Q11. What is the difference between classical and state space control methods?
Classical control is suitable for single input single output systems whereas state space methods can be useful for linear and non linear time invariant and time varying , multi input multi output systems also.
Q12. In a PID Controller what is the role of P, I and D values ?
Kp is proportional gain which gives a large correction for a large error. Immediate response to errors. It does not affect gain or phase margin.
Ki is Integral Gain which allows the steady state error to be zero & reduces peak overshoot.
It is used to increase the open loop gain at low frequencies.
Kd is the derivative gain which reduces oscillation, makes the system more stable. It also introduces phase lag, increases speed of response, decreases gain margin & amplifies noise. It may result in instability.
Q13. What is sliding control ?
The basic idea is that the control signal changes abruptly on the basis of the state of the system. A control system of this type is also referred to as a variable structure system.
Q14. What is the difference between linear & nonlinear analysis?
One can obtain closed form solutions for linear systems. This is seldom the case for nonlinear systems.
Q15. How does the sampling time affect the response of a continuous time system ?
The D/A converter between the discrete & continuous time process implements holding of the calculated control signal during the time step (sampling interval). This holding implies that the control signal is time delayed by approx h/2. The delay influences the stability of the control loop. Suppose we have tuned a continuous time PID & apply these PID parameters on a discrete time PID Controller, then the control loop will get reduced stability because of the approx delay of h/s. As a rule of thumb the stability reduction is small & tolerable if the time delay is less than 1/10th of the response time of the control system as it would have been with a continuous time controller or a controller having small sampling time.
h/2 < Tr/10
h < Tr/5 ;
h < 1/ 5wb
The response time here is 63% rise time which can be read off from the setpoint step response. For a system having dominating time constant T, the response time is approx equal this time constant. If the bandwidth of the control system is wb ( assuming PID parameters have been found using continuous time PID controller) the response time of the control system can be estimated by
Tr 1/wb
Q16. How are poles and eigenvalues related?
Poles and eigenvalues are equal for most systems not having pole zero cancellation.
Q17. How to handle a nonlinear model ?
If the model is nonlinear it must be linearised before calculating the poles or eigenvalues.
Q18. What is the relation between the transfer function and state space model ?
When we convert the state space model to the transfer function, we observe that the denominator of the transfer function is the determinant of (sI-A).
d(s) = |sI-A|
Q19 What is the condition for stability of a continuous time system & for a discrete time system from its pole locations?
The system is stable if all poles are in the Left hand plane i.e. d(s) = 0 are with all negative real parts. A sampled or discrete time system is stable if all the poles of the closed loop transfer function lie within the unit circle of the z-plane.
In terms of linear systems, we recognize that the stability requirement may be defined in terms of the location of the poles of the closed-loop transfer function.
A necessary and sufficient condition that a feedback system be stable is that all the poles of the system transfer function have negative real parts.
Q19a. What is the Nyquist Criterion?
The Nyquist criterion is a graphical method and deals with the loop gain transfer function, i.e., the open-loop transfer function.
We can determine the stability of the closed-loop system by locating the zeros of 1 + G(s). This result is of prime importance in the following development. For the moment, let us assume that 1 + G(s) is known in factored form so that we have Obviously, if 1 + G(s) were known in factored form, there would be no need for the use of the Nyquist criterion, since we could simply observe whether any of the zeros of 1 + G(s) [which are the poles of Y( s ) /R( s ) ]l,i e in the right half of the s plane. In fact, the primary reason for using the Nyquist criterion is to avoid this factoring. Although it is convenient to think of 1 + G(s) in factored form at this time, no actual use is made of that form.
A c c o r d ing to Nyquist plot stability evaluation methods, a system with no open-loop right half-plane (RHP) poles should have no clockwise (CW) encirclements of the -1 point for stability, and a system with open-loop RHP poles should have as many counter-clockwise (CCW) encirclements as there are open-loop RHP poles
Q19b. What are the properties of Bode plots?
Any information that can be wrenched out of the Bode plots of the loop gain is critically important for two reasons. First, the Bode plots are a natural place to judge the properties of the feedback loop. When the magnitude of the loop gain is large, positive feedback properties such as good disturbance rejection and good sensitivity reduction are obtained. When the magnitude of the loop gain is small, these properties are not enhanced. The work of this chapter completes the missing information about transient response that can be read from the loop gain Bode plots.
Second, it is the Bode plot that we are able to manipulate directly using series compensation techniques. It is important to be able to establish the qualities of the Bode plots that produce positive qualities in a control system because only then can the Bode plots be manipulated to attain the desired qualities.
Q20. How do we map the s-plane to the z-plane
For s = +jw , we have
z = esT = e +jw
For < 0 we have |z| < 1 i.e. the left half of the s-plane corresponds to the area within the unit circle in z-plane.
Q21 What is meant by system observability and controllability?
A system is completely controllable if there exists a control input u(t) that can transfer any initial state x(0) to any other desired location x in a finite time. For
X’ = AX + Bu
Y = CX
The system is controllable if the rank of the controllability matrix is n
Pc = rank [ B AB ...... A n-1 B] = n
A system is completely observable if there exists a finite time T such that the initial state x(0) can be determined from the observation history y(t) given the control u(t).
The system is observable if the determinant of the observability matrix Po is non zero.
Po = C
CA
.
.
CA n-1
Q22. What is Optimal Control?
Optimal control is concerned with the design need to minimize a performance index. Exp
J = ∫▒[x^T Qxdt]dt
In Engg systems the bigger the control input, the bigger the energy consumed. To consider the input in the performance index we often define
J = ∫_0^∞▒〖[x^T 〗 Qx+u^T Ru]dt
Where R is a positive definite matrix. Consider full state feedback control
u = -Kx
The performance index can be written as
J = ∫_0^∞▒〖[x^T 〗 Qx+x^T K^T RKx]dt
Q23. What is meant by system transfer function?
The TF of a linear time-invariant system is defined to be the ratio of the Laplace Transform of the output variable to the LT of the input variable under the assumption that all initial conditions are zero.
G(s) = Y(s)/U(s)
Q24. What is meant by system bandwidth? What are the gain and phase margins obtained for the closed loop PID control?
It is a range of frequencies for which system gain is more than -3dB. 20 log(0.7) Such gain is considered adequate for good transmission of signal.
Q25. What is a time invariant system?
If the coefficients of the describing differential equations are constants then the system is time invariant else it is time varying.
Q25a. What are the different types of numerical solver routines available and which one was used here and why?
The numerical solver routine chosen depends on 3 factors
Nature of the model being simulated
Accuracy required in the simulated data
Computing effort available for the simulation study
Common solvers are Fixed step methods Euler’s Method where accuracy is poor, RK4 method and multistep methods like Adams Bashforth and Adam’s Moulton. With single step methods calculation can proceed from any known stage. State x(k) is calculated based on knowledge of state x(k-1).
Multistep methods use the stored values of 2 or more previously calculated states or derivatives in order to compute the derivative approximation for the current time step. They cannot be self starting since the computation cannot proceed from the initial state alone. They also have a strategy to increase or decrease the step size depending on the difference between the predicted and corrected x(k) value. Such variable time step methods are useful if the simulated system possesses local time constants that differ by several orders of magnitude or if there is little prior knowledge about the system response.
Q26. How does a rotary encoder function?
It uses a patterned spinning disc attached to the motor shaft marked with a large number of radial lines like the spokes of a wheel. There are opaque and transparent sectors. As the disk rotates transparent sectors allow light & form square wave pulses. An optical switch like a photodiode or LED generates an electrical pulse whenever one of these lines passes through its field of view. An electronic control circuit counts the pulses to determine the angle by which the shaft has turned. In its simplest form it can only measure change in angle relative to some base point. Another sensor can be added to determine when the shaft passes its 0 position.
Rotation direction is determined by adding a second sensor placed at a different angle around the shaft from the initial sensor. This is called quadrature encoder. Encoders have a 3rd output channel called 0 or reference signal which supplies a single pulse per revolution. This is used for determination of reference position called index pulse.
A quadrature encoder supports 2 output channels (A&B) to sense position, velocity & direction of rotation. Using 2 code tracks with sectors positioned 90 deg out of phase the 2 output channels of the quadrature encoder indicate both position & direction of rotation. If A leads B then it is rotating in one direction & if B leads A then in reverse.
In order to decode the o/p from the quadrature encoders, 2 sequential samples are used. Sample from Time n & n+1. These 2 sets of data points create a 4 bit word that is used for a 16 entry look up table. The table has 4 each gotos for count up, count down, no change, error(overspeed). The no of pulses in a fixed time interval are counted to estimate the velocity & acceleration of the encoder. Once the no of pulses in a fixed time interval is measured, the angular velocity can be calculated using
Velocity = Encoder Pulses x 60/Fixed time interval
Acceleration = Encoder pulses_n – Encoder Pulses_n-1/Pulses per Rev/(Fixed Time Interval )2
Inputs are affected by noise, so each input must be filtered
QEA ------------>Filter---------------
Index-------- Filter----------------
QEB--------Filter------------------
The filtered phase edges are counted by a position counter usually 16 bit up/down counter. To establish a reference point for position & speed measurements the counter can be reset by using an index signal.
Single ended incremental quadrature encodes – A & B signals are referenced to ground so there is 1 wire per signal.
Differential encoder – There are 2 lines per signal A &B – A, A’, B, B’. This is called push pull as all 4 lines supply a known voltage ( either 0 or Vcc). Differential encoders are used in noisy environment as taking differential measurements protects the integrity of the signal.
What is the purpose of the index signal?
The index signal provides the position of the motor & typically a single pulse is generated for every 360 degree of shaft rotation.
Q27. What are the advantages & disadvantages of hydraulic systems as compared to electrical systems.
Heat generated by internal losses is a basic limitation of any machine. Hydraulic components are superior in that the fluid carries away the heat generated.
Hydraulic fluid acts as a lubricant.
There is no phenomenon comparable to saturation & losses in magnetic materials of electrical machines. The torque developed in a electric motor is proportional to current & limited by magnetic saturation. The torque developed in hydraulic actuators is proportional to pressure difference & limited by safe stress levels.
Electric motors are simple lag device from applied voltage to speed. Hydraulic actuators are quadratic resonance from flow to speed with high natural frequency. Hydraulic actuators have high speed of response with fast start , stop & speed reversals.
Hydraulic actuators have high stiffness. This results in great positional stiffness and less position error.
High power to weight ratio
Disadvantages i) Hydraulic power is not so readily available as that of electrical power
ii) Contamination & leakage of hydraulic fluid
Q28. Define delay time, Rise Time & Settling Time
Delay Time : The time required for the response to reach half the final value the 1st time
Rise Time : Time taken for the response to rise from 10% to 90% of final value.
Settling Time : Time taken for the response to reach and stay within 2-5% of its final value.
Q29. What are the features of a servovalve & how is it different from proportional valves?
Single stage servovalves consist of a torque motor which is directly attached to & positions a spool valve. 2 Stage servovalves have a hydraulic preamplifier (first stage) which multiplies the force output of the torque motor to a level sufficient to overcome flow forces, stiction forces etc. Flapper, jet pipe & spool valves act as 1st stage valve while second stage is almost always spool type. 2 stage servovalves may be classified as spool position, load pressure & load flow feedback.
Q30. What is Update Rate?
When it comes to specmanship, the frequency with which the servo is updated ranks high among those specifications that are not well understood. Is 2 milliseconds OK? What advantage is there in going to 200 microseconds or 20 microseconds? First, we must be clear on what is being updated. Is it the position command, the position feedback, the position loop, the velocity loop or the current (which equates to torque) loop? Let’s start by talking about the position command. There are a variety of algorithms used within the control to generate a path for each axis. These algorithms can be linear/circular interpolation, leader/follower, cam following, registration, etc. The result of these calculations is that each axis command register gets a new number every time the calculation is made. If one were to plot the position of the command register for one of these axes as a function of time, it would look like a flight of stairs where the height of the step is the distance that axis must move in one update period. As an example, if the update time for that calculation is one millisecond and the axis is being commanded at 600 inches per minute (IPM), this becomes 10 inches per second or .01 inches every millisecond. The riser on the staircase will be .01 inches and the run will be one millisecond. The ideal servo command would be the straight line generated if the update time became infinitesimally small. The disturbance from the ideal servo command is a saw tooth wave with a height of .01 inches and a frequency of 1000 Hertz. How will the servo react to this command disturbance? For the majority of industrial machinery, the servo bandwidth will be in the range of 2 to 10 Hertz. The highest bandwidth would be most apt to respond to this disturbance, so let’s consider a 10 Hertz bandwidth. Once the bandwidth of a servo is exceeded, the output rolls off at a 40 DB per decade rate or a ratio of 100 to 1 for each decade increase in frequency. Since the 1000 Hertz disturbance is two decades from the bandwidth, the roll off will be 10,000 to 1. This would reduce a .01 inch disturbance to .000001 inches or 1 micro-inch. Hardly a problem. Unless you are dealing with some extremely high velocities and some pretty hot servos, it appears that a 1 millisecond update suffices for command generation. There is one caveat, however. If you are using the command register to trigger some event, be sure the 1 millisecond delay is not of consequence. It is usually advisable to trigger events off the actual feedback as will be discussed next. When using encoders as the feedback device, the position feedback register is updated every time a pulse comes along, so it stays in sync with the actual machine position. When using resolvers, most digitizing techniques update the feedback register once per carrier frequency cycle. Thus if the analog sine wave used to drive the resolver is 1 Kilohertz, the feedback register will be upgraded once per millisecond. This is why most suppliers are using carrier frequencies in the 2,000 to 10,000 Hertz range now days. This assures updates in the 500 to 100 microsecond arena. There are some resolver-to-digital (R/D) converters that employ techniques to overcome this problem. Also when triggering events off the position, don’t forget that there will be a servo error unless feedforward techniques or PID is used. Also, since these features are sometimes switched in and out, it is safer to simply trigger events off the feedback, rather than the command. How often the servo loop itself is updated is another concern. And, there are three loops to worry about. Updating a loop means subtracting the feedback from the command to generate the error and manipulating that error with digital or analog lead/lag filters, PID, notch filters, feedforward, or any other loop feature that may be available. There has been a rough rule of thumb that the update rate of a servo loop should be at least 10 times the bandwidth. For a position loop with a 10 Hertz bandwidth, this suggest an update frequency of 100/second or 10 milliseconds per update. For a velocity loop with a 100 Hertz bandwidth, the update would be every one millisecond. For a current loop bandwidth of 1,000 Hertz, the update time would be 100 microseconds. This helps explain why the PWM chopping frequencies are often in the 10 Kilohertz range for those who are doing the current loop digitally. It also explains why many are doing the current loop in an analog fashion due to the high burden on the computer time when the loop has to be served 10,000 times per second. Usually, in the specification, only one update time is given. That number usually means the time elapsed between position loop closure calculations. And, with today’s technology, it is not hard to achieve 1 millisecond, which most vendors are meeting or beating. With industrial machinery, this is adequate, especially if the current (torque) loop is analog. Those vendors who are in the 100 microsecond arena are either playing the specmanship game or are doing the current loop digitally (or maybe a bit of both). As computers continue to get faster, the update times will get shorter and shorter, but it will be hard to see any further significant changes in servo performance with industrial machines.
Classical control is suitable for single input single output systems whereas state space methods can be useful for linear and non linear time invariant and time varying , multi input multi output systems also.
Q12. In a PID Controller what is the role of P, I and D values ?
Kp is proportional gain which gives a large correction for a large error. Immediate response to errors. It does not affect gain or phase margin.
Ki is Integral Gain which allows the steady state error to be zero & reduces peak overshoot.
It is used to increase the open loop gain at low frequencies.
Kd is the derivative gain which reduces oscillation, makes the system more stable. It also introduces phase lag, increases speed of response, decreases gain margin & amplifies noise. It may result in instability.
Q13. What is sliding control ?
The basic idea is that the control signal changes abruptly on the basis of the state of the system. A control system of this type is also referred to as a variable structure system.
Q14. What is the difference between linear & nonlinear analysis?
One can obtain closed form solutions for linear systems. This is seldom the case for nonlinear systems.
Q15. How does the sampling time affect the response of a continuous time system ?
The D/A converter between the discrete & continuous time process implements holding of the calculated control signal during the time step (sampling interval). This holding implies that the control signal is time delayed by approx h/2. The delay influences the stability of the control loop. Suppose we have tuned a continuous time PID & apply these PID parameters on a discrete time PID Controller, then the control loop will get reduced stability because of the approx delay of h/s. As a rule of thumb the stability reduction is small & tolerable if the time delay is less than 1/10th of the response time of the control system as it would have been with a continuous time controller or a controller having small sampling time.
h/2 < Tr/10
h < Tr/5 ;
h < 1/ 5wb
The response time here is 63% rise time which can be read off from the setpoint step response. For a system having dominating time constant T, the response time is approx equal this time constant. If the bandwidth of the control system is wb ( assuming PID parameters have been found using continuous time PID controller) the response time of the control system can be estimated by
Tr 1/wb
Q16. How are poles and eigenvalues related?
Poles and eigenvalues are equal for most systems not having pole zero cancellation.
Q17. How to handle a nonlinear model ?
If the model is nonlinear it must be linearised before calculating the poles or eigenvalues.
Q18. What is the relation between the transfer function and state space model ?
When we convert the state space model to the transfer function, we observe that the denominator of the transfer function is the determinant of (sI-A).
d(s) = |sI-A|
Q19 What is the condition for stability of a continuous time system & for a discrete time system from its pole locations?
The system is stable if all poles are in the Left hand plane i.e. d(s) = 0 are with all negative real parts. A sampled or discrete time system is stable if all the poles of the closed loop transfer function lie within the unit circle of the z-plane.
In terms of linear systems, we recognize that the stability requirement may be defined in terms of the location of the poles of the closed-loop transfer function.
A necessary and sufficient condition that a feedback system be stable is that all the poles of the system transfer function have negative real parts.
Q19a. What is the Nyquist Criterion?
The Nyquist criterion is a graphical method and deals with the loop gain transfer function, i.e., the open-loop transfer function.
We can determine the stability of the closed-loop system by locating the zeros of 1 + G(s). This result is of prime importance in the following development. For the moment, let us assume that 1 + G(s) is known in factored form so that we have Obviously, if 1 + G(s) were known in factored form, there would be no need for the use of the Nyquist criterion, since we could simply observe whether any of the zeros of 1 + G(s) [which are the poles of Y( s ) /R( s ) ]l,i e in the right half of the s plane. In fact, the primary reason for using the Nyquist criterion is to avoid this factoring. Although it is convenient to think of 1 + G(s) in factored form at this time, no actual use is made of that form.
A c c o r d ing to Nyquist plot stability evaluation methods, a system with no open-loop right half-plane (RHP) poles should have no clockwise (CW) encirclements of the -1 point for stability, and a system with open-loop RHP poles should have as many counter-clockwise (CCW) encirclements as there are open-loop RHP poles
Q19b. What are the properties of Bode plots?
Any information that can be wrenched out of the Bode plots of the loop gain is critically important for two reasons. First, the Bode plots are a natural place to judge the properties of the feedback loop. When the magnitude of the loop gain is large, positive feedback properties such as good disturbance rejection and good sensitivity reduction are obtained. When the magnitude of the loop gain is small, these properties are not enhanced. The work of this chapter completes the missing information about transient response that can be read from the loop gain Bode plots.
Second, it is the Bode plot that we are able to manipulate directly using series compensation techniques. It is important to be able to establish the qualities of the Bode plots that produce positive qualities in a control system because only then can the Bode plots be manipulated to attain the desired qualities.
Q20. How do we map the s-plane to the z-plane
For s = +jw , we have
z = esT = e +jw
For < 0 we have |z| < 1 i.e. the left half of the s-plane corresponds to the area within the unit circle in z-plane.
Q21 What is meant by system observability and controllability?
A system is completely controllable if there exists a control input u(t) that can transfer any initial state x(0) to any other desired location x in a finite time. For
X’ = AX + Bu
Y = CX
The system is controllable if the rank of the controllability matrix is n
Pc = rank [ B AB ...... A n-1 B] = n
A system is completely observable if there exists a finite time T such that the initial state x(0) can be determined from the observation history y(t) given the control u(t).
The system is observable if the determinant of the observability matrix Po is non zero.
Po = C
CA
.
.
CA n-1
Q22. What is Optimal Control?
Optimal control is concerned with the design need to minimize a performance index. Exp
J = ∫▒[x^T Qxdt]dt
In Engg systems the bigger the control input, the bigger the energy consumed. To consider the input in the performance index we often define
J = ∫_0^∞▒〖[x^T 〗 Qx+u^T Ru]dt
Where R is a positive definite matrix. Consider full state feedback control
u = -Kx
The performance index can be written as
J = ∫_0^∞▒〖[x^T 〗 Qx+x^T K^T RKx]dt
Q23. What is meant by system transfer function?
The TF of a linear time-invariant system is defined to be the ratio of the Laplace Transform of the output variable to the LT of the input variable under the assumption that all initial conditions are zero.
G(s) = Y(s)/U(s)
Q24. What is meant by system bandwidth? What are the gain and phase margins obtained for the closed loop PID control?
It is a range of frequencies for which system gain is more than -3dB. 20 log(0.7) Such gain is considered adequate for good transmission of signal.
Q25. What is a time invariant system?
If the coefficients of the describing differential equations are constants then the system is time invariant else it is time varying.
Q25a. What are the different types of numerical solver routines available and which one was used here and why?
The numerical solver routine chosen depends on 3 factors
Nature of the model being simulated
Accuracy required in the simulated data
Computing effort available for the simulation study
Common solvers are Fixed step methods Euler’s Method where accuracy is poor, RK4 method and multistep methods like Adams Bashforth and Adam’s Moulton. With single step methods calculation can proceed from any known stage. State x(k) is calculated based on knowledge of state x(k-1).
Multistep methods use the stored values of 2 or more previously calculated states or derivatives in order to compute the derivative approximation for the current time step. They cannot be self starting since the computation cannot proceed from the initial state alone. They also have a strategy to increase or decrease the step size depending on the difference between the predicted and corrected x(k) value. Such variable time step methods are useful if the simulated system possesses local time constants that differ by several orders of magnitude or if there is little prior knowledge about the system response.
Q26. How does a rotary encoder function?
It uses a patterned spinning disc attached to the motor shaft marked with a large number of radial lines like the spokes of a wheel. There are opaque and transparent sectors. As the disk rotates transparent sectors allow light & form square wave pulses. An optical switch like a photodiode or LED generates an electrical pulse whenever one of these lines passes through its field of view. An electronic control circuit counts the pulses to determine the angle by which the shaft has turned. In its simplest form it can only measure change in angle relative to some base point. Another sensor can be added to determine when the shaft passes its 0 position.
Rotation direction is determined by adding a second sensor placed at a different angle around the shaft from the initial sensor. This is called quadrature encoder. Encoders have a 3rd output channel called 0 or reference signal which supplies a single pulse per revolution. This is used for determination of reference position called index pulse.
A quadrature encoder supports 2 output channels (A&B) to sense position, velocity & direction of rotation. Using 2 code tracks with sectors positioned 90 deg out of phase the 2 output channels of the quadrature encoder indicate both position & direction of rotation. If A leads B then it is rotating in one direction & if B leads A then in reverse.
In order to decode the o/p from the quadrature encoders, 2 sequential samples are used. Sample from Time n & n+1. These 2 sets of data points create a 4 bit word that is used for a 16 entry look up table. The table has 4 each gotos for count up, count down, no change, error(overspeed). The no of pulses in a fixed time interval are counted to estimate the velocity & acceleration of the encoder. Once the no of pulses in a fixed time interval is measured, the angular velocity can be calculated using
Velocity = Encoder Pulses x 60/Fixed time interval
Acceleration = Encoder pulses_n – Encoder Pulses_n-1/Pulses per Rev/(Fixed Time Interval )2
Inputs are affected by noise, so each input must be filtered
QEA ------------>Filter---------------
Index-------- Filter----------------
QEB--------Filter------------------
The filtered phase edges are counted by a position counter usually 16 bit up/down counter. To establish a reference point for position & speed measurements the counter can be reset by using an index signal.
Single ended incremental quadrature encodes – A & B signals are referenced to ground so there is 1 wire per signal.
Differential encoder – There are 2 lines per signal A &B – A, A’, B, B’. This is called push pull as all 4 lines supply a known voltage ( either 0 or Vcc). Differential encoders are used in noisy environment as taking differential measurements protects the integrity of the signal.
What is the purpose of the index signal?
The index signal provides the position of the motor & typically a single pulse is generated for every 360 degree of shaft rotation.
Q27. What are the advantages & disadvantages of hydraulic systems as compared to electrical systems.
Heat generated by internal losses is a basic limitation of any machine. Hydraulic components are superior in that the fluid carries away the heat generated.
Hydraulic fluid acts as a lubricant.
There is no phenomenon comparable to saturation & losses in magnetic materials of electrical machines. The torque developed in a electric motor is proportional to current & limited by magnetic saturation. The torque developed in hydraulic actuators is proportional to pressure difference & limited by safe stress levels.
Electric motors are simple lag device from applied voltage to speed. Hydraulic actuators are quadratic resonance from flow to speed with high natural frequency. Hydraulic actuators have high speed of response with fast start , stop & speed reversals.
Hydraulic actuators have high stiffness. This results in great positional stiffness and less position error.
High power to weight ratio
Disadvantages i) Hydraulic power is not so readily available as that of electrical power
ii) Contamination & leakage of hydraulic fluid
Q28. Define delay time, Rise Time & Settling Time
Delay Time : The time required for the response to reach half the final value the 1st time
Rise Time : Time taken for the response to rise from 10% to 90% of final value.
Settling Time : Time taken for the response to reach and stay within 2-5% of its final value.
Q29. What are the features of a servovalve & how is it different from proportional valves?
Single stage servovalves consist of a torque motor which is directly attached to & positions a spool valve. 2 Stage servovalves have a hydraulic preamplifier (first stage) which multiplies the force output of the torque motor to a level sufficient to overcome flow forces, stiction forces etc. Flapper, jet pipe & spool valves act as 1st stage valve while second stage is almost always spool type. 2 stage servovalves may be classified as spool position, load pressure & load flow feedback.
Q30. What is Update Rate?
When it comes to specmanship, the frequency with which the servo is updated ranks high among those specifications that are not well understood. Is 2 milliseconds OK? What advantage is there in going to 200 microseconds or 20 microseconds? First, we must be clear on what is being updated. Is it the position command, the position feedback, the position loop, the velocity loop or the current (which equates to torque) loop? Let’s start by talking about the position command. There are a variety of algorithms used within the control to generate a path for each axis. These algorithms can be linear/circular interpolation, leader/follower, cam following, registration, etc. The result of these calculations is that each axis command register gets a new number every time the calculation is made. If one were to plot the position of the command register for one of these axes as a function of time, it would look like a flight of stairs where the height of the step is the distance that axis must move in one update period. As an example, if the update time for that calculation is one millisecond and the axis is being commanded at 600 inches per minute (IPM), this becomes 10 inches per second or .01 inches every millisecond. The riser on the staircase will be .01 inches and the run will be one millisecond. The ideal servo command would be the straight line generated if the update time became infinitesimally small. The disturbance from the ideal servo command is a saw tooth wave with a height of .01 inches and a frequency of 1000 Hertz. How will the servo react to this command disturbance? For the majority of industrial machinery, the servo bandwidth will be in the range of 2 to 10 Hertz. The highest bandwidth would be most apt to respond to this disturbance, so let’s consider a 10 Hertz bandwidth. Once the bandwidth of a servo is exceeded, the output rolls off at a 40 DB per decade rate or a ratio of 100 to 1 for each decade increase in frequency. Since the 1000 Hertz disturbance is two decades from the bandwidth, the roll off will be 10,000 to 1. This would reduce a .01 inch disturbance to .000001 inches or 1 micro-inch. Hardly a problem. Unless you are dealing with some extremely high velocities and some pretty hot servos, it appears that a 1 millisecond update suffices for command generation. There is one caveat, however. If you are using the command register to trigger some event, be sure the 1 millisecond delay is not of consequence. It is usually advisable to trigger events off the actual feedback as will be discussed next. When using encoders as the feedback device, the position feedback register is updated every time a pulse comes along, so it stays in sync with the actual machine position. When using resolvers, most digitizing techniques update the feedback register once per carrier frequency cycle. Thus if the analog sine wave used to drive the resolver is 1 Kilohertz, the feedback register will be upgraded once per millisecond. This is why most suppliers are using carrier frequencies in the 2,000 to 10,000 Hertz range now days. This assures updates in the 500 to 100 microsecond arena. There are some resolver-to-digital (R/D) converters that employ techniques to overcome this problem. Also when triggering events off the position, don’t forget that there will be a servo error unless feedforward techniques or PID is used. Also, since these features are sometimes switched in and out, it is safer to simply trigger events off the feedback, rather than the command. How often the servo loop itself is updated is another concern. And, there are three loops to worry about. Updating a loop means subtracting the feedback from the command to generate the error and manipulating that error with digital or analog lead/lag filters, PID, notch filters, feedforward, or any other loop feature that may be available. There has been a rough rule of thumb that the update rate of a servo loop should be at least 10 times the bandwidth. For a position loop with a 10 Hertz bandwidth, this suggest an update frequency of 100/second or 10 milliseconds per update. For a velocity loop with a 100 Hertz bandwidth, the update would be every one millisecond. For a current loop bandwidth of 1,000 Hertz, the update time would be 100 microseconds. This helps explain why the PWM chopping frequencies are often in the 10 Kilohertz range for those who are doing the current loop digitally. It also explains why many are doing the current loop in an analog fashion due to the high burden on the computer time when the loop has to be served 10,000 times per second. Usually, in the specification, only one update time is given. That number usually means the time elapsed between position loop closure calculations. And, with today’s technology, it is not hard to achieve 1 millisecond, which most vendors are meeting or beating. With industrial machinery, this is adequate, especially if the current (torque) loop is analog. Those vendors who are in the 100 microsecond arena are either playing the specmanship game or are doing the current loop digitally (or maybe a bit of both). As computers continue to get faster, the update times will get shorter and shorter, but it will be hard to see any further significant changes in servo performance with industrial machines.
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