![Electrical/Mechanical Analogies
• Mechanical systems, like electrical networks, have 3 passive,
linear components:
– Two of them (spring and mass) are energy-storage elements; one of
them, the viscous damper, dissipates energy.
– The two energy-storage elements are analogous to the two electrical
energy-storage elements, the inductor and capacitor. The energy
dissipater is analogous to electrical resistance.
• Displacement ‘x’ is analogous to current I
• Force ‘f’ is analogous to voltage ‘v’
• Impedance (Z=V/I) is therefore Z=F/X
• Since, [Sum of Impedances] I(s) = [Sum of applied voltages]
• Hence, [Sum of Impedances] X(s) = [Sum of applied forces]](https://image.slidesharecdn.com/me314-week03-modeling2-170919164046/85/ME-314-Control-Engineering-Week-03-04-18-320.jpg)
![Use Analogy
[(𝑀1 𝑠2 + 𝑓𝑣1
+ 𝑓𝑣3
𝑠 + (𝐾1+𝐾2)] 𝑋1 𝑠 − (𝐾2+𝑓𝑣3
𝑠)𝑋2 𝑠 = 𝐹 𝑠
−(𝐾2+𝑓𝑣3
𝑠)𝑋1 𝑠 + [(𝑀2 𝑠2
+ 𝑓𝑣2
+ 𝑓𝑣3
𝑠 + (𝐾1+𝐾2)]𝑋2 𝑠 = 0
Loop 1
Loop 2](https://image.slidesharecdn.com/me314-week03-modeling2-170919164046/85/ME-314-Control-Engineering-Week-03-04-24-320.jpg)
ME-314- Control Engineering - Week 03-04
- 1. ME-314 Control Engineering Dr. Bilal A. Siddiqui Mechanical Engineering DHA Suffa University and Brian Douglas
- 2. Mathematic Modeling of Dynamical Systems • Dynamics of many systems can be written in terms of differential equations – Mechanical, thermal, electrical, economic, biological systems etc. • We said that these D.E.’s can be derived using basic physical laws • All systems we will study will be ‘causal’, i.e. the system’s response at any time ‘t’ depends only on past and not future inputs • We should always be aware about – The scope of our assumptions (how valid they are and when are they valid?) – Simplicity is desired but oversimplification is also dangerous • Recall transfer functions: – It is the ratio of Laplace Transform of output to Laplace Transform of input, when initial conditions are zero. – We assume • Zero initial conditions • Linearity • Time Invariance
- 3. Similarities in Mechanical and Electrical Systems • 3 basic components in mech systems: – Mass – Spring – Damper • 3 basic components in electric systems: – Resistance – Capacitor – Inductor • Basic form of differential equations is the same. • Therefore, learning to model one type of system easily leads to modeling method for the other. • Also, electrical and mechanical systems can be easily cascaded in block diagrams due to this similarity. • In fact, many other types of systems have similar forms • We will begin with electrical systems. This will make modeling mech easier!
- 4. Modeling Electrical Systems (Nise) • Current (i) is the rate of flow of charge (q) 𝑖 𝑡 = 𝑑𝑞 𝑡 𝑑𝑡 ⇒ 𝑞 𝑡 = 𝑖 𝑡 𝑑𝑡 • Taking Laplace transform 𝐼 𝑠 = 𝑠𝑄 𝑠 ⇒ 𝑄 𝑠 = 1 𝑠 𝐼(𝑠) • Impedance (complex resistance) is defined as 𝑧 = v i ⇒ Z s = V s I s • It’s mathematical inverse is called admittance.
- 5. Passive Electric Components • Resistor: 𝑣 𝑡 = 𝑖 𝑡 𝑅 = 𝑅 𝑑𝑞 𝑡 𝑑𝑡 ⇒ 𝑉 𝑠 = 𝐼 𝑠 𝑅 𝑖 𝑡 = 𝑣 𝑡 𝑅 ⇒ 𝐼 𝑠 = 𝑉 𝑠 𝑅 • Capacitor: 𝑣 𝑡 = 1 𝐶 𝑞 𝑡 = 1 𝐶 𝑖 𝑡 𝑑𝑡 ⇒ 𝑉 𝑠 = 1 𝐶𝑠 𝐼 𝑠 𝑖 𝑡 = 𝐶 𝑑𝑣 𝑡 𝑑𝑡 ⇒ 𝐼 𝑠 = 𝐶𝑠𝑉(𝑠) • Inductor: 𝑣 𝑡 = 𝐿 𝑑𝑖 𝑡 𝑑𝑡 = 𝐿 𝑑2 𝑞 𝑡 𝑑𝑡2 ⇒ 𝑉 𝑠 = 𝐿𝑠𝐼 𝑠 𝑖 𝑡 = 1 𝐿 𝑣 𝑡 𝑑𝑡 ⇒ 𝐼 𝑠 = 1 𝐿𝑠 𝑉 𝑠
- 6. Summary of Passive Components • We will combine these elements in complex networks using Kirchoff’s Laws – Current and Voltages in a loop sum to zeros
- 7. Single Loop RLC Circuit • Find transfer function of Vc(s) to input V(s) ⇒
- 8. Simplifying the Procedure • Let us look at this in another way. – Resistor: 𝑉𝑅 𝑠 = 𝐼 𝑠 𝑅 – Capacitor: 𝑉𝐶 𝑠 = 1 𝐶𝑠 𝐼 𝑠 – Inductor: 𝑉𝐿 𝑠 = 𝐿𝑠𝐼 𝑠 • Let’s define impedance (similar to resistance) as 𝑍 𝑠 = 𝑉 𝑠 𝐼 𝑠 • Unlike resistance, impedance is also applicable to capacitors & inductors. • It represents information about dynamic behavior of components. c
- 9. Solving Multi-Loop Electric Circuits • For multiple loops and loads, use the following recipe. – Replace passive element values with their impedances. – Replace all sources and time variables with their Laplace transform. – Assume a transform current and a current direction in each mesh. – Write Kirchhoff’s voltage law around each mesh. – Solve the simultaneous equations for the output. – Form the transfer function.
- 10. Multi-loop Example • Find the transfer function I2(s) / V (s) • Solving for Loop 1 and Loop 2 • There are various ways to solve this. 𝑅1 + 𝐿𝑠 −𝐿𝑠 −𝐿𝑠 𝐿𝑠 + 𝑅2 + 1/𝐶𝑠 𝐼1 𝑠 𝐼2 𝑠 = 𝑉 𝑠 0 • This will yield the following transfer function (1) (2)
- 11. Summarizing the Method • Let us look at the pattern in the last example • This form will help us write such equations rapidly • Mechanical equations of motion (covered next) have the same form. So, this form is very useful!
- 12. Another Example
- 13. Loop 1
- 14. Loop 2 Loop 3
- 16. Class Quiz
- 17. Mechanical Systems (Translational) • Many concepts applied to electrical networks can also be applied to mechanical systems via analogies. • This will also allow us to model hydraulic/pneumatic/thermal systems.
- 18. Electrical/Mechanical Analogies • Mechanical systems, like electrical networks, have 3 passive, linear components: – Two of them (spring and mass) are energy-storage elements; one of them, the viscous damper, dissipates energy. – The two energy-storage elements are analogous to the two electrical energy-storage elements, the inductor and capacitor. The energy dissipater is analogous to electrical resistance. • Displacement ‘x’ is analogous to current I • Force ‘f’ is analogous to voltage ‘v’ • Impedance (Z=V/I) is therefore Z=F/X • Since, [Sum of Impedances] I(s) = [Sum of applied voltages] • Hence, [Sum of Impedances] X(s) = [Sum of applied forces]
- 20. Recall Spring Mass Damper System
- 21. Multiple Degrees of Freedom Mechanical Systems • Many mechanical systems are similar to multiple-loop and multiple-node electrical networks, where more than one simultaneous differential equation is required to describe the system. • In mechanical systems, the number of equations of motion required is equal to the number of linearly independent motions. • Linear independence implies that a point of motion in a system can still move if all other points of motion are held still. • Another name for the number of linearly independent motions is the number of degrees of freedom. • In a two-loop electrical network, each loop current depends on the other loop current, but if we open-circuit just one of the loops, the other current can still exist if there is a voltage source in that loop. • Similarly, in a mechanical system with two degrees of freedom, one point of motion can be held still while the other point of motion moves under the influence of an applied force.
- 22. 2 DOF Transfer Function • System has two degrees of freedom, since each mass can be moved in the horizontal direction while the other is held still. • 2 simultaneous equations of motion will be required to describe system. • The two equations come from free-body diagrams of each mass. • Forces on M1 are due to (a) its own motion and (b) motion of M2 transmitted to M1 through the system.
- 23. Force Analysis a. Hold M2 still, move M1 to right b. Hold M1 still, move M2 to right c. combined a. Hold M1 still, move M2 to right b. Hold M2 still, move M1 to right c. combined Forces on M1 Forces on M2
- 24. Use Analogy [(𝑀1 𝑠2 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + (𝐾1+𝐾2)] 𝑋1 𝑠 − (𝐾2+𝑓𝑣3 𝑠)𝑋2 𝑠 = 𝐹 𝑠 −(𝐾2+𝑓𝑣3 𝑠)𝑋1 𝑠 + [(𝑀2 𝑠2 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + (𝐾1+𝐾2)]𝑋2 𝑠 = 0 Loop 1 Loop 2
- 25. 3 DOF Translational System
- 26. Loop 1
- 27. Loop 2
- 28. Loop 3
- 29. Homework 2a
- 30. Mechanical Systems (Rotational) • Torque replaces force ; angular displacement replaces translational displacement.
- 31. An Example Physical system Schematic of the system
- 32. Loop 1
- 33. Loop 2
- 34. 3 DOF Rotational System
- 35. Homework 2b
- 38. Generalizing for Gears • From the following observation
- 39. Example of Geared System
- 40. An important note • Intuitively, we will start forming two simultaneous equations for each inertia. • But, the inertias are not independent (we cannot stop the motion of one and study the motion of the other), since they are tied together by the gears. • Thus, there is only one degree of freedom and hence one equation of motion !
- 41. • Reflecting the torque and impedances
- 42. Gear Train 10/2/2017Spring 312 ( Cpoyrights Bilal Siddiqui 2015 42 • Max gear ratio of only about 10 : 1 can be obtained with one pair of gears. • Greater ratios can be obtained in less space and with fewer dynamic problems by compounding additional pairs of gears. • A two-stage compound gear train can obtain a train value of up to 100 : 1. • Consider a pinion 2 driving a gear 3. The speed of the driven gear is 𝑛3 = 𝑁2 𝑁3 𝑛2 = 𝑑2 𝑑3 𝑛2 • This relation is valid regardless of the type of gears meshed. • Consider the more complex train shown above. • The train ratio is simply 𝑛6 = − 𝑁2 𝑁3 𝑁3 𝑁4 𝑁5 𝑁6 𝑛2 • In other words, 𝑛 𝐿 = 𝑒. 𝑛 𝐹 • Where, 𝑒 = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑡𝑜𝑜𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑛 𝑡𝑜𝑜𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 • nL is the speed of the last gear and nF is the speed of the first gear in the train.
- 43. More on Gear Trains • Making a train also helps reduce the package size • Sometimes, idler gears are necessary to provide support to one of the gears. • Used in many applications, like automobiles, rotorcraft etc 10/2/2017 43
- 44. Example Find the train ratio 10/2/2017 44
- 45. Gear Trains
- 46. Example with Gear Train
- 47. Gear Train System (continued) • Therefore, the transfer function will be
- 48. Homework 2c
- 49. Modeling Electromechanical Systems • Systems that are hybrid of electrical and mechanical systems
- 50. DC Motor – An Electromechanial System • Mechanical output (θ), electrical input (V)
- 51. Schematic of DC Motor
- 52. How DC Motor works? • A rotating circuit called the armature, through which current ia flows, passes through the fixed magnetic field at right angles and feels a torque which turns the rotor. This torque is 𝑻 𝒎 𝒔 = 𝑲 𝒕 𝑰 𝒂 𝒔 ⇒ 𝑰 𝒂 𝒔 = 𝑻 𝒎 𝒔 /𝑲 𝒕 • When a conductor moves through a magnetic field, it develops a voltage, called back emf eb, proportional to the speed. 𝒗 𝒃 𝒕 = 𝑲 𝒃 𝒅𝜽 𝒎 𝒅𝒕 ⇒ 𝑽 𝒃 𝒔 = 𝑲 𝒃 𝒔𝜽 𝒎 𝒔 • Write the loop equation for the electric circuit 𝑬 𝒂 𝒔 = 𝑹 𝒂 + 𝑳 𝒂 𝒔 𝑰 𝒂 + 𝑽 𝒃 𝒔 • We can write 𝑬 𝒂 𝒔 = 𝑹 𝒂+𝑳 𝒂 𝒔 𝑲 𝒕 𝑻 𝒎(𝒔) + 𝑲 𝒃 𝒔𝜽 𝒎 𝒔
- 53. Mechanical Loading of the Rotor • We must find 𝑇 𝑚 𝑠 𝜃 𝑚 𝑠 if we want to obtain the transfer function 𝜃 𝑚 𝑠 𝐸 𝑎 𝑠 • Jm is the equivalent inertia at the armature and includes both the armature inertia, and (as we will see later) the load inertia reflected to the armature. 𝑇 𝑚 𝑠 = 𝐽 𝑚 𝑠2 + 𝐷 𝑚 𝑠 𝜃 𝑚 𝑠 • Back-substituting in previous equation 𝐸 𝑎 𝑠 = 𝑹 𝒂+𝑳 𝒂 𝒔 𝑲 𝒕 𝐽 𝑚 𝑠2 + 𝐷 𝑚 𝑠 𝜃 𝑚 𝑠 + 𝐾𝑏 𝑠𝜃 𝑚 𝑠 • Assuming that 𝐿 𝑎 ≪ 𝑅 𝑎 The motor transfer function can simply be written as 𝜽 𝒎 𝒔 𝑬 𝒂 𝒔 = 𝑲 𝒔 + 𝜶
- 54. Assignment 2d • What happens when the motor is connected to a load through a gear? Write the transfer function.