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Explain the difference between open- and closed-loop control. (15pts)
Closed loop control systems differ from open loop control systems in the fact that the closed loop system provides feedback information. This feedback information is used as a comparison comparison between the desired result and the actual result. An open loop system does not have this feedback, so the output is is not compared with the input. This type of control is good for systems that are more basic and do not require a “check” on the output value. Conversely, closed loop control is better for systems that have continuous variability variability or unpredictable unpredictable perturbations. An example of this is the thermostat device. It must check the actual room temperature te mperature and compare it to the setpoint.
2.) The automatic control system for the temperature of a bath of a liquid consists of a reference
voltage fed into a differential amplifier. This is connected to a relay which then switches on or off the electrical power to a heater in the liquid. Negative feedback is provided by a measurement system which feeds a voltage into the differential amplifier. Sketch a block diagram of the system, explain the role of each component (subsystem), and describe each signal. (20pts)
The reference input could be a knob that allows the user to select the desired setpoint. This signal is an analog voltage reading that can be manipulated to allow it to be used in the control system. It reaches the differential amplifier which will amplify the difference between the setpoint voltage and the feedback voltage. This will then be sent as an actuating signal that allows the relay to logically logically decide whether to allow the plant to operate. This could be in the form of a simple simple binary signal. The plant will then operate as instructed and heat the bath. The bath temperature will then be read by a sensor which will convert the measured temperature to a voltage which will then be fed back into the differential amplifier to be used for comparison of the setpoint temperature versus the measured temperature. This is an idealized diagram as it does not have any perturbations or error included.
3.) Explain the significance of the following information given in the specification of the transducers. (In other words, define these types of errors and explain what the actual value can be if the measured value is, let's say, M ) (15pts)
(a) A capacitive linear displacement transducer. Non-Linearity and hysteresis: ± 0.01% of full range Non-Linearity error stems from the chosen characteristic need to quantify a system as linear. This is easier for engineers to deal with, so a certain amount of error is induced when this assumption is made. This is generally one of the most common types of error as signal reading is a very important part of any meaningful system. Hysterisis error is a symmetric error that comes from the dynamic lag in a system that is turning on or turning off. This lag is a result of the signal and can either overshoot or undershoot the desired value or measured value. This error is very common and has ways of being compensated for. The actual value could be: [M*0.999 , M*1.001]
(b) A resistance strain gauge force measurement transducer. Temperature sensitivity: ± 1% of the full range over normal environmental temperatures When working with metals, especially tiny pieces that are used to measure strain in elements, it is important to remember that the metal itself will vary with the ambient temperature. If the room temp of the sample is 23ºC vs 25ºC, there could be a small error in the reading. This is called the temperature sensitivity, and is described as a percentage of the FR value. This error has been mostly compensated for by use of different manufacturing techniques and the use of other sensor circuits (half bridge, for example). The actual value could be: [M*0.99 , M*1.01]
(c) Gyroscope for angular velocity measurement Repeatability: ± 0.01% of full range Repeatability is a natural error that is experienced using the same measurement procedure, the same observer, the same measuring instrument, used under the same conditions, in the same location and in a short period of time. Even if all of these criteria for repeatability are met, there will still be some inherent error. This error should be very small as repeatability is maximized under the above conditions.
The actual value could be: [M*0.999 , M*1.001]
4.) A platinum resistance temperature detector has a resistance of 100.00 Ω at 0˚C, 138.50 Ω at
100˚ C, and 175.83 Ω at 200˚ C. What will be the non-linearity error in ˚C at 100˚C if the detector is assumed to have a linear relationship between 0 and 200˚C? (20pts) First, write the equation that describes the linear relationship used in the detector model: Resistance= [(175.83 – 100) / 200 * Temp] + 100 Then plug in the desired value of 100°C [(175.83 – 100) / 200 * 100°C] + 100 = 137.915 Ω Find the real difference between the linear value and the measured value Actual – Linear = NL error 138.50 Ω - 137.915 Ω = 0.585 Ω
5.) A thermocouple element when taken from a liquid at 50°C and plunged into a liquid at 100° C at time t = 0 gave the following e.m.f. values. Determine the 95% response time. Use linear interpolation inside the intervals. Hint: You must consider 95% of the difference between the final steady state value and the initial steady state value. (Answer: 75s) (20pts) Time (s) 0 20 40 60 80 100 120 e.m.f (mV) 2.5 3.8 4.5 4.8 4.9 5.0 5.0
The final steady-state value is, through observation of the results, 5.0. This is reached at 100 s. The total time interval that will be examined is 100 s. To determine the 95% response time, first we need to know what 95% of the final steady state value minus the initial steady state value is. (5 – 2.5 )* 0.95 + 2.5 = 4.875 mV And preforming a interpolation to solve for the 95% response time: (4.875 – 4.8) / (4.9 – 4.8) = (x – 60) / (80 – 60) [(4.875 – 4.8) / (4.9 – 4.8)] * (80 – 60) + 60 = 75 seconds
6.) What will be the change in resistance of an electrical resistance strain gauge with a gauge factor
of 2.1 and resistance 50 Ω if it is subject to a strain of 0.001? (10pts) ε*R*G=Δ
0.001 * 50 Ω * 2.1 = 0.105 Ω
