Ch 06 electrical and electromechanical systems

10/27/2013 System Dynamics 6.01 Electrical and Electromechanical Systems System Dynamics 6.02 Electrical and Electromechanical Systems §1.Electrical Elements - Voltage and current are the primary variables used to describe a circuit’s behavior - Current: the flow of electrons, the time rate of change of electrons passing through a defined area Electrical and Eletromechanical Systems 𝑖= 𝑑𝑄 , 𝑑𝑡 𝑄= 𝑖𝑑𝑡 𝑖: current, 𝐴 𝑄: charge, 𝐶 Electrons are negatively charged, the positive direction of current flow is opposite to that of the electron flow HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.03 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements - Voltage: energy is required to move a charge between two points in a circuit The work per unit charge required to this is called voltage Unit of voltage: 𝑉 ≡ 𝐽/𝐶 - Ohm’s Law 𝑣 = 𝑖𝑅 𝑖: current, 𝐴 𝑅: resistance, Ω battery-lightbulb circuit HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.05 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements Electrical quantities HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.04 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements 1.Active and passive Elements - Elements that provide energy are sources, and elements that dissipate energy are loads - Circuit elements may be classified as active or passive • Passive elements: resistors, capacitors, and inductors are not sources of energy • Active elements: energy sources that drive the system - Several types of energy sources • Chemical: batteries • Mechanical: generators • Thermal: thermocouples • Optical: solar cells - Active elements are modeled as either ideal voltage sources or ideal current sources HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.06 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements - Power 𝑃: work done per unit time, so the power generated by an active element, or the power dissipated or stored by a passive element, can be calculated as 𝑤𝑜𝑟𝑘 𝑤𝑜𝑟𝑘 𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 = = = = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 × 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡𝑖𝑚𝑒 𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑡𝑖𝑚𝑒 𝑣 𝑃 = 𝑖𝑣 = 𝑖 𝑅 = 𝑅 𝑃: power, 𝑊 ≡ 𝐽/𝑠 𝑖: current, 𝐴 𝑅: resistance, Ω 𝑣: voltage, 𝑉 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/27/2013 System Dynamics 6.07 Electrical and Electromechanical Systems §1.Electrical Elements 2.Modeling Circuits - The dynamics of physical systems result from the transfer, loss, and storage of mass or energy - Basic laws used to model electrical systems • Conservation of charge: Kirchhoff ’s current law • Conservation of energy: Kirchhoff ’s voltage law System Dynamics 6.08 Electrical and Electromechanical Systems §1.Electrical Elements 3.Resistances - Parallel resistances Kirchhoff’s Current law gives 𝑖 = 𝑖1 + 𝑖2 with 𝑖1 = 𝑣𝑠 /𝑅1, 𝑖2 = 𝑣𝑠 /𝑅2 We have 1 1 1 𝑖= + 𝑣 = 𝑣, ≡ + 𝑅1 𝑅2 𝑠 𝑅 𝑠 𝑅 𝑅1 𝑅2 and 𝑅2 𝑖, 𝑅1 + 𝑅2 The current-divider rule 𝑖1 = currents entering the node = currents leaving the node System Dynamics 6.09 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements9 - Series Resistances Kirchhoff’s voltage law gives 𝑣𝑠 − 𝑣1 − 𝑣2 = 𝑣𝑠 − 𝑖𝑅1 − 𝑖𝑅2 = or 𝑣𝑠 = 𝑅1 + 𝑅2 𝑖 Current in the loop 1 1 𝑖= 𝑣 = 𝑣, ≡ 𝑅1 + 𝑅2 𝑠 𝑅 𝑠 𝑅 𝑅1 + 𝑅2 and 𝑅1 𝑅2 𝑣1 = 𝑅1 𝑖 = 𝑣, 𝑣2 = 𝑅2 𝑖 = 𝑣 𝑅1 + 𝑅2 𝑠 𝑅1 + 𝑅2 𝑠 𝑣1 𝑅1 = 𝑣2 𝑅2 HCM City Univ of Technology, Faculty of Mechanical Engineering 6.11 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements 4.Capacitance - Capacitance: the ability of a device to store charge for a given voltage difference across the element - A capacitor is designed to store charge Charge on a capacitor: 𝑄 = 𝑖𝑑𝑡 - Voltage across the capacitor 𝑄 1 𝑣= = 𝑖𝑑𝑡 = 𝐶 𝐶 𝐶 𝑡 𝑖𝑑𝑡 + 𝑄0 𝐶 𝐶: capacitance, 𝐹 ≡ 𝐶/𝑉 𝑖: current across the capacitor, 𝐴 𝑄0 : the initial charge on the capacitor at time 𝑡 = 0, 𝐶 𝑣= 𝐶 𝑖𝑑𝑡 𝑑𝑣 𝑖=𝐶 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.10 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements - Nonlinear Resistances • Not all resistance elements have the linear voltage-current relation 𝑣 = 𝑖𝑅 • An example of a specific diode’s voltage-current relationship found from experiments is 𝑖 = 0.16(𝑒 0.12𝑣 − 1) For low voltages, we can approximate this curve with a straight line whose slope equals the derivative 𝑑𝑖/𝑑𝑣 at 𝑣 = 𝑑𝑖 = 0.16(0.12𝑒 0.12𝑣 ) = 0.0192 𝑑𝑣 𝑣=0 𝑣=0 Thus, for small voltages, 𝑖 = 0.0192𝑣, and the resistance is 𝑅 = 𝑣/𝑖 = 1/0.0192 = 52Ω The voltage-divider rule System Dynamics 𝑅1 𝑖 𝑅1 + 𝑅2 𝑖1 𝑅2 = 𝑖2 𝑅1 the sum of all the voltage drops around the loop is equal to zero HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑖2 = Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.12 Nguyen Tan Tien Electrical and Electromechanical Systems §1.Electrical Elements 6.Power and Energy - The power dissipated by or stored by an electrical element 𝑃 = 𝑖𝑣 𝑃: power, 𝑊 ≡ 𝐽/𝑠 𝑖: current, 𝐴 𝑣: voltage, 𝑉 - Capacitors and inductors store electrical energy as stored charge and in a magnetic field • The energy 𝐸 stored in a capacitor 𝑑𝑣 𝐸 = 𝑃𝑑𝑡 = 𝑖𝑣𝑑𝑡 = 𝐶 𝑣𝑑𝑡 = 𝐶 𝑣𝑑𝑣 = 𝐶𝑣 𝑑𝑡 • The energy 𝐸 stored in an inductor 𝑑𝑖 𝐸 = 𝑃𝑑𝑡 = 𝑖𝑣𝑑𝑡 = 𝑖 𝐿 𝑑𝑡 = 𝐿 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑖𝑑𝑖 = 𝐿𝑖 2 Nguyen Tan Tien 10/27/2013 System Dynamics 6.13 Electrical and Electromechanical Systems System Dynamics 6.14 Electrical and Electromechanical Systems §1.Electrical Elements Summarizes the voltage-current relations and the energy expressions for resistance, capacitance, and inductance elements §2.Circuit Examples - Ex.6.2.1 Current and Power in a Resistance Circuit For the circuit shown in the figure, the applied voltage is 𝑣𝑠 = 6𝑉 and the resistance is 𝑅 = 10Ω Determine the current and the power that must be produced by the power supply Solution The current is found from 𝑣𝑠 𝑖= = = 0.6𝐴 𝑅 10 The power is computed from 𝑣𝑠2 62 𝑃= = = 3.6𝑊 𝑅 10 Note that we can also compute the power from 𝑃 = 𝑖𝑣𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.15 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.2 A Summing Circuit The figure shows a circuit for summing the voltages 𝑣1 and 𝑣2 to produce 𝑣3 Derive the expression for 𝑣3 as a function of 𝑣1 and 𝑣2 , for the case where 𝑅1 = 𝑅2 = 10Ω and 𝑅3 = 20Ω Solution The voltage-current relations 𝑣1 − 𝑣3 𝑣2 − 𝑣3 𝑣3 𝑖1 = , 𝑖2 = , 𝑖3 = 𝑅1 𝑅2 𝑅3 Kirchoff’s current law gives 𝑖3 = 𝑖1 + 𝑖2 We obtain 𝑣3 𝑣1 − 𝑣3 𝑣2 − 𝑣3 = + ⟹ 𝑣3 = 0.4(𝑣1 + 𝑣2 ) 𝑅3 𝑅1 𝑅2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.17 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples Resistors 𝑅3 and 𝑅4 are in series 𝑅𝑠 = + = 8Ω Resistors 𝑅𝑠 and 𝑅2 are in parallel 1 40 = + = ⟹ 𝑅𝑝 = Ω 𝑅𝑝 10 40 §2.Circuit Examples - Ex.6.2.3 6.16 Electrical and Electromechanical Systems Application of the Voltage-Divider Rule Consider the circuit shown in the figure Obtain the voltage 𝑣0 as a function of the applied voltage 𝑣𝑠 by applying the voltage-divider rule Use the values 𝑅1 = 5Ω , 𝑅2 = 10Ω, 𝑅3 = 6Ω, and 𝑅4 = 2Ω Solution Let 𝑣𝐴 be the voltage at the node as shown The voltage-divider rule applied to resistors 𝑅3 and 𝑅4 gives 𝑅4 𝑣0 = 𝑣 = 𝑣 = 𝑣 𝑅3 + 𝑅4 𝐴 + 𝐴 𝐴 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.18 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples - Potentiometer: a resistance with a sliding electrical pick-off Apply the voltage-divider rule 40/9 𝑣𝐴 = 𝑣 = 𝑣 + 40/9 𝑠 17 𝑠 1 ⟹ 𝑣0 = 𝑣𝐴 = 𝑣 = 𝑣 4 17 𝑠 17 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien • The resistance 𝑅1 between the sliding contact and ground is a function of the distance 𝑥 of the contact from the end of the potentiometer • Potentiometers, commonly called pots, are used as linear and angular position sensors Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/27/2013 System Dynamics 6.19 Electrical and Electromechanical Systems Đ2.Circuit Examples Rotational potentiometer System Dynamics 6.21 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.5 Maximum PowerTransferin a Speaker-AmplifierSystem A common example of an electrical system is an amplifier and a speaker The load is the speaker, which requires current from the amplifier in order to produce sound The resistance 𝑅𝐿 is that of the load The source supplies a voltage 𝑣𝑠 and a current 𝑖𝑆 , and has its own internal resistance 𝑅𝑆 For optimum efficiency, we want to maximize the power supplied to the speaker, for given values of 𝑣𝑆 and 𝑅𝑆 Determine the value of 𝑅𝐿 to maximize the power transfer to the load HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.23 6.20 Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.4 Potentometer Assuming the potentiometer resistance 𝑅1 is proportional to 𝑥 Derive the expression for the output voltage 𝑣0 as a function of 𝑥 Solution The length of the pot is 𝐿 and its total resistance is 𝑅1 + 𝑅2 From the voltage-divider rule 𝑅1 𝑣𝑜 = 𝑉 𝑅1 + 𝑅2 The resistance 𝑅1 proportional to 𝑥: 𝑅1 = 𝑅1 + 𝑅2 𝑥/𝐿 Substituting the above equation gives 𝑥 𝑣𝑜 = 𝑉 = 𝐾𝑥 𝐿 𝐾 = 𝑉/𝐿: the gain of the pot • Voltage relation for rotational pot 𝜃 𝑣𝑜 = 𝑉 = 𝐾𝜃 𝜃𝑚𝑎𝑥 • Voltage relation for linear pot 𝑥 𝑣𝑜 = 𝑉 = 𝐾𝑥 𝐿 𝐾: the gain of the pot HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien Electrical and Electromechanical Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.22 Electrical and Electromechanical Systems §2.Circuit Examples Solution From Kirchhoff’s voltage law 𝑣𝑠 − 𝑖𝑠 𝑅𝑠 − 𝑖𝑠 𝑅𝐿 = From the voltage-divider rule 𝑅𝐿 𝑣𝐿 = 𝑣 𝑅𝑆 + 𝑅𝐿 𝑆 The power consumed by the load 𝑃𝐿 = 𝑖𝑆2 𝑅𝐿 = 𝑣𝐿2 /𝑅𝐿 Using the relation between 𝑣𝐿 and 𝑣𝑆 we can express 𝑃𝐿 in terms of 𝑣𝑆 as 𝑅𝐿 𝑃𝐿 = 𝑣2 𝑅𝑆 + 𝑅𝐿 𝑆 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.24 §2.Circuit Examples To maximize 𝑃𝐿 for a fixed value of 𝑣𝑆 , we must choose 𝑅𝐿 to maximize the ratio 𝑅𝐿 𝑟= 𝑅𝑆 + 𝑅𝐿 The maximum of 𝑟 occurs where 𝑑𝑟 =0 𝑑𝑅𝐿 or 𝑑𝑟 (𝑅𝑆 + 𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑆 + 𝑅𝐿 ) = =0 𝑑𝑅𝐿 (𝑅𝑆 + 𝑅𝐿 )4 ⟹ 𝑅𝐿 = 𝑅𝑆 This result for a resistance circuit is a special case of the more general result known as impedance matching §2.Circuit Examples - Ex.6.2.6 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Nguyen Tan Tien Electrical and Electromechanical Systems A Feedback Amplifier Early in the twentieth century engineers struggled to design vacuum-tube amplifiers whose gain remained constant at a predictable value The gain is the ratio of the output voltage to the input voltage The vacuum-tube gain 𝐺 can be made large but is somewhat unpredictable and unreliable due to heat effects and manufacturing variations A solution to the problem is shown in the figure Derive the input-output relation for 𝑣𝑜 as a function of 𝑣𝑖 Investigate the case where the gain 𝐺 is very large Nguyen Tan Tien 10/27/2013 System Dynamics 6.25 Electrical and Electromechanical Systems §2.Circuit Examples Solution Solve for 𝑣𝑜 𝐺 𝑣 + 𝐺𝑅2 /(𝑅1 + 𝑅2 ) 𝑖 If 𝐺𝑅2 /(𝑅1 + 𝑅2 ) ≫ 1, then 𝑅1 + 𝑅2 𝑣𝑜 ≈ 𝑣𝑖 𝑅2 𝑣𝑜 = System Dynamics 6.27 6.26 Electrical and Electromechanical Systems §2.Circuit Examples 1.Loop Currents Sometimes the circuit equations can be simplified by using the concept of a loop current, which is a current identified with a specific loop in the circuit Use of loop currents usually reduces the number of unknowns to be found, although when deriving the circuit equations you must be careful to use the proper algebraic sum for each element - Example 6.2.7 Analysis with Loop Currents Given the values of the voltages and the resistances for the circuit in the figure (a) Solve for the currents 𝑖1 , 𝑖2 and 𝑖3 passing through the three resistors (b) Use the loopcurrent method to solve for the currents The input voltage to the amplifier 𝑅2 𝑣𝑖 − 𝑣 𝑅1 + 𝑅2 𝑜 Thus the amplifier’s output 𝑅2 𝑣𝑜 = 𝐺 𝑣𝑖 − 𝑣 𝑅1 + 𝑅2 𝑜 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien Electrical and Electromechanical Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.28 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples Solution a.Applying Kirchhoff’s voltage law to the left-hand loop gives 𝑣1 − 𝑅1𝑖1 − 𝑅2 𝑖2 = For the right-hand loop 𝑣2 − 𝑅2 𝑖2 + 𝑅3 𝑖3 = From conservation of charge 𝑖1 = 𝑖2 + 𝑖3 Solving for the solution 𝑅2 + 𝑅3 𝑣1 − 𝑅2 𝑣2 𝑖1 = 𝑅1𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 𝑅3 𝑣1 + 𝑅1𝑣2 𝑖2 = 𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 𝑅2 𝑣1 − (𝑅1 + 𝑅2 )𝑣2 𝑖3 = 𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 §2.Circuit Examples b.Define the loop currents 𝑖𝐴 and 𝑖𝐵 positive clockwise Note - voltage drop 𝑅2 𝑖𝐴 due to 𝑖𝐴 - voltage increase 𝑅2 𝑖𝐵 due to 𝑖𝐵 Applying Kirchhoff’s voltage law to the left-hand loop gives 𝑣1 − 𝑅1𝑖𝐴 − 𝑅2 𝑖𝐴 + 𝑅2 𝑖𝐵 = 𝑣2 + 𝑅3 𝑖𝐵 + 𝑅2 𝑖𝐵 − 𝑅2 𝑖𝐴 = The solution 𝑅2 + 𝑅3 𝑣1 − 𝑅2 𝑣2 𝑖𝐴 = 𝑅1 𝑅2 + 𝑅1𝑅3 + 𝑅2 𝑅3 𝑅2 𝑣1 − (𝑅1 + 𝑅2 )𝑣2 𝑖𝐵 = 𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3 ⟹ 𝑖1 = 𝑖𝐴 , 𝑖3 = 𝑖𝐵 , 𝑖2 = 𝑖𝐴 + 𝑖𝐵 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.29 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.30 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples 2.Capacitance and Inductance in Circuits - Ex.6.2.8 Series RC Circuit The resistor and capacitor in the circuit shown in the figure are said to be in series because the same current flows through them Assume that the supply voltage 𝑣𝑆 is known Obtain the model of the capacitor voltage 𝑣1 §2.Circuit Examples Solution From Kirchhoff’s voltage law 𝑣𝑆 − 𝑅𝑖 − 𝑣1 = For the capacitor 𝑡 𝑄0 𝑣1 = 𝑖𝑑𝑡 + 𝐶 𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Differentiate this with respect to 𝑡 to obtain 𝑑𝑣1 = 𝑖 𝑑𝑡 𝐶 Then substitute for 𝑖 from the first equation 𝑑𝑣1 𝑑𝑣1 = 𝑣 − 𝑣1 ⟹ 𝑅𝐶 + 𝑣1 = 𝑣𝑆 𝑑𝑡 𝑅𝐶 𝑆 𝑑𝑡 This the required model Nguyen Tan Tien 10/27/2013 System Dynamics 6.31 Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.9 Pulse Response of a Series RC Circuit A rectangular pulse input is a positive step function that lasts a duration 𝐷 One way of producing a step voltage input is to use a switch like that shown in the figure The battery voltage 𝑉 is constant and the switch is initially closed at point 𝐵 At 𝑡 = the switch is suddenly moved from point 𝐵 to point 𝐴 Then at 𝑡 = 𝐷 the switch is suddenly moved back to point 𝐵 Obtain the expression for the capacitor voltage 𝑣1 (𝑡) assuming that 𝑣1 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.33 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples The solution of this equation for 𝑡 ≥ 𝐷 is simply the free response with the initial condition 𝑣1 (𝐷) 𝑡−𝐷 𝐷 𝑡−𝐷 𝑣1 𝑡 = 𝑣1 𝐷 𝑒 − 𝑅𝐶 = 𝑉(1 − 𝑒 −𝑅𝐶 )𝑒 − 𝑅𝐶 System Dynamics 6.32 Electrical and Electromechanical Systems §2.Circuit Examples Solution When at 𝑡 = the switch is suddenly moved from point 𝐵 to point 𝐴, the circuit is identical to that shown in the figure with 𝑣𝑆 = 𝑉, and its model is 𝑑𝑣1 𝑅𝐶 + 𝑣1 = 𝑉 𝑑𝑡 The input voltage is a step function of magnitude 𝑉 The solution is the forced response 𝑣1 𝑡 = 𝑉(1 − 𝑒 −𝑡/𝑅𝐶 ) When the switch is moved back to point 𝐵 at time 𝑡 = 𝐷, the circuit is equivalent to that shown in the figure, whose model is equation the same above with 𝑉 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.34 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.10 Series 𝑅𝐶𝐿 Circuit The resistor, inductor, and capacitor in the circuit shown in the figure are in series because the same current flows through them Obtain the model of the capacitor voltage 𝑣1 with the supply voltage 𝑣𝑆 as the input Solution From Kirchhoff’s voltage law 𝑑𝑖 𝑣𝑆 − 𝑅𝑖 − 𝐿 − 𝑣1 = 𝑑𝑡 For the capacitor 𝑡 𝑑𝑣1 𝑣1 = 𝑖𝑑𝑡 ⟹ 𝑖 = 𝐶 𝐶 𝑑𝑡 Pulse response of a series 𝑅𝐶 circuit HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.35 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples Substitute this for 𝑖 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics §2.Circuit Examples - Ex.6.2.11 𝑑𝑣1 𝑑 𝑣1 𝑣𝑆 − 𝑅𝐶 − 𝐿𝐶 − 𝑣1 = 𝑑𝑡 𝑑𝑡 It can be expressed in the following form 𝑑 𝑣1 𝑑𝑣1 𝐿𝐶 + 𝑅𝐶 + 𝑣1 = 𝑣𝑆 𝑑𝑡 𝑑𝑡 6.36 Nguyen Tan Tien Electrical and Electromechanical Systems Parallel 𝑅𝐿 Circuit 𝑅 and 𝐶 in the circuit shown in the figure are said to be in parallel because they have the same voltage 𝑣1 Given supply current 𝑖𝑆 , Obtain the model of the current 𝑖2 passing through the inductor Solution The currents 𝑖1 and 𝑖2 are defined in the figure Then 𝑑𝑖2 𝑣1 = 𝐿 = 𝑅𝑖1 𝑑𝑡 From conservation of charge, 𝑖1 + 𝑖2 = 𝑖𝑆 , we obtain 𝑑𝑖2 𝐿 𝑑𝑖2 𝐿 = 𝑅(𝑖𝑆 − 𝑖2 ) ⟹ + 𝑖2 = 𝑖𝑆 𝑑𝑡 𝑅 𝑑𝑡 This is the required model HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/27/2013 System Dynamics 6.37 Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.12 Analysis of a Telegraph Line Figure shows a circuit representation of a telegraph line The resistance 𝑅 is the line resistance and 𝐿 is the inductance of the solenoid that activates the receiver’s clicker The switch represents the operator’s key Assume that when sending a “dot,” the key is closed for 0.1𝑠 Using the values 𝑅 = 20Ω,𝐿 = 4𝐻, obtain the expression for the current 𝑖(𝑡) passing through the solenoid HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.39 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples - Ex.6.2.13 An 𝑅𝐿𝐶 Circuit with Two Input Voltages The 𝑅𝐿𝐶 circuit shown in the figure has two input voltages Obtain the differential equation model for the current 𝑖3 Solution Applying Kirchhoff’s voltage law to the left-hand loop gives 𝑑𝑖3 𝑣1 − 𝑅𝑖1 − 𝐿 =0 𝑑𝑡 For the right-hand loop 𝑑𝑖3 𝑣2 − 𝑖2 𝑑𝑡 − 𝐿 =0 𝐶 𝑑𝑡 Differentiate this equation with respect to 𝑡 𝑑𝑣2 𝑑 𝑖3 − 𝑖2 − 𝐿 = 𝑑𝑡 𝐶 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.41 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.38 Electrical and Electromechanical Systems §2.Circuit Examples Solution From the voltage law we have 𝑑𝑖 𝐿 + 𝑅𝑖 = 𝑣𝑖 (𝑡) 𝑑𝑡 𝑣𝑖 (𝑡): the input voltage due to the switch and the 12𝑉 supply The voltage 𝑣𝑖 (𝑡) can be modeled as an impulsive input 𝑣𝑖 𝑡 = 1.2𝛿(𝑡) The Laplace transform of equation with 𝑖(0) = gives 1.2 0.3 4𝑠 + 20 𝐼 𝑠 = 1.2 ⟹ 𝐼 𝑠 = = 4𝑠 + 20 𝑠 + This gives the solution 𝑖 𝑡 = 0.3𝑒 −5𝑡 Note that this solution gives 𝑖(0+) = 0.3, whereas 𝑖(0) = The difference is due to the impulsive input HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.40 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples From conservation of charge 𝑖3 = 𝑖1 + 𝑖2 Solve for 𝑖1 𝑑𝑖3 𝑖1 = 𝑣 −𝐿 𝑅 𝑑𝑡 Then 𝑑𝑣2 𝑑 𝑖3 − 𝑖3 − 𝑖1 − 𝐿 = 𝑑𝑡 𝐶 𝑑𝑡 𝑑𝑣2 1 𝑑𝑖3 𝑑 𝑖3 ⟹ − 𝑖3 + 𝑣 −𝐿 −𝐿 = 𝑑𝑡 𝐶 𝐶 𝑅 𝑑𝑡 𝑑𝑡 Rearrange this equation to obtain the answer 𝑑 𝑖3 𝑑𝑖3 𝑑𝑣2 𝐿𝑅𝐶 + 𝐿 + 𝑅𝑖3 = 𝑣1 + 𝑅𝐶 𝑑𝑡 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.42 Nguyen Tan Tien Electrical and Electromechanical Systems §2.Circuit Examples 3.State Variable Models of Circuits The presence of several current and voltage variables in a circuit can sometimes lead to difficulty in identifying the appropriate variables to use for expressing the circuit model ⟹ Use of state variables can often reduce this confusion - Ex.6.2.14 State-Variable Model of a Series 𝑅𝐿𝐶 Circuit Consider the series 𝑅𝐿𝐶 circuit Choose a suitable set of state variables, and obtain the state variable model of the circuit in matrix form The input is the voltage 𝑣𝑠 §2.Circuit Examples Solution The energy stored in the capacitor is 𝐶𝑣12 /2 and the energy stored in the inductor is 𝐿𝑖 /2 ⟹ state variables: 𝑣1 and 𝑖 From Kirchhoff’s voltage law 𝑑𝑖 𝑑𝑖 1 𝑅 𝑣𝑠 − 𝑅𝑖 − 𝐿 − 𝑣1 = ⟹ = 𝑣 − 𝑣 − 𝑖 𝑑𝑡 𝑑𝑡 𝐿 𝑠 𝐿 𝐿 From the capacitor relation 𝑑𝑣1 𝑣1 = 𝑖𝑑𝑡 ⟹ = 𝑖 𝐶 𝑑𝑡 𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 𝑑𝑖 𝑅 − 𝑑𝑡 = 𝐿 𝑑𝑣1 𝐶 𝑑𝑡 Rewrite in matrix form 1 − 𝐿 𝑖 + 𝐿 𝑣𝑠 𝑣1 0 Nguyen Tan Tien 10/27/2013 System Dynamics 6.43 Electrical and Electromechanical Systems §3.Impedance and Amplifier - Ex.6.3.1 Coupled 𝑅𝐶 Loops Determine the transfer function 𝑣𝑜 (𝑠)/𝑣𝑠 (𝑠) of the circuit Solution The energy in this circuit is stored in the two capacitors The energy stored in a capacitor is expressed by 𝐶𝑣 /2 ⟹ the state variables are the voltages 𝑣1 and 𝑣𝑜 The capacitance relations are 𝑑𝑣𝑜 𝑑𝑣1 = 𝑖3 , = 𝑖2 𝑑𝑡 𝐶 𝑑𝑡 𝐶 For the right-hand loop 𝑣1 − 𝑣𝑜 𝑑𝑣𝑜 𝑖3 = ⟹ = (𝑣 − 𝑣𝑜 ) 𝑅 𝑑𝑡 𝑅𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.45 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics §3.Impedance and Amplifier For the left-hand loop 6.44 Electrical and Electromechanical Systems 𝑣𝑠 − 𝑣𝑜 𝑅 From conservation of charge 𝑖2 = 𝑖1 − 𝑖3 = (𝑣𝑠 − 2𝑣1 + 𝑣𝑜 ) 𝑅 𝑖1 = 𝑑𝑣1 = (𝑣 − 2𝑣1 + 𝑣𝑜 ) 𝑑𝑡 𝑅𝐶 𝑠 Transform these equations for zero initial conditions to obtain 𝑠𝑉𝑜 𝑠 = 𝑉 𝑠 − 𝑉𝑜 𝑠 𝑅𝐶 1 𝑠𝑉1 𝑠 = [𝑉 𝑠 − 2𝑉1 (𝑠) + 𝑉𝑜 (𝑠)] 𝑅𝐶 𝑠 𝑉𝑜 (𝑠) ⟹ = 𝑉𝑠 (𝑠) 𝑅 2𝐶 𝑠 + 3𝑅𝐶𝑠 + ⟹ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.46 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier 1.Impedance - A resistance resists or “impedes” the flow of current The corresponding relation is 𝑣/𝑖 = 𝑅 Capacitance and inductance elements also impede the flow of current - In electrical systems an impedance is a generalization of the resistance concept and is defined as the ratio of a voltage transform 𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a current source - Standard symbol for impedance 𝑉(𝑠) 𝑍(𝑠) ≡ 𝐼(𝑠) §3.Impedance and Amplifier - The impedance of a resistor is its resistance 𝑍 𝑠 =𝑅 - For a capacitor 𝑡 𝐼(𝑠) 𝑣 𝑡 = 𝑖𝑑𝑡 ⟹ 𝑉 𝑠 = 𝐶 𝐶 𝑠 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.47 Nguyen Tan Tien Electrical and Electromechanical Systems The impedance of a capacitor 𝑍 𝑠 = 𝐶𝑠 - For an inductor 𝑑𝑖 𝑣 𝑡 = 𝐿 ⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠 𝑑𝑡 The impedance of a inductor 𝑍 𝑠 = 𝐿𝑠 System Dynamics 6.48 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier 2.Series and Parallel Impedances - The concept of impedance is useful because the impedances of individual elements can be combined with series and parallel laws to find the impedance at any point in the system - The laws for combining series or parallel impedances are extensions to the dynamic case of the laws governing series and parallel resistance elements §3.Impedance and Amplifier - Series Impedances • Two impedances are in series if they have the same current If so, the total impedance is the sum of the individual impedances 𝑍 𝑠 = 𝑍1 𝑠 + 𝑍2 (𝑠) • Example: a resistor 𝑅 and capacitor 𝐶 in series have the equivalent impedance 𝑅𝐶𝑠 + 𝑍 𝑠 =𝑅+ = 𝐶𝑠 𝐶𝑠 𝑉(𝑠) 𝑅𝐶𝑠 + ⟹ ≡𝑍 𝑠 = 𝐼(𝑠) 𝐶𝑠 and the differential equation model is 𝑑𝑣 𝑑𝑖 𝐶 = 𝑅𝐶 + 𝑖(𝑡) 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 10/27/2013 System Dynamics 6.49 Electrical and Electromechanical Systems System Dynamics 6.50 Electrical and Electromechanical Systems Đ3.Impedance and Amplifier - Parallel Impedances Two impedances are in parallel if they have the same voltage difference across them Their impedances combine by the reciprocal rule 1 = + 𝑍(𝑠) 𝑍1 (𝑠) 𝑍2 (𝑠) • Example: a resistor 𝑅 and capacitor 𝐶 in parallel have the equivalent impedance 1 𝑉(𝑠) 𝑅 = + ⟹ ≡𝑍 𝑠 = 𝑍(𝑠) 1/𝐶𝑠 𝑅 𝐼(𝑠) 𝑅𝐶𝑠 + and the differential equation model is 𝑑𝑣 𝑅𝐶 + 𝑣 = 𝑅𝑖(𝑡) 𝑑𝑡 §3.Impedance and Amplifier - Ex.6.3.2 Circuit Analysis Using Impedance For the circuit shown in the figure, determine the transfer function between the input voltage 𝑣𝑠 and the output voltage 𝑣𝑜 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.51 Nguyen Tan Tien Electrical and Electromechanical Systems Solution The equivalent impedance 𝑍(𝑠) for 𝑅 and 𝐶 1 𝑅 = + ⟹𝑍 𝑠 = 𝑍(𝑠) 1/𝐶𝑠 𝑅 𝑅𝐶𝑠 + In this representation we may think of the impedance as a simple resistance, provided we express the relations in Laplace transform notation System Dynamics 6.52 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier Kirchhoff’s voltage law gives 𝑉𝑠 𝑠 − 𝑅1 𝐼 𝑠 − 𝑍 𝑠 𝐼 𝑠 = The output voltage is related to the current by 𝑉𝑜 𝑠 = 𝑍 𝑠 𝐼(𝑠) Eliminating 𝐼(𝑠) from these two relations gives 𝑉𝑜 (𝑠) 𝑉𝑠 𝑠 = 𝑅1 − 𝑉𝑜 𝑠 = 𝑍(𝑠) which yields the desired transfer function 𝑉𝑜 (𝑠) 𝑍(𝑠) 𝑅 = = 𝑉𝑠 (𝑠) 𝑅1 + 𝑍(𝑠) 𝑅𝑅1 𝐶𝑠 + 𝑅 + 𝑅1 This network is a first-order system whose time constant is 𝑅𝑅1 𝜏= 𝑅 + 𝑅1 §3.Impedance and Amplifier 3.Isolation Amplifier - A voltage-isolation amplifier is designed to produce an output voltage that is proportional to the input voltage - Such an amplifier may be considered to be a voltage source • if it does not affect the behavior of the source circuit that is attached to the amplifier input terminals, and • if the amplifier is capable of providing the voltage independently of the particular circuit (the “load”) attached to amplifier output terminals HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.53 §3.Impedance and Amplifier - Consider the system 𝑉𝑠 𝑠 − 𝐼1 𝑠 𝑍𝑠 𝑠 − 𝑉𝑖 𝑠 = 0, Nguyen Tan Tien Electrical and Electromechanical Systems 𝐼1 𝑠 = 𝑉𝑖 (𝑠)/𝑍𝑖 (𝑠) • 𝑍𝑖 (𝑠) is large ⟶ 𝑖1 is small: the amplifier does not affect the current 𝑖1 ⟶ the amplifier does not affect the source circuit • 𝑍𝑖 (𝑠) is large 𝑍𝑖 𝑠 𝑉𝑖 𝑠 = 𝑉 (𝑠) ≈ 𝑉𝑠 (𝑠) 𝑍𝑖 𝑠 + 𝑍𝑠 𝑠 𝑠 ⟹ A voltage-isolation amplifier must have a high input impedance HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 6.54 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier - Denote the amplifier’s voltage gain as 𝐺 This means that the amplifier’s output voltage 𝑣𝑜 is 𝑣𝑜 = 𝐺𝑣𝑖 For the load circuit 𝐺𝑉𝑖 𝑠 − 𝐼𝑜 𝑠 𝑍𝑜 𝑠 − 𝐼𝑜 𝑠 𝑍𝐿 𝑠 = ⟹ 𝐼𝑜 𝑠 = 𝐺𝑉𝑖 (𝑠) 𝑍𝑜 𝑠 + 𝑍𝐿 (𝑠) 𝑍𝐿 𝑠 𝐺𝑉𝑜 (𝑠) 𝑍𝑜 𝑠 + 𝑍𝐿 𝑠 • 𝑍𝑜 (𝑠) is small ⟶ 𝑣𝐿 ≈ 𝐺𝑣𝑜 : the voltage 𝑣𝐿 delivered to the load is independent of the load ⟹ 𝑉𝐿 𝑠 = 𝑍𝐿 𝑠 𝐼𝑜 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/27/2013 System Dynamics 6.55 Electrical and Electromechanical Systems System Dynamics 6.56 Electrical and Electromechanical Systems §3.Impedance and Amplifier 4.Operational Amplifier (Op Amp) - An operational amplifier is a high-gain linear amplifier - The op-amp approaches the ideal amplifier of the analog designer’s dreams because it has such ideal characteristics • Very high open-loop gain: 𝐺 = 100,000 + • Very high input resistance: 𝑅𝑖𝑛 > 𝑀Ω • Low output resistance: 𝑅𝑜𝑢𝑡 = 50 ÷ 75 Ω - The output voltage 𝑉𝑜𝑢𝑡 = 𝐺 𝑣2 − 𝑣1 𝑣𝑜𝑢𝑡 : output voltage, 𝑉 𝐺: open-loop gain 𝑣1 : inverting input, 𝑉 𝑣2 : non-inverting input, 𝑉 0.8 × (−𝑉𝑐𝑐 ) ≈ −𝑣𝑠𝑎𝑡 ≤ + 0.8 ì (+ ) Đ3.Impedance and Amplifier - Ex.6.3.3 Op-Amp Multiplier Determine the relation between the input voltage 𝑣𝑖 and the output voltage 𝑣𝑜 of the op-amp circuit Assume that the op amp has the following properties • very large gain 𝐺 • 𝑣𝑜 = −𝐺𝑣1 • very large input impedance Solution The current drawn by the op amp is very small 𝑖3 ≈ 0, the input terminal pair can be represented as an open circuit 𝑣𝑖 − 𝑣1 𝑣1 − 𝑣𝑜 𝑖1 = , 𝑖2 = 𝑅𝑖 𝑅𝑓 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.57 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier From conservation of charge, 𝑖1 = 𝑖2 + 𝑖3 ≈ 𝑖2 𝑣𝑖 − 𝑣1 𝑣1 − 𝑣𝑜 ⟹ = 𝑅𝑖 𝑅𝑓 System Dynamics 6.58 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier 5.Generation Op-Amp Input-Output Relation Substitute 𝑣1 = −𝑣𝑜 /𝐺 into the equation 𝑣𝑖 𝑣𝑜 𝑣𝑜 𝑣𝑜 + =− − 𝑅𝑖 𝑅𝑖 𝐺 𝑅𝑓 𝐺 𝑅𝑓 With very large 𝐺 𝑣𝑖 𝑣𝑜 =− 𝑅𝑖 𝑅𝑓 𝑅𝑓 ⟹ 𝑣𝑜 = − 𝑣𝑖 𝑅𝑖 This circuit can be used to multiply a voltage by the factor 𝑅𝑓 /𝑅𝑖 , and is called an op-amp multiplier Note that this circuit inverts the sign of the input voltage HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier The impedance 𝑍𝑖 (𝑠) of the input elements is defined such that 𝑉𝑖 𝑠 − 𝑉1 𝑠 = 𝑍𝑖 𝑠 𝐼1 (𝑠) For the feedback elements 𝑉1 𝑠 − 𝑉𝑜 𝑠 = 𝑍𝑓 𝑠 𝐼2 (𝑠) The high internal impedance: 𝑖3 ≈ ⟹ 𝑖1 ≈ 𝑖2 The amplifier relation: 𝑣𝑜 = −𝐺𝑣1 ⟹ 𝑉𝑜 𝑠 = −𝐺𝑉1 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.60 Nguyen Tan Tien Electrical and Electromechanical Systems §3.Impedance and Amplifier The gain of this multiplier is 𝑅𝑓 /𝑅𝑖 , with a sign reversal 𝑍𝑓 𝑠 𝑅𝑓 𝑉𝑜 (𝑠) ≈− =− 𝑉1 (𝑠) 𝑍𝑖 𝑠 𝑅𝑖 𝑍𝑓 𝑠 𝑍𝑖 𝑠 ⟹ 𝑉1 𝑠 = 𝑉 𝑠 + 𝑉 (𝑠) 𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 𝑖 𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 𝑜 𝑍𝑓 𝑠 𝑉𝑜 𝑠 𝐺 𝑍𝑖 (𝑠) ⟹ =− ,𝐻 𝑠 = 𝑉1 𝑠 𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 + 𝐺𝐻 𝑠 𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 To eliminate the sign reversal, using an inverter, which is a multiplier having equal resistances Very large 𝐺 ⟹ |𝐺𝐻(𝑠)| ≫ 𝑍𝑓 𝑠 𝑅𝑓 𝑉𝑜 (𝑠) ≈− =− 𝑉1 (𝑠) 𝑍𝑖 𝑠 𝑅𝑖 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10 10/27/2013 System Dynamics 6.61 Electrical and Electromechanical Systems System Dynamics 6.62 Electrical and Electromechanical Systems §3.Impedance and Amplifier - Ex.6.3.4 Integration with Op Amps Determine the transfer function 𝑉𝑜 (𝑠)/𝑉𝑖 (𝑠) of the circuit §3.Impedance and Amplifier - Ex.6.3.5 Differentiation with Op Amps Design an op-amp circuit that differentiates the input voltage Solution The impedance of a capacitor is 1/𝐶𝑠 ⟹ the transfer function of the circuit is found from 𝑍𝑓 𝑠 𝑉𝑜 (𝑠) =− =− 𝑉𝑖 (𝑠) 𝑍𝑖 𝑠 𝑅𝐶𝑠 In the time domain, the circuit model 𝑡 𝑣𝑜 = − 𝑣 𝑑𝑡 𝑅𝐶 𝑖 Solution A differentiator can be created by interchanging the resistance and capacitance in the integrator circuit 𝑍𝑖 𝑠 = 1/𝐶𝑠, 𝑍𝑓 𝑠 = 𝑅 The circuit is called an op-amp integrator, and is used in many devices for computing, signal generation, and control HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.63 Nguyen Tan Tien Electrical and Electromechanical Systems The input-output relation for this ideal differentiator 𝑍𝑓 𝑠 𝑉𝑜 (𝑠) =− = −𝑅𝐶𝑠 𝑉𝑖 (𝑠) 𝑍𝑖 𝑠 The model in the time domain 𝑑𝑣𝑖 (𝑡) 𝑣𝑜 (𝑡) = −𝑅𝐶 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.64 Nguyen Tan Tien Electrical and Electromechanical Systems §4.Electric Motors - Electromechanical systems consist of an electrical subsystem and a mechanical subsystem with mass and possibly elasticity and damping - In some devices, such as motors and speakers, the mass is driven by a force generated by the electrical subsystem In other devices, such as microphones, the motion of the mass generates a voltage or current in the electrical subsystem ⟹ we must apply electrical principles and Newton’s laws to develop a model of an electromechanical system - Often the forces and torques are generated electromagnetically, but other methods are used as well; for example, piezoelectric devices contain crystals that generate forces when a voltage is applied to them §4.Electric Motors 1.Magnetic Coupling - The majority of electromechanical devices utilize a magnetic field - Two basic principles apply to a conductor, such as a wire, carrying a current within a magnetic field • a force is exerted on the conductor by the field 𝑓 = 𝐵𝐿𝑖 𝑓: force, 𝑁 𝐵: the flux density of the field, 𝑊𝑏/𝑚2 𝐿: the length of the conductor, 𝑚 𝑖: current, 𝐴 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.65 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.66 Đ4.Electric Motors if the conductor moves relative to the field, the field induces a voltage in the conductor that opposes the voltage producing the current 𝑣𝑏 = 𝐵𝐿𝑣 𝑣𝑏 : the voltage induced in the conductor, 𝑉 𝐵: the flux density of the field, 𝑊𝑏/𝑚2 𝐿: the length of the conductor, 𝑚 𝑣: conductor velocity in the field, 𝑚/𝑠 §4.Electric Motors HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Electrical and Electromechanical Systems - The circuit represents the electrical behavior of the conductor and the mass 𝑚 represents the mass of the conductor and any attached mass The power generated by the circuit: 𝑣𝑏 𝑖 The power applied to the mass 𝑚 by the force 𝑓: 𝑓𝑣 Neglecting any energy loss due to resistance in the conductor or friction or damping acting on the mass 𝑣𝑏 𝑖 = 𝑓𝑣 = 𝐵𝐿𝑖𝑣 In addition, from Newton’s law 𝑚𝑣 = 𝑓 Therefore 𝑚𝑣 = 𝐵𝐿𝑖 Nguyen Tan Tien 11 10/27/2013 System Dynamics 6.67 Electrical and Electromechanical Systems §4.Electric Motors 2.The D’Arsonval Meter §4.Electric Motors - Ex.6.4.1 The current to be measured is passed through a coil to which a pointer is attached The coil is positioned within a magnetic field and is wrapped around an iron core to strengthen the effects of the field The core thus acts like an inductor The interaction between the current and the field produces a torque that tends to rotate the coil and pointer This rotation is opposed by a torsional spring of stiffness 𝑘 𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics System Dynamics 6.69 Nguyen Tan Tien Electrical and Electromechanical Systems 6.68 Electrical and Electromechanical Systems A Model of the D'Arsonval Meter Derive a model of a D’Arsonval meter in terms of the coil angular displacement 𝜃 and the coil current 𝑖 The input is the applied voltage 𝑣𝑖 Discuss the case where there are 𝑛 coils around the core Solution The torque acting on both sides of the coil 𝐿 𝑇 = 𝑓𝑟 = 2𝐵 𝑖 𝑟 = (𝐵𝐿𝑟)𝑖 The equation of motion of the core/coil unit is 𝑑2𝜃 𝑑𝜃 𝐼 +𝑐 + 𝑘 𝑇 𝜃 = 𝑇 = (𝐵𝐿𝑟)𝑖 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.70 Nguyen Tan Tien Electrical and Electromechanical Systems §4.Electric Motors The rotation of the coil induces a voltage 𝑣𝑏 in the coil that is proportional to the coil’s linear velocity 𝑣 such that 𝑣𝑏 = 𝐵𝐿𝑣 The linear velocity is related to the coil’s angular velocity 𝜃 by 𝑣 = 𝑟𝜃 Thus 𝑑𝜃 𝑣𝑏 = 𝐵𝐿𝑣 = 𝐵𝐿𝑟 𝑑𝑡 The coil circuit is represented in the figure, where 𝑅 represents the resistance of the wire in the coil Kirchhoff’s voltage law applied to the coil circuit gives 𝑑𝑖 𝑣𝑖 − 𝐿 − 𝑅𝑖 − 𝑣𝑏 = 𝑑𝑡 𝑑𝑖 𝑑𝜃 ⟹ 𝐿 + 𝑅𝑖 + 𝐵𝐿𝑟 = 𝑣𝑖 𝑑𝑡 𝑑𝑡 §4.Electric Motors If there are 𝑛 coils • the resulting torque expression: 𝑇 = 𝑛(𝐵𝐿𝑟)𝑖 • the induced voltage expression: 𝑣𝑏 = 𝑛𝐵𝐿𝑟𝜃 the model is 𝑑2𝜃 𝑑𝜃 𝐼 +𝑐 + 𝑘 𝑇 𝜃 = 𝑇 = 𝑛(𝐵𝐿𝑟)𝑖 𝑑𝑡 𝑑𝑡 𝑑𝑖 𝑑𝜃 𝐿 + 𝑅𝑖 + 𝑛𝐵𝐿𝑟 = 𝑣𝑖 𝑑𝑡 𝑑𝑡 If the applied constant voltage 𝑣𝑖 , the system will reach a steady-state in which the pointer comes to rest At steadystate, 𝜃 = 𝑑𝑖/𝑑𝑡 = 0, the above equations become 𝑣𝑖 𝑛𝐵𝐿𝑟𝑖 𝑛𝐵𝐿𝑟𝑣𝑖 𝑖= , 𝜃= = 𝑅 𝑘𝑇 𝑅𝑘 𝑇 This equation can be used to calibrate the device by relating the pointer displacement 𝜃 to either the measured current 𝑖 or voltage 𝑣𝑖 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.71 Nguyen Tan Tien Electrical and Electromechanical Systems §4.Electric Motors 3.DC Motors - There are many types of electric motors, but the two main categories are direct current (dc) motors and alternating current (ac) motors - Within the dc motor category there are the armature-controlled motor and the field-controlled motor - Basic motor elements: stator, rotor, armature, commutator • Stator: stationary and provides the magnetic field • Rotor: an iron core that is supported by bearings • Armature: the coils are attached to the rotor • Commutator: a pair of electrically conducting, spring-loaded carbon sticks (called brushes) that slide on the armature HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 6.72 Nguyen Tan Tien Electrical and Electromechanical Systems §4.Electric Motors 4.Model of an Armature Controlled DC Motors Model - input: the armature voltage 𝑣𝑎 - output: the armature current 𝑖𝑎 motor speed 𝜔 The torque on the armature 𝑇 = 𝑛𝐵𝐿𝑖𝑎 𝑟 = 𝑛𝐵𝐿𝑟 𝑖𝑎 = 𝐾𝑇 𝑖𝑎 𝐾𝑇 ≡ 𝑛𝐵𝐿𝑟: the motor’s torque constant given by the manufacturer of motor The voltage in the armature, back emf, is proportional to the speed 𝑣𝑏 = 𝑛𝐵𝐿𝑣 = 𝑛𝐵𝐿𝑟 𝜔 = 𝐾𝑏 𝜔 𝐾𝑏 ≡ 𝑛𝐵𝐿𝑟: the motor’s back emf constant or the voltage constant Note that 𝐾𝑏 ≡ 𝐾𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 12 10/27/2013 System Dynamics 6.73 Electrical and Electromechanical Systems System Dynamics 6.74 Electrical and Electromechanical Systems §4.Electric Motors The back emf is a voltage drop in the armature circuit Thus, Kirchhoff’s voltage law gives 𝑑𝑖𝑎 𝑣𝑎 − 𝑅𝑎 𝑖𝑎 − 𝐿𝑎 − 𝐾𝑏 𝜔 = 𝑑𝑡 Apply Newton’s law 𝑑𝜔 𝐼 = 𝑇 − 𝑐𝜔 − 𝑇𝐿 𝑑𝑡 𝑑𝜔 ⟹𝐼 = 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿 𝑑𝑡 The above two equations constitute the system model §4.Electric Motors Motor Transfer Functions 𝐼𝑎 (𝑠) 𝐼𝑠 + 𝑐 𝐼𝑎 (𝑠) 𝐾𝑏 = , = 𝑉𝑎 (𝑠) 𝐷(𝑠) 𝑇𝐿 (𝑠) 𝐷(𝑠) Ω(𝑠) 𝐾𝑇 Ω(𝑠) 𝐿𝑎 𝑠 + 𝑅𝑎 = , = 𝑉𝑎 (𝑠) 𝐷(𝑠) 𝑇𝐿 (𝑠) 𝐷(𝑠) with 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 State-Variable Form of the Motor Model 𝑑𝑖𝑎 = (𝑣𝑎 − 𝑅𝑎 𝑖𝑎 − 𝐾𝑏 𝜔) 𝑑𝑡 𝐿𝑎 𝑑𝜔 = (𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿 ) 𝑑𝑡 𝐼 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.75 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.76 Nguyen Tan Tien Electrical and Electromechanical Systems §4.Electric Motors 5.Field Controlled Motors - Require two power supplies • for the armature circuit • for the field circuit - Requite a control circuit to maintain a constant armature current in the presence of the back emf, which varies with motor speed and field strength - In general, the field strength 𝐵(𝑖𝑓 ) is a nonlinear function of the field current 𝑖𝑓 The torque on the armature 𝑇 = 𝑛𝐵 𝑖𝑓 𝐿𝑖𝑎 𝑟 = 𝑛𝐿𝑟𝑖𝑎 𝐵 𝑖𝑓 = 𝑇(𝑖𝑓 ) Often the linear approximation 𝑇 − 𝑇𝑟 ≈ 𝐾𝑇 (𝑖𝑓 − 𝑖𝑓𝑟 ) is used 𝑇𝑟 ,𝑖𝑓𝑟 : torque, current at a reference operating equilibrium 𝐾𝑇 : the slope of the 𝑇(𝑖𝑓 ) curve at the reference condition §4.Electric Motors - Ex.6.4.2 Model of a Field-Controlled dc Motor Develop a model of the field-controlled motor Solution The voltage 𝑣𝑓 is applied to the field circuit, whose inductance and resistance are 𝐿𝑓 and 𝑅𝑓 No back emf exists in the field circuit because it does not move within the field, and Kirchhoff’s voltage law applied to the field circuit gives 𝑑𝑖𝑓 𝑣𝑓 = 𝑅𝑓 𝑖𝑓 + 𝐿𝑓 𝑑𝑡 For the inertia 𝐼 𝑑𝜔 𝐼 = 𝑇 − 𝑐𝜔 − 𝑇𝐿 = 𝐾𝑇 𝑖𝑓 − 𝑐𝜔 − 𝑇𝐿 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.77 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance 1.Steady-State Motor Response The steady-state operating conditions can be obtained by applying the final value theorem, 𝑥 ∞ = lim 𝑠𝑋(𝑠), to the 𝑇𝐹s 𝑠→0 • The steady-state value of 𝑖𝑎 to applied voltage 𝑉𝑎 (𝐼𝑠 + 𝑐) 𝑉𝑎 𝑐𝑉𝑎 𝑠 ⟹ 𝑖 = lim 𝑠𝐼 𝑠 = 𝐼𝑎 𝑠 = 𝑎 𝑎 𝑠→0 𝐷(𝑠) 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 where 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 to applied torque 𝑇𝐿 𝐾𝑏 𝑇𝐿 𝐾𝑏 𝑇𝐿 𝑠 ⟹ 𝑖 = lim 𝑠𝐼 𝑠 = 𝐼𝑎 𝑠 = 𝑎 𝑎 𝑠→0 𝐷(𝑠) 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 𝑐𝑉𝑎 + 𝐾𝑏 𝑇𝐿 ⟹ 𝑖𝑎 = 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 6.78 Nguyen Tan Tien Electrical and Electromechanical Systems Đ5.Analysis of Motor Performance The steady-state value of 𝜔 to applied voltage 𝑉𝑎 𝐾𝑇 𝑉𝑎 𝐾𝑇 𝑉𝑎 𝑠 ⟹ 𝜔 = lim 𝑠Ω 𝑠 = Ω(𝑠) = 𝑠→0 𝐷(𝑠) 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 where 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 to applied torque 𝑇𝐿 𝐿𝑎 𝑠 + 𝑅𝑎 𝑇𝐿 𝑠 ⟹ 𝜔 = lim 𝑠Ω 𝑠 Ω 𝑠 =− 𝑠→0 𝐷 𝑠 −𝑅𝑎 𝑇𝐿 = 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 𝐾𝑇 𝑉𝑎 − 𝑅𝑎 𝑇𝐿 ⟹𝜔= 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 13 10/27/2013 System Dynamics 6.79 Electrical and Electromechanical Systems System Dynamics 6.80 Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Ex.6.5.1 No-Load Speed and Stall Torque Motor: 𝐾𝑇 = 𝐾𝑏 = 0.05𝑁𝑚/𝐴 , 𝑐 = 10−4 𝑁𝑚𝑠/𝐴 , 𝑅𝑎 = 0.5Ω , 𝜔𝑚𝑎𝑥 = 3000𝑟𝑝𝑚, 𝑖𝑚𝑎𝑥 = 30𝐴 Compute the no-load speed, the no-load current, and the stall torque Determine whether the motor can be used with an applied voltage of 𝑣𝑎 = 10𝑉 Solution 𝑐𝑉𝑎 + 𝐾𝑏 𝑇𝐿 10−4 × 10 + 0.05𝑇𝐿 𝑖𝑎 = = = 0.392 + 19.61𝑇𝐿 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 10−4 × 0.5 + 0.05 × 0.05 𝐾𝑇 𝑉𝑎 − 𝑅𝑎 𝑇𝐿 0.05 × 10 − 0.5𝑇𝐿 ω= = = 196.1 − 196.1𝑇𝐿 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 10−4 × 0.5 + 0.05 × 0.05 No load, 𝑇𝐿 = 𝑖𝑎 = 0.392𝐴 < 𝑖𝑚𝑎𝑥 𝜔 = 196.1𝑟𝑎𝑑 = 1872𝑟𝑝𝑚 < 𝜔𝑚𝑎𝑥 Stall torque, 𝜔 = 𝑇𝐿 = 1𝑁𝑚 𝑖𝑎 = 20𝐴 < 𝑖𝑚𝑎𝑥 §5.Analysis of Motor Performance 2.Motor Dynamics Response Ex.6.5.2 Response of an Armature-Controlled dc Motor The parameter values for a certain motor are 𝐾𝑇 = 𝐾𝑏 = 0.05𝑁𝑚/𝐴 𝑐 = 10−4 𝑁𝑚𝑠/𝐴 𝑅𝑎 = 0.5Ω 𝐿𝑎 = × 10−3 𝐻 𝐼 = × 10−5𝑘𝑔𝑚2 where 𝐼 includes the inertia of the armature and that of the load The load torque 𝑇𝐿 is zero Obtain the step response of 𝑖𝑎 (𝑡) and 𝜔(𝑡) if the applied voltage is 𝑣𝑎 = 10𝑉 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.81 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.82 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance Solution Substituting the given parameter values into motor’s 𝑇𝐹s 𝐼𝑎 (𝑠) × 10−5 𝑠 + 10−4 = 𝑉𝑎 (𝑠) 18 × 10−8 𝑠 + 4.52 × 10−5 𝑠 + 2.55 × 10−3 Ω(𝑠) 0.05 = 𝑉𝑎 (𝑠) 18 × 10−8 𝑠 + 4.52 × 10−5 𝑠 + 2.55 × 10−3 If 𝑣𝑎 is a step function of magnitude 10𝑉 × 103𝑠 + 5.555 × 104 𝐶1 𝐶2 𝐶3 𝐼𝑎 𝑠 = = + + 𝑠 𝑠 + 165.52 𝑠 + 85.59 𝑠 𝑠 + 165.52 𝑠 + 85.59 2.777 × 10 𝐷1 𝐷2 𝐷3 Ω 𝑠 = = + + 𝑠 𝑠 + 165.52 𝑠 + 85.59 𝑠 𝑠 + 165.52 𝑠 + 85.59 Evaluating the partial-fraction coefficients 𝑖𝑎 𝑡 = 0.39 − 61𝑒 −165.52𝑡 + 61.74𝑒 −85.59𝑡 𝜔 𝑡 = 196.1 + 210𝑒 −165.52𝑡 − 406𝑒 −85.59𝑡 §5.Analysis of Motor Performance Step response of an armature-controlled dc motor HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.83 Nguyen Tan Tien Electrical and Electromechanical Systems - The large overshoot in 𝑖𝑎 , which is caused by the numerator dynamics The plot shows that the steady-state calculation of 𝑖𝑎 = 0.39𝐴 greatly underestimates the maximum required current, which is approximately 15𝐴 - In practice, a pure step input is impossible, and thus the required current will not be as high as 15𝐴 The real input would take some time to reach 10𝑉 System Dynamics 6.84 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance 3.The Effect of Armature Inductance - The denominator of motor transfer functions 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 If setting 𝐿𝑎 = 0, the model reduces to a first-order model - Another reason 𝐿𝑎 is sometimes neglected is that it is difficult to calculate or to measure - With 𝐿𝑎 = the characteristic equation and the motor differential equations are still only second order and thus are manageable Be careful in using the approximation 𝐿𝑎 = §5.Analysis of Motor Performance 4.Determining Motor Parameters - Motor parameter values can be applied from • the motor manufacturer • the experiment - The voltage and torque constants 𝐾𝑏 , 𝐾𝑇 : cannot be calculated from 𝑛𝐵𝐿𝑟 because the value of the magnetic field parameter 𝐵 might be difficult to determine - The armature inertia 𝑖𝑎 can be • calculated from the formula for the inertia of a cylinder • measured by suspending it with a metal wire and measuring the torsional oscillation frequency 𝑓𝑛 𝐻𝑧 as the armature twists on the wire, then 𝑖𝑎 = 𝑘 𝑇 /(2𝜋𝑓𝑛 )2 𝑘 𝑇 : the torsional spring constant of the wire HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 14 10/27/2013 System Dynamics 6.85 Electrical and Electromechanical Systems System Dynamics 6.86 Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Some parameters can be measured with static (steady-state) tests • By slowly increasing the load torque 𝑇𝐿 until the motor stalls and measuring the resulting stall current, we can compute 𝐾𝑇 from 𝐾𝑇 = 𝑇𝐿 /𝑖𝑎 • Knowing the voltage 𝑉𝑎 , we can compute the armature resistance from 𝑅𝑎 = 𝑉𝑎 /𝑖𝑎 • By measuring the no-load speed 𝜔 and the resulting current 𝑖𝑎 , and knowing 𝑉𝑎 , 𝑅𝑎 , and 𝐾𝑇 , we can compute 𝑐 from the steady-state relations with 𝑇𝐿 = - Because 𝑐 is difficult to determine precisely, its value is rarely reported by motor manufacturers However, in motors with good bearings the damping can be slight and is often taken to be zero §5.Analysis of Motor Performance - The inductance 𝐿𝑎 is • difficult to determine ⟹ special instruments such as an impedance meter can be used to measure 𝐿𝑎 • often assumed to be very small and therefore is often taken to be zero This is sometimes a good approximation, but not always - Units for 𝐾𝑇 and 𝐾𝑏 𝑇 𝑁𝑚 𝐾𝑇 = = 𝑖𝑎 𝐴 𝑣𝑏 𝑉 𝑊 𝑁𝑚/𝑠 𝑁𝑚 𝐾𝑏 = = = 𝑉𝑠 = 𝑠 = 𝑠= 𝜔 1/𝑠 𝐴 𝐴 𝐴 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.87 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance 5.The Trapezoidal Profile and Motion Control HCM City Univ of Technology, Faculty of Mechanical Engineering 6.89 Nguyen Tan Tien Electrical and Electromechanical Systems Đ5.Analysis of Motor Performance Average required current, 𝑖𝑟𝑚𝑠 , and motor torque, 𝑇𝑟𝑚𝑠 The average usually stated is the rms average, which stands for root mean square For torque, it is calculated as follows 𝑇𝑟𝑚𝑠 = 𝑡𝑓 𝑡𝑓 𝑇 𝑡 𝑑𝑡 , 𝑖𝑟𝑚𝑠 = 𝑇𝑟𝑚𝑠 𝐾𝑇 • Maximum speed error: the maximum difference between the desired speed given by the profile and the actual speed • Average speed error: the average speed error is commonly computed as the rms value • Displacement error: the difference between the specified displacement and the actual displacement • System response time: the system must be able to respond fast enough to follow the profile - Note: not all these criteria are important for every application HCM City Univ of Technology, Faculty of Mechanical Engineering 6.88 Electrical and Electromechanical Systems §5.Analysis of Motor Performance 6.Motor and Amplifier Performance - In evaluating the performance of a motion-control system, the following are important • Energy consumption per cycle: the sum of the energy loss in the motor resistance and in the damping - The trapezoidal speed profile shown in the figure is used in many motion-control applications The constant-speed phase of the profile is called the slew phase - Aplications • Speed control: conveyor of assembly line • Position control: robot arm moving a specified angle System Dynamics System Dynamics Nguyen Tan Tien Nguyen Tan Tien 𝑡𝑓 𝐸= 𝑅𝑖 𝑡 𝑑𝑡 + 𝑡𝑓 𝑐𝜔2 𝑡 𝑑𝑡 𝑡𝑓 : the duration of the cycle 𝜔: the speed of the system at the location of the damping • Maximum required current, 𝑖𝑚𝑎𝑥 , and maximum motor torque, 𝑇𝑚𝑎𝑥 • Maximum required motor speed, 𝜔𝑚𝑎𝑥 • Maximum required voltage, 𝑣𝑚𝑎𝑥 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.90 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance - In the following, assume that the damping constant 𝑐 is zero - Obtain the following motor model 𝑑𝑖 𝑣 = 𝑅𝑖 + 𝐿 + 𝐾𝑏 𝜔 𝑑𝑡 𝑑𝜔 𝐼 = 𝐾𝑇 𝑖 − 𝑇𝑑 𝑑𝑡 𝜔: the motor speed, 𝑟𝑎𝑑/𝑠 𝜔𝐿 = 𝜔/𝑁: the load speed, 𝑟𝑎𝑑/𝑠 𝑁: the speed reduction ratio 𝑇𝑑 = 𝑇𝐿 /𝑁: the torque 𝑇𝑑 opposing the motor torque 𝑇 is due to the torque 𝑇𝐿 acting on the load, 𝑁𝑚 𝐼: the inertia includes the motor inertia 𝐼𝑚 and the load inertia 𝐼𝐿 reflected to the motor shaft, 𝑘𝑔𝑚2 𝐼 = 𝐼𝑚 + 𝐼𝐿 /𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 15 10/27/2013 System Dynamics 6.91 Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Energy lost With 𝑐 = 0, the expression for the energy loss per cycle 𝑡𝑓 𝐸= 𝑅𝑖 𝑡 𝑑𝑡 = 𝑅 𝑡𝑓 𝑑𝑡 Assuming that 𝑇𝑑 is constant, and expanding this expression 𝑅𝑖 𝑡𝑓 2𝑅𝐼𝑇𝑑 𝑡𝑓 𝑅𝑇𝑑2 𝑡𝑓 𝐸= 𝜔 𝑑𝑡 + 𝜔 𝑑𝑡 + 𝑑𝑡 𝐾𝑇 𝐾𝑇2 𝐾𝑇 With the assumption that the motion begins and ends at rest so that 𝜔(0) = 𝜔(𝑡𝑓 ), the second integral is zero 𝑅𝑇𝑑2 𝑡𝑓 𝑅𝐼2 𝑡𝑓 𝐸= 𝜔 𝑑𝑡 + 𝐾𝑇 𝐾𝑇2 Note that the first term on the right depends on the profile, while the second term depends on the disturbance torque 𝑇𝑑 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 6.93 Electrical and Electromechanical Systems §5.Analysis of Motor Performance ≤ 𝑡 ≤ 𝑡1 , 𝜔 = 𝜔𝑚𝑎𝑥 /𝑡1 = 𝜃𝑓 /𝑡1 𝑡2 𝑡1 ≤ 𝑡 ≤ 𝑡2 , 𝜔 = 𝑡2 ≤ 𝑡 ≤ 𝑡𝑓 , 𝜔 = −𝜔𝑚𝑎𝑥 /𝑡1 = −𝜃𝑓 /𝑡1 𝑡2 𝑡𝑓 ⟹ 𝑡1 𝜔2 𝑑𝑡 = 𝜃𝑓 𝑡1 𝑡2 𝑡2 𝑑𝑡 + 𝑡𝑓 0𝑑𝑡 + − 2 𝜃𝑓 𝑡1 𝑡2 = 𝜃𝑓 𝑡1 𝑡2 𝑡1 + 𝑡𝑓 − 𝑡2 = 𝜃𝑓 𝑡1 𝑡2 6.92 Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Maximum Motor Speed 𝐼𝜔 + 𝑇𝑑 𝐾𝑇 System Dynamics 𝑑𝑡 2𝑡1 Evaluate the integral in previous equation for the trapezoidal profile, which is specified in terms of the motor speed 𝜔(𝑡) The total angular displacement is the area under the trapezoidal profile shown in the figure Assuming that 𝜃(0) = 0, and because 𝑡𝑓 − 𝑡2 = 𝑡1 , we have 𝑡𝑓 𝜃 𝑡𝑓 = 𝜃𝑓 = 𝜔𝑑𝑡 = 𝜔𝑚𝑎𝑥𝑡1 + 𝜔𝑚𝑎𝑥 𝑡2 − 𝑡1 = 𝜔𝑚𝑎𝑥𝑡2 𝜃𝑓 ⟹ 𝜔𝑚𝑎𝑥 = 𝑡2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.94 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Maximum Motor Torque Maximum required acceleration for the trapezoidal profile 𝜃𝑓 𝜔𝑚𝑎𝑥 𝛼𝑚𝑎𝑥 = = 𝑡1 𝑡1 𝑡2 The motor torque 𝑑𝜔 𝑇 = 𝐾𝑇 𝑖 = 𝐼 + 𝑇𝑑 = 𝐼𝛼 + 𝑇𝑑 𝑑𝑡 The maximum required motor torque 𝜃𝑓 𝜔𝑚𝑎𝑥 𝑇𝑚𝑎𝑥 = 𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 = 𝐼 + 𝑇𝑑 = 𝐼 + 𝑇𝑑 𝑡1 𝑡1 𝑡2 Since 𝑡𝑓 − 𝑡2 = 𝑡1 𝐸= 𝑅𝐼2 𝐾𝑇2 𝜃𝑓 𝑡1 𝑡2 2𝑡1 + 𝑅𝑇𝑑2 𝑡𝑓 𝑅 2𝐼2 𝜃𝑓2 = + 𝑇𝑑2 𝑡𝑓 𝐾𝑇2 𝐾𝑇 𝑡1 𝑡22 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 6.95 §5.Analysis of Motor Performance - RMS Motor Torque 𝑡𝑓 2 𝑇𝑟𝑚𝑠 = 𝑇 𝑡 𝑑𝑡 = 𝑡𝑓 𝑡𝑓 ⟹ 𝑇𝑟𝑚𝑠 = 𝑡𝑓 𝑡1 Electrical and Electromechanical Systems 𝑡𝑓 6.96 Electrical and Electromechanical Systems (𝐼𝛼 + 𝑇𝑑 )2 𝑑𝑡 𝑡2 (𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 )2 𝑑𝑡 + (0 + 𝑇𝑑 )2 𝑑𝑡 𝑡1 𝑡𝑓 𝑇𝑟𝑚𝑠 = System Dynamics Nguyen Tan Tien §5.Analysis of Motor Performance + This reduces to HCM City Univ of Technology, Faculty of Mechanical Engineering (−𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 )2 𝑑𝑡 𝑡2 2𝐼2 𝛼𝑚𝑎𝑥 𝑡1 + 𝑇𝑑2 (𝑡1 + 𝑡2 ) 𝑡𝑓 Since 𝛼𝑚𝑎𝑥 = 𝜃𝑓 /𝑡1 𝑡2 and 𝑡1 + 𝑡2 = 𝑡𝑓 𝑇𝑟𝑚𝑠 = 2𝐼2 𝜃𝑓2 𝑡𝑓 𝑡1 𝑡22 + 𝑇𝑑2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 16 10/27/2013 System Dynamics 6.97 Electrical and Electromechanical Systems System Dynamics 6.98 Electrical and Electromechanical Systems §5.Analysis of Motor Performance - Amplifier Requirements Using the motor current equation 𝑖 = 𝑇/𝐾𝑇 , the maximum current and the rms current required 𝑇𝑚𝑎𝑥 𝑇𝑟𝑚𝑠 𝑖𝑚𝑎𝑥 = , 𝑖𝑟𝑚𝑠 = 𝐾𝑇 𝐾𝑇 The motor voltage equation 𝑑𝑖 𝑣 𝐿 𝑑𝑖 𝐾𝑏 𝑣 = 𝑅𝑖 + 𝐿 + 𝐾𝑏 𝜔 ⟹ = 𝑖 + + 𝜔 𝑑𝑡 𝑅 𝑅 𝑑𝑡 𝑅 If the electrical time constant 𝐿/𝑅 is very small, neglect the second term on the right to obtain 𝑣 𝐾𝑏 =𝑖+ 𝜔 𝑅 𝑅 The maximum voltage required is given approximately by 𝑣𝑚𝑎𝑥 = 𝑅𝑖𝑚𝑎𝑥 + 𝐾𝑏 𝜔𝑚𝑎𝑥 §5.Analysis of Motor Performance - Ex.6.5.3 Calculating Motor-Amplifier Requirements The trapezoidal profile requirements for a specific application are given in the following table, along with the load and motor data HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.99 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance Solution The total inertia 𝐼 𝐼𝐿 × 10−3 𝐼 = 𝐼𝑚 + = 10−3 + = × 10−3 𝑘𝑔𝑚2 𝑁 22 The reduction ratio is 𝑁 = 2, the required motor displacement is 𝑁𝜃𝐿𝑓 = 10𝜋 = 20𝜋 𝑟𝑎𝑑, and the load torque as felt at the motor shaft is 𝑇𝑑 = 𝑇𝐿 /𝑁 = 0.1/2 = 0.05 𝑁𝑚 The motor’s energy consumption per cycle 2(4 × 10−6)(20𝜋)2 𝐸= + (0.05)20.6 = 22𝐽/𝑐𝑦𝑐𝑙𝑒 (0.3)2 0.2(0.4)2 The power consumption is 22/𝑡𝑓 = 37𝐽/𝑠 = 37𝑊 The maximum speed for the trapezoidal profile 𝜔𝑚𝑎𝑥 = 𝜃𝑓 /𝑡2 = 50𝜋 𝑟𝑎𝑑/𝑠 = 1500 𝑟𝑝𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.101 Nguyen Tan Tien Electrical and Electromechanical Systems Determine the motor and amplifier requirements System Dynamics 6.100 Nguyen Tan Tien Electrical and Electromechanical Systems §5.Analysis of Motor Performance The rms torque 𝑇𝑟𝑚𝑠 = 2(4 × 10−6 )(20𝜋)2 + (0.05)2 = 1.28𝑁𝑚 0.6(0.2)(0.4)2 The maximum required torque × 10−3 (20𝜋) 𝑇𝑚𝑎𝑥 = + 0.05 = 1.57 + 0.05 = 1.62𝑁𝑚 0.2(0.4) Note that the load torque contributes little to 𝑇𝑚𝑎𝑥 Most of the required torque is needed to accelerate the inertia The amplifier requirements 1.62 𝑖𝑚𝑎𝑥 = 5.4𝐴 0.3 1.28 𝑖𝑟𝑚𝑠 = = 4.27𝐴 0.3 𝑣𝑚𝑎𝑥 = 5.4 + 0.3 50𝜋 = 10.8 + 47.1 = 57.9𝑉 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.102 Nguyen Tan Tien Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices 1.Tachometer Equipment used to measure linear and rotational velocity Consider the circuit equation for an armature-controlled motor 𝑑𝑖𝑎 𝑣𝑎 − 𝑖𝑎 𝑅𝑎 − 𝐿𝑎 − 𝐾𝑏 𝜔 = 𝑑𝑡 With the tachometer • there is no applied voltage 𝑣𝑎 = • at steady-state 𝑑𝑖𝑎 /𝑑𝑡 = ⟹ −𝑖𝑎 𝑅𝑎 − 𝐾𝑏 𝜔 = Denote the voltage 𝑖𝑎 𝑅𝑎 across the resistor by 𝑣𝑡 𝑣𝑡 = 𝐾𝑏 𝜔 Measure the voltage 𝑣𝑡 ⟹ Determine the velocity 𝜔 §6.Sensors and Electroacoustic Devices 2.Accelerometer Apply the Newton’s law 𝑘 𝑘 𝑚𝑥 = −𝑐 𝑥 − 𝑧 − 𝑥 − 𝑧 − 𝑥 − 𝑧 2 or 𝑚𝑦 + 𝑐𝑦 + 𝑘𝑦 = −𝑚𝑧 The transfer function between the input 𝑧 and the output 𝑦 𝑌(𝑠) −𝑚𝑠 = 𝑍(𝑠) 𝑚𝑠 + 𝑐𝑠 + 𝑘 The device can be used - as a vibrometer to measure the amplitude of a sinusoidal displacement 𝑧 = 𝐴𝑠𝑖𝑛𝜔𝑡, or - as an accelerometer to measure the amplitude of the acceleration 𝑧 = −𝐴𝜔2 sin𝜔𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 17 10/27/2013 System Dynamics 6.103 Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices 3.Strain Gauge Accelerometers System Dynamics 6.105 Nguyen Tan Tien Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices - Ex.6.6.1 An Electromagnetic Speaker Develop a model of the electromagnetic speaker shown in the figure, and obtain the transfer function relating the diaphragm displacement 𝑥 to the applied voltage 𝑣 Solution Consider the simplified model 𝑚: the mass of the diaphragm and the coil 𝑘: spring constant 𝑘 𝑐: damping constant 𝑓: the magnetic force, 𝑓 = 𝑛𝐵𝐿𝑖 = 𝐾𝑓𝑖 𝐾𝑓 ≡ 𝑛𝐵𝐿 𝑛: the number of turns in the coil HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.107 6.104 Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices 5.Electro Acoustic Devices - Speakers and microphones are common examples of a class of electromechanical devices called electroacoustic • Speaker electrical energy ⟹ mechanical energy • Microphone mechanical energy ⟹ electrical energy - A radio amplifier produces a current in a coil that is attached to a diaphragm in the cone This causes the coil and diaphragm to move relative to the permanent magnet The motion of the diaphragm produces air pressure waves, which is sound 4.Piezoelectric Devices HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien Electrical and Electromechanical Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.106 Nguyen Tan Tien Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices From Newton’s law 𝑑2𝑦 𝑑𝑥 𝑚 = −𝑐 − 𝑘𝑥 + 𝐾𝑓 𝑖 𝑑𝑡 𝑑𝑡 Consider the electrical subsystem 𝐿: the coil’s inductance 𝑅: the coil’s resistance The coil experiences a back emf, 𝐾𝑏 𝑥 , because it is a current conductor moving in a magnetic field From Kirchhoff’s voltage law 𝑑𝑖 𝑑𝑥 𝑣 = 𝐿 + 𝑅𝑖 + 𝐾𝑏 𝑑𝑡 𝑑𝑡 𝑣: the voltage signal from the amplifier HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.108 Nguyen Tan Tien Electrical and Electromechanical Systems §6.Sensors and Electroacoustic Devices Transforming the equation and solving for 𝑋(𝑠) gives 𝐾𝑓 𝑋 𝑠 = 𝐼(𝑠) 𝑚𝑠 + 𝑐𝑠 + 𝑘 Transforming the equation and solving for 𝐼(𝑠) gives 𝐼 𝑠 = [𝑉 𝑠 − 𝐾𝑏 𝑠𝑋(𝑠)] 𝐿𝑠 + 𝑅 Eliminating 𝐼(𝑠) from the above two equations, obtain the desired transfer function 𝐾𝑓 𝑋(𝑠) = 𝑉(𝑠) 𝑚𝐿𝑠 + 𝑐𝐿 + 𝑚𝑅 𝑠 + 𝑘𝐿 + 𝑐𝑅 + 𝐾𝑓 𝐾𝑏 𝑠 + 𝑘𝑅 §7.Matlab Applications Step Response from Transfer Functions 𝐼𝑎 (𝑠) 𝐼𝑠 + 𝑐 = 𝑉(𝑠) 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 Ω(𝑠) 𝐾𝑇 = 𝑉(𝑠) 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 The motor parameters Matlab % Program motor_par.m (Motor parameters in SI units) global KT Kb La Ra Im cm KT = 0.05;Kb = KT; La = 2e-3;Ra = 0.5; Im = 9e-5;cm = 1e-4 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 18 10/27/2013 System Dynamics 6.109 Electrical and Electromechanical Systems System Dynamics 6.110 Electrical and Electromechanical Systems §7.Matlab Applications LTI models based on the motor transfer functions Matlab % Program motor_tf.m (TFs for voltage input) I = Im; c = cm; current=tf([I,c],[La*I,Ra*I+c*La,c*Ra+Kb*KT]); %current speed=tf(KT,[La*I,Ra*I+c*La,c*Ra+Kb*KT]); %speed Computes and plots the step response for an input of 10𝑉 Matlab % Program motor_step.m (Motor step response) motor_par motor_tf [current, tc] = step(current); [speed, ts] = step(speed); subplot(2,1,1),plot(tc,10*current), xlabel('t (s)'),ylabel('Current (A)') subplot(2,1,2),plot(ts,10*speed), xlabel('t (s)'),ylabel('Speed (rad/s)') §7.Matlab Applications Modified Step Response A more realistic model of a suddenly applied voltage input is 𝑣(𝑡) = 10(1 − 𝑒 −𝑡/𝜏 ), where 𝜏 is no larger than the system’s dominant time constant Computes motor response to this modified step input for 𝜏 = 0.01𝑠 Matlab % Program motor_mod (Response with modified step) motor_par motor_tf mod_step % Program mod_step.m % Motor simulation with modified step input t = (0:0.0001:0.07); v = 10*(1-exp(-t/0.01)); ia = lsim(current,v,t); plot(t,ia,t,v) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.111 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.112 Nguyen Tan Tien Electrical and Electromechanical Systems §7.Matlab Applications Step Response from State Variable Model 𝑑𝑖𝑎 𝑣 − 𝑖𝑎 𝑅𝑎 − 𝐾𝑏 𝜔 = 𝑑𝑡 𝐿𝑎 𝑑𝜔 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝑑 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿 /𝑁 = = 𝑑𝑡 𝐼 𝐼 𝑁: the gear ratio The appropriate state and input matrices 𝑣 𝑖 𝒙 = 𝑎 ,𝒖 = 𝑇 𝜔 𝐿 𝑅𝑎 𝐾𝑏 − − 𝐿𝑎 𝐿𝑎 𝐿 0 𝑨= ,𝑩 = 𝑎 ,𝑪 = ,𝑫 = 𝐾𝑇 𝑐 0 0 − − 𝐼 𝐼 𝑁𝐼 §7.Matlab Applications The parameter values for the load Matlab % load_par.m Load parameters IL = 0; cL = 0; N = 1; TL = 0; Inertia, damping, the matrices and the state space model Matlab % motor_mat.m Motor state matrices I = Im +IL/N^2; c = cm + cL/N^2; A = [-Ra/La,-Kb/La;KT/I,-c/I;]; B = [1/La,0;0, -1/(N*I)]; C = [1,0;0,1]; D = [0,0;0,0]; sysmotor = ss(A,B,C,D); HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 6.113 Nguyen Tan Tien Electrical and Electromechanical Systems System Dynamics 6.114 Nguyen Tan Tien Electrical and Electromechanical Systems §7.Matlab Applications Computes and plots the response due to a step voltage of magnitude 10𝑣, with the load torque 𝑇𝐿 = Matlab % state_step.m (Motor step response with state model) motor_par load_par motor_mat [y, t] = step(sysmotor); subplot(2,1,1),plot(t,10*y(:,1)), xlabel('t (s)'),ylabel('Current (A)') subplot(2,1,2),plot(t,10*y(:,2)), xlabel('t (s)'),ylabel('Speed (rad/s)') §7.Matlab Applications Trapezoidal Response The applied voltage is the following trapezoidal function, which is shown in the figure for the case where 𝑣𝑚𝑎𝑥 = 20𝑉, 𝑡1 = 0.3𝑠, 𝑡2 = 0.9𝑠, 𝑡𝑓 = 1.2𝑠, 𝑡3 = 1.5𝑠 𝑣𝑚𝑎𝑥 𝑡 ≤ 𝑡 ≤ 𝑡1 𝑡1 𝑣𝑚𝑎𝑥 𝑡1 < 𝑡 < 𝑡2 𝑣 𝑡 = 𝑣𝑚𝑎𝑥 (𝑡𝑓 − 𝑡) 𝑡2 ≤ 𝑡 ≤ 𝑡𝑓 𝑡1 𝑡𝑓 < 𝑡 ≤ 𝑡3 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 19 10/27/2013 System Dynamics 6.115 Electrical and Electromechanical Systems System Dynamics 6.116 Electrical and Electromechanical Systems §7.Matlab Applications Creates the voltage array 𝑣 Matlab % Program trapezoid.m (Trapezoidal voltage profile) t1 = 0.3; t2 = 0.9; tfinal = 1.2; t3 = 1.5; v_max = 20; dt = t3/1000; t = (0:dt:t3); for k = 1:1001 if t(k)