Practise Test Questions Flashcards
In a control system, define and provide examples of the following terms: system, plant and controller. Explain what are the differences between plant and controller.
A system can be defined as a connected set of elements which transforms an input signal into an output signal and they are usually modelled as dynamical systems, differential equations or difference equations.
A plant is just a system that has to be controlled, that means its output has to fulfil some criterion or reach some desired value. Usually, the plant is given and we cannot modify it, we can only act on or modify its state through its inputs.
A controller is a designed system or function that generates an action signal or control for the plant. The control signal generated by the controller should be such that it drives the output of the plant to the reference signal.
The main difference between a plant and a controller is that we cannot modify the plant, since it is given, but we can design and implement a controller to control the plant. The controller itself can be viewed as a system which can be modified.
You are designing a controller for the forward speed of a robot and you try three different approaches: 1) an open-loop controller, 2) a proportional controller (P), and 3) a proportional integral controller (PI). You try the three controllers on the robot with a flat floor where the surface changes from a smooth floor to a carpet (i.e. a perturbation for the controller). When you set the desired (reference) velocity to 1 m/s you obtain the velocity profiles (a), (b), (c) in Figure 1. Please associate the plots (a), (b), and (c) to the corresponding controller 1), 2) and 3). Justify your answer.
https://ulster-my.sharepoint.com/:i:/g/personal/mcerlean-a6_ulster_ac_uk/EaDUj1EKhv5IqJCpgltYBLABqL1Fl7gs9ZNTTnDa90Lixg?e=8S1O8r
The open loop controller 1) corresponds to figure c). Although the robot reaches the target velocity, compared to the other figures the time it takes to reach that velocity is considerably longer. Moreover, as the perturbation kicks in the drop on the velocity is around 20%, indicating there is no corrective action on the perturbation, i.e. there is no feedback signal towards the controller. The proportional controller corresponds 2) to figure b). A proportional controller, depending on the system to control, can have a steady state error, which is the case for the robot as it doesn’t reach the target velocity. The effect of the perturbation in this case is smaller than for the open-loop controller, but it still exists. The response time of the controller, i.e. how fast it reaches the steady state, is shorter than in the open-loop controller. Finally, the PI controller 3) corresponds to plot a). The PI controller eliminates the steady state error, and therefore, the robot reaches the reference velocity. Although in general PI controllers are slower than P controllers, in this case the response time, i.e. the time the system takes to reach the steady state, is not very different from the P controller. Another clue indicating that figure (c) corresponds to a PI controller is that the perturbation is being eliminated after a while, and, although the robot slows down once it reaches the carpet the speed tends to grow towards the reference.
Ultrasound sensors are time-of-flight sensors commonly used in robotics. They measure the time an ultrasound wave takes to go forth and back from the sensors to the obstacle. Given the measured time t, the distance is obtained as d=v*t/2, where v is the speed of sound (approximately 340 m/s). If a sonar sensor has a blanking time of 0.0009 seconds, and provides readings with a frequency of 5 Hertz, what is the range of distances (minimum and maximum distances) the sensor can read? Explain how you obtain these values.
The blanking time is the time the sensor needs to wait for the device generating the sound wave (the metallic sheet) to stop moving (due to the inertia) so that the sensor can go from emitting ultrasound to detecting the ultrasound. During that time the sensor cannot perceive any returning ultrasound wave, i.e. the ultrasound sensor is “blind” to near objects.
The minimum distance is determined by the blanking time. The time of flight of the ultrasound wave is equal to the blanking time, 0.0009 seconds. Therefore, the minimum distance is
d = v*t/2 = 340*0.0009/2 = 0.153 m. This means the sensor cannot detect objects closer than 15.3 cm.
The frequency of the sensor is directly related to the time between two consecutive readings, and follows the relation f=1/T, i.e T=1/f, where T is the period measured in seconds and f is the frequency measured in Hz. Because the frequency is 5Hz, the period between two consecutive readings will be 1/5=0.2 seconds.
The maximum distance is determined by the frequency of the sensor. Therefore, the maximum distance is
d = v*t/2 = 340*0.2/2 = 34 m.
This means the sensor cannot detect objects farther than 34 m. Therefore, the range of distances the sensor can read is from 0.153 m to 34 m
Classify robotic sensors? Explain each with an example?
Robotic sensors can be classified as:
Proprioceptive vs. Exteroceptive
Proprioceptive: Measure an internal robot variable; battery level, motor temperature, encoders. . .
Exteroceptive: Measure external (environmental) variables; distance sensors, compass, bumpers,
cameras. . .
Active vs. Passive.
Active: These sensors measure the effect of something they put in the environment: laser scanners, encoders, Kinect. .
Passive: Measure some variable that is in the environment: compass, camera. . .
Give any definition of a robot? Discuss practical applications of robots with examples?
(SPA) A goal-oriented machine that can sense, plan, and act
Industrial Applications:
* Welding, painting, cutting, packaging. . .
* Pick and place: Baxter, Kuka. . .
* Benefits: Productivity, costs. .
Applications:
* Medical robots: surgical, rehabilitation, prosthetics. . .
* Social robots: Companion, caretaker. . .
* Aerial robots: Observation, delivery, surveillance. . .
* Mining, construction,
You are working in the robotics lab and just got your brand-new robot with ultrasound sensors and a laser scanner. According to the specifications of the robot the resolution of the sonars is 1mm (0.001m) and the resolution of the laser scanner is 1cm (0.01m). Your colleague is a bit puzzled because the laser scanner is much more expensive but its resolution is ten times worse than that of the sonars’.
(d). You implement a gap detector which consists on subtracting one reading from the previous one. Two consecutive readings are r1=2m and r2=5m. Given the above covariance (0.0625), give a confidence interval such that the difference between the two readings will be with 99% probability within that range.
(d). Since we are assuming the readings follow a Gaussian distribution, the difference between two readings will follow such distribution. When computing the distance between two measurements the average distance will just be difference between the averages, that means the average distance in this case will be:
2-5 = -3m or 5-2=3m
depending on how we compute the difference. The covariance of the difference is the sum of the two covariances, which for this case are the same, and therefore
0.0625+ 0.0625= 0.125
The standard deviation is the square root of 0.125=0.3535. Again, for a Gaussian variable, the range of distances such that there is 99% certainty the difference is within the range can be computed as the mean or average plus/minus three times the standard deviation, that gives a range between 3-30.3535= 1.9395 and 3+30.3535= 4.0605 meters.
What are the input, output and parameters of the Split & Merge Algorithm?
Split:
* Input: List of points (x i , y i ) i = 1, · · · , n.
* Output: List of segments (pairs of points)
* Parameters: Threshold distance d th
Merge:
* Input: List of segments (pairs of points)
* Output: List of segments (pairs of points)
* Parameters: Threshold angle θ th
You are working in the robotics lab and just got your brand-new robot with ultrasound sensors and a laser scanner. According to the specifications of the robot the resolution of the sonars is 1mm (0.001m) and the resolution of the laser scanner is 1cm (0.01m). Your colleague is a bit puzzled because the laser scanner is much more expensive but its resolution is ten times worse than that of the sonars’.
(a). Given the speed of sound 340 m/s and the speed of light 3*10^8 m/s, calculate the time resolution of the hardware necessary to achieve the above resolutions. Use these values to explain the difference in price to your colleague.
(a). Both sensors considered are time of flight sensors, and they measure the distance based on the measure of the time the sound or the light takes to go forth and back from the sensor to an obstacle.
Once the sensor has received the corresponding wave back after a specific time “t”, the distance can be computed as d=vt/2, where v is the speed of sound for the ultra-sound sensor and the speed of light for the laser sensor. This formula is also used to compute the relations between the distance resolution of the sensor and the necessary time resolution of the hardware involved in measuring the time of flight. If we solve this equation for the time variable t, we have: t = (2d) /v, where we can substitute the measure distance “d” by the resolution corresponding to each sensor. If we
compute the time resolution for the sonar, we obtain:
t=(2 * 0.001) / 340 = 0.000006 = 6 x 10^(-6) s
Doing the same calculation for the laser we obtain:
t=(2 * 0.01) / 3x10^8=6x10^(-11) s.
Therefore the laser scanner needs to have a much higher time resolution than the sonar sensors, which means the electronic circuitry for the laser will be more expensive. If we convert these times into frequencies we can see that the frequency of the hardware needed to read a sonar sensor is of the order of 100 KHz, while the frequency to read a laser sensor is of the order of the GHz.
You are working in the robotics lab and just got your brand-new robot with ultrasound sensors and a laser scanner. According to the specifications of the robot the resolution of the sonars is 1mm (0.001m) and the resolution of the laser scanner is 1cm (0.01m). Your colleague is a bit puzzled because the laser scanner is much more expensive but its resolution is ten times worse than that of the sonars’.
(b). You make the standard assumption that the noise in the laser sensor follows a Gaussian distribution. You ask your colleague to run an experiment to calculate the covariance of the Gaussian (calibrate the laser). Describe the experiment he should perform to obtain the covariance of the readings. [6 marks]
(b). To calibrate the laser scanner the robot should be placed in a static environment, while the robot itself is not moving. With the help of a program a high number of laser scans should be collected and stored in a data file. Ideally, since the environment is static and the robot is not moving the readings provided by the laser should be identical, but since real sensors have noise, there will be some variation in the distance measured by each laser beam. For each of the laser beams the covariance can be obtained, and then one of the covariance values can be used as the value to characterise the error of the sensor. Alternatively the average of all of them can be used, which would provide a more accurate result.
You are working in the robotics lab and just got your brand-new robot with ultrasound sensors and a laser scanner. According to the specifications of the robot the resolution of the sonars is 1mm (0.001m) and the resolution of the laser scanner is 1cm (0.01m). Your colleague is a bit puzzled because the laser scanner is much more expensive but its resolution is ten times worse than that of the sonars’.
(c). Once your colleague runs the experiment he tells you that the covariance of the readings is σ^2=0.01 and the standard deviation is σ=0.1. If the laser scanner gives a 15m reading, what would be the interval in which the real distance is with 99% probability. Explain your answer?
(c). Since we assume the readings of the laser scanner follow a normal or Gaussian distribution, we can take the measured distance as the mean of that distribution and use the covariance to define the uncertainty. For the Gaussian distribution the probability of a value being between the mean plus/minus three times the standard deviation is 99%. This means the real distance will be bounded by:
15 - (3 * 0.1) = 14.7
15 + (3 * 0.1) = 15.3
with 99% probability
