2.0000 + 0.0000i 0.0000 + 1.4142i 0.0000 - 1.4142i >> 4./min(abs(real(roots(p)))) 4.8038e+016 kb=3; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Zero real part for the complex roots Turbine Speed 3 ----------------------------- s3+ 2s2+2s+4 Nut Angle

----------------------------- >> p=[1 2 2 4.1]; >> roots(p) ans = -2.0165 + 0.0000i 0.0082 + 1.4259i 0.0082 - 1.4259i kb=3.1; num=kb; den=[1 2 2 (1+kb)]; sys = tf(num,den); step(sys) grid Turbine Speed 3.1 ----------------------------- s3+ 2s2+2s+4.1 Nut Angle