Help with waveform control: frequency, phase and amplitude

[hutchphd]

Also never be too hasty to go down the digital path. Analog could be cheaper and easier.

For completeness I mention Mr Hewlett's wonderful master's thesis invention with the nonlinear (light bulb) feedback.

https://people.ohio.edu/postr/bapix/HP200CD.htm

For harmonic waveforms at least, an oscilloscope and a knob to turn is often a good solution

 

[artis]

pardon , @Averagesupernova I meant that with changing frequency the output amplitude will vary with different loads and even with the same load.

 

[sophiecentaur]

@sophiecentaur a bit skeptical and critical as usual , welcome :)

Only critical about the lack of real direction in the thread. It seems to me that the OP is after a box with two terminals (or more?) that will provide any required volts and current to suit any load that may be chosen. Is not the normal approach for the 'consumer' appliance to present a load that can be fed from a general purpose Power Supply (usually a nominal constant Voltage)? The internal power regulator would expect to do the 'clever stuff' to deal with the peculiar requirements of any devices that it feeds inside the box. To invent the sort of device that the OP seems to be after would be needlessly hard and would be sure to fall short in some way. It would be like the marvellous medicines that are advertised on the net to sort out your arthritis, impotence, shortness of breath and dyspepsia, all in the same bottle. But you would still need another bottle to deal with your sore throat or piles. One job, done properly and efficiently, is surely the essence of good design.

 

[Averagesupernova]

It seems to me that the OP is after a box with two terminals (or more?) that will provide any required volts and current to suit any load that may be chosen.

It seems that way. Design experience teaches that there are trade-offs in the real world. Convenience of use and max number of applications possible do not generally occur simultaneously.

 

[artis]

you know when I come to think of it I actually just need voltage (amplitude) and phase shift feedback, because after all i am using an amplifier as my source not a mechanical generator where an increase in load results in a slight decrease in rpm/frequency which then needs to be adjusted for , in my case as long as the amplifier supply power doesn't sag my frequency should stay constant and as set by the amplifier input signal, at least I think so...?

So basically depending on the load output voltage will change , current will be then determined by the load and also will be phase shift so if I can keep voltage constant and compensate for phase shift if it gets too large then I think I'm okay.So the logic goes like this. I set a fixed frequency and the feedback control maintains fixed voltage at that frequency , if I set a new frequency value the frequency rises and the feedback then sets the voltage again or adjusts it given that my voltage value isn't changed.
Additionally I should adjust for phase shift if necessary.

 

[anorlunda]

I still don't understand what you'll adjust phase shift for. Also, you never answered the question, "Phase shift of what relative to what?"

 

[hutchphd]

It would be like the marvellous medicines that are advertised on the net to sort out your arthritis, impotence, shortness of breath and dyspepsia, all in the same bottle.

Can I get a citation here?

 

[artis]

The phase shift I simply thought to be the current either lagging or leading , say I attach a very inductive load and the current starts to lag , then the idea is to simply adjust the signal waveform driving the amplifier so that the phase shift is minimized .

Ok to be honest I haven't put all of this together and maybe I'm just making a fool of myself here , but at least in theory shouldn't that be possible?

Also was I right in saying that since I'm driving my load with an amplifier even under varying load conditions where the output voltage increases/decreases the frequency should stay fixed as it it set by the input signal?

 

[sophiecentaur]

say I attach a very inductive load and the current starts to lag

It doesn't "start to lag" because the word Lag can only apply to the steady state. You connect the load and then you wait until it's all settled down. Only then can you measure a meaningful Lag.

the frequency should stay fixed as it it set by the input signal?

One way (the only one) that you can cause the output frequency to be different from the input frequency (say, lower) is to separate the source from the output with a steadily increasing distance (increasing time lag as the ambulance speeds away from you). Then you have the Doppler effect at work. There is no way that the output frequency of a static setup can be different from the input frequency. Ask yourself at which point in the circuit would the voltage be going up and down at a different rate from the input voltage? Amplitude continuity is always maintained across an interface.
There are signal processors that can achieve this, though - the easiest to describe is the old tape loop with several read heads on a wheel which rotates faster of slower than the tape speed, the outputs from the heads are switched to select the head that happens to be scanning over the tape. This is really just the Doppler shift at work. Other (digital) methods are available but none of this is relevant to the simple source / load system that's been suggested.

 

[sophiecentaur]

Can I get a citation here?

You can actually write your own and someone, somewhere will read and believe it.

 

[anorlunda]

The phase shift I simply thought to be the current either lagging or leading , say I attach a very inductive load and the current starts to lag , then the idea is to simply adjust the signal waveform driving the amplifier so that the phase shift is minimized .

OK to be honest I haven't put all of this together and maybe I'm just making a fool of myself here , but at least in theory shouldn't that be possible?

No. In theory it is not possible. That's what we have been trying to tell you.

If you connect a pure resistive load, the phase between V and I will be zero.
If you connect a pure inductive load, the phase between V and I will by 90 degrees.
If you connect a pure capacitive load, the phase between V and I will be 90 degrees the other direction.

The voltage source can do nothing to change that. There is nothing to adjust. The signal source can't make an inductor into a resistor.

 

[artis]

@sophiecentaur , thanks that's what I thought , an amplifier will have a fixed frequency just a varying output amplitude under a changing load impedance , a mechanical generator would change it's frequency because with more load the rpm would decrease and need to be adjusted.@anorlunda hey, I messed up terribly and only now I understood that when I said phase shift I was actually thinking just controlling the shape of the output waveform itself not the waveform of the voltage vs current as it would change under a different type of load (inductive/capacitive or reactive)
when you said that an inductor would never perform as a resistor I though damn sure makes sense ,

so surely I would not be able to undo the phase shift present within an inductor , what I wanted to is to simply simulate a capacitor added to an inductor , although I assume that would need to be done the old fashioned way as to connect a physical cap in the output much like power stations switching on capacitor banks to compensate for overly lagging current , am I now thinking correctly?

So I guess what I am asking is , how could one simulate the added capacitor effect to an inductive load simply by having an amplifier and an inductive load without a real capacitor , is that even possible ?IIRC a capacitor added to an inductor like an induction motor doesn't actually decrease the total lag of current or reactive power it just forms a loop where these reactive currents circulate between the cap and motor instead of them circulating from the motor back to grid etc ?

 

[anorlunda]

I'm afraid your misconception is more basic. Consider the following circuit.

Let Z be an impedance constructed of any combination of R, L and C you please, series and/or parallel connections, as many passive R, L, C components as you please. The voltage measured across Z is V, and the current through Z is I.

Let S be a source. It could be anything, a mains plug, a signal generator, a battery , or whatever. We'll exclude DC, but S is a periodic signal of any shape.

Z and S are connected only via the two terminals (black dots).

Now, if you allow V, I and Z to be complex numbers, Ohm's Law determines the relationship $\frac{V}{I}=Z$ between voltage and current. Nothing that S does can change that.

If S is not a sine wave, then it can be represented as a Fourier Series which has several harmonic frequencies. $\frac{V}{I}=Z$ applies separately to each harmonic, and the total is the sum of all harmonics. Changing the fundamental frequency of S can change Z, but still $\frac{V}{I}=Z$ . The relationship between V and I depends on Z, not S. Changing the waveform of S, changing the phase angle of S relative to some external reference does not change $\frac{V}{I}=Z$.

So, what you say your objective is, to change the relationship between I and V using adjustments in S is theoretically impossible. That can be done only by changing the contents of the box Z. I hope that I'm making the message clear.

 

[artis]

yes, you made this clear , thanks for the explanation.
but my question about the classical case of an capacitor connected to an induction motor is true isn't it ? where you basically control how far in the grid the "Z" imposed phase shift goes , that's like PFC correction in smps , one cannot change the phase shift of the mains input rectifier or the power of the switches being taken mostly from the peaks of the rectified sine due to stored charge in capacitors but the PFC circuit changes the properties of this connected Z load as to make it more resistive in nature ?maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

 

[Baluncore]

maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

Phase is independent of voltage in a linear system.
Now you are just confusing things by writing one long misinterpretable sentence.
Sit down with a book on AC theory and write out the equations.
You will be able to ask more sensible questions once you understand the concepts.

 

[sophiecentaur]

maybe a little different question , but then increasing voltage through an inductive or reactive load both increases the real power flowing in the circuit as well as reactive but the proportion by which both increase is determined by the phase angle and at some phase angle (45?) both real and reactive increase at the same rate ?

I was contemplating an answer to this when I read @Baluncore 's post, above - which I totally endorse. The above quote is a great demonstration of how 'home brewed' personal theories can lead you into more nonsensical questions and answers.

All elementary 'AC Theory' is based on linear generators and Loads so changing the supply volts cannot do anything to change the IMPEDANCES in the circuit.

You really should read some basic texts and do some graft to get a hold of this topic, instead of trying to get there by Q and A which, as PF often points out, is not a way to learn anything technical. Just being apologetic about asking bad questions is not excuse, either, if you really want to find out stuff.

 

[anorlunda]

I think we are going in circles. Thread closed.

@artis , take the advice given by @Baluncore and @sophiecentaur