Solutions Manual INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 5th edition 2018 SOLUTIONS MANUAL David G. Alciatore, PhD, PE Department of Mechanical Engineering Colorado State University Fort Collins, CO 80523 Introduction to Mechatronics and Measurement Systems 1 Solutions Manual This manual contains solutions to the end-of-chapter problems in the fifth edition of "Introduction to Mechatronics and Measurement Systems." Only a few of the open-ended problems that do not have a unique answer are left for your creative solutions. More information, including an example course outline, a suggested laboratory syllabus, Mathcad/Matlab files for examples in the book, and other supplemental material are provided on the book website at: mechatronics.colostate.edu We have class-tested the textbook for many years, and it should be relatively free from errors. However, if you notice any errors or have suggestions or advice concerning the textbook's content or approach, please feel free to contact me via e-mail at David.Alciatore@colostate.edu. I will post corrections for reported errors on the book website. Thank you for choosing my book. I hope it helps you provide your students with an enjoyable and fruitful learning experience in the exciting cross-disciplinary subject of mechatronics. 2 Introduction to Mechatronics and Measurement Systems Solutions Manual 2.1 D = 0.06408 in = 0.001628 m. 2 D –6 A = ---------- = 2.082 10 4 = 1.7 x 10-8 m, L = 1000 m L R = ------- = 8.2 A 2.2 4 (a) R 1 = 21 10 20% so 168k R 1 252k (b) R 2 = 07 10 20% so 5.6k R 2 8.4k (c) R s = R 1 + R 2 = 217k 20% so 174k R s 260k (d) R1 R2 R p = ------------------R1 + R2 3 R 1MIN R 2MIN - = 5.43k R pMIN = ------------------------------R 1MIN + R 2 MIN R 1MAX R 2MAX - = 8.14k R pMAX = --------------------------------R 1 MAX + R 2 MAX 2.3 2 R 1 = 10 10 , R 2 = 25 10 1 2 1 R1 R2 10 10 25 10 ------------------------------------------------------------------- = 20 10 1 R = = 2 1 R1 + R2 10 10 + 25 10 a = 2 = red, b = 0 = black, c = 1 = brown, d = gold 2.4 In series, the trim pot will add an adjustable value ranging from 0 to its maximum value to the original resistor value depending on the trim setting. When in parallel, the trim pot could be 0 perhaps causing a short. Furthermore, the trim value will not be additive with the fixed resistor. 2.5 When the last connection is made, a spark occurs at the point of connection as the completed circuit is formed. This spark could ignite gases produced in the battery. The negative terminal of the battery is connected to the frame of the car, which serves as a ground reference throughout the vehicle. Introduction to Mechatronics and Measurement Systems 3 Solutions Manual 2.6 No, as long as you are consistent in your application, you will obtain correct answers. If you assume the wrong current direction, the result will be negative. 2.7 Place two 100 resistors in parallel and you immediately have a 50 resistance. 2.8 Put two 50 resistors in series: 5050 2.9 Put a 100 resistor in series with the parallel combination of two 100 resistors: 100100100100100 2.10 From KCL, I s = I 1 + I 2 + I 3 Vs Vs Vs Vs so from Ohm’s Law -------- = ------ + ------ + -----R eq R1 R2 R3 R1 R2 R3 1 1 1 1 Therefore, -------- = ------ + ------ + ------ so R eq = ---------------------------------------------------R eq R1 R2 R3 R2 R3 + R1 R3 + R1 R2 2.11 Is Is From Ohm’s Law and Question 2.10, V = -------- = ---------------------------------------------------R eq R2 R3 + R1 R3 + R1 R2 ---------------------------------------------------R1 R2 R3 and for one resistor, V = I 1 R 1 R2 R3 Therefore, I 1 = ---------------------------------------------------- I s R 2 R 3 + R 1 R 3 + R 1 R 2 2.12 R1 R2 R1 R2 lim ------------------- = ------------- = R 2 R1 R 1 R 1 + R 2 2.13 dV 1 dV 2 dV I = C eq ------- = C 1 ---------- = C 2 ---------dt dt dt From KVL, V = V1 + V2 so dV dV dV ------- = ---------1- + ---------2dt dt dt 4 Introduction to Mechatronics and Measurement Systems Solutions Manual Therefore, C1 C2 II I 1 1 1 ------= ------ + ------ so -------- = ------ + ------ or C eq = ------------------C1 + C2 C eq C1 C2 C eq C1 C2 2.14 V = V1 = V2 dV 1 dV 2 dV dV I 1 = C 1 ---------- = C 1 ------- and I 2 = C 2 ---------- = C 2 ------dt dt dt dt From KCL, dV dV dV I = I 1 + I 2 = C 1 ------- + C 2 ------- = ------- C 1 + C 2 dt dt dt dV Since I = C eq ------dt C eq = C 1 + C 2 2.15 I = I1 = I 2 From KVL, dI dI dI V = V 1 + V 2 = L 1 ----- + L 2 ----- = ----- L 1 + L 2 dt dt dt dI Since V = L eq ----dt L eq = L 1 + L 2 2.16 dI 1 dI 2 dI V = L ----- = L 1 ------- = L 2 ------dt dt dt From KCL, I = I 1 + I 2 V V V Therefore, ---- = ------ + -----L L1 L2 so dI dI dI ----- = -------1 + -------2 dt dt dt L1 L2 1 1 1 so --- = ------ + ------ or L = -----------------L L1 L2 L1 + L2 2.17 V o = 1V , regardless of the resistance value. 2.18 40 From Voltage Division, V o = ------------------ 5 – 15 = – 8V 10 + 40 Introduction to Mechatronics and Measurement Systems 5 Solutions Manual 2.19 Combining R2 and R3 in parallel, R2 R3 23 R 23 = ------------------- = ------------ = 1.2k R2 + R3 2+3 and combining this with R1 in series, R 123 = R 1 + R 23 = 2.2k (a) Using Ohm’s Law, V in 5V I 1 = ---------- = ---------- = 2.27mA R 123 2.2k (b) Using current division, R2 2 I 3 = ------------------- I 1 = --- 2.27mA = 0.909mA 5 R2 + R3 (c) Since R2 and R3 are in parallel, and since Vin divides between R1 and R23, R 23 1.2 V 3 = V 23 = --------------------- V in = ------- 5V = 2.73V R 1 + R 23 2.2 2.20 (a) From Ohm’s Law, V out – V 1 14.2V – 10V I 4 = ----------------------- = ------------------------------- = 0.7mA R 24 6k (b) V 5 = V 6 = V 56 = V out – V 2 = 14.2V – 20V = – 5.8V (a) R 45 = R 4 + R 5 = 5k 2.21 R 3 R 45 R 345 = --------------------- = 1.875k R 3 + R 45 R 2345 = R 2 + R 345 = 3.875k R 1 R 2345 R eq = -------------------------- = 0.795k R 1 + R 2345 (b) 6 R 345 V A = ----------------------- V s = 4.84V R 2 + R 345 Introduction to Mechatronics and Measurement Systems Solutions Manual (c) VA I 345 = ---------- = 2.59mA R 345 R3 I 5 = --------------------- I 345 = 0.97mA R 3 + R 45 2.22 This circuit is identical to the circuit in Question 2.21. Only the resistance values are different: (a) R 45 = R 4 + R 5 = 4k R 3 R 45 R 345 = --------------------- = 2.222k R 3 + R 45 R 2345 = R 2 + R 345 = 6.222k R 1 R 2345 R eq = -------------------------- = 1.514k R 1 + R 2345 (b) R 345 V A = ----------------------- V s = 3.57V R 2 + R 345 (c) VA I 345 = ---------- = 1.61mA R 345 R3 I 5 = --------------------- I 345 = 0.89mA R 3 + R 45 2.23 Using superposition, R2 V R21 = ------------------- V 1 = 0.909V R1 + R2 R1 V R22 = ------------------- i 1 = 9.09V R1 + R2 V R2 = V R21 + V R22 = 10.0V 2.24 R4 R5 R 45 = ------------------- = 0.5k R4 + R5 V1 – V2 I = ------------------- = – 0.5mA R1 + R2 Introduction to Mechatronics and Measurement Systems 7 Solutions Manual R 45 V A = --------------------- V 1 – V 2 = – 0.238 V R 3 + R 45 2.25 R 45 = R 4 + R 5 = 9k R 3 R 45 R 345 = --------------------- = 2.25k R 3 + R 45 R 2345 = R 2 + R 345 = 4.25k R 1 R 2345 R eq = -------------------------- = 0.81k R 1 + R 2345 2.26 Using loop currents, the KVL equations for each loop are: V 1 – I out R 1 = 0 V2 – I5 R5 – I3 R3 – V1 = 0 – I6 R6 + I5 R5 = 0 I 3 R 3 – I 24 R 4 – I 24 R 2 = 0 and using selected KCL node equations, the unknown currents are related according to: I out = I 2 + I 3 + I V1 I V1 = I out – I 5 + I 6 I 3 = I 5 + I 6 – I 24 This is now 7 equations in 7 unknowns, which can be solved for Iout and I6. The output voltage is then given by: V out = V 2 – I 6 R 6 2.27 Applying Ohm’s Law to resistor combination R24 gives: V out – V 1 4.2V I 4 = ----------------------- = ------------ = 0.7mA R 24 6k The voltage across R5 is: V 5 = V 6 = V 56 = V + – V - = V out – V 2 = – 5.8V 2.28 8 It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 M. Introduction to Mechatronics and Measurement Systems Solutions Manual 2.29 2.30 Since the input impedance of the oscilloscope is 1 M, the impedance of the source will be in parallel, and the oscilloscope impedance will affect the measured voltage. Draw a sketch of the equivalent circuit to convince yourself. R2 R3 R 23 = ------------------R2 + R3 R 23 V out = --------------------- V in R 1 + R 23 (a) R 23 = 9.90k , V out = 0.995V in (b) R 23 = 333k , V out = 1.00V in When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good. 2.31 R2 V out = ------------------- V in R1 + R2 (a) 10 V out = ------------- V in = 0.995V in 10.05 (b) 500 V out = ---------------- V in = 0.9999V in 500.05 For a larger load impedance, the output impedance of the source less error. 2.32 The theoretical value of the voltage is: R 1 V theor = -------------- V s = --- V s R+R 2 The equivalent resistance of the parallel combination of the resistor and the voltmeter input impedance is: R 5R5 ---------------= --- R R + 5R 6 And the measured voltage across this resistance is: V meas 5--R 5 6 = ----------------- V s = ------ V s 11 5 R + --- R 6 Therefore, the percent error in the measurement is: V meas – V theor ----------------------------------- = – 9% V theor Introduction to Mechatronics and Measurement Systems 9 Solutions Manual 2.33 It will depend on the supply; check the specifications before answering. 2.34 With the voltage source shorted, all three resistors are in parallel, so, from Question 2.10: R1 R2 R3 R TH = ---------------------------------------------------R2 R3 + R1 R3 + R1 R2 2.35 V in = 5 45 Combining R2 and L in series and the result in parallel with C gives: R 2 + Z L Z C Z R2 LC = ------------------------------------- = 1860.52 – 60.25 = 923.22 – 1615.30j R2 + ZL + ZC Using voltage division, Z R2 LC -V V C = -------------------------R 1 + Z R2 LC in where R 1 + Z R2 LC = 1000 + 923.22 – 1615.30j = 2511.57 – 40.02 so 1860.52 – 60.25 V C = -------------------------------------------- 5 45 = 3.70 24.8 = 3.70 0.433rad 2511.57 – 40.02 Therefore, V C t = 3.70 cos 3000t + 0.433 V 2.36 With steady state dc Vs, C is open circuit. So V C = V s = 10V so V R1 = 0V and V R2 = V s = 10V 2.37 (a) In steady state dc, C is open circuit and L is short circuit. So Vs I = ------------------- = 0.025mA R1 + R2 (b) = 6 6 –j – 10 10 Z C = -------- = ----------- j = -------- – 90 C 5 5 Z LR2 = Z L + R 2 = jL + R 2 = 10 + 20j = 10 0.036 10 Introduction to Mechatronics and Measurement Systems Solutions Manual Z C Z LR2 Z CLR2 = -----------------------= 91040 – 28550j = 95410 – 17.4 Z C + Z LR2 Z eq = R 1 + Z CLR2 = 191040 – 28550j = 193200 – 8.50 Vs I s = -------- = 0.0259 8.50 mA Z eq ZC I = ------------------------ I s = 0.954 – 17.44 I s = 0.0247 – 8.94 mA Z C + Z LR2 So I t = 24.7 cos t – 0.156 A 2.38 (a) rad = -------- , f = ------ = 0.5Hz sec 2 A pp = 2A = 4.0 , dc offset = 0 (b) rad = 2 -------- , f = ------ = 1Hz sec 2 A pp = 2A = 2 , dc offset = 10.0 (c) rad = 2 -------- , f = ------ = 1Hz sec 2 A pp = 2A = 6.0 , dc offset = 0 (d) rad = 0 -------- , f = ------ = 0Hz sec 2 A pp = 2A = 0 , dc offset = sin + cos = – 1 2 2.39 V rms P = ----------- = 100W R 2.40 V pp V rms = --------- 2 = 35.36V 2 Introduction to Mechatronics and Measurement Systems 11 Solutions Manual 2 V rms P = ----------- = 12.5W R 2.41 Vm = 2V rms = 169.7V 2.42 For V rms = 120V , V m = 2V rms = 169.7V , and f = 60 Hz, V t = V m sin 2f + = 169.7 sin 120t + 2.43 From Ohm’s Law, 5V – 2V 3V I = --------------------- = ------R R Since 10mA I 100mA , 3V 10mA ------- 100mA R giving 3V 3V ---------------- R --------------- or 30 R 300 100mA 10mA 2 V For a resistor, P = ------ , so the smallest allowable resistance would need a power rating of R at least: 2 3V P = --------------- = 0.3W 30 so a 1/2 W resistor should be specified. The largest allowable resistance would need a power rating of at least: 2 3V P = --------------- = 0.03W 300 so a 1/4 W resistor would provide more than enough capacity. 2.44 Using KVL and KCL gives: V 1 = I R1 R 1 V 1 = I 1 – I R1 R 2 + I 1 – I R1 – I 2 R 3 V 3 – V 2 = I 1 – I R1 – I 2 R 3 – I 2 R 4 12 Introduction to Mechatronics and Measurement Systems Solutions Manual The first loop equation gives: V1 I R1 = ------ = 10mA R1 Using this in the other two loop equations gives: 10 = I 1 – 10m 2k + I 1 – 10m – I 2 3k 10 – 5 = I 1 – 10m – I 2 3k – I 2 4k or 5k I 1 – 3k I 2 = 60 3k I 1 – 7k I 2 = 35 Solving these equations gives: I 1 = 12.12mA and I 2 = 0.1923mA 2.45 (a) V out = I 2 R 4 – V 2 = – 4.23V (b) P 1 = I 1 V 1 = 121mW , P 2 = I 2 V 2 = 0.962mW , P 3 = – I 2 V 3 = – 1.92mW Using KVL and KCL gives: V 1 = I R1 R 1 V 1 = I 1 – I R1 R 2 + I 1 – I R1 – I 2 R 3 V 3 – V 2 = I 1 – I R1 – I 2 R 3 – I 2 R 4 The first loop equation gives: V1 I R1 = ------ = 10mA R1 Using this in the other two loop equations gives: 10 = I 1 – 10m 2k + I 1 – 10m – I 2 2k 10 – 5 = I 1 – 10m – I 2 2k – I 2 1k or 4k I 1 – 2k I 2 = 50 2k I 1 – 3k I 2 = 25 Introduction to Mechatronics and Measurement Systems 13 Solutions Manual Solving these equations gives: I 1 = 12.5mA and I 2 = 0mA 2.46 (a) V out = I 2 R 4 – V 2 = – 5V (b) P 1 = I 1 V 1 = 125mW , P 2 = I 2 V 2 = 0mW , P 3 = – I 2 V 3 = 0mW P avg Vm Im 1 = --- V t I t dt = -------------- sin t + V sin t + I dt T T T T 0 0 Using the product formula trigonometric identity, T P avg Vm Im = -------------- cos V – I – cos 2t + V + I dt 2T 0 Therefore, Vm Im Vm Im P avg = -------------- cos V – I = -------------- cos 2 2 T 2.47 I rms = 2 --1- I 2m sin t + I dt T 0 Using the double angle trigonometric identity, 2T I rms = Im 1 ----- --- – cos 2 t + I dt T 2 0 Therefore, 2 I rms = 2.48 I I m T ----- --- = ----mT 2 2 R2 R3 R 23 = ------------------- = 5k R2 + R3 R 23 1 V o = --------------------- V i = --- sin 2t 2 R 1 + R 23 This is a sin wave with half the amplitude of the input with a period of 1s. 14 Introduction to Mechatronics and Measurement Systems Solutions Manual 2.49 No. A transformer requires a time varying flux to induce a voltage in the secondary coil. 2.50 V Np ------ = ------p = 120V ------------- = 5 Ns Vs 24V 2.51 RL = Ri = 8 for maximum power 2.52 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed. Introduction to Mechatronics and Measurement Systems 15 Solutions Manual 3.1 For Vi > 0, Vo = 0 For Vi < 0, Vo = Vi The resulting waveform consists only of the negative "humps" of the original cosine wave. Each hump has a duration of 0.5s and there is a 0.5s gap between each hump. 3.2 For Vi > 0.7V, Vo = 0.7V For Vi < 0.7V, Vo = Vi The resulting waveform consists only of the negative "humps" (below 0.7V) of the original cosine wave. The positive "humps" (above 0.7V) are clipped off and held constant at 0.7V. 3.3 (a) output passes the positive humps only (b) output passes the negative humps only (c) output passes the positive humps only (d) output passes the negative humps only (e) output passes the positive humps and scales the negative humps by 1/2 (f) output passes the full wave (a) output passes the positive humps above -0.7V only, with the negative humps clipped at -0.7V (b) output passes the negative humps below 0.7V only, with the positive humps clipped at 0.7V (c) output passes the positive humps above 0.7V only, with the negative humps clipped at 0.7V (d) output passes the negative humps below -0.7V only, with the positive humps clipped at -0.7V (e) output passes the positive humps above -0.7V only, and scales the negative humps below -0.7V by 1/2 (f) output passes the full wave 3.4 3.5 16 When the diode is forward biased, the output voltage is -0.7V, so the output signal is chopped off at -0.7V instead of 0V. Introduction to Mechatronics and Measurement Systems Solutions Manual 3.6 When the switch is closed and the circuit is in steady state, the current through the load is constant, and the diode is reverse biased (i.e., there is no diode current). When the switch is opened, the inductor generates a forward voltage to oppose a decrease in current. Now the diode forms a circuit with the load, allowing the current to dissipate through the resistor. If there were no diode, and the switched were opened, because the current would attempt to decrease instantaneously, the inductor would generate a very large voltage which would create an arc (current through air) across the switch contacts. 3.7 Forward bias (Vin > Vout + 0.7V) is required for charging. "Leaking" causes voltage decay (i.e., Vout decreases slowly). 3.8 See the data sheet for a LM7815C voltage regulator. 3.9 (a) For Vi > 0.5V, Vo = 0.5V. For Vi < 0.5V, Vo = Vi. The resulting waveform is the original sine wave with the top halves of the positive "humps" (above 0.5 V) clipped off. (b) For Vi < 0.5V, Vo = 0.5V. For Vi > 0.5V, Vo = Vi. The resulting waveform is the original sine wave with the bottom of the negative "humps" (below 0.5 V) clipped off. 3.10 Using superposition, Ohm’s Law, and current division, 1V 2 I 1left = ----------------- = ------R 5R 2R + ---2 1 1 I 2 left = – --- I 1 left = – ------2 5R 1 1 1V 3 I 4left = --- I 1 left = ------- , I 4 right = ----------------- = ------2 5R 2R 5R R + ------3 2R 2 I 2right = ----------------- I 4right = ------R + 2R 5R 1 I 1right = I 4right – I 2right = ------5R Introduction to Mechatronics and Measurement Systems 17 Solutions Manual 3 I 1 = I 1left + I 1right = ------- 0 5R 1 I 2 = I 2left + I 2right = ------- 0 5R I3 = 0 4 I 4 = I 4left + I 4right = ------- 0 5R 1 V diode = 1V – I 4 R = --- V 0 5 3.11 With I2=I3=0, I1 and I4 are equal. The current (I=I1=I4) is: 1V + 1V 2 I = --------------------- = ------- V 2R + R 3R and the voltage of node A relative to node B is: 1 V AB = 1V – I 2R = – 1V + I R = – --- V 3 Therefore, the voltage polarity on the left diode is incorrect. 3.12 When the left diode is forward biased and the right diode is reverse biased, V out = V H and when the right diode is forward biased and the left diode is reverse biased, V out = V L When both diodes are reverse biased, RL V out = ------------------ V i Ri + RL Therefore, the output is a scaled version of the input chopped off below VL and above VH. 3.13 1 For Vin > 0, V out = --- V in = 5 sin t 2 For Vin < 0, V out = V in = 10 sin t The positive "bumps" of the resulting waveform are half the amplitude (5 vs. 10) of the original, and the lower bumps are the same. 18 Introduction to Mechatronics and Measurement Systems Solutions Manual 3.14 In steady state dc, the capacitor is equivalent to an open circuit. Therefore, the steady state current through the capacitor is 0 and the steady state voltage across the capacitor is Vout. (a) For Vs=10V, the diode is forward biased and is equivalent to a short circuit. Therefore, the equivalent resistance of the two horizontal resistors is R/2 and from voltage division, R 2 V capacitor = V out = -------------- V s = --- V s = 6.66V R 3 ---- + R 2 (b) For Vs=10V, the diode is reverse biased and is equivalent to an open circuit. Therefore, the circuit simplifies to two series resistors and R 1 V capacitor = V out = -------------- V s = --- V s = – 5 V R+R 2 3.15 There are three possible states of the diodes. When only the left diode is forward biased, V out = V H . When only the right diode is forward biased, V out = V L . When both diodes are reverse biased, V L V out V H . In this case, the circuit is a voltage divider and RL 1 V out = ------------------ V in = --- V in Ri + RL 2 The upper limit of this state is when V out = V H = 5V corresponding to V in = 10V . Vout remains at 5V when Vin increases above 10V. The lower limit of the double reverse biased state is when V out = V L = – 5 V corresponding to V in = – 10 V . Vout remains at 5V when Vin decreases below 10V. It is not possible for both diodes to be reverse biased at the same time in this circuit. The resulting output signal Vout is a sin wave scaled by 1/2 with the peaks clipped off at 10V . 3.16 3.17 (a) output passes first (positive) hump only (b) output is 5.1V over the whole input cycle Use a resistor in series with the LED where: 5V – V LED I = --------------------------R Introduction to Mechatronics and Measurement Systems 19 Solutions Manual The required resistance value is 5V – V LED R -------------------------------I max 3.18 (a) 5V R --------------- = 100 50mA (b) 5V – 2V R -------------------------- = 60 50mA V E = 5V – V LED – V CE = 2.8V V in saturation = V E + V BE = 2.8V + 0.7V = 3.5V (a) For the LED to be ON, the transistor must be in saturation and V B = V LED + V BE = 1V + 0.7V = 1.7V When the LED is off, IB = 0 and 1 V B = --- V in 2 So for the LED to be ON, V in 2V B = 3.4V (b) When the transistor is fully saturated, V B = V LED + 0.7V = 1.7V and V C = V LED + 0.2V = 1.2V and 5V – V C 3.8V I C = --------------------- = ------------- = 11.5mA 330 330 Assume 1 I B = --------- I C = 0.115mA 100 If I1 is the current through the horizontal 1k resistor and I2 is the current through the right 1k resistor, then VB I 1 = I 2 + I B = ------- + 0.115mA = 1.815mA 1k and V in = V B + 1k I 1 = 3.52V 20 Introduction to Mechatronics and Measurement Systems Solutions Manual 3.19 When the transistor is in full saturation, V CE = 0.2V and V BE = 0.7V and IC I out = I B + I C = --------- + I C = 1.01I C 100 In the collector-to-emitter circuit, V out = V s – I C R C – V CE = I out R out giving 5V – I C 1k – 0.2V = 1.01I C 1k Now we can solve for the collector and emitter currents: 4.8V I C = ------------- = 2.39mA 2.01k and I out = 1.01I C = 2.41mA Therefore, V out = I out R out = 2.41mA 1k = 2.41V and the minimum required input voltage is: V in = V out + V BE + I B R B = 2.41V + 0.7V + 0.239mA 1k = 3.13V 3.20 1: a resistor (e.g., 1k) to limit the base current while ensuring the transistor is in full saturation 2: 24 Vdc capable of at least 1A of current 3: power diode capable of carrying at least 1A for flyback protection 4: ground 3.21 The transistor begins to saturate at approximate Vin=0.88V, where: IC 4.869 = ----- = ------------- = 90.2 IB 0.054 VBE = 0.73V VCE = 0.16V We usually assume these values are 100, 0.7V, and 0.2 V at the verge of saturation. Introduction to Mechatronics and Measurement Systems 21 Solutions Manual 3.22 A voltage source (e.g., 5V) and current limiting series resistor (e.g., 330 ) is required on the LED side. On the phototransistor side, a pull-up resistor (e.g, 1k) and a voltage source (e.g., 5V) is required on the collector lead and ground is required on the emitter lead. 3.23 From the figure, the approximate "ON" values for the drain-to-source voltage and current are: V ds 0.25V and I ds 48mA so the "ON" resistance is: V ds R ON = -------- 5.2 I ds 3.24 1: nothing required 2: 24 Vdc capable of at least 1A of current 3: power diode capable of carrying at least 1A for flyback protection 4: ground 3.25 The type of BJT required is an npn and an additional resistor must be added in series with the open collector output to pull up the voltage enough to bias the BE junction of the BJT. 3.26 The upper FET is a p-channel enhancement mode MOSFET and the lower is an n-channel enhancement mode MOSFET. When Vin = 5V, the upper MOSFET doesn’t conduct but the bottom one does, so Vout = 0V. When Vin = 0V, the upper MOSFET conducts but the bottom one doesn’t, so Vout = Vcc. 3.27 The requirements are that 2 I d cont 10A and P d max I on R on = 100R on IRF530N is a good choice 3.28 22 (a) cutoff (b) ohmic (c) saturation (d) cutoff Introduction to Mechatronics and Measurement Systems Solutions Manual 4.1 4.2 (a) linear (b) nonlinear (c) linear (d) linear (e) nonlinear (f) nonlinear (g) linear The Fourier Series is: F t = 5 sin 2t and the fundamental frequency is 0 f 0 = ------ = 1Hz 2 4.3 Since Vrms = 120V and f = 60Hz, the Fourier Series is: Ft = 2V rms sin 2ft = 169.7 sin 120t and the fundamental frequency is 60Hz. 4.4 We need to prove: A n cos n 0 t + B n sin n 0 t = C n cos n 0 t + n We can use the following trig identity: cos a + b = cos a cos b – sin a sin b where: a = n 0 t An cos b = -----Cn Bn – sin b = -----Cn b = n Equations 4.8 and 4.9 can now be verified: Bn 2 An 2 2 2 sin b + cos b = ------ + ------ = 1 C n C n Therefore, Cn = 2 2 An + Bn Introduction to Mechatronics and Measurement Systems 23 Solutions Manual Also, Bn – -----Cn Bn sin b tan n = tan b = ---------------- = --------- = – -----cos b An An -----Cn Therefore, –1 Bn –1 Bn n = tan – ------ = – tan ------ A n A n T --2 T 4.5 2 2 A n = --- F t cos n 0 t dt = --T T 0 T cos n0 t dt – cos n0 t dt 0 T --2 So T --2 2 T A n = ------------- sin n 0 t 0 – sin n 0 t T n 0 T --- 2 2 But 0 = ------ , so T 1 A n = ------ sin n – sin 2n + sin n n But sine of any multiple of is 0, so An = 0 24 Introduction to Mechatronics and Measurement Systems Solutions Manual 4.6 Using MathCAD: t 0 0.01 3.0 2 4 10 n m 2 4 50 Va( t ) Vb( t ) Vc( t ) sin ( 2 t ) 1 2 sin ( 2 t ) 2 2 1 sin ( 2 t ) 2 1 n 2 m cos ( 2 n t ) ( n 1) ( n 1) cos ( 2 m t ) ( m 1) ( m 1) 1 1.5 1 0.5 Va( t ) Vb( t ) 0.5 0 0 0.5 0 1 2 0.5 3 0 1 t 2 3 t 1.5 1 Vc( t ) 0.5 0 0.5 0 1 2 3 t 4.7 (a) fH 1- 1 – ----- 2 = 6 + --------------------- 10 – 6 = 7.17Hz 1 So the bandwidth is: 0 Hz to 7.17 Hz Introduction to Mechatronics and Measurement Systems 25 Solutions Manual (b) T = 1s rad 1 f o = --- = 1Hz , 0 = 2f 0 = 2 -------sec T n (rad/sec) f (Hz) Ain Aout/Ain Aout = (Aout/Ain)Ain 1 0 1 4/ 1 4/ 2 0 3 4/3 1 4/3 3 0 5 4/5 1 4/5 4 0 7 4/7 3/4 3/7 5 0 9 4/9 1/4 1/9 >5 (2n-1)0 (2n-1) 4/f 0 0 4 4 4 3 1 V out = --- sin 2t + ------ sin 6t + ------ sin 10t + ------ sin 14t + ------ sin 18t 3 5 7 9 (c) Using MathCAD: t 0 0.01 2 Vout( t ) 4 4 sin ( 6 t ) 3 sin ( 2 t ) 4 sin ( 10 t ) 5 3 sin ( 14 t ) 7 2 2.0 Vout( t ) 0 2 0 1 t 4.8 26 (a) 1 1 - = 1000 rad ------- c = -------- = ----------------------------------------------3 – 6 RC sec 1 10 1 10 (b) T = 1s, f = 1Hz, 0 = 2 Introduction to Mechatronics and Measurement Systems 2 1 sin ( 18 t ) 9 Solutions Manual A out n sin n t Ain n ---------A in Ft = n=1 where 4 A in n = ---------------------- 2n – 1 A out 1 ---------- n = --------------------------A in 2 n 1 + ------ c n = 2n – 1 2 (c) Using MathCAD: n 1 100 c t 0 0.01 2 1000 n 1 4 F ( t) ( 2 n 1 ) n 1 ( 2 n 1 ) 2 sin t n n 2 c 1 F( t ) 0 1 0 1 2 t 4.9 (a) 1 1 - = 10 rad ------- c = -------- = ----------------------------------------------------3 –6 RC sec 100 10 1 10 (b) T = 1s, f = 1Hz, 0 = 2 Introduction to Mechatronics and Measurement Systems 27 Solutions Manual A out n sin n t Ain n ---------A in Ft = n=1 where 4 A in n = ---------------------- 2n – 1 A out 1 ---------- n = --------------------------A in 2 n 1 + ------ c n = 2n – 1 2 (c) Using MathCAD: n 1 100 c 10 n ( 2 n 1) 2 t 0 0.01 2 F ( t) n 4 ( 2 n 1) sin t n 2 n 1 c 1 1 0.5 F ( t) 0 0.5 1 0 0.5 1 1.5 2 t 4.10 1 rad rad L = ------- -------- = 0.707 -------- = 0.113Hz sec sec 2 1 rad rad H = 3 + 1.5 1 – ------- -------- = 3.44 -------- = 0.547Hz sec 2 sec L bandwidth H 28 Introduction to Mechatronics and Measurement Systems Solutions Manual 4.11 4.12 (a) rad rad 1 -------- 5 -------sec sec (b) and (c) n Ain Aout 1 1 1 2 2 2 3 3 3 4 4 1.33 5 5 0 R V o = --------------------- V i 1 R + ---------jC Vo jRC ------ = ----------------------Vi jRC + 1 1 To find the cut-off frequency, set the amplitude ratio magnitude to ------- : 2 Vo 1 RC ------ = --------------------------------- = ------Vi 2 RC 2 + 1 Solving for the frequency gives 1 c = -------RC Using this expression gives: ---- Vo c ------ = ---------------------------Vi 2 ----- +1 c Introduction to Mechatronics and Measurement Systems 29 Solutions Manual and now we can plot the frequency response in terms of the dimensionless frequency ratio r = ------ : c r 0 0.01 5.0 r Ar r r 2 1 1 Ar r 0.5 0 0 2 4 6 r 4.13 Vo RC = arg ------ = 1 – 1 + RCj = 0 – atan ------------ = – atan RC Vi 1 1 Using c = -------- and r = ------ = RC , RC c = – atan r 0 0 20 r 40 60 68.199 80 0 0.5 1 0 30 Introduction to Mechatronics and Measurement Systems 1.5 r 2 2.5 2.5 Solutions Manual 4.14 The answer is in Figure 4.4 (see the "0, 30, 50" curve). 4.15 Using MathCAD: 1 20 n n t 0 0.01 2 ( 2 n 1 ) 2 4 F ( t) ( 2 n 1 ) sin t n n ATTn 1 ATT1 .25 ATT2 4 F low( t ) ( 2 n 1 ) .25 ATT3 .25 ATT sin t n n n ATTn 1 ATT17 .25 ATT18 4 F high( t ) ( 2 n 1 ) .25 ATT20 .25 ATT sin t n n n 1 2 1 F( t ) F low( t ) 0 0 1 2 0 1 1 2 0 1 t 2 t 2 1 F high ( t ) 0 1 2 0 1 2 t Introduction to Mechatronics and Measurement Systems 31 Solutions Manual 4.16 Using MathCAD: n low 1 50 n high 1 1.1 10 low_pass ( n ) e 0.1 ( 2 n 1) high_pass ( n ) 1 e ( 2 n 1) 1 0.8 0.6 low_pass nlow 0.4 0.2 0 10 20 30 40 6 8 50 nlow 1 0.9 high_pass nhigh 0.8 0.7 0.6 2 4 nhigh 4.17 When the mass is in static equilibrium, M in g = kX out so g X out = --- M in k and the static sensitivity is g K = --k 32 Introduction to Mechatronics and Measurement Systems 10 Solutions Manual 4.18 We generally assume that the displayed voltage is the gain times the input voltage. This assumption will be in error if the oscilloscope is dc coupled and some of the frequencies in the signal exceed the bandwidth of the oscilloscope. 4.19 KVL gives: 1 IR + ---- q = V in C where q is the charge on the capacitor. Putting this in standard form gives: RC q· + q = CV in where q is the dependent variable, the time constant () is RC, and the sensitivity (K) is C. Using the general solution for a 1st order system, t – -- q t = CA i 1 – e Therefore, the step response output voltage (which is the voltage across the capacitor) is t – -- 1 V out t = ---- q t = A i 1 – e C 4.20 Applying KVL around the flyback loop gives: L d + IR = 0 dt Putting this in standard first-order-system form gives: L- d --+I = 0 Rdt so the time constant is: = L/R. The root of the characteristic equation is s=-1/, so the equation for current is: I t = Ce t – The current is Iss at t=0, so C=Iss, giving: I t = I ss e 4.21 t – - The rate of change of internal energy is equal to the rate of heat transfer: d· --- E = Q in dt in Introduction to Mechatronics and Measurement Systems 33 Solutions Manual so dT out mc ------------- = hA T in – T out dt 1 Defining C t = mc (thermal capacitance) and R t = ------- (thermal resistance) and hA converting into standard form gives: dT out C t R t ------------- + T out = 1 T in dt mc where the time constant is = R t C t = ------- and the sensitivity is K=1. hA 4.22 Plotting the data Xout(t) shows a steady state asymptote of approximately 5 indicating that: KA in = 5 X out t Plotting Z t = ln 1 – ----------------- shows a near linear relation indicating that the system KA in can be modeled as 1st order. Z 0.5 0 ‐0.5 0 0.2 0.4 0.6 0.8 1 1.2 ‐1 ‐1.5 Z ‐2 Linear (Z) ‐2.5 ‐3 ‐3.5 ‐4 y = ‐3.772x + 0.145 ‐4.5 The slope of the line is approximately 3.772, indicating a time constant of = 1 / 3.772 = 0.265 sec 4.23 34 The damped natural frequency is always smaller if there is damping in the system. Introduction to Mechatronics and Measurement Systems Solutions Manual 4.24 (a) mechanical rotary (applied torque, torsion spring, rotary damper, and rotary inertia): ·· · J + B + k = ext (b) electrical (voltage source and series resistor, inductor, and capacitor): 1 1 Lq·· + Rq· + ---- q = V s or LI· + RI + ---- I dt = V s C C (c) hydraulic (pump with inlet in reservoir, long pipe with friction loss and fluid inertia, and tank): 1 dQ I ------- + RQ + ---- V = P C dt where V = Q dt 4.25 Given the differential equation: x· out + x out = Kx in Applying the procedure results in: s + 1 X out s = KX in s X out s K G s = ------------------ = ------------------X in s s + 1 K G j = -----------------1 + j X out K ------------ = G j = --------------------------X in 2 1 + = arg G j = 0 – atan = – atan 4.26 rad F 0 = 20N = 0.75 -------Sec n = b k- = 1.095 , = ------= 0.685 , = --------------- = 0.456 r n m 2 km X0 1 -----------= ------------------------------------------------- = 1.219 F0 k 2 2 2 1 – r + 4 2 r Introduction to Mechatronics and Measurement Systems 35 Solutions Manual X0 F0 X 0 = ------------ ----- = 2.032m F 0 k k – 1 2 = – tan ----------------- = – 49.6 = – 0.866rad 1 ----– r r Therefore, the steady state response is: x t = X 0 sin t + = 2.032 sin 0.75t – 0.866 m 4.27 The equation of motion is mx··out + bx· out + kx out = kx in Taking the Laplace transform of both sides gives the transfer function: X out s k G s = ------------------ = -----------------------------2 X in s ms + bs + k Now k G j = --------------------------------------- k – m2 + bj X out k ------------ = G j = ---------------------------------------------------X in k – m2 2 + b 2 –1 b = – tan -------------------2- k – m So the steady state response is X out x out t = X in ------------- sin t + X in 36 Introduction to Mechatronics and Measurement Systems Solutions Manual Using MathCAD, m 0.10 k 100000 b 10 k X in X out( ) k 2 m 2 ( ) ( b ) X in 0.05 angle k 2 X out( 10 ) 0.05 ( 10 ) 0.057 deg X out( 1000 ) 0.5 ( 1000 ) 90 deg X out( 10000 ) 5.05 10 2 m b 4 ( 10000 ) 179.421 deg Only the first input results in an acceptable output. 4.28 The response would be the same since there is no "g" in the equations. The only difference would be the initial "equilibrium position," which would be at the unstretched length of the spring. One method to determine the mass is to measure the natural frequency with a spring of known stiffness and calculate: k m = -----2n 4.29 xh t = e – n t A cos d t + B sin d t xp t = C x t = xh t + xp t F0 x = ----- gives C = 0 k x(0) = 0 gives A = -C x· 0 = 0 gives B = 0 Therefore, F0 – t x t = ----- 1 – e n cos d t k Introduction to Mechatronics and Measurement Systems 37 Solutions Manual 4.30 The natural frequencies and damping constants can be used to predict the results. To model the tire, the mass of the wheel, stiffness of the tire, and position of the spindle (new variable) would also need to be included. The input force or displacement would then be at the tire-road interface. 4.31 The volume in the tank is 2 D V = ---------- h 4 The pressure at the bottom of the tank is P = gh = h Solving the volume expression for h and substituting into the pressure expression gives 1 4 P = ---------2- V = ---- V C D so 2 D C = ---------4 4.32 Q Since F = ma , a = x·· and x· = ---- , A Q· PA = LA ---- A L P = ------- Q· = IQ· A 4.33 Element [flow] analogies: F in v in V in I in k 1 v in – v m C 1 I in – I m m vm L Im b1 vm R1 Im k2 vm C2 Im Analogous free body diagram equations [KVL]: V in = V C1 V C1 = V R1 + V C2 + V L 38 Introduction to Mechatronics and Measurement Systems Solutions Manual The resulting analogous electrical circuit follows: L + Vin R1 C1 C2 4.34 Element [flow] analogies: F1 v1 V1 I1 m v1 L I1 k1 v1 C1 I1 b1 v1 – v2 R1 I1 – I2 k2 v1 – v2 C2 I1 – I2 F2 –v2 V2 –I2 Analogous free body diagram equations [KVL]: V 1 – V C2 – V R1 – V C1 = V L VR1 + VC – V2 = 0 2 The resulting analogous electrical circuit follows: R1 C2 C1 + L V2 + V1 4.35 Hydraulic elements are direct analogies to electrical elements. The capacitors are replaced by tanks, and the resistor and inductor are replaced by a long pipe with flow resistance and inertance. Introduction to Mechatronics and Measurement Systems 39 Solutions Manual 4.36 Element [flow] analogies: V1 I1 F1 v1 L I1 m v1 C1 I1 k1 v1 R I1 – I2 b v1 – v2 C2 I1 – I2 k2 v1 – v2 V2 –I2 F2 –v2 Analogous KVL [free body diagram] equations: F 2 + F k1 + F 2 – F 1 = 0 F 1 – F 2 – F k1 + F b – F k2 = 0 F2 – Fb – Fk2 = 0 The resulting analogous mechanical system follows: k1 F1 m1 F2 b k2 4.37 Element [flow] analogies: F in v 1 V in I 1 b1 v1 – v2 R1 I1 – I2 k1 v1 – v2 C1 I1 – I2 m v2 L I2 k2 v2 – v3 C2 I2 – I3 b2 v3 R2 I3 Analogous free body diagram equations [KVL]: V in = V R1 + V C 1 V R1 + V C = V L + V C2 1 40 Introduction to Mechatronics and Measurement Systems Solutions Manual V C2 = V R2 The resulting analogous electrical circuit follows: I2 I1 L (I2-I3) R1 + (I1-I2) Vin I3 C2 R2 C1 4.38 k z – x + b z· – x· – m 1 g sgn x· = m 1 x·· k z – x + b z· – x· – F 1 = – m 2 ·· z ·· – rF 1 = I 2 where · ·· z = y – r , z· = y· – r , ·· z = y·· – r Introduction to Mechatronics and Measurement Systems 41 Solutions Manual 5.1 The power dissipated by each resistor is 2 2 2 2 V in V out GAIN V in 25GAIN ------- = 25 ------ and ---------= ------------------------------------ = -----------------------R R RF RF RF To be able to use 1/4 W resistors, the following must be true: 25 ------ 0.25 or R 100 R 2 25GAIN ------------------------ 0.25 or R F GAIN 2 100 RF (a) for GAIN=1, RF > 100 (b) for GAIN=10, RF > 10k (a) R4 V + = V - = ------------------- V out R3 + R4 5.2 V+ I = ------R2 R2 R3 + R4 V out = ------------------------------- I R4 (b) V + = V - = V out V out V+ I 1 = I 2 = ------- = ---------R2 R2 V + + I 1 R 1 = V out + I 3 R 3 so R1 R1 I 3 = ------ I 1 = ------------- V out R3 R2 R3 R1 1 I = I 1 + I 3 = V out ------ + ------------- R2 R2 R 3 so R2 R3 V out = ------------------- I R1 + R3 42 Introduction to Mechatronics and Measurement Systems Solutions Manual 5.3 With RF replaced by a short, the op amp circuit becomes a buffer so the gain is 1. 5.4 (a) R2 V - = V + = ------------------- V 1 = 5V R1 + R2 V out = V - + V 2 – I 3 R 3 but I3 = 0, so V out = V - + V 2 = 10V (b) 5.5 same as (a) V+ = V- = Vi R3 V 4 = 1 + ------ V + R 2 R2 + R3 V4 I 4 = ------ = ------------------- V i R2 R4 R4 5.6 If VA denotes the voltage at the output of the first op amp, V A = V - = V + = 0V and from Ohm’s Law, the current from voltage source V1 is V1 V1 – VA I 1 = -------------------- = -----R R If VB denotes the voltage at the inverting input of the second op amp, VB = V- = V+ = V2 and from Ohm’s Law, V 2 – V out V2 VB V B – V out I 4 = ------- = ------ and I 2 = ------------------------ = ----------------------R R R R where I4 is the current through the vertical resistor and I2 is the current through the feedback resistor of the second op amp. From this, V out = V 2 – I 2 R Now applying Ohm’s Law to the resistor between the op amps gives V2 VA – VB I 3 = --------------------- = – -----R R Introduction to Mechatronics and Measurement Systems 43 Solutions Manual where I3 is the current through the resistor. From KCL, the current out of the first op amp is V2 V1 1 I out 1 = I 3 – I 1 = – ------ – ------ = – ---- V 1 + V 2 R R R The negative sign indicates that the current is actually into the op amp. From KCL, V2 V2 2 I 2 = I 3 – I 4 = – ------ – ------ = – ---- V 2 R R R Therefore, 2V 2 V out = V 2 – – ---------- R = 3V 2 R 5.7 V o V i because of positive feedback. 5.8 Applying Ohm’s Law to both resistors gives V1 V2 I 1 = ------ and I 2 = -----R1 R2 From KCL, IF = I1 + I2 Since V o + I F R F = 0 , V1 V2 V o = – R F ------ + ------ R1 R2 For R1 = R2 = RF = R, Vo = – V1 + V2 5.9 Applying Ohm’s Law to both resistors gives V1 – V3 V2 – V3 I 1 = ------------------- and I 2 = ------------------R1 R2 From KCL, IF = I1 + I2 44 Introduction to Mechatronics and Measurement Systems Solutions Manual Since V o + I F R F = V 3 , V1 – V3 V2 – V3 V o = V 3 – R F ------------------- + ------------------- R1 R2 For R1 = R2 = RF = R, V o = V 3 – V 1 – V 3 + V 2 – V 3 = 3V 3 – V 1 + V 2 5.10 From voltage division, RF V - = V + = ------------------- V 2 R F + R 2 From Ohm’s Law, V1 – V- I 1 = -----------------------R1 The output voltage can be found with: RF V – -----------------1 R F + R 2 V 2 R F V o = V - – I 1 R F = ------------------- V 2 – -------------------------------------------------- R F R F + R 2 R1 Simplifying gives R1 V2 – V1 RF + R2 – RF V2 V o = ---------------------------------------------------------------------------R1 RF + R2 RF V2 RF + R1 – V1 RF + R2 V o = --------------------------------------------------------------------R1 RF + R2 RF For R1=R2=R, RF V o = ------ V 2 – V 1 R 5.11 RF RF V outin = – ------ V in and V outref = 1 + ------ V R R ref RF RF V out = V out in + V out ref = – ------ V in + 1 + ------ V ref R R Introduction to Mechatronics and Measurement Systems 45 Solutions Manual 5.12 Using superposition, R4 V o1 = – ------ V 3 R3 R5 R4 V o2 = 1 + ------ ------------------- V 4 R 3 R 3 + R 5 R5 R4 R4 V o = V o1 + V o2 = – ------ V 3 + 1 + ------ ------------------- V 4 R3 R 3 R 3 + R 5 5.13 Using MathCAD: The input can be expressed as: T 2 100 1 100 Vin( t) sin ( t) 0.1 Assuming RC=1, the output of the integrator will be: 1 Vout ( t) cos ( t) 0.1 t 3 210 0 3 210 Vout ( t ) 3 410 3 610 3 810 0 0.02 0.04 t 5.14 V+ = V- = 0 dI L 1 V i = L -------- so I L = --- V i dt L dt Vo = V- + IR R 46 Introduction to Mechatronics and Measurement Systems Solutions Manual R but IR = IL, so V o = ---- V i dt L 5.15 From Ohm’s Law, the input currents can be related to the circuit voltages with: V+ I + = – ------R2 and V in – V - V I - = -------------------- – -----R1 Rs If the input voltages and currents are assumed to be equal (I+ = I-), equating these expressions, setting Vin=0, and dividing through by the voltage (V+ = V-) gives: 11 1 ----= ------ + ----R2 R1 Rs which gives: R1 Rs R 2 = -----------------R1 + Rs 5.16 (a) RF V o = – ------ V i = – 2V i R (b) 1 V o = – -------- V i dt = – V i dt RC Introduction to Mechatronics and Measurement Systems 47 Solutions Manual (c) RF V o = – ------ V 1 + V 2 = – 4V i R (d) V - = V + = 0V From Ohm’s Law, Vi Vi V1 V2 I 1 = ------ = ------ and I 2 = --------- = --------5k 5k 10k 10k From KCL, 1 1 I F = I 1 + I 2 = V i ------ + --------- 5k 10k but from Ohm’s Law, 0 – Vo I F = --------------5k so 1 3 V o = – 5kI F = – V i 1 + --- = – --- V i 2 2 5.17 comparator + + 5V + 330 Vin LED 48 Introduction to Mechatronics and Measurement Systems Solutions Manual 5.18 330 LED + + 5V + Vin 5.19 open collector comparator The limit on the feedback resistor current is: V out 10V I F = ---------- = ---------- 10mA RF RF Therefore, 10V R F --------------- = 1k 10mA 5.20 RF V outmax = 13.6V and V out = – ------ V in = – 2V in R so: V outmax - = 6.8V V inmax = -----------------2 5.21 RF closed loop gain = ------ = 10 R so the fall-off frequency is 105Hz. 5.22 The amplifier will saturate (reach the minimum swing voltage limit) as the integrated dc component grows. Introduction to Mechatronics and Measurement Systems 49 Solutions Manual 6.1 11111111111111112 = 216 1 = 65,535 6.2 (a) 128 = 100000002 since 27 = 128 (b) 127 = 11111112 since (27 1) = 127 (a) 128 = 8016 since 8(16) = 128 (b) 127 = 7F16 since 7(16) + 15 = 127 6.3 6.4 (a) 1 1 1101 + 1001 (b) 13 + 9 10110 22 1101 13 - 1001 - 9 0100 (c) 4 1101 x 1001 13 x 9 1101 27 0000 9 0000 117 1101 1110101 (d) 50 111 111 7 + 111 + 7 1110 14 Introduction to Mechatronics and Measurement Systems Solutions Manual (e) 111 x 111 7 x 7 11 1111 111 49 111 111 110001 6.5 (a) A B C (b) A C B 6.6 A AA = A A 6.7 (a) The output (C) is high (5V) iff both inputs are low (0V). C = AB = A + B A B C 0 0 1 0 1 0 1 0 0 1 1 0 Introduction to Mechatronics and Measurement Systems 51 Solutions Manual (b) (c) C = AB AB = AB A B C 0 0 0 0 1 0 1 0 0 1 1 1 C = AABBAB = AAB + BAB = A A + B + B A + B C = AB + BA = A B (d) A B C 0 0 0 0 1 1 1 0 1 1 1 0 A B C 0 0 0 0 1 1 1 0 1 1 1 1 C = AB = A + B 6.8 A B C 6.9 A B C 52 Introduction to Mechatronics and Measurement Systems Solutions Manual 6.10 A B C D E F 0 0 0 0 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 A B C D E F 6.11 (a) 1 0 + 1 0 + 1 + 0 1 + 0 11+11+01 1+1+0 1 (b) A B + A A + B AB + A + B AA + 1 A1 A 6.12 X = AB + BC + BC + C = ABBC + BC + C = BC A + 1 + C = C B + 1 = C Introduction to Mechatronics and Measurement Systems 53 Solutions Manual 6.13 A + A B = A + B Multiplying (ANDing) both sides by A gives: AA+AAB = AA+AB Simplifying, gives: A+0 = A+AB A = A 1 + B A = A Thus, the identity is valid. 6.14 A + B A + B = AA + AB + BA + BB = A + A B + B = A + A = A 6.15 Equation 6.21: A + B A + C AA + AC + BA + BC A1 + C + BA + C A1 + BA + C A + BA + BC A 1 + B + BC A 1 + BC A + BC 54 A + B A + C A + BC A B C A+B A+C BC 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 Introduction to Mechatronics and Measurement Systems Solutions Manual 6.16 6.17 Equation 6.22: A B A+B AB A + B + AB 0 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 Equation 6.23: A B C AB BC BC AB + BC + BC AB + C 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 A B C AB AC BC AB + AC + BC AB + BC 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 0 1 1 6.18 Introduction to Mechatronics and Measurement Systems 55 Solutions Manual 6.19 (a) A B C AB BC BC AB + BC + BC AB + C 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 0 0 0 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 1 1 1 1 1 0 1 1 (b) A B C ABC A+B+C 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 0 (c) 56 A B C AB BC BC AB + BC + BC AB + C 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 0 1 1 Introduction to Mechatronics and Measurement Systems Solutions Manual 6.20 6.21 A B A B A+B AB 0 0 1 1 1 1 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 X = AP + BP P A B X 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 The circuit is called a multiplexer because P allows one of two (multiple) inputs to pass through to the output. 6.22 X = AB + A A + B X = AB + A + B X = A A + 1 = AA = A equivalent circuit: one wire connecting input A to output X! 6.23 Y = AD + A + B C For the unallowed code CD=11, the output (Y) would be: Y = A + A + B = A + B In this state, the alarm would go off when windows or doors are disturbed or when motion is detected. This state is the same as state 2 (CD = 10). 6.24 The simplified Boolean expression is: X = B C + A Introduction to Mechatronics and Measurement Systems 57 Solutions Manual Using DeMorgan’s Laws, the all-AND representation is: B C A = B C A and the all-OR representation is: B + C + A The original expression contains 1 AND operations, 1 OR operation, and one inversion, requiring 3 ICs. The all-AND version contains 2 ANDs and 2 inversions, requiring 2 ICs. The all-OR version contains 2 ORs and 4 inversions, also requiring 2 ICs. 6.25 Y = A D + A + B C Y = A + D + A + B + C A 6.26 B C D Segment c is OFF only for the digit 2, so the output of the logic circuit must be: X = DCBA or more simply X = CBA 6.27 X = ABC + A + B C = ABC A + B C = ABCABC 6.28 X = AA C + C + BC = A1 + BC = A + BC = A + B + C = A + B + C 6.29 The required Boolean expression is: X = C A + B which can be implemented with the following logic circuit: A B C 58 Introduction to Mechatronics and Measurement Systems X Solutions Manual The complete truth table (including the sub expression A + B ) is: 6.30 A B C A + B X 0 0 0 1 0 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 0 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 0 0 X = PAB + PAB + PAB + PAB X = PA B + B + PB A + A X = PA + PB 6.31 For the two expressions for S: A B AB AB AB + AB A+B A+B A + BA + B 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 1 0 0 For the two expressions for C: 6.32 A B AB A+B A+B A+B A + B A + B A + B 0 0 0 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 1 1 1 1 From the logic circuit: C = AB which is the sum-of-products result. And, using DeMorgan’s Law, S = A + B AB = A + B A + B which is the product-of-sums result. Introduction to Mechatronics and Measurement Systems 59 Solutions Manual 6.33 Sum of products circuit: A B S C Product of sums circuit: A B S C 6.34 Ci-1 Ai Bi Si Ci 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Si = Ci – 1 Ai Bi + Ci – 1 Ai Bi + Ci – 1 Ai Bi + Ci – 1 Ai Bi Ci = Ci – 1 Ai Bi + Ci – 1 Ai Bi + Ci – 1 Ai Bi + Ci – 1 Ai Bi 60 Introduction to Mechatronics and Measurement Systems Solutions Manual 6.35 preset clear toggle set toggle set J reset reset K CK Q 6.36 T Preset Clear Q Q 1 1 Q0 Q0 1 1 Q0 Q0 0 1 1 Q0 Q0 1 1 1 Q0 Q0 0 0 1 1 0 1 1 0 0 1 Introduction to Mechatronics and Measurement Systems 61 Solutions Manual 6.37 CK D Preset Clear Q Q x 1 1 Q0 Q0 0 1 1 0 1 1 1 1 1 0 0 x 1 1 Q0 Q0 1 x 1 1 Q0 Q0 x x 0 1 1 0 x x 1 0 0 1 6.38 62 Introduction to Mechatronics and Measurement Systems Solutions Manual 6.39 6.40 6.41 There is a delay between the release at contact "B" (which can exhibit bounce) and the connection at contact "A" (which can exhibit bounce). There are also small but important switching delays in the NAND gates. See Figure 6.7. The debounce circuit is an RS flipflop with inverters at each input (to effectively eliminate the internal inverters). 6.42 S1 S2 D Q CK D Q CK Q S4 S3 D Q CK Q D Q CK Q Q E Introduction to Mechatronics and Measurement Systems 63 Solutions Manual 6.43 The P0P1P2P3 values change on the negative edge of each clock pulse as follows: 0000 (after reset pulsed low), 1000 (after 1st bit clock pulse), 0100 (after 2nd bit clock pulse), 1010 (after 3rd bit clock pulse), 1101 (after 4th bit clock pulse). 6.44 The Q0Q1Q2Q3 values (where Q3 = Sout) change on the negative edge of each clock pulse as follows: 0000 (after reset pulsed low), 1011 (after load line pulsed high), 0101 (after 1st bit clock pulse), 0010 (after 2nd bit clock pulse), 0001 (after 3rd bit clock pulse), 0000 (after 4th bit clock pulse). 6.45 1 J Q CK 1 K Q 6.46 5V sensor D Q CK CL Q 5V NC button 64 Introduction to Mechatronics and Measurement Systems 1k Solutions Manual 6.47 5V NO NC 5V S D Q CK Q C 7474 counter B oneshot O 7409 7490 74124 Note - With this design, the counter could be negative- or positive-edge triggered. If the counter is negative-edge triggered (as shown), the one-shot is actually not required. B S Q O C Introduction to Mechatronics and Measurement Systems 65 Solutions Manual 6.48 totem pole TTL 5V sinking current with low output totem pole TTL sourcing current with high output opencollector TTL 5V sinking current with high output TTL can sink more current than it can source, so the sourcing option wouldn’t be as bright. 6.49 5V 1k 74LS00 4011B 6.50 The CMOS LOW can sink only 1mA per gate which is enough to drive only two LS TTL inputs (which require 0.36 mA per gate). 6.51 c = QD QC QB QA e = QD + QC + QB + QA QD + QC + QB + QA QD + QC + QB + QA QD + QC + QB + QA 6.52 66 See info in TTL Data Book. Introduction to Mechatronics and Measurement Systems Solutions Manual 6.53 The input is the same: some sort of clock signal. Three of the four outputs look the same (the 3 least significant bits), but in the decade counter the most significant bit resets all the bits on the count of 10. The binary counter will continue to increment bits until 16 is reached. In the binary counter the output code provides 16 combinations, but in the decimal counter the output code provides 10 different output combinations. 6.54 The output goes high when the signal increases through 4V and low when it decreases through 1V. t 6.55 V CAPACITOR – -- = V cc 1 – e where = RC But when t = T, VCAPACITOR = 2/3 Vcc, so T T – -------- – ------- 2 --- = 1 – e RC so e RC = 1 --3 3 and T = RC ln 3 1.1RC 6.56 See a Linear Circuits data and/or applications book. 6.57 The time to discharge from 2/3Vcc to 1/3Vcc is the same as the time to charge from 1/3Vcc to 2/3Vcc. From the section, the time to charge to 2/3Vcc is: 1 t b = – R 2 C ln --- 3 and the time to charge to 1/3Vcc is: 2 t a = – R 2 C ln --- 3 Therefore, the elapsed time would be: 2 1 23 T 2 = t b – t a = R 2 C ln --- – ln --- = R 2 C ln ---------- = R 2 C ln 2 3 3 1 3 6.58 Ideally, making R1=0 would make T1 = T2, which would result in a perfectly symmetric square wave. However, with R1 shorted, there would no longer be any resistance in the transistor collector-emitter circuit which could result in excessive current to be sunk by the 555 when the base goes high, and this could result in damage. If the capacitor has a partial charge initially, the first square-wave pulse width will be off slightly, but all subsequent pulses will be consistent. Introduction to Mechatronics and Measurement Systems 67 Solutions Manual 6.59 If the count is updating immediately during the negative edge of signal L, it is possible that individual bits are latched either before or after the actual transition, depending on the exact timing of the counter outputs. This unlikely, but possible, scenario can be prevented by blocking the input pulses during the latch period, when L is high. This could be done by ANDing the input pulse line (I) with the inversion of the latch signal (L), and attach this output to the counter. 6.60 Assume that the digital event is a digital pulse that can be applied to the input of the counter. Cascade 3 74LS90's and connect the output of the third to an LED. Refer to the IC spec sheet for the appropriate wiring. 6.61 With the solution in 6.47, bounce with the SPDT switch has no effect. This can be verified with a timing diagram. If the button were pressed immediately after the switch, while bounce were still occurring, the stored value would be uncertain, but timing this fast would not be detectable (or repeatable) by a human anyway. If the button exhibited bounce, the circuit would have a problem. Positive and negative edges would occur during the bounce, which would result in premature latching during the button press, and re-latching during the button release. 6.62 See "Case Study 2" in Chapter 11. 6.63 See "Case Study 2" in Chapter 11. 68 Introduction to Mechatronics and Measurement Systems Solutions Manual 7.1 With the code provided, when the target LED turns on (after the 1st countdown to zero), it never goes off. A more graceful solution would be to reset the counter and LEDs when the button is pressed again after the decrement to zero. If a "start" label is inserted above the "movlw target" line, we would just need to replace "goto begin" with "goto start." Then, the target LED would turn off and the countdown would start over again. 7.2 PIC16F84 1 1k RA1 2 RA3 RA0 3 RA4 OSC1 MCLR OSC2 4 5V RA2 5 6 7 8 9 Vss Vdd RB0 RB7 RB1 RB6 RB2 RB5 RB3 RB4 18 330 17 LED 16 15 4 MHz 22 pF 14 22 pF 13 12 11 10 5V 0.1 F list p=16f84 include <p16F84.inc> ; Define counter variable locations c1 equ 0x0c c2 equ 0x0d c3 equ 0x0e ; Initializes PORTA to output all zeros bcf STATUS, RP0 ; select bank 0 clrf PORTA ; initialize all pin values to zero bsf STATUS, RP0 ; select bank 1 clrf TRISA ; designate all PORTA pins as outputs bcf STATUS, RP0 ; select bank 0 Introduction to Mechatronics and Measurement Systems 69 Solutions Manual ; Main program loop start bsf PORTA, 0 call pause bcf PORTA, 0 call pause ; turn on the LED connected to RA0 ; pause for 1 second ; turn off the LED connected to RA0 ; pause for 1 second goto start ; Subroutine to pause for approximately 1 second pause ; Initialize counter variables movlw 0x00 movwf c1 movlw 0x00 movwf c2 movlw 0xFA movwf c3 loop incfsz c1, F goto loop incfsz c2, F goto loop incfsz c3, F goto loop return ; end of subroutine end 70 ; end of instructions Introduction to Mechatronics and Measurement Systems Solutions Manual 7.3 PIC16F84 1 1k RA3 RA0 3 RA4 OSC1 MCLR OSC2 5 6 7 8 NO RA1 2 4 5V RA2 9 Vss Vdd RB0 RB7 RB1 RB6 RB2 RB5 RB3 RB4 18 17 LED 330 16 15 4 MHz 22 pF 14 22 pF 13 12 11 10 5V 0.1 F 5V ’ Program to turn an LED on and off at 1 Hz while a pushbutton switch ’ is being held down ’ Define variable names for the I/O pins my_button Var PORTB.0 my_led Var PORTA.0 begin: While (my_button == 1) ’ Turn on the LED High my_led ’ while the switch is held down ’ Wait for 1/2 sec Pause 500 ’ Turn off the LED Low my_led ’ Wait for 1/2 sec Pause 500 Wend Introduction to Mechatronics and Measurement Systems 71 Solutions Manual 7.4 Goto begin ’ continue End ’ end of instructions ’ Program to perform the functionality of the Pot statement ’ assumed variables: ’ pin: I/O pin identifier ’ scale: byte variable containing maximum time constant ’ var: byte variable containing measured time constant ’ Charge the capacitor High pin Pause scale ’ Discharge the capacitor and measure the discharge time var = 0 Low pin Input pin While (pin == 1) ’ while the capacitor is not discharged Pause 1 var = var + 1 Wend 7.5 ’ Subroutine to perform a software debounce on pin RB0 debounce: ’ Define pin RB0 as an input pin Var PORTB.0 Input pin ’ Wait for the pushbutton switch to be pressed (1st bounce) loop: If (pin == 0) Then loop ’ Wait 10 milliseconds for the switch bounce to settle Pause 10 ’ Wait for the pushbutton switch to be released (1st bounce) loop2: If (pin == 1) Then loop2 ’ Wait 10 milliseconds for the switch bounce to settle Pause 10 ’ End of subroutine Return 72 Introduction to Mechatronics and Measurement Systems Solutions Manual 7.6 ’ Using a 20MHz oscillator (PULSIN gives number of 2us increments) sensor Var PORTA.0 high_width Var WORD low_width Var WORD period Var WORD rpm Var WORD start: ’ Time entire pulse period PULSIN sensor, 1, high_width PULSIN sensor, 0, low_width period = high_width/2 + low_width/2 ’ Convert period to units of 10 mmin (10 milli-minutes) period = period / 60 / 100 ’ Calculate and display rpm (assuming rpm is in the approximate 10 to 10000 range) If (period > 0) Then rpm = 10000 / period Gosub display_rpm Endif Goto start 7.7 ’ Subroutine to perform a simulated D/A conversion, holding a voltage for approximately ’ 1 second D_to_A: ’ Define variables: ’ digital_value: predefined byte variable indicating the relative voltage value pin Var PORTA.0 ’ output pin i Var BYTE ’ counter variable used in For loop ’ Maintain filtered PWM signal for approximately 1 second For i = 1 To 1000 ’ Hold the pin high for the portion of a second based on the digital value High pin ’ Pause for the appropriate number of 1/255 (approximately) increments Pause (4*digital_value) ’ Hold the pin low for the remainder of the second Low pin ’ Pause for the appropriate number of 1/255 (approximately) increments Pause (4 * (255 digital_value)) Next i ’ End of subroutine Return Introduction to Mechatronics and Measurement Systems 73 Solutions Manual 7.8 The polling loop continues to run after the alarm as been activated, and the if the door and windows are closed the alarm will go off. An improved design would branch off to another section of code when the alarm is activated that would wait for some sort of alarm reset signal before deactivating the alarm. 7.9 ’ PicBasic Pro program to perform the control functions of the security system example ’ using interrupts ’ Define variables for I/O port pins door_or_window Var PORTB.0 motion Var PORTB.1 c Var PORTB.2 d Var PORTB.3 alarm Var PORTA.0 ’ signal A ’ signal B ’ signal C ’ signal D ’ signal Y ’ Define constants for use in IF comparisons OPEN Con 1 ’ to indicate that a door OR window is open DETECTED Con 1 ’ to indicate that motion is detected ’ Initialize interrupts OPTION_REG = $7F On Interrupt Goto myint INTCON = $88 ’ enable PORTB pull-ups ’ enable interrupts on RB4 through RB7 ’ Main loop waiting for sensors to change value (i.e., wait for interrupts) always: Low alarm ’ keep the alarm low until a sensor changes value Goto always ’ continue ’ Interrupt service routine that runs until sensors return to inactive states Disable ’ disable interrupts during the interrupt service routine myint: While ((door_or_window == OPEN) Or (motion == DETECTED)) If ((c == 0) And (d == 1)) Then ’ operating state 1 (occupants sleeping) If (door_or_window == OPEN) Then High alarm Else Low alarm Endif Else If ((c == 1) And (d == 0)) Then ’ operating state 2 (occupants away) If ((door_or_window == OPEN) Or (motion == DETECTED)) Then High alarm Else Low alarm Endif 74 Introduction to Mechatronics and Measurement Systems Solutions Manual Else ’ operating state 3 or NA (alarm disabled) Low alarm Endif Endif Wend INTCON.1 = 0 ’ clear the interrupt flag Resume ’ end of interrupt service routine Enable ’ allow interrupts again End 7.10 // Declare all global variables const int switch_1=1; // first combination switch const int switch_2=2; // second combination switch const int switch_3=3; // third combination switch const int enter_button=4; // combination enter key const int green_led=5; // green LED indicating a valid combination const int red_led=6; // red LED indicating an invalid combination const int speaker=7; // speaker signal for sounding an alarm const int motor=8; // signal to bias the motor power transistor const int a=9; // bit 0 for the 7447 BCD input const int b=10; // bit 1 for the 7447 BCD input const int c=11; // bit 2 for the 7447 BCD input const int d=12; // bit 3 for the 7447 BCD input byte combination; // stores the valid combination in the 3 LSBs byte number_invalid; // counter used to keep track of the number of bad // combinations // Initializations void setup() { // Define pin I/O status pinMode(switch_1, INPUT); pinMode(switch_2, INPUT); pinMode(switch_3, INPUT); pinMode(enter_button, INPUT); pinMode(green_led, OUTPUT); pinMode(red_led=6, OUTPUT); pinMode(speaker=7, OUTPUT); pinMode(motor=8, OUTPUT); pinMode(a=9, OUTPUT); pinMode(b=10, OUTPUT); pinMode(c=11, OUTPUT); pinMode(d=12, OUTPUT); // Initialize the valid combination and turn off all output functions combination = B101; // valid combination (switch 3:on, switch // 2:off, switch 1:on) // Make sure the LEDs and motor are off and initialize the display Introduction to Mechatronics and Measurement Systems 75 Solutions Manual digitalWrite(green_led, LOW); digitalWrite(red_led, LOW); digitalWrite(motor, LOW); digitalWrite(a, LOW); digitalWrite(b, LOW); digitalWrite(c, LOW); digitalWrite(d, LOW); // Initialize invalid combo counter number_invalid = 0; } // Main polling loop void loop() { // Wait for the pushbutton switch to be pressed while (enter_button == 0); // Read switches and compare their states to the valid combination if ( (digitalRead(switch_1) == bitRead(combination, 0) && (digitalRead(switch_2) == bitRead(combination, 1) && (digitalRead(switch_2) == bitRead(combination, 1)) { // Turn on the green LED digitalWrite(green_led, HIGH); // Turn on the motor digitalWrite(motor, HIGH); // Reset the combination attempt counter number_invalid = 0 } else { // Turn on the red LED digitalWrite(red_led, HIGH); // Sound the alarm tone (speaker, 80, 100); // Increment the combination attempt counter and check for overflow number_invalid = number_invalid + 1 if (number_invalid > 9) number_invalid = 0; } // Update the invalid combination attempt counter digit display a = bitRead(number_invalid, 0); b = bitRead(number_invalid, 1); c = bitRead(number_invalid, 2); d = bitRead(number_invalid, 3); // Wait for the pushbutton switch to be released 76 Introduction to Mechatronics and Measurement Systems Solutions Manual while (enter_button == 1); // Turn off the LEDs and the motor digitalWrite(green_led, LOW); digitalWrite(red_led, LOW); digitalWrite(motor, LOW); } 7.11 // Displays the scaled resistance value of a potentiometer on an LCD. #include <LiquidCrystal.h> // Define variables, pin assignments, and constants byte value; // scaled potentiometer value short percentage; // displayed potentiometer percentage value const int pot_pin=1; // potentiometer pin LiquidCrystal lcd(2, 3, 4, 5, 6, 7); void setup() { lcd.begin (16, 2); } void loop() { Pot pot_pin, SCALE, value ’ read the potentiometer value value = analogRead(pot_pin); // Scale value from 0-1023 to 0-100 range percentage = map(value, 0, 1023, 0, 100); // Display the percentage value on LCD display lcd.clear(); lcd.print("pot value = "); lcd.print(percentage); } 7.12 See 7.14 solution with: #include <LiquidCrystal.h> // Initialize arrays of the two 5-digit numbers byte digits_prev[5]; byte digits[5]; // Initialize other variables byte i; // for loop counter byte display[5];’ //number being displayed LiquidCrystal lcd(2, 3, 4, 5, 6, 7); Introduction to Mechatronics and Measurement Systems 77 Solutions Manual void setup() { lcd.begin (16, 2); } void loop() { // When enter key is pressed, initialize the digits for the next number to be entered for (i=0; i<=4; i++) { digits_prev[i] = digits[i]; digits[i] = 10; // to indicate a missing digit (for numbers with < 5 digits) } // Display the numbers on the LCD display lcd.clear(); for (i=0; i<=4; i++) display[i] = digits_prev[i]; display_digits(); lcd.setCursor(0, 1); // move to 2nd line of display for (i=0; i<=4; i++) display[i] = digits[i]; display_digits(); } // Function to display the digits of a number stored in an array of five elements void display_digits(void) { for (i=0; i<=4; i++) if (display[i] != 10) // skip missing digits lcd.print(display[i]); } 78 Introduction to Mechatronics and Measurement Systems Solutions Manual 7.13 20x2 LCD character display Vss Vcc Vee RS 1 2 3 4 R/W 5 E 6 DB4 11 DB5 12 DB6 13 DB7 14 5V 5V 20k pot PIC16F84 1 1k 5V 1k 5V RA2 RA1 2 RA3 RA0 3 RA4 OSC1 MCLR OSC2 4 5 6 5k pot 7 8 0.1 F 9 Vss Vdd RB0 RB7 RB1 RB6 RB2 RB5 RB3 RB4 18 17 16 15 4 MHz 22 pF 14 22 pF 13 12 5V 11 10 0.1 F ’ Displays the scaled resistance value of a potentiometer on an LCD. ’ Define variables, pin assignments, and constants value Var BYTE ’ scaled potentiometer value percentage Var WORD ’ displayed potentiometer percentage value pot_pin Var PORTB.0 ’ pin to which the potentiometer and series capacitor are ’ attached (RB0) SCALE Con 200 ’ value for Pot statement scale factor Introduction to Mechatronics and Measurement Systems 79 Solutions Manual loop: Pot pot_pin, SCALE, value ’ read the potentiometer value percentage = (value * 100) / 255 ’ convert to percentage value ’ Display the percentage value on LCD display Lcdout $FE, 1, "pot value = ", DEC percentage, " %" Goto loop ’ continue to sample and display the potentiometer value End 7.14 Use a combination of Figures 7.11 and 7.13 along with the associated code. Use byte array variables called "digits_prev" and "digits" to store the digits of the entered numbers, where "digits_prev" contains the digits of the previous number entered and "digits" contains the digits of the current number being entered. Add appropriate processing statements to the keypad code to keep track of and store the digits of the current number. Here are excerpts of code needed in the implementation: ’ Initialize arrays of the two 5-digit numbers digits_prev Var BYTE[5] digits Var BYTE[5] ’ Initialize other variables i Var BYTE display Var BYTE[5] ’ For loop counter ’ number being displayed ’ When enter key is pressed, initialize the digits for the next number to be entered For i = 0 To 4 digits_prev[i] = digits[i] digits[i] = 10 ’ to indicate a missing digit (for numbers with < 5 digits) Next i ’ Display the numbers on the LCD display Lcdout $FE, 1 ’ clear the display For i = 0 to 4 display[i] = digits_prev[i] Next i Gosub display_digits Lcdout $FE, $C0 ’ go to the next line of the display For i = 0 to 4 display[i] = digits[i] Next i Gosub display_digits ’ Subroutine to display the digits of a number stored in an array of five elements display_digits: 80 Introduction to Mechatronics and Measurement Systems Solutions Manual For i = 0 To 4 If (display[i] != 10) ’ skip missing digits Lcdout DEC display[i] Endif Next i Return 7.15 Pin RA4 is an open-collector output. The two possible states are open-circuit and ground. The pull-up resistor results in a logic high signal (5V) at Vee when RA4 is in the opencircuit state. Vee is at logic low (0V) when RA4 is grounded. 7.16 See the figure in the solution of Question 6.47 for the "hardware solution." A "software solution" would look something like: B S state count Var Var Var Var PORTB.0 PORTB.1 BYTE BYTE ’ pin attached to the bounce-free, NO button ’ pin attached to the SPDT switch ’ state of the switch when the button is pressed ’ variable used to track the ’ Reset the counter variable count = 0 ’ Main loop loop: ’ Wait for the button to be pressed, and store the state of the switch While (B == 0) : Wend state = S ’ Wait for the button to be released, and increment the count if appropriate While (B == 1) : Wend If (state = 1) Then count = count + 1 Endif Goto loop 7.17 If the button were not bounce-free, we would just need to add pause statements to allow the bounce to settle after each while loop (e.g., Pause 10). 7.18 See Design Example 7.1 for driving the display and see Question 7.5 for how to debounce the inputs (or use the Button statement). Here are some code excerpts that might be useful in the implementation: my_count first second Var Var Var BYTE BYTE BYTE ’ current count (0 to 99) ’ first digit (tens place) ’ second digit (ones place) Introduction to Mechatronics and Measurement Systems 81 Solutions Manual inc dec reset Var Var Var UP DOWN PORTB.0 PORTB.1 PORTB.2 ’ pin attached to "increment" button ’ pin attached to "decrement" button ’ pin attached to "reset" button Con Con 0 1 ’ Process the input If (inc == DOWN) Then my_count = my_count + 1 If (my_count > 99) Then my_count = 0 Endif Endif If (dec == DOWN) Then If (my_count > 0) Then my_count = my_count 1 Endif Endif If (reset == DOWN) Then my_count = 0 Endif Endif ’ Determine the digits first = my_count / 10 second = my_count (10*first) 7.19 Go to www.microchip.com, navigate to the 8-bit, 16-series PIC Microcontrollers, and sort by "Memory Type" (for "FLASH"), then select the appropriate model from the table. 7.20 See the comments and program flow in the "poweramp.bas" code in Threaded Design Example A.4. 7.21 See the comments and program flow in the "stepper.bas" code in Threaded Design Example B.2. 7.22 See the comments and program flow in the "move," "move_steps," and "step_motor" subroutines in the "stepper.bas" code in Threaded Design Example B.2. 7.23 See the comments and program flow in the "speed" and "get_AD_value" subroutines in the "stepper.bas" code in Threaded Design Example B.2. 7.24 See the comments and program flow in the "position" and "get_encoder" subroutines in the "master PIC code" (dc_motor.bas) in Threaded Design Example C.3. 82 Introduction to Mechatronics and Measurement Systems Solutions Manual 7.25 See the comments and program flow in the "slave PIC code" (dc_enc.bas) in Threaded Design Example C.3. Introduction to Mechatronics and Measurement Systems 83 Solutions Manual 8.1 A digital computer or microprocessor uses digital or discrete data, that is, data that are simply strings of 1's and 0's that have no time correspondence. We have to add some type of time coding to make sense of the data. Therefore we have to design interfaces that will change (convert) analog information into a discretized form that will be compatible with a computer. Again, additional code must be included to provide the time references. 8.2 > 2*(15 kHz) = 30 kHz 8.3 84 (a) 1 sample per minute would probably suffice, so fs = 1/60 Hz (b) f s 2 120MHz = 240MHz (c) f s 2 20kHz = 40MHz Introduction to Mechatronics and Measurement Systems Solutions Manual 8.4 Using MathCAD: 0 f max max 1 max ( 2 ) f max 0.318 T 2 2 0 T 6.283 V( t ) t 0.33 t 0 2 T fs t 1.571 sin ( 2 t ) sin ( t ) 0 1 t f s 0.637 t 0.67 0.33 2 f max fs t 0.67 2 T t2 0 t 2 T 2 0 t 10 t 2 T 10 2 1.5 1 V t 0.33 0.5 V t 0.67 V t2 0 V t 10 0.5 1 1.5 2 0 2 4 6 8 10 12 t 0.33 t 0.67 t 2 t 10 Introduction to Mechatronics and Measurement Systems 85 Solutions Manual 8.5 Using MathCAD: a 2 t.5 0 0.5 40 b 0.9 a t1 0 1 40 F( t) 2 cos ab t sin 2 t10 0 10 40 ab 2 t 1 F t .5 0 1 0 10 20 30 40 t .5 1 F t1 0 1 0 10 20 30 40 30 40 t1 1 F t 10 0 1 0 10 20 t 10 86 Introduction to Mechatronics and Measurement Systems Solutions Manual 8.6 From equation 8.2, to prevent aliasing, f s 2f max = 2a For a higher fidelity representation, a much higher sampling rate is required. 8.7 Using MathCAD, Introduction to Mechatronics and Measurement Systems 87 Solutions Manual 8.8 88 5V – – 5V Q = ----------------------------- = 2.44mV 4096 Introduction to Mechatronics and Measurement Systems Solutions Manual 8.9 5V – – 5V N = ----------------------------- = 2000 0.005V An 11 bit A/D converter would suffice since 211 = 2048. A 12-bit converter would be the minimal acceptable standard size available. 8.10 N = 28 = 256 Q = 10V / N = 0.0391 V The digital state number for a given voltage V between 0 and 10 is the truncated value of V/Q. The code is the binary equivalent of the state number. (a) 0/Q=0 corresponding to state 0: 00000000 (b) 1/Q=25.6 corresponding to state 25: 00011001 (c) 5/Q=127.9 corresponding to state 127: 01111111 (d) 7.5/Q=191.8 corresponding to state 191: 10111111 8.11 1byte samples bits 10sec 5000 ------------------- 12 ------------------ ------------- = 75000bytes sec sample 8bits 8.12 B0 = G2 G1 G0 + G2 G1 G0 but it is clear in the truth table that G 1 G 0 = G 1 , G 2 G 1 = G 2 , and G 2 G 1 = G 1 , so B0 = G1 G0 + G2 B1 = G2 G1 G0 + G2 G1 G0 = G1 G0 = G1 Also, 8.13 bit scale fraction bit value cumulative voltage 5 1/2 1 -5V + 1/2(10 V) = 0 V 4 1/4 0 0V 3 1/8 1 1.25 V 2 1/16 1 1.875 V 1 1/32 1 2.1875 V digital output = 10111 8.14 more memory will be required Introduction to Mechatronics and Measurement Systems 89 Solutions Manual 8.15 See www.microchip.com resolution: 12 bits architecture: successive approximation 8.16 See www.national.com 8.17 1 1 V out1 = – --- V 1 = – V o = – --- V s 2 8 8.18 1 1 V out2 = – --- V 2 = – 2V o = – --- V s 2 4 8.19 1 1 V out3 = – --- V 3 = – 4V o = – --- V s 2 2 8.20 The low end of the range (at 0000) would be -10V. The increment between states would be: 1 1 5 – ------ 10V – – 10V = – ------ 20V = – --- V 16 16 4 So the value at 0001 would be: 5 1 – 10V – --- V = – 11 --- V 4 4 The value at 1111, would be: 15 3 – 10V – ------ 20V = – 28 --- V 16 4 8.21 The standard sampling rate for high-fidelity audio recordings is: f s = 44kHz 2 20kHz The sample interval corresponding to this frequency is: 1 ms t = --- = 0.023 ---------------fs sampe For a total time of T=45min, the total required number of samples is: T---= 118800000samples t 90 Introduction to Mechatronics and Measurement Systems Solutions Manual Assuming stereo audio without compression, and using the standard high-fidelity resolution of 16 bits per sample, the total number of memory required is: bits T 1byte 1kB 1MB ----- 16 ------------------ 2channels -------------- -------------------------- ------------------- = 453MB 8bits 1024bytes 1024kB t sample Introduction to Mechatronics and Measurement Systems 91 Solutions Manual 9.1 SPDT 5V NO DPST NO SPST NC NO 9.2 5V NC NO digital signal 9.3 NO 5V reset 9.4 V1 V2 92 Introduction to Mechatronics and Measurement Systems circuit 1 circuit 2 Solutions Manual 9.5 AC power switch 1 switch 2 light + 9.6 Regardless of the polarity of the voltage in each secondary coil, the current flows through an upper diode, down through the resistor, then through a lower diode back to the coil. Therefore, the voltage polarities do change across the resistors, and Vout = Vleft Vright, where Vleft and Vright are the secondary coil voltages. When the core is to the left, Vleft is larger and Vout>0. When the core is to the right, Vright is larger and Vout<0. When the core is centered, both secondaries have the same voltage and Vout=0. 9.7 The excitation frequency (fex) should be much larger than the maximum core displacement frequency (fmax) to prevent aliasing and to result in a high-fidelity representation. The lowpass filter cut-off frequency (flow_pass) should be between fex and fmax to filter out the high frequency of the excitation but pass the lower frequency displacement signal. 9.8 During the transition from 3 (0011) to 4 (0100), any of the following 8 codes could result: 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111. 9.9 From Table 9.1, the all four bits change value between decimal code 7 (0111) and 8 (1000), so during this transition, any bit can have either value (0 or 1), so the maximum count uncertainty is the full range. 9.10 1rev - ------------------1line - 0.18 360 ----------- -----------------------= ------------- rev 1000lines 2pulses pulse 9.11 ’ Declare signal and count variables A Var PORTB.0 B Var PORTB.1 count Var WORD ’ Store the initial states of signals and initialize count A_prev = A B_prev = B count = 0 Introduction to Mechatronics and Measurement Systems 93 Solutions Manual ’ Polling loop to monitor signal negative edges and update count loop: If ((A == 0) And (A_prev == 1) And (B == 1)) Then count = count + 1 EndIf A_prev = A If ((B == 0) And (B_prev ==1) And (A == 1)) Then count = count - 1 EndIf B_prev = B Goto loop 9.12 Checking row 7 (B3B2B1B0=0111, G3G2G1G0=0100) gives: 0=0 1 = 01 1 = 10 1 = 10 Checking row 8 (B3B2B1B0=1000, G3G2G1G0=1100) gives: 1=1 0 = 11 0 = 00 0 = 00 2 9.13 D A = ---------4 D dA = 2 ---- dD 4 dD dL but ------- = – ------- , so D L D dL 2 ---- – D ------- 4 L dA ------- = ------------------------------------ = – 2 dL ------2 A L D--------4 94 Introduction to Mechatronics and Measurement Systems Solutions Manual 2 9.14 D 2 A = ---------- = 0.0491in 4 P = ---- = 10 190psi A 6 E steel = 30 10 psi –4 –6 = --- = 3.40 10 = 340 10 = 340 E R R 0.01 120F = ---------------- = -------------------------= 0.245 –4 3.40 10 9.15 For a metal foil strain gage with F=2 and =0.3, Equation 9.11 gives 2 = 1 + 2(0.3) + PZ where PZ is the piezoresistive term, which works out to be 0.4. Therefore, the change in length term (1) provides 50% (1/2) of the effect, the change is area term (0.6) provides 30% (0.6/2) of the effect, and PZ accounts for the remaining 20% (0.4/2). 9.16 9 E = 200 10 Pa F G = 2.115 D = 0.010m 3 P = 50 10 N R = 120 2 D –5 2 A = ---------- = 7.854 10 m 4 P 9 = ---- = 0.637 10 Pa A R = F G R = 0.808 = --- = 0.00318 E R 1 = R + R = 120.808 R 2 = R 3 = R 4 = 120 R2 R1 V o = V ex ------------------- – ------------------- = 0.00168V ex R1 + R4 R2 + R3 9.17 From Equation 9.21, with Vout=0 and R2=R3, R1 1 0 = V ex ------------------- – --- R 1 + R 4 2 Therefore, R1 ------------------ = 1 --R1 + R4 2 Introduction to Mechatronics and Measurement Systems 95 Solutions Manual which gives: R 4 = R1 So the potentiometer must be adjusted to the exact resistance of the strain gage to balance the bridge. 9.18 9.19 Combination of Figures 9.24b and 9.25. V ex 10V I --------- = ---------------------- = 14.2mA 2 350 2R V ex V --------- = 5V 2 P = IV = 71.4mW 9.20 0.001 2R' 0.001R G so R' ------------- R G 2 but R' = L 0.050 ---- , so m 0.001 L --------------------- 120m = 1.2m 2 0.050 96 Introduction to Mechatronics and Measurement Systems Solutions Manual 9.21 Using MathCAD: Type J Coefficents: c0 0.0488683 c3 c1 19873.1 c4 c2 218615 c5 7 1.1569210 8 2.6491810 9 2.0184410 Temperature/Voltage Relationship: 5 ci V i T( V) i= 0 Approximate Sensitivity: T( 0 ) 0.049 T( 0.03) 546.224 0.03 0 DVDT T( 0.03) DVDT 5.492 10 5 T( 0 ) The sensitivity is approximately 0.055 mV / deg C. 9.22 Using MathCAD: Type J Coefficents: 7 7.8025610 c0 0.100861 c3 c1 25727.9 c4 9 9.2474910 c2 767346 c5 11 6.9768810 c6 c7 13 2.6619210 14 3.9407810 Temperature/Voltage Relationship: 7 ci V i T( V) i= 0 Approximate Sensitivity: T( 0 ) 0.101 DVDT T( 0.015) 302.478 0.015 0 T( 0.015) T( 0.010) 213.286 DVDT 4.961 10 5 T( 0 ) The sensitivity is approximately 0.050 mV / deg C. Introduction to Mechatronics and Measurement Systems 97 Solutions Manual 9.23 Using MathCAD: Type J Coefficents: c0 0.0488683 c3 c1 19873.1 c4 c2 218615 c5 7 1.1569210 8 2.6491810 9 2.0184410 Temperature/Voltage Relationship: 5 ci V i T( V) i= 0 V 0 V 200 root ( T( V) 200 V) V 200 0.011 T V 200 200 So V 200 0 = 11mV 9.24 V T 100 = 30mV 2 3 4 5 100 = a 0 + a 1 V 100 0 + a 2 V 100 0 + a 3 V 100 0 + a 4 V 100 0 + a 5 V 100 0 V 100 0 = 5.26mV V T 0 = V T 100 + V 100 0 = 35.26mV T V T 0 = 636.6 C 9.25 V T 11 = 30mV 2 3 4 5 11 = a 0 + a 1 V 11 0 + a 2 V 11 0 + a 3 V 11 0 + a 4 V 11 0 + a 5 V 11 0 V 11 0 = 0.559mV V T 0 = V T 11 + V 11 0 = 30.559mV T V T 0 = 556 C 98 Introduction to Mechatronics and Measurement Systems Solutions Manual 9.26 Using MathCAD: Type J Coefficents: c0 0.0488683 c3 c1 19873.1 c4 c2 218615 c5 7 1.1569210 8 2.6491810 9 2.0184410 Temperature/Voltage Relationship: 5 ci V i T( V) i= 0 V 0 V 120 root ( T( V) V 10 root ( T( V) V V 120 V 10 120 V) V 120 6.356 10 V 10 5.084 10 10 V) V 5.848 10 4 3 T V 120 120 T V 10 10 3 So the change in voltage would be 5.85 mV. 9.27 From Equation 9.62: 1 2 2 1 2 -----2- s + ------ s + 1 X r s = – -----2- s X i s n n n so the transfer function is: 1 2 – -----2- s Xr s n G s = ------------- = ------------------------------------Xi s 2 1- 2 --------s + -s + 1 2 n n Substituting s=j gives: - 2 ---- n G j = -----------------------------------------------2 – ------ + 2 ------ j + 1 n n Introduction to Mechatronics and Measurement Systems 99 Solutions Manual Therefore, the amplitude magnitude is: 2 ---- n Xr ------ = G j = --------------------------------------------------------------Xi - 2 2 ----- 2 1 – ----+ 2 n n 9.28 n = k- = --m N 5000 ---m- = 316.2 rad ---------------------sec 0.05kg 100 ------ = ------------- = 0.316 n 316.2 - 2 ---- n = 0.1 b 30 = --------------- = ----------------------------------- = 0.949 2 km 2 5000 0.05 (a) rad 2 x··in actual = X in = 5mm 100 ------- sec 2 m 4 mm = 5 10 ---------2- = 50 ---------2sec sec m m sec x··in actual = 50 ---------2- 9.81 ---------------- = 5.1g g sec (b) 1 H a = ------------------------------------------------------------------ = 1.08 2 2 2 2 + 4 ------ 1 – ------ n n 1 1 m 2 X r = -----2- H a X in = ----------------------------2- 1.08 50 ---------2- n sec 316.2 rad ------- sec –4 X r = 5.4 10 m = 0.54mm (c) Since H a is assumed to be 1 for the accelerometer device for all ’s, m 2 x··in measured = n X r 1 = 54 ---------2sec 100 Introduction to Mechatronics and Measurement Systems Solutions Manual (d) 2 ----n –1 – 1 2 0.949 0.316 = – tan ----------------------2- = – tan ---------------------------------------- = – 33.7 = – 0.588rad 1 – 0.1 - 1 – ---- n x r t = X r sin t + = 0.54 sin 100t – 0.588 mm 9.29 m = 1kg n = N k = 2 ---m k- = 1.414 rad ---------s m Ns b = 2 ------m rad = 1.25 -------s r = ------ = 0.884 n X i = 0.010m b = --------------- = 0.707 2 km 2 Xr r ------ = -----------------------------------------------= 0.616 Xi 2 2 2 2 1 – r + 4 r – 1 2 r = – tan --------------2- = – 80.1 = – 1.398 rad 1 – r Xr X r = X i ------ = 0.00616m X i Therefore, the steady state output displacement response is: x o t = x i t + x r t = X i sin t + X r sin t + x o t = 10 sin 1.25t + 6.16 sin 1.25t – 1.40 mm Introduction to Mechatronics and Measurement Systems 101 Solutions Manual 10.1 Use a combination of a power transistor switch circuit and a diode clamp. 10.2 Electric motors and solenoids create changing magnetic fields which induce voltages in nearby unshielded circuits. 10.3 P = T s 1 – ------------ max At maximum power, 1 = --- max 2 so the maximum power is: P max 10.4 1 --- max 1 1 1 2 = P --- max = --- max T s 1 – ---------------- = --- max T s max 2 4 2 At the maximum no-load speed, the motor torque is 0 so V in 10V rad max = -------- = ----------------------------------------------- = 833rpm = 87.3 -------ke 12V 1000rpm sec V in I s = -------- = 6.67A R V in T s = k t -------- = 0.8Nm R P max 10.5 102 max ----------- max max max 2 = P ------------ = ------------ T s 1 – ------------ = ------------ T s = 17.45W 2 2 max 4 See the documentation on the LMD18200. Introduction to Mechatronics and Measurement Systems Solutions Manual 10.6 RESET STEP CW/CCW B0 B1 1 10.7 B1 B0 1 2 3 4 step 2 1 B0 B1 B1 1 0 0 1 0 0 1 2 1 0 1 0 1 0 1 0 1 3 0 1 1 1 0 0 1 1 0 4 0 1 0 1 1 1 0 1 0 1 1 0 0 This checks out with both Table 10.1 and Equation 10.18. For 2, sum of products (SOP) gives: 2 = B1 B0 + B1 B0 and product of sums gives: 2 = B1 + B0 B1 + B0 = B1 B0 + B0 B1 which is the same as the SOP result. 10.8 Search the Internet. Introduction to Mechatronics and Measurement Systems 103 Solutions Manual x 0.001 d = ------- = ------------- = 0.02rad r 0.05 10.9 m = 3 d = 0.06rad = 3.44 Therefore, the minimum required number of steps per revolution is 360 N = ----------- = 105 m To achieve the maximum speed, cm 0.10 ------v max s rad d = ----------- = ------------------ = 2 -------r 0.05cm s rad m = 3 d = 6 -------s Therefore, the required step rate is m steps ---------= 100 ------------ m s 10.10 104 (a) servo motor (b) ac induction motor (c) series dc (d) ac induction (e) servo motor (f) series dc (g) stepper motor (h) synchronous motor (i) dc motor (j) ac induction motor (k) ac motor (l) ac motor Introduction to Mechatronics and Measurement Systems Solutions Manual 10.11 The load speed is related to the motor speed with: l = rg m where the gear box reduction ratio is: M r g = ----N (a) J = Jr + Jl rg 2 The maximum acceleration occurs at start-up where: Ts Ts max = ----- = --------------------2J Jr + Jl rg (b) At steady state, m T m = r g T l = r g k l and T m = T s – T s ------------ mmax l = rg m Combining these equations gives: m 2 kr g m = T s – T s ------------ mmax so Ts Ts m = ------------------------------ and l = r g m = -------------------------------Ts Ts 2 kr g + ------------kr g + ----------------- mmax rg m max (c) Designating the torque on the motor (rotor) side of the gear box as Tgm and the torque on the load side of the gear box as Tgl, the equations of motion for the motor rotor and load can be written as: T m – T gm = J r m and T gl – T l = J l l where the gear box torques are related by: T gm = r g T gl and the load and motor angular accelerations are related by: l = rg m Introduction to Mechatronics and Measurement Systems 105 Solutions Manual Therefore, the motor equation of motion can be written as: Tm – rg Tl + Jl rg m = Jr m This can be written as: T m – r g T l = J eff m where Jeff is the total effective inertia seen by the motor, given by: 2 J eff = r g J l + J r Designating the motor speed m as , the motor and load torques are given by: T m = T s 1 – ------------ max and T l = k l = kr g and the motor equation can now be written as: d T s 1 – ------------ – r g kr g = J eff ------ max dt which can be written as: d J eff ------- + b = T s dt where: Ts 2 b = ------------ + kr g max The particular (steady state) solution to the equation of motion is: Ts ss = ----b and the total general solution for the initial condition (0) = 0 is: b – ------- t J t = ss 1 – e eff Therefore, when the motor is at 95% of its steady state speed, b 0.95 ss 106 – ------- t J = ss 1 – e eff Introduction to Mechatronics and Measurement Systems Solutions Manual so the time required for to reach this speed is: 2 rg Jl + Jr J eff t = – -------- ln 1 – 0.95 = 2.996 --------------------------b Ts ----------- + kr 2g max 10.12 If the inner diameter of the cylinder (i.e., the diameter of the piston face) is D, the diameter of the piston rod is d, and the fluid pressure is P, then the force to extend the cylinder (pressure on the face of the piston) is: 2 D F = PA face = ---------4 and the force to retract the cylinder (pressure on rod side of the piston) is: 2 F = PA face – rod D – d = -----------------------4 2 d 2 10.13 A = --------- = 0.785in 4 2 F = PA = 1000psi 0.785in = 785lb 2 2 d 10mm 2 10.14 A = --------- = --------------------------- = 78.5mm 4 4 F 2000N P = ---- = ----------------------2- = 25.5MPa A 78.5mm 10.15 Search manufacturer catalogs or the Internet. 10.16 Required components: pressure regulator (e.g., 1500 psi), pneumatic cylinder (doubleacting or spring return), valve. The required cylinder area is: F 100 2 2 A = --- = ------------ in = 0.0667in p 1500 Therefore, the required cylinder diameter is: D = 4A- = 0.29in ----- Introduction to Mechatronics and Measurement Systems 107 Solutions Manual 11.1 Analytically, a running average of the three most recent derivative calculations involves the following: ei – 2 – ei – 3 D i – 2 = --------------------------t ei – 1 – ei – 2 D i – 1 = --------------------------t ei – ei – 1 D i = --------------------t ei – ei – 3 Di – 2 + Di – 1 + Di D avg = ------------------------------------------- = --------------------3 3t The last equation can be implemented in code based on the either of the expressions. Here’s the code using the first expression: ’ before loop Di2 = 0 : Di1 = 0 ’ modified derivative calc inside loop Di = (error - error_previous) / DT derivative = (Di2 + Di1 + Di) / 3 ’ update previous two derivative calcs for next loop cycle Di2 = Di1 : Di1 = Di 11.2 See Section 1.1 and Internet Link 7.14 for some examples. 11.3 See Section 1.1 and Internet Link 7.14 for some examples. 108 Introduction to Mechatronics and Measurement Systems Solutions Manual 11.4 penny nickel quarter Shigh Smiddle Slow C B L R X (quarter) Y (nickel) Z (penny) S_low Var PORTB.0 S_mid Var PORTB.1 S_high Var PORTB.2 X_q Var BIT Y_n Var BIT Z_p Var BIT loop: ’ Clear out coin identifiers X_q = 0 : Y_n = 0 : Z_p = 0 ’ Wait for S_low to go low (i.e. wait for a coin). While (S_low == 1) : Wend ’ wait for S_low to go low ’ Check for middle and high sensors while low sensor is active While (S_low == 0) If (S_mid == 0) Then While (S_mid == 0) If (S_high == 0) Then Introduction to Mechatronics and Measurement Systems 109 Solutions Manual X_q = 1 Goto done End If Wend Y_n = 1 Goto done EndIf Wend Z_p = 1 done: ’ The correct coin is identified at this point ............ use the coin info .......... Goto loop ’ process next coin 110 Introduction to Mechatronics and Measurement Systems Solutions Manual A.1 a. 100,000,000 kg = 100,000,000,000 g = 100 x 109 g = 100 Gg b. 0.000000025 m = 25 x 10-9 m = 25 nm c. 16.9 x 10-10 s = 169 x 10-9 s = 169 ns A.2 Do Class Discussion Items A.4 and A.5 in class. A.3 The stress is given by: max h FL --2 FL = ---------------- = 6 ---------21 3 wh ------ wh 12 Using MathCAD: ( F Lw h ) 6 F L wh 2 ddF ( F Lw h ) 6 L d ( F Lw h ) 2 dF h w ddw ( F Lw h ) 6 F L d ( F Lw h ) 2 2 dw h w ddL ( F Lw h ) 6 F d ( F Lw h ) 2 dL h w ddh ( F Lw h ) 12 F L d ( F Lw h ) 3 dh h w F 12520 N F 10N w 11.8cm w 0.5mm L 0.95 m L 0.5mm h 12.1cm h 0.5mm ( F Lw h ) 41.307MPa E( F Lw h ) ddF ( F Lw h ) F ddL ( F Lw h ) L ddw ( F Lw h ) w ddh ( F Lw h ) h 5 E( F Lw h ) 5.711 10 Pa 2 2 Erms( F Lw h ) ( ddF ( F Lw h ) F ) ( ddL ( F Lw h ) L ) 2 2 ( ddw ( F Lw h ) w ) ( ddh ( F Lw h ) h ) 5 Erms( F Lw h ) 3.857 10 Pa Introduction to Mechatronics and Measurement Systems 111