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.
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...
If you have to change the voltages across a capacitor, you must have current flowing through this capacitor.
If just one switch is open, no current can pass through the serial connection.
Remember Kirchhoffs current law ( KCL ).
The capacitors are coupled in series. All the switches must be closed before the charges can get away ( rearrange ).
None of the suggestions have all switches closed, so D) is correct because the voltages V1, V2 are unchanged ( no change in the location of the charges ).
Don't bother about the...
I think the man is called Mason, and sorry: I think you are your method is wrong.
Now, if you number the points from x to y: 1 . . 6, the transfer function from 1 to 2 seems to be 1. But later you will have a problem with the W4 from 1 to 3. You must move this W4 connection from 1-3 to 1-2. You...
I recommend a core with an actual airgap.
The inductance depends on the amount of magnetic energy in the core, created at some current. This energy is proportional to H*B. If you sketch a B(H) hysterisis curve as for the magnetic material, the magnetic losses in the core will be proportional to...
The burning of L1 may be due to:
1) Too big conductive losses in the coil.
2) Too big magnetic ( hysteresis ) losses in the core.
In case of 2), use another core with an airgap, thereby reducing the magnetic losses.
Most of the variation of magnetic energy in the core will take place in the...
Why do you want to measure? What's the purpose?
Say you going to make a digital controller for this motor. Having measured L and R, you will have to z-transform a transfer function, taking calculation time delay in the controller into consideration, and so on. If that's the case, there is...
The equation
2s2 + 4s + 8 = 0
has two complex roots: s = -1 ± j√3.
If the characteristic equation were to have a damping ratio = 1, it should have two real roots at the same location, for example.
s1 = -1.2 , s2 = -1.2.
In this case the characteristic equation could be written:
s2 + 2.4s +...
Ii = Io = Vi/(R - j/ωC)
Vo = Io * ( -j/ωC )
Phase shift = Φ , where Φ is calculated from Vo/Vi = xxxx∠Φ ( result in polar notation )
You may find an easier way, but this is the "basic" method.
You don't need to show in this form.
Using Mason you should get:
h(s)/H(s) = Kp * Kb / ( 2s + 4 + Kp ) →
h(s) = H(s) * Kp * Kb / ( 2s + 4 + Kp )
( if you wish )
Figure 1 shows two closed loops: An inner and an outer loop.
Use Mason's rule to calculate the transfer function as for the inner loop.
Insert this inner transfer function in the outer loop and use Mason again to calculate h(s)/H(s) for the outer loop ( the transfer function for the outer...
Read Faradays law again:
dΨn/dt does not induce a current, but a voltage.
Say that the secondary winding is unloaded, no current at all will pass the secondary winding.
Yes, but:
The DFT coefficients of all harmonic frequecies gives the differences in phases, one signal to another.
Think of a Bode plot where the phase varies as a function of frequency. Likewise the phases of different harmonics in a DFT ( or FFT ) depend on the order of the harmonic.
Say you do a Fast Fourier Transform (FFT) at some signal, some complex coefficient to some harmonic could be: 3 + j4.
This means that this harmonic has a phase referred to the Fourier interval = θ.
Tan(θ) = 4/3 → θ ≈ 53°.
The absolute value of the coefficient = 5, so this coefficient could...