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Suppose I have a simple RL-circuit with known resistance R and, inductance L with a voltage source where I can apply any input voltage (via an arbitrary waveform generator). What I want now is a specific shaped current through the coil (I need the b-field), for example the convolution of a rectangle signal with a cosine (i.e. half cycle of a cosine). What voltage do I have to apply to get the desired current through the coil.

I'm pretty sure it is possible to calculate it with the response function and several Fourier-transformations, but I'm unfortunately not an expert in this area.

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  • \$\begingroup\$ You say "specific frequency" and this normally implies a sinewave. Then you appear to be hinting that maybe with harmonics present on the signal you can shape the current waveform (and hence the flux density) to what you desire. If you can draw what your looking for and post a link to it here someone with more reputation can edit your question to incorporate the picture. \$\endgroup\$
    – Andy aka
    Oct 5, 2013 at 11:05
  • \$\begingroup\$ yeah you're right silly me... I have a arbitrary waveform-generator, so I can apply any voltages necessary. I'll edit my question \$\endgroup\$
    – wa4557
    Oct 5, 2013 at 11:59

4 Answers 4

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Given the context, I assume you have this:

schematic

simulate this circuit – Schematic created using CircuitLab

There is a series current \$i(t) \$ and so, by KVL:

$$v_1(t) = Ri(t) + L \frac{di(t)}{dt} $$

What voltage do I have to apply to get the desired current through the coil.

Since the desired current is just \$i(t)\$ and is given, the equation above is all you need assuming the time derivative of the desired current exists.

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  • \$\begingroup\$ yeah exactly, so there is no phase shift, between the applied voltage and the current through the coil? So lets assume I apply just a single rectangle, will I get a single rectangle of current? \$\endgroup\$
    – wa4557
    Oct 5, 2013 at 13:10
  • \$\begingroup\$ @user1943296, you're mixing contexts together that shouldn't be mixed. If the voltage source is sinusoidal, then you can properly talk of phase shifts because, in that case, the voltage and current are sinusoidal. For the general case, the applied voltage and resulting current do not have the same form so the notion of phase shift isn't meaningful (unless your talking about the phase shift of the sinusoidal components of the Fourier decomposition). What exactly do you mean by "apply just a single rectangle"? \$\endgroup\$ Oct 5, 2013 at 13:15
  • \$\begingroup\$ yeah I see what you mean... With single rectangle I mean that I want one single rectangle for the current. Does that mean, that I have to apply two delta-functions as voltage? (assuming the resistance is zero...) \$\endgroup\$
    – wa4557
    Oct 5, 2013 at 13:22
  • \$\begingroup\$ @user1943296, that's correct. The more rapidly the inductor current must change, the greater the inductor voltage must be. \$\endgroup\$ Oct 5, 2013 at 13:32
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Theoretically, it could be done by using the governing equation \$V_i = L\frac{dI}{dt} + RI\$ and Laplace or Fourier transforming both sides. Usually that method is used for finding a particular solution to the differential equation, but since you're dealing with an algebraic expression there's no reason the procedure can't be inverted to solve for \$V_i\$ in terms of \$I\$.

Unfortunately, for all but the simplest types of waveforms, the expression that you get after manipulating the equation is unlikely to be easy to inverse transform back into the time domain. It is probably easiest to numerically calculate the required input waveform directly from the governing equation point-by-point by substituting in values for \$\frac{dI}{dt}\$ and \$I\$.

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  • \$\begingroup\$ There is no need to solve differential equations, because the current is known. So, only the first derivative has to be find, which is easy enough. So, no Laplace or Fourier transformations are needed. \$\endgroup\$
    – johnfound
    Oct 5, 2013 at 12:42
  • \$\begingroup\$ @johnfound The advantage of using the Laplace transform in this situation would be with inductor current waveforms which have points that are not differentiable. It may be possible to obtain via inverse Laplace transform an analytic expression for the input waveform directly in terms of step functions, while in the time domain the equation has to be re-evaluated every time the inductor current has a point that is not differentiable. \$\endgroup\$
    – Bitrex
    Oct 5, 2013 at 13:45
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If you want a raised cosine current through an inductor then: -

  • generate a raised cosine voltage waveform (a signal) and...
  • feed it in to a voltage to current converter (simple circuit) with...
  • feedback from the resistor (or another small resistor) and this will...
  • ensure the current through the resistor looks like your signal and...
  • because R and L are in series, this current is also the current through the inductor

If this makes sense then to make a VtoI you'll need to tell what sort of indutance and current amplitudes you want. Or maybe you can take it from here?

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  1. The voltage on L is the first derivative of the current. So, if know what current you need, then find the first derivative and you will have the voltage on the coil.

  2. The voltage on the resistance is by Ohm's law: Ur=Ir*R, where Ir = IL (because they are serially connected.

  3. The sum of these two voltages will give you the input voltage you need to apply in order to get the needed current.

Another approach is to simply make a model of the needed schematic and play with it:

schematic

simulate this circuit – Schematic created using CircuitLab

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