Recent content by Hesch

Yes. Say that the original values are: N = Norg , I = Iorg. In the approximated model the values become: Napp = 1 , Iapp = Iorg * Norg. The reason for this approximation is that the coil is formed like the thread on a screw, which make calculations difficult.

Well, no emf is induced in wire, due to some magneting field is passing through the wire ( except for some Eddy voltages and -currents ). The formula to be used is: E = dΨ/dt , Ψ is the magnetic flux through some loop formed by a wire. A straight wire doesn't form a loop. Now, if you connect...

Of course the currents through two resistors in series will be the same, but this common current will change, when an inner resistance in the battery is inserted.

You have chosen the right circular current path to be clockwise, and that's ok except +48Ileft then becomes -48Ileft. The reason why you have to choose directions ( arrows ) before setting up equations, is that you must respect your chosen directions! Otherwise you will get a wrong result.

Don't use the node law to define directions. Just define. You are free to choose directions. If your choise - as for an arrow direction - turns out to be opposite to the physical positive current direction, the resulting current will just be negative. You can use the KVL law, as you have...

You must choose a current through all the resistors in series. The currents are the same, you know. Let's say you choose 2A, then the resistor values will be: 1. (12V - 8V) / 2A = 2Ω 2. (8V - 6V) / 2A = 1Ω and so on. But other values can be used. Say that you choose a current = 2mA, the...

You can do a lot with these transforms. Say you have a sattelite photo of a milititary airport and you want to know how many jet fighters of which type is parked in this airport, you can employ a lot of people with magnifying glasses to count these planes. But also you could stuff the photo...

Well, I have recently done some experiments, transforming some shapes like the letters 'E' and 'F'. Then I calculate the transfer function from 'F' to 'E': FFT(H) = FFT('E') / FFT('F'). Now, if I transform the letter 'O' and calculate: IFFT( FFT('O') * FFT(H) ), will I get a 'Q' ??

A FFT will give you a complex value for each harmonic in your set of discrete data points. Say that the FFT value of the 4. harmonic is ( 0.3 + 0.5i ), you may interprete it as 0.3*cos(4ωt) + 0.5*sin(4ωt) . . . if I understand you correct.

Ka and Kb have the same value, but the units are [Nm/A] and [Vs] respectively. There are people that can prove it.

Ok, then use these types as input/output. Setting La = 0, you will get a 2. order transfer function.

Of course the input and output type matters. If you change the type of input or output, the motors behaviour will change. In the model, choosing ω(s) as output, the transfer function will be H(s). Choosing θ(s) as output, the transfer function will be H(s)/s: θ(s) = ω(s)/s.

Well, the transfer function is a differential equation. Using Laplace transform it may be written: H(s) = output(s)/input(s) When expressing the transfer function by its Laplace transform, it becomes much easier to calculate at controller. I have some questions: - Is it a DC-motor ? - What...

Well, yes, but a line imager only sees a line. Say that there is a disturbance in the image, e.g. an unwanted reflection near the edge of the ball and on that line, the line imager will be completely confused: The position of the ball will not be measured correct. In a spatial image most of...

The idea of using a camera will work, but there are some things to be considered: - The images must be of good quality ( e.g. white ball, black background, good illumination ). - The image must be at least 1000 x 1000 pixels - The lens will distort the image ( lens distortion ) so that the...