(Solution Manual) Feedback Control of Dynamic Systems

Franklin,Powell,Emami-Naeini, 7

Engineering

Solution Manual

7th Edition

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Solution Manual for Feedback Control of Dynamic Systems, The downloadable file consists of 397 pages, and the total size of the file is 7.36 MB. Purchase now only for highly discounted price.

Author(s): Franklin, Powell, Emami-Naeini

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Problem 1.01PP Draw a component block diagram for each of the following feedback control systems, (a) The manual steering system of an automobile (b) Drebbel’s incubator (c) The water level controlled by a float and valve (d) Watt’s steam engine with fly-bail governor In each case, indicate the location of the elements listed below and give the units associated with each signal. In each case, indicate the location of the elements listed below and give the units associated with each signal. • The process • The process desired output signal • The sensor • The actuator • The actuator output signal • The controller • The controller output signal • The reference signal • The error signal Notice that in a number of cases the same physical device may perform more than one of these functions. Step-by-step solution step 1 of 4 (A). Manual Steering Systems Step 2 of 4 (B) Step 3 of 4 (c) Set water leveKRef Input) (CorMleror ■ Actuator o/p) Ftootvatve Floating Of floatWater Tank (process)Water level (Sensor+contioller+Actualor) "Water Level Controller Step 4 of 4 (D) Fly-Ball Governor Problem 1.02PP Identify the physical principles and describe the operation of the thermostat in your home or office. Step-by-step solution step 1 of 1 Thermostat is used as the sensing element for controlling the room temperature. It performs the task of automatic reduction of error to zero, irrespectiTe o f the situation created by disturbance. Working Principle: It contains a fluid which is able to e ^ a n d or contract due to temperature change, which causes the sni^ - action of a switch that makes switching- ON or OFF ofthe heat source. Problem 1.03PP A machine for making paper is diagrammed in Fig. There are two main parameters under feedback control: the density of fibers as controlled by the consistency of the thick stock that flows from the headbox onto the wire, and the moisture content of the final product that comes out of the dryere. Stock from the machine chest is diluted by white water returning from under the wire as controlled by a control valve {CV).A meter supplies a reading of the consistency. At the “dry end” of the machine, there is a moisture sensor. Draw a block diagram and identify the nine components listed in Problem part (d) for the following: (a) Control of consistency (b) Control of moisture Figure A papermaking machine Figure A papermaking machine Watt’s steam engine with fly-ball governor In each case, indicate the location of the elements listed below and give the units associated with each signal. • The process • The process desired output signal • The sensor • The actuator • The actuator output signal • The controller • The controller output signal • The reference signal • The error signal Step-by-step solution step 1 of 13 (a) Step 2 of 13 Figure 1 shows the general block diagram to understand the process of the system. The nine components listed below are identified in figure 1. • Process • Process desired output signal • Sensor • Actuator • Actuator output signal • Controller • Controller output signal • Reference signal • Error signal Actuating signalActuator ou^ut signal Figure 1 step 3 of 13 Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of consistency. ErrorStep 4 of 13 ^ Actuator Paper consistency Sensor Figure 2 Step 5 of 13 Compare figure 1 and figure 2. The comparison shows the working of the paper making machine > A /ifh the mainr ^nmnnnonfc in fho nrr»rocc nf nr*ntml nf nrinciotonnw Step 6 of 13 The input reference signal is given to the controller and here the valve acts as an actuator for the system. The valve initiates the mixing process and the consistency of the paper is monitored according to time delay. The consistency checking meter checks the output of the process and sends the signal to error detector. Here, the original required consistency Is checked with the output received and checks for any variation in the output. The corresponding required adjustments may be done in the controller to reduce the error rate, to improve the total output of the process. Step 7 of 13 Thus, the control of consistency is explained and the major components are identified. Step 8 of 13 (b) Step 9 of 13 Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of moisture. Step 10 of 13 Paper m oisture Sensor Figures step 11 of 13 Compare figure 1 and figure 3. The comparison shows the working of the paper making machine with the major components in the process of control of moisture. Step 12 of 13 Here, Dryer remains as a main process and the moisture is checked in the moisture checking meter (sensor). The total output is compared in the error detector and required adjustments are done by the controller to improve the accuracy. Step 13 of 13 Thus, the control of moisture is explained and the major components are identified. Problem 1.04PP Many variables in the human body are under feedback control. For each of the following controlled variables, draw a block diagram showing the process being controlled, the sensor that measures the variable, the actuator that causes it to Increase and/or decrease, the information path that completes the feedback path, and the disturbances that upset the variable. You may need to consult an encyclopedia or textbook on human physiology for information on this problem. (a) Blood pressure (b) Blood sugar concentration (c) Heart rate (c) Heart rate (d) Eye-pointing angle (e) Eye-pupil diameter Step-by-step solution step 1 of 2 (A) . (B) . (C) . Block Diagram for Blood Pre ssure. Blood Sugar Concentration and Heart Bate Control Step 2 of 2 (D ). rm * 12^T_*2?2!E!S. Block Diagram of Eye-Pointing Angle and Eye - P i^ il Diameter Problem 1.05PP Draw a block diagram of the components for temperature control in a refrigerator or automobile air-conditioning system. Step-by-step solution step 1 of 2 Figure 1 shows the general block diagram to understand the temperature control of refrigerator and in the automobile air-conditioning system. ActuMor Process and in the automobile air-conditioning system. Figure 1 Step 2 of 2 The required temperature signal is set in the thermostat, and then the controller actuates the compressor for cooling process. The temperature sensor measures the temperature and compares the required reference temperature with the measured temperature in the comparator. When the required temperature is achieved, the controller stops the compressor and maintains the temperature. The temperature sensor checks the temperature of the system periodically and actuates the controller if there is difference in the required reference temperature. This process is continued to maintain the required temperature of the system thereby controlling the ON/OFF input to the compressor. Thus, the temperature control of refrigerator and in the automobile air-conditioning system is explained. Problem 1.06PP Draw a block diagram of the components for an elevator-position control. Indicate how you would measure the position of the elevator car. Consider a combined coarse and fine measurement system. What accuracies do you suggest for each sensor? Your system should be able to correct for the fact that in elevators for tall buildings there is significant cable stretch as a function of cab load. Step-by-step solution step 1 of 2 The comoonents of elevator oosition control are shown in fiouret. Step 1 of 2 The components of elevator position control are shown in figure!. Controller Actuator Process Floor Sensor Figure I step 2 of 2 The input is given to the logic controller through comparator. The logic controller drives the transmission mechanism of the elevator (Motor or hydraulic). The elevator moves up and down and reaches the required floor position. When the corresponding floor button is pressed, the controller reduces the speed of the motor to stop the elevator in the respective floor. The electro switch acts as a sensor for coarse measurement to measure the floor level. Accuracy level can be fixed in the sensor considering the cable stretch due to cab load. This sensor enables to locate the elevator in the respective floor accurately without any deviation in the measurements. This complete mechanism forms a closed loop as in figure 1. The error detector compares the reference input to the output of the closed loop to ensure the exact location of the elevator. Hence, the block diagram of the elevator mechanism considering negligence is explained. Problem 1.07PP Feedback control requires being able to sense the variable being controlled. Because electrical signals can be transmitted, amplified, and processed easily, often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. Describe a sensor that would give an electrical output proportional to the following; (a) Temperature (b) Pressure (c) Liquid level (d) EJpwj^fJinJjiri ainnn a ninp (nr hlnnH alnnn an artprv^ (d) Flow of liquid along a pipe (or blood along an artery) (e) Linear position (t) Rotational position (g) Linear velocity (h) Rotational speed (i) Translational acceleration (j) Torque Step-by-step solution step 1 of 2 (A) . TEMP M ATURE : Thermocoi^lc. (B) . PRESSURE : Pressure gauge. (C) . U Q UlL? LEVEL : Bourdon tube and LVDT s combination. (Py FLOW OF LIQUID ALONG PIPE FORCE : Any pres sure sensor can be used. LINEAR POSITION : LVDT lin e a r Variable Differential Transformer). Step 2 of 2 ROTATIONAL POSiriON : Potentiometer. (G). UNEAR VELOCITY : Speedometer. 0 ^ . ROTATIONAL SPEED : Tachometer. ( 1 \ TRANSLATIONAL ACCLERATION : LVDT ( J y TORQUE: Combination of Gear and Tachometer. Problem 1.08PP Each of the variables listed in Problem can be brought under feedback control. Describe an actuator that could accept an electrical input and be used to control the variables listed. Give the units of the actuator output signal. Problem Feedback control requires being able to sense the variable being controlled. Because electrical signals can be transmitted, amplified, and processed easily, often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. Describe a sensor that would give an electrical output proportional to the following; (a) Temperature (b) Pressure (b) Pressure (c) Liquid level (d) Flow of liquid along a pipe (or blood along an artery) (e) Linear position (t) Rotational position (g) Linear velocity (h) Rotational speed (i) Translational acceleration (j) Torque Step-by-step solution step 1 of 1 ^ An actuator amplifies the signal taken from the sensor. Any Electronic A n^lifier can be used to do such. The Operational Amplifier is the most commonly used Actuator. e .- ' OPAMP OP AMP Units o f the Actuator output signal are current (Anq^ere) or Voltage (Volts) Problem 1.09PP Feedback in Biology (a) Negative Feedback in Biology. When a person is under long-term stress {say, a couple of weeks before an exam!), hypothalamus (in the brain) secretes a hormone called Corticotropin Releasing Factor (CRF) which binds to a receptor in the pituitary gland stimulating It to produce Adrenocorticotropic hormone (ACTH), which in turn stimulates the adrenal cortex (outer part of the adrenal glands) to release the stress hormone Glucocorticoid (GC). This In turn shuts down (turns off the stress response) for both CRF and ACTH production by negative feedback via the bloodstream until GC returns to Its normal level. Draw a block diagram of this closed-loop system. (b) Positive Feedback in Biology. This happens in some unique circumstances. Consider the birth process of a baby. Pressure from the head of the baby going through the birth canal causes (b) Positive Feedback in Biology. This happens in some unique circumstances. Consider the birth process of a baby. Pressure from the head of the baby going through the birth canal causes contractions via secretion of a hormone called oxytocin which causes more pressure which in turn intensifies contractions. Once the baby is born, the system goes back to normal (negative feedback). Draw a block diagram of this closed-loop system. Step-by-step solution step 1 of 9 (a) Step 2 of 9 The block diagram to show an example of closed loop negative feedback in a biological system Is shown in figure 1. Step 3 of 9 Long Nofinal Gluco oofticoids(CD)— ^L^Hypolhalam c«L|^^^Pituitary Adrenal GCs glands M i1Glands I Stress response - off ^Stress response • off Negatice feedback _______ Figure 1 Step 4 of 9 Here, ‘GD’ is the Glucocorticiods, ‘CRF’ is the Corticotrophin Releasing Factor, and ‘ACTH’ is the Adrenocorticotropic hormone. Step 5 of 9 Thus the block diagram of closed loop negative feedback In a biological system is drawn and shown in figure 1. Step 6 of 9 (b) Step 7 of 9 The block diagram to show an example of closed loop positive feedback in a biological system Is shown in figure 2. Step 8 of 9 Oj^tocin Desired , Baby Birth 1 ^ f 1 ^ I s Pressure Pressure y ? y head canal \ n (Contractions) Positive feedback Figure 2 Step 9 of 9 Thus the block diagram is drawn to show the closed loop positive feedback in a biological system. Problem 2.01 PP Write the differential equations for the mechanical systems shown in Fig.. For Fig.(a) and (b), state whether you think the system will eventually decay so that it has no motion at all, given that there are nonzero initial conditions for both masses and there is no input; give a reason for your answer. Figure Mechanical systems Nofnctioa Nofrictkn (I) h (I) I— w v —^ —I Step-by-step solution step 1 of 6 (a) Refer to Figure 2.41 (a) in the text book for the block diagram of a mechanical system. Draw the free body diagram of mass m ,. - *1*1- ^ *1 *^ *a*a Write the differential equation describing the system. m,x, -- ^ x , -k^x^-k^x^ Step 2 of 6 Draw the free body diagram of mass . t z Write the differential equation describing the system. m^x^ ^-k^x^-k^x^-^k^x^-¥k^y + *3 (*1 - ■»!)+*3 - .y ) = 0 Thus, the differential equations describing the system are. (x,-JCj) s 0 There is friction that affects the motion of both the masses. Thus, the system decays to zero motion for both the masses. Step 3 of 6 (b) Refer to Figure 2.41 (b) in the text book for the block diagram of a mechanical system. Draw the free body diagram of mass ■ Write the differential equation describing the system. ( x j- x , ) = 0 Step 4 of 6 Draw the free body diagram of mass . *2 *1 Write the differential equation describing the system. m^x^ --b ^ x ^ -k ^ x ^ -k ^ x ^ + k ^ A l|X 2 4'i^(x2~Xi)4'A^A^ = 0 Thus, the differential equations describing the system are. + it,x, + (xj - ) * 0 Although friction affects only the motion of the left mass directly, continuing motion of the right mass excites the left mass, and that interaction continues until all motion damps out. Step 5 of 6 (c) Refer to Figure 2.41 (c) in the text book for the block diagram of a mechanical system. Draw the free body diagram of mass m ,. | _ x . ^*1 - *1*1 - *2 * 1 *► tjX j Write the differential equation describing the system. m,x, *-i^x, -ik,x, -A^x, +i^Xj+i\x^ =0 m,jt, + 4| (x, - X,) + V i + *1 (*i - * 2) = 0 Step 6 of 6 Draw the free body diagram of mass . iW j s -k^x^ - A ^ X j - A iX j + ^X| + + F + i|X i+ i| ( x i- x i) + * 2 (X 2 -x i) = F Thus, the differential equations describing the system are. n\Xy + 6 |(x,-X j)+ A |X ,+ A ^(x, - X j) * 0 m ,x ,+ i|X ,+ A ^ (i,-X |)+ < :,(x ,-X |) = f Problem 2.02PP Write the differential equation for the mechanical system shown in Fig. State whether you think the system wili eventuaily decay so that it has nomotion at ail, given that there are nonzero initial conditions for both masses, and give a reason for your answer. Figure Mechanical system M < 1 Noftictioo Step-by-step solution Step-by-step solution step 1 of 4 Refer to Figure 2.40 in the textbook. Draw the free body diagram of left side mass, m j. K ^ ,J— , m^l U X i - x ^ Figure 1 Step 2 of 4 Write the equation of motion for left side mass, m |. m, + JCj (jcj - Xj * 0 »Hj^+(Ar|+Ar,)aii-iC, x ,+ i,{ x ,~ x ,) = 0 (1) Draw the free body diagram of right side mass, m |. Figure 2m , Step 3 of 4 Write the equation of motion for right side mass, m2 ■ m ,X 2+JC 2(xj-x,)+ 6 ^ (x j- xj)+JC,;^ = 0 i ^ ( i , - i i ) = 0 (2) Therefore, the differential equations for mechanical system are. + /T j)x,-Jir2 - JITjX,+(AT, + Jirj)x2 + ( ij - x ,)s 0 step 4 of 4 The relative motion between x, and ji^ , decays to zero due to the damper. However, the two masses continue oscillating together without decay, since there is no friction opposing that motion and also no flexure of the end springs, and these are the essentials to maintain oscillation of the two masses. Thus, the system will finally decay, as the relative motion between x, and x^. decays to zero. Problem 2.03PP Write the equations of motion for the double-pendulum system shown in Fig. Assume that the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length /, and the springs are attached three-fourths of the way down. Step-by-step solution step 1 of 6 Consider the circuit diagram. r m m Double pendulum system Step 2 of 6 ^ Re-draw the circuit diagram. ^ / I ^ / /1 m Here there are two degrees of freedom Step 3 of 6 ^ At any instant, let one of them be displayed by 6| and other by 6,. Step 4 of 6 Writing about point ‘0’ the moment equation is. Jt(sinO2-sinO,)cos0sm /^d, Similarly, write the moment equation for other pendulu -mg/sin 6,- ^ ^ / j *(sin 6,-sine|)cos 8 2 =m/^e. Step 5 of 6 Assume the angles are small, then sin0| s 0 |, cos9|S ]and cos02 sl.T he above equations are modified as. * (e,-e,)= m /= 9 , Therefore, the above equation is moment equilibrium about the pivot point of the left pendulum. Step 6 of 6 ^ Write the equation for moment equilibrium about the pivot point of the right pendulum. * ( 6 ,- 6 ,) - 0 Therefore, the above equation is moment equilibrium about the pivot point of the right pendulum. Problem 2.04PP Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length / suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec? (The inertia I of a thin stick about an end point is Assume that 0 is small enough that sin Q ^ Q.) Why do you think grandfather clocks are typically about 6 ft high? Step-by-step solution step 1 of 3 Consider the following equation for the sum of all external moments about the center of mass of Consider the following equation for the sum of all external moments about the center of mass of a body. M ^ I a ......(1) Where. / is the body’s mass moment of inertia about its center of mass, a is the angular acceleration of the body. The coordinates and forces relationship is shown in Figure 1. o Figure 1 Step 2 of 3 From Figure 1, the equation of moment about point O is given below. M o = -m g n ^ s m 0 — ........( 2) Substitute for /p in equation (2). -m g = ^ m /*tf+ mg X ^ sin = 0 f l + ^ 8 i n f l = 0 ......(3)2i Consider ^is very small in equation (3). g + ^ e = o (4) Consider the following general equation. ^+24^a>e+ai‘ff= 0 (5) Compare equation (4)and equation (5). Step 3 of 3 ^ Consider the following equation for the period for a swing. r=— ....(7) Substitute equation (6)in equation (7). 2/r - 2 > r /- = - ......(8) Substitute 2 for / and 9.81 for g in equation (8). r = 2 ;r jI 2x2 ’'V 3 x9.8I = 2.31 = 2sec Consider the following equation for the value of length. /= -? § - (9) Substitute 9.81 for g in equation (9). 3x9.81/ * - 8^* * 0.3727 m Thus, the vaiue of length and period for a swing are [0.3727m]a Hence, the 1 second for a swing from one side to the other. This pendulum is shorter because the period is faster. But if the period had been 2 second, the pendulum length is 1.5 meters. Problem 2.05PP For the car suspension discussed in Example, plot the position of the car and the wheel after the car hits a “unit bump”(that is. ris a unit step) using Matlab. Assume that m1 = 10 kg, m2 = 250 kg, Kw= 500,000 N/m, and Ks = 10,000 N/m. Find the value of b that you would prefer if you were a passenger in the car. Example A Two-Mass System: Suspension Model Figure 1 shows an automobile suspension system. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel. A system consisting of one of the four wheel suspensions is usually referred to as a quarter-car model. The system can be approximated by the simplified system shown in Fig. 2 where two spring constants and a damping coefficient are defined. Assume that the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20 kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20 kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied weight, we find that Mv ~ 1,000,000 N/m. By using the step response data in Fig. 3(b) and qualitatively observing that the car?s response to a step change matches the damping coefficient curve for ^ = 0.7 in the figure, we conclude that b = 9800 N?sec/m. Solution. The system can be approximated by the simplified system shown in Fig. 2. The coordinates of the two masses, x and y, with the reference directions as shown, are the displacements of the masses from their equilibrium conditions. The equilibrium positions are offset from the springs? unstretched positions because of the force of gravity. The shock absorber is represented in the schematic diagram by a dashpot symbol with friction constant b. The magnitude of the force from the shock absorber is assumed to be proportional to the rate of change of the relative displacement of the two masses?that is, the force = The force of gravity could be included in the free-body diagram; however, its effect is to produce a constant offset of X and y. By defining x and y to be the distance from the equilibrium position, the need to include the gravity forces is eliminated. The force from the car suspension acts on both masses in proportion to their relative displacement with spring constant ks. Figure 2.6 shows the free-body diagram of each mass. Note that the forces from the spring on the two masses are equal in magnitude but act in opposite directions, which is also the case for the damper. A positive displacement y of mass m2 will result in a force from the spring on m2 in the direction shown and a force from the spring on m1 in the direction shown. However, a positive displacement x of mass m1 will result in a force from the spring ks on m1 in the opposite direction to that drawn in Fig. 4, as indicated by the minus X term for the spring force. The lower spring kw represents the tire compressibility, for which there is insufficient damping (velocity-dependent force) to warrant including a dashpot in the model. The force from this spring is proportional to the distance the tire is compressed and the nominal equilibrium force would be that required to support m1 and m2 against gravity. By defining x to be the distance from equilibrium, a force will result if either the road surface has a bump (r changes from its equilibrium value of zero) or the wheel bounces (x changes). The motion of the simplified car over a bumpy road will result in a value of r(t) that is not constant. As previously noted, there is a constant force of gravity acting on each mass; however, this force has been omitted, as have been the equal and opposite forces from the springs. Gravitational forces can always be omitted from vertical-spring mass systems (1) if the position coordinates are defined from the equilibrium position that results when gravity is acting, and (2) if the spring forces used in the analysis are actually the perturbation in spring forces from those forces acting at equilibrium. Figure 1 Automobile suspension Figure 2 The quarter-car model? T Figure 3 Responses of second-order systems versus ^ ; (a)impulse responses; (b) step responses Figure 4 Free-body diagrams for suspension system Ki-i 0 ■ r ) ili^-r) Applying Eq. (1) to each mass and noting that some forces on each mass are in the negative (down) direction yields the system of equations Some rearranging results in F=m a, (2.1) 4 ( J - i ) + t ,( y - i) - * ,( i- r ) = « iii , (18) = (19) The most common source of error in writing equations for systems like these are sign emors. The method for keeping the signs straight in the preceding development entailed mentally picturing the displacement of the masses and drawing the resulting force in the direction that the displacement would produce. Once you have obtained the equations for a system, a check on the signs for systems that are obviously stable from physical reasoning can be quickly carried out. As we will see when we study stability in Section 6 of Chapter 3, a stable system always has the same signs on similar variables. For this system, Eq. (2) shows that the signs on the and X terms are all positive, as they must be for stability. Likewise, the signs on the and /term s are all positive in Eq. (2). 1 + — ( t - W + . ^ d - y ) + ^ i = ^ r . (1 1 0 ) Ml Ml Mi 9 + - ^ 0 - t ) + ^ ( y - x > = o . M2 M2Mi (2.11) The transfer function is obtained in a similar manner as before for zero initial conditions. Substitutino s for d/dt in the differential eouations vields J^X(I)+ 1 — (Xd) - X(D) + .^(X(I) - X(D) + ^ X (.) = ^ « (I),Ml Dll Jfl] Mi ^ n s ) + s — m ) - x m + — o r ( s ) - x m = a which, after some algebra and rearranging to eliminate X(s), yields the transfer function I'd ) ^ ( » + > ) * * + ( s r + ^ ) * " + ( ^ r + ^ + f e ) ‘" + ( ^ ) ' + i ^ (112) To determine numerical values, we subtract the mass of the four wheels from the total car mass of 1580 kg and divide by 4 to find that m2 = 375 kg. The wheel mass was measured directly to be m1 = 20 kg. Therefore, the transfer function with the numerical values is Y(s) 131e06(5 + 13.3) *d) “ J* + (516.1)s“ + (5.685«04)j“ + (I J(n < 0 ^ + l.733«Or 0.13) We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find the response of the car body to inputs resulting from the car motion over a bumpy road. Step-by-step solution step 1 of 3 Consider the transfer function. i '( 4 '»!'»! 4 j Ja l 'U A A 1 m i«j Where. is the mass. kg is the car deflection. k^ is the applied weight. b is the damping. Write the MATLAB program, m l = 10; m2 = 250; kw = 500000; ks= 10000; Bd = [1000 3000 4000 5000]; t = 0:0.01:2; for i = 1:4 b = Bd(i): A=[0 1 0 0;-( ks/ml + kw/ml ) -b/ml ks/ml b/m l; 0 0 0 1; ks/m2 b/m2 -ks/m2 -b/m2 ]; B=[0; kw/ml :0 ;0 ]; C=[1 0 0 0 ; 00 1 0]; D=0; y=step(A.B.C.D,1,t): subplot(2.2,i): plot(t, y(:,1),t, y(:,2)): legend(Wheer,'Car’): title = sprintf('Response with b = % 4.1f, b ); end Step 2 of 3 The output for the MATLAB code is given in figure 1. Step 3 of 3 Thus, the smallest overshoot is with b is 5,000, and the jump in car position is the fastest with this damping value. From figure 1, ^ s 3,000 is the best compromise. Therefore, overshoot for lower values, and the system gets too fast for larger values. Problem 2.06PP Write the equations of motion for a body of mass M suspended from a fixed point by a spring with a constant k. Carefully define where the body’s displacement is zero. Step-by-step solution step 1 of 3 Displacement is Zero when forces are balanced Hence Ti-k xT No displecement (di^koement) Step 2 of 3 ^ e n M g=-la, ^x&ere S =dt =Acceleration due to gravity. Then, The displacement be comes Zero. It is shown in the figure where displacement is zero. Step 3 of 3 From the above figure, kx+Mg = 0 I .', kx = -hK I Problem 2.07PP Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a changeable damping, b(u1). It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force, u2. in opposite directions on the wheel axle and the car body. (a) Modily the equations of motion in Example to include such control inputs. (b) Is the resulting system linear? (c) Is it possible to use the force u2 to completely replace the springs and shock absorber? Is this a good idea? Example A Two-Mass System: Suspension Model a good idea? Example A Two-Mass System: Suspension Model Figure 1 shows an automobile suspension system. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel. A system consisting of one of the four wheel suspensions is usually referred to as a quarter-car model. The system can be approximated by the simplified system shown In Fig. 2 where two spring constants and a damping coefficient are defined. Assume that the model is for a car with a mass of 1580 kg, including the four wheels, which have a mass of 20 kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection, we find that ks = 130,000 N/m. Measuring the wheel?s deflection for the same applied weight, we find that Mv ~ 1,000,000 N/m. By using the step response data in Fig. 3(b) and qualitatively observing that the car?s response to a step change matches the damping coefficient curve for ^ = 0.7 in the figure, we conclude that b = 9800 N?sec/m. Solution. The system can be approximated by the simplified system shown in Fig. 2. The coordinates of the two masses, x and y, with the reference directions as shown, are the displacements of the masses from their equilibrium conditions. The equilibrium positions are offset from the springs? unstretched positions because of the force of gravity. The shock absorber is represented in the schematic diagram by a dashpot symbol with friction constant b. The magnitude of the force from the shock absorber is assumed to be proportional to the rate of change of the relative displacement of the two masses?that is, the force = The force of gravity could be included In the free-body diagram; however, its effect is to produce a constant offset of X and y. By defining x and y to be the distance from the equilibrium position, the need to include the gravity forces is eliminated. The force from the car suspension acts on both masses in proportion to their relative displacement with spring constant ks. Figure 2.6 shows the free-body diagram of each mass. Note that the forces from the spring on the two masses are equal in magnitude but act in opposite directions, which is also the case for the damper. A positive displacement y of mass m2 will result in a force from the spring on m2 in the direction shown and a force from the spring on m1 in the direction shown. However, a positive displacement x of mass m1 will result in a force from the spring ks on m1 in the opposite direction to that drawn in Fig. 4, as indicated by the minus X term for the spring force. The lower spring kw represents the tire compressibility, for which there is insufficient damping (velocity-dependent force) to warrant including a dashpot in the model. The force from this spring is proportional to the distance the tire Is compressed and the nominal equilibrium force would be that required to support m1 and m2 against gravity. By defining x to be the distance from equilibrium, a force will result if either the road surface has a bump (r changes from its equilibrium value of zero) or the wheel bounces (x changes). The motion of the simplified car over a bumpy road will result in a value of r(t) that is not constant. As previously noted, there is a constant force of gravity acting on each mass; however, this force has been omitted, as have been the equal and opposite forces from the springs. Gravitational forces can always be omitted from vertical-spring mass systems (1) if the position coordinates are defined from the equilibrium position that results when gravity is acting, and (2) if the spring forces used in the analysis are actually the perturbation in spring forces from those forces acting at equilibrium. Figure 1 Automobile suspension Figure 2 The quarter-car model? U Figure 3 Responses of second-order systems versus ^ ; (a)impulse responses; (b) step responses 3 ^ z ms Figure 4 Free-body diagrams for suspension system lO-i IJi-H h^-ii Applying Eq. (1) to each mass and noting that some forces on each mass are in the negative (down) direction yields the system of equations Some rearranging results in F = m a , (2.1) 4 ( J - i ) + t , ( y - x ) - t , ( i - r ) = « > i * , (2.8) - f c O '- J ) - * ( » - « (2.9) The most common source of error in writing equations for systems like these are sign emors. The method for keeping the signs straight In the preceding development entailed mentally picturing the displacement of the masses and drawing the resulting force in the direction that the displacement would produce. Once you have obtained the equations for a system, a check on the signs for systems that are obviously stable from physical reasoning can be quickly carried out. As we will see when we study stability In Section 6 of Chapter 3, a stable system always has ? tKa oinne- nn X terms are all positive, as they must be for stability. Likewise, the signs on the and /term s are all positive In Eq. (2). y +A ( t - y , + i. ( x - y ) + t x = ^ r . (2.10)Ml Mi M i — 0 - t ) + ^ (y -x )= O . M 2 M 2Mi (2.11) The transfer function is obtained In a similar manner as before for zero initial conditions. Substituting s for d/dt in the differential equations yields j^ x c i)+ - r m + ^ ( X ( j) - Y(sn+ ^ x w =Mi M i M i M i + » — ()'(») - * ( ! ) ) + — ()'(j) - * ( » ) ) = a which, after some algebra and rearranging to eliminate X(s), yields the transfer function X M _____________________ ^ ( » + > ) «(») (2.12) To determine numerical values, we subtract the mass of the four wheels from the total car mass of 1580 kg and divide by 4 to find that m2 = 375 kg. The wheel mass was measured directly to be m l = 20 kg. Therefore, the transfer function with the numerical values is r(s) 131e06(s + 13.3) S(j) “ J* + (516.1)s“ + (5.685«04)j2 + (Ume06)s + l.733«07' 0.13) We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find the response of the car body to inputs resulting from the car motion over a bumpy road. Step-by-step solution step 1 of 3 (A) The FBD shows the a<iditioa of the variable force, \3 2 and shows 6 as in the FBD of figure 2.5, However, here 6. Is a function of the control variable ui. The forces below are drawn in the direction that would result from a po sitive displac ement of ^ O " - * ) - ■ K '. M ,y = - K , ( y - x ) - i{ u ^ ) { y - i ) +Uj Step 2 of 3 The system is linear with respect to because it is additive. Step 3 of 3 (C) It is technically possible. However it would take very high forces and thus a lot of power and is therefore not done. Problem 2.08PP In many mechanical positioning systems there is flexibility between one part of the system and another. An example is shown in Fig. 1 where there is flexibility of the solar panels. Figure 2 depicts such a situation, where a force u is applied to the mass M and another mass m is connected to it. The coupling between the objects is often modeled by a spring constant k with a damping coefficient b, although the actual situation is usually much more complicated than this, (a) Write the equations of motion governing this system. (b) Find the transfer function between the control input u and the output y. Figure 1 Communications satellite Source: Courtesy Space Systems/Loral (SSL) Figure 2 Schematic of a system with flexibility Step-by-step solution step 1 of 7 Sketch the schematic o f a sjrstem with flexibility. Step 2 of 7 ^ Sketch the free body diagram of the given figure. iibc-y) b ( f - y )U Step 3 of 7 > X K x-y) Step 4 of 7 ^ From the free body diagrams, we get mx = - J t( x - - i ( x - j^) = u + ^ (x - + A ( i - J> ) Simplify further. X H — X H — X ------- y-------- y = 0 m m m m - k A . _ jt A . 1 — X -------- x + V H -------- v + — V = — u M M M M M Thus, the equation of motion governing the system is obtained as -k^b.k A . - X H — X H — X ------- y-------- y = 0 m m m m - k A . . ^ A . 1 — X -------- x + V H -------- v + — V = — u M M M M M Step 5 of 7 ^ (b) Find ^ Limlace tra n sf^ n o f the equations -k\.k A . . *X H — X H — X ----y------ y = 0 m m m m - k A . - it ^ A . 1 M M M M M s ^ x + - X + - s x - - r - - s r = 0 — X - — sX + + — Y + — ysY = — U M M M M M Rewrite in matrix form (m ^ + b s + k -(b s + k ) ^0^ [ -(As + Jt) Ms^ + bs + k ) [ Y ; [ u ) Step 6 of 7 Solve using the Cramer’s rule, detl( m ^ + b s + k 0^ T - ( t e + t ) u ] + As + it - (As + it) det 7 =-(A s+it) Afs^+As+it + As + it {^m s^ + As + itJ^Afe^ + As + itj - (As + it)^ Step 7 of 7 Simplify further. y ms^ + As + it ^ + bs +k^ (^M s^ +bs + k'j - (As + k ^ + As + it U mMs + (« + M')bs + (M + m) ks^ Problem 2.09PP Modify the equation of motion for the cruise control in Example, Eq., so that it has a control law; that is, let u = K (vr-v), (2.108) where vr= reference speed. (2.109) /C= constant. (2.110) This is a “proportionarcontrol law in which the difference between \^rand the actual speed is used as a signal to speed the engine up or slow it down. Revise the equations of motion with vr as the input and v as the output and find the transfer function. Assume that/n = 1500 kg and b = 70 N -sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a value of Kthat you think would result in a control system In which the actual speed converges as quickly as possible to the reference speed with no objectionable behavior. 70 N -sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a value of Kthat you think would result in a control system In which the actual speed converges as quickly as possible to the reference speed with no objectionable behavior. Eq. b uV H — V = —. m m EX A M PLE 2.1 A Simple System; Cruise Control Model 1. Write die equedon of motion for the ipeed m l forward motion of lhecaribownmFig.2.1 aasnmiiig that the en^ne imparta a force ■ ae abown. Take the Laplace transform of the reauiting differential eqoadon and find the transfer function between the input« and the output v. 2. Use Matlab to 6nd the response of the velocity of the car for the case in which the inpntjumpe from being « = Oat time f = 0 to a constant Hgui«2.1 Cruise control model « B 500 N thereafter. Assume that the car a visoous drag coeflkient, b B 50 N*secAn.I « is 1000 kg and Solution 1. Equations of motion: For sunpBdty we asanme that the rotmkNinl inertia o f the wheeb b negligible and that there b ftiction retarding the motion of the car that b proportional to the car’s speed with a proportionality constant, The car can then be approximated for modeling poipoeea using the ftee-body diagram men m Fig. 2.2, which defines coordinmes, shows all forces acting on the body (heavy lineaX and imficates the accdetaCion (dashed line). The coordinate of the car’a positioo X u the distaoce from the refotenoe line sbo«m and b chosen ao that positive isto the r i ^ Note that in this case the meitial acceleration u amply the second derivative of x (that is, a * x) because the car position b measured wkh respect to « inertial reference frame. The equation of motion b found mung Eq. (2.1). The friction force acts opposite to the direction motiMi; therefore k b drawn rqiposhe to the direction o f positive motion m l eiketcd as a negative fioroe in Eq. (2.1). Theresukb (2.2) ^ b . M X + — X = — . 03) For the case of the automotive cruise control where the variable of interest b the speed, v ( ^ ) , the equbion of motion becomes * “ 0.4)V + —V B —. Hk sobtion of wch an equation win be cowered in detail in Q upter 3; however, the essence b that you aaaume a aolotionoftfae form v s pvea an input of the form ■ b Then, tince v b the differential equation chi be written aa^ 03) Rgurt2,2 r > c o body diagram for t:: 0 6 ) a 7 ) S te p re sp o n se wtth H atlabThe c* term C H iceb out, and we find that V - !■ yg _ m « . . + 1 ' For reasons that will become efetf in Quqker 3, this b often written using capital letters to signify that k b the‘Itunsforro'’of the solution, or v w _ i * + h' This expresskm of die dUferential equation (2.4) b called the tranti’ er function and wiD be used extensively m later chapters. Note that, in essence, we have aubstkuled i fof dfdt in Eq. (Z4X Thb transfer Auh tion serves as a mwh model that relates the ch’s velocity to the forces propelling the car, that b , inputs from the accelerator pedal. Transfer functions of a system will be used in later chapters to design feedback corXrollers such as a cruise control device found in many modern cars. X Time reaponae: The dynamics ofn system can be prescribed to Matlab in terms of its transfer function as can be seen in the Matlab statemenb below that implemeiits E<^ (2.7). The step fimetion in Madab calcn- bles die time response of a Ikictf system to ■ unit step input Bwuuie the system b finear. the output for thb case can be multiplied by the magnitude of the input mep to derive a step re 9onae of any ampUtude. Equivaleikly. ^ can be multiplied by die magnitude of the input step. % setsupthem odetod«flnetlN transfer ftincUon 9S • (1/1000]/(s -f 5<V1000); % defines the transfer function from Eq. (2.7) with the numbers filled l a step(S 00*9s); % plots the step response for u • 500. calculate and plot the time reaponie o f velocity for an input step with n 500>N magnitude. The Mep responae b shown m Fig. 2.3. Rgure2.3 Response of the car velodtytoastepinti Step-by-step solution step 1 of 7 Consider the equation of motion. V + — V ® — IT Where, u is the input V is the output m is the mass b is the friction force Substitute JC (v^-v) foru. m m . b K K V + — V * — V , ------- Vm m m . b K K V + — V + — V = — V-m m m A block diagram of the scheme is shown in Figure 1. Step 2 of 7 1u m . b . K Step 3 of 7 Figure 1 Step 4 of 7 Consider the transfer function of the closed loop system. y(s) K (x ) b K S + — + —m m Write the inputs for MATLAB. K m denI m mj Step 5 of 7 Consider K=100,200,1000,5000 Write the MATLAB program, m = 1500; b = 70; k = [100 200 1000 5000]; hold on t=0:0.2:50; for i=1:length(k) K=k(i); num =K/m; den = [1 b/m+K/m]; sys=tf(num,den); y = step(sys,t); plot(t,y) grid end hold off Step 6 of 7 The output for the MATLAB code is given in Figure 2. Step 7 of 7 Thus, the larger the K is the better the performance with no objectionable behavior for any of the cases. Increasing K also results in the need for higher acceleration is less obvious. In Figure 2, for K - 1,000 , there is a response in 5 seconds and the steady state error is 5%. Problem 2.1 OPP Determine the dynamic equations for lateral motion of the robot in Fig. 1. Assume it has three wheels with a single, steerable wheel in the front where the controller has direct control of the rate of change of the steering angle, Usteer, with geometry as shown in Fig. 2. Assume the robot is going in approximately a straight line and its angular deviation from that straight line is very small. Also assume that the robot is traveling at a constant speed, Vo. The dynamic equations relating the lateral velocity of the center of the robot as a result of commands in Usteer are desired. Figure 1 Robot for delivery of hospital supplies. Source: AP Images Figure 2 Model for robot motion Step-by-step solution step 1 of 3 Refer FIGURE 2.46 in the textbook. Consider the following equation for the sum of all external moments about the center of mass of a body. M ^ I a .......(1) Where. / is the body’s mass moment of inertia about its center of mass. a is the angular acceleration of the body. Consider the following equation for the time rate of change of the steering wheel angle. S, - U ^ .......(2) Where. is the control input. The turning rate change with respect to x axis is shown in Figure 1. Step 2 of 3 Consider the following equation for the carts turningrate of change with respect to x axis. sin 9, Where. is nonzero. L is the length of the wheel, F^is the constant speed. Consider is small and rearrange the above equation. (3) The lateral motion as a function of y is shown in Figure 2. Step 3 of 3 Consider the following equation for the actual change in the carts lateral position. .....(4) Take differentiation on both sides in equation (4). y = ......(5) Substitute equation (3)in equation (5). .. v^e. Take differentiation on both sides, y »-2— *• ......(o) Substitute equation (1 )in equation (6). (7) Thus, there is no dynamics come into equation (7). Therefore, there was no need to invoke equation (1). Problem 2.11 PP A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig.: 10^ i+ = i-= 0. Figure Circuit for Problem Find the transfer function of the simple amplification circuit shown using this model. Step-by-step solution step 1 of 2 > Refer to Figure 2.43 in the circuit. The output voltage is, r ^ = — [ r , - r . ] o ) 5 + r * The currents at inverting and noninverting terminals is, I . - L - 0 Apply KirchhofTs current law at node V_. y - V y - V—— — s f i - = 0 R, « , ) K R, \ R,R^ ) R^R, R„+R^ V = V^+ ^ r , .....(2)R^+ R, - R^ + R, - Step 2 of 2 The voltage at noninverting input terminal is, > ; = o v Substitute O Vfor y and ------^ f o r y in equation (1). v ^ = — ( o —“ 5+ it^ R^+R/ Rm*Rf J - 10’ * / y <«’ R . y s + lRi, + R ^ “ s+lJ^+Rf “ s+lRj^+R^ s+lR^,+R^ K^LJ 2L_5^1=.ioL_L^-^ j + + S + lR^+Ry " ^ r (s + I)( R ,* R ,)+ IO ^ R ,) iQt R , •“[ (/i„ + Ji/)(j + l) J + “ »'-[(*+i)(^+«,)+ioX]— io ’«A ________ -1 0 ’ i?, K. "(*+i)(J?.+«^)+ioX [R^+R/J *'* » + i+ io ’ f — 5s— 1 Therefore, the transfer function of the circuit, is K. j + 1 + 10’ Problem 2.12PP Show that the op-amp connection shown in Fig. 1 results in Vout = Vin if the op-amp is ideal. Give the transfer function if the op-amp has the nonideal transfer function of Problem. Figure 1 Circuit A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. 2: A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. 2: 10^ i+ = i-= 0 . Figure 2 Circuit Find the transfer function of the simple amplification circuit shown using this model. Step-by-step solution step 1 of 3 Refer to the circuit diagram in Figure 2.38 in the textbook. The voltage at the non-inverting terminal of the op-amp is, v ,-V ^ The voltages at the inverting and non-inverting terminals of the op-amp are same for ideal op- amp. v. = n The inverting node of the op-amp is directly connected to output node. So, y^=y- Substitute for v_ in the equation. y ^ = y u Hence, it is proved. Step 2 of 3 The simplified op-amp circuit is shown in Figure 1. Step 3 of 3 From Figure 1, the output voltage is, = .4 (v .-v .) Substitute for and for v_ in the equation. y ^ + A r ^ = A V ^ ( l + A )V ^ = AVi. y « A K. 1 + .4 Therefore, the transfer function. of non-ideal op-amp isA \ + A Problem 2.13PP A common connection for a motor power amplifier is shown in Fig. The idea is to have the motor current follow the input voltage, and the connection is calied a current amplifier. Assume that the sense resistor rs is very smail compared with the feedback resistor R. and find the transfer function from Vin to la. Also show the transfer function when Rf= Figure Op-amp circuit for Problem Step-by-step solution step 1 of 3 A common connection for a motor power an^lifier is given. Sketch the given figure. Step 2 of 3 At node A, write the voltage equation. V L -0 ^ R At node B, write the voltage equation, with R^-^ R , 0 - f% 0 - V . , R R, ^ r + r/ ‘ step 3 of 3 ^ The d3mamics o f the motor is modeled with negligible inductance as J^9^ + We can rewrite as 4, + * 4 = A'^.as + At the ou^ut, &om the equation , 0 - r . 0 - r . ,, R 5, j,e ,+ b e , = K,i^ And the motor equation Find To V „ = l^R, + I^R^^■ K, J^s -I - b y — 0 V —0 V — 0Substitute in the equation ------ 1 - — = 0 and sinqjlify.R^ Rj R R^ 5 ,1 .* ■ • J , s ^ b ) R This e^^ression shows that, in the steady state, when s — > 0 , the current is proportional to the ii^ut voltage. We know that the current anq^lifier has no feedback from the output voltage, in which And thus, we get U _ R y > . Thus, the transfer fimction is obtained as ^ - - M l U R Problem 2.14PP An op-amp connection with feedback to both the negative and the positive terminals is shown in Fig. 1. If the op-amp has the nonideal transfer function given in Problem, give the maximum value possible for the positive feedback ratio, ' ’ = 7Tir' in terms of the negative feedback ratio, N = for the circuit to remain stable. Figure 1 Op-amp circuit A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. 2: 10^ i+ = i-= 0 . Figure 2 Circuit for Problem Find the transfer function of the simple amplification circuit shown using this model. Step-by-step solution step 1 of 4 Clearly — r+ R E „ E,Step 2 of 4 • [ r„ r,J L e,+e„J r ]V. rt-R Step 3 of 4 If the Op - Amp is non ideal then. A(say) M l , I ,r + R jj R (+ R k R ,^V „ = A E ,+ R „ -AR, E ,+ E „ I, R ,+ R ,. R +rJ Step 4 of 4 In the limiting case we put denominator = 0 R*,A rAWe get 1+ R,4-Rj^ R+r 1+NA=PA 1=(P-N)10’ 's + 1 s + U C P - Al)10’ Poleisat[(P-N )10’ -l]. Equating it to Zero for limiting case, ,1+N.lO’We have P S - 10' Max value is |N+10'^ P roblem 2.15PP Write the dynamic equations and find the transfer functions for the circuits shown in Fig.. (a) Passive lead circuit (b) Active iead circuit (c) Active lag circuit (d) Passive notch circuit Figure (a) Passive iead; (b) active lead; (c) active lag; and (d) passive notch circuits Figure (a) Passive iead; (b) active lead; (c) active lag; and (d) passive notch circuits —W ^ ^ V V V - j— -o*U 0 -i—v v \ I t vvnJ - o K/2> ^ 2 C S-------------- A A o S tep-by-step s o lu tio n (a) Refer to Figure 2.48 (a) in the textbook for passive lead circuit. Write the KirchhofTs current law equation at node y . (ri)Q -+ |> - = C B + -^ii (1) Therefore, the dynamic equation is Apply Laplace transform on both sides. > { i) = c » i( i) + — »(») / X C l + — ^ u(s)U Therefore, the transfer function. Z i £ li ii(s ) (b) Refer to Figure 2.48 (b) in the textbook for active lead circuit. Draw the circuit diagram with node voltages. R H ^ Figure 1 Write the KirchhofTs current law at node y . From the circuit, y - y 0 - r . Substitute K i + ' ^ f ^ f o r y in equation (2). Therefore, the dynamic equation is“ R, “....(3) —+ - ^ V ■*/[r, - “ R, “ Apply Laplace transform on both sides of equation (3). y ^ ( s ) = - c s K . ( s ) - ± y ^ ( s ) HI f k W K 4-i f e ] /i,C ^ s J - L ^ - L ] U c Therefore, the transfer function.U f ) : * r R.C K, U C R,C) (C) Refer to Figure 2.48 (c) in the textbook for active lag circuit. Draw the circuit diagram with node voltages. R^ R, -VcM t Write the KirchhofTs cument law equation at inverting terminal. Z . - i i . K Write the KirchhofTs cument law equation at node y . Substitute - Z l k for y . K S, d l[ 1 1 ,“ - ) K, - ^ y ^ - ^ - c ^ K - c v „ ~ v , = o c v , . + ^ = - c ^ v , - — v , —^ v , «• 0 0*>0M00W c r „ + ^ = - c - ^ r „ - — r i + - ^ ^ i r „ (4) - R, ^ J !|j ^ Therefore, the dynamic equation is c v ^ , y ^ = ^ ^ v , . ± y^ “ I t , K . It,. Apply Laplace transform on both sides. n . ( 4 K. ■^ C s +til Therefore, the transfer function.U f ) :/J,Cs+til step 8 of 1 1 A (<i) Refer to Figure 2.48 (d) in the textbook for passive notch circuit. Draw the circuit diagram with node voltages. Write the KirchhofTs cument law equation at node P,. Apply Laplace transform. C t lK W - r . w ) + ^ + c i ( F ; w - F „ ( » ) ) = 0 = a K „ W + c iK , ,( » ) Write the KirchhofTs cument law equation at node P,. R d t' ' R Apply Laplace transform. + 2 C s K (i) + — 2 < i_ i = 0 R ' ' R Step 10 Of 1 1 From the circuit diagram. Apply Laplace transform. step 11 of 1 1 Substitute p;(«) and ^ j( ^ ) in the equation. f _ « ^ + ____ I____ ] ( . 2 ^ + 2 2R + 2R^Cs)K . U ) R C V 2RO + 2 2R+2R‘ s-H lJ Therefore, the transfer function.U f ) :f , ____ !____ ] 2 ^ 0 + 2 2R+2R‘ C sJ R C V 2RCS+2 2R+2R‘