Test the phase shift / relation

Hi,

I'm building a rack for 6 nice old preamps. I added phase switch and output transformers. Now I'd like to test the phase shift / relation to eliminate mistakes at the wiring. There must be the same phase relation on output of the rack like on the input of the preamp if phase switch is turning off. But how can I test this? I have multimeter (Fluke), sinus/square-generator and an oscilloscope. Can somebody help?

barthman.de said:

.. an oscilloscope...

Click to expand...

Dual trace?

moamps said:

barthman.de said:

.. an oscilloscope...

Click to expand...

Dual trace?

Click to expand...

Two channels: A / B; A+B so I think dual trace. I have not used it often, so I have no practical knowledge with it. It is a Hameg HM203-6.

Don't over think this and make this seem harder than it is. Take a normal or mult from the input signal into one input channel of a mixer, and the output to test into another input. Assign them both to a bus and adjust for similar level. If they cancel out at the bus they are opposite polarity, if they get 6 dB louder they are same polarity.


JR

barthman.de said:

....It is a Hameg HM203-6....

Click to expand...

Here is the procedure:

- connect the sine generator (1Vpp/1kHz) to the +input of the preamp,
- connect  -input of the preamp to the ground,
- set the gain of the preamp to minimum (0dB),
- connect CH1 probe to the sine generator output (or +input of the preamp),
- connect CH2 probe to the +output of the preamp,
- connect -output of the preamp to the ground (if you are using transformer at the preamp's out),
- set the scope inputs to AC, 1V/div, noninvert, timebase to 1ms/div or so, two channel chop mode
- using trace vertical position knobs move one trace up, second down to distinct the waveforms clearly,
- observe the waveforms, if both are positive or negative for the same time position, they are in phase

P.S. you can use the XY mode, if the resulting trace is placed in first and third quadrant, signals are in phase 

JohnRoberts said:

Don't over think this and make this seem harder than it is. Take a normal or mult from the input signal into one input channel of a mixer, and the output to test into another input. Assign them both to a bus and adjust for similar level. If they cancel out at the bus they are opposite polarity, if they get 6 dB louder they are same polarity.


JR

Click to expand...

Thank you for helping me to remember this easy way of testing this.  ;D This is a good idea.

moamps said:

Here is the procedure:

- connect the sine generator (1Vpp/1kHz) to the +input of the preamp,
- connect  -input of the preamp to the ground,
- set the gain of the preamp to minimum (0dB),
- connect CH1 probe to the sine generator output (or +input of the preamp),
- connect CH2 probe to the +output of the preamp,
- connect -output of the preamp to the ground (if you are using transformer at the preamp's out),
- set the scope inputs to AC, 1V/div, noninvert, timebase to 1ms/div or so, two channel chop mode
- using trace vertical position knobs move one trace up, second down to distinct the waveforms clearly,
- observe the waveforms, if both are positive or negative for the same time position, they are in phase

P.S. you can use the XY mode, if the resulting trace is placed in first and third quadrant, signals are in phase

Click to expand...

OK. Works fine:

In Phase:



Out of phase:



Thank you.

Not to be overly pedantic but the proper terminology is same "polarity" not phase.

Phase implies a fraction of a cycle offset, and for a sine wave opposite polarity looks like 180' of phase shift, but it isn't.

JR

PS: It sure is easier to say in phase and you won't confuse many people, just trying to be accurate.

JohnRoberts said:

Not to be overly pedantic but the proper terminology is same "polarity" not phase.

Phase implies a fraction of a cycle offset, and for a sine wave opposite polarity looks like 180' of phase shift, but it isn't.

JR

PS: It sure is easier to say in phase and you won't confuse many people, just trying to be accurate.

Click to expand...

Yes you are right. And all because so many manufacturer print "phase bottom" in their manuals or on front panels. Polarity switch would be better.  I do it too :'(

barthman.de said:

JohnRoberts said:

Not to be overly pedantic but the proper terminology is same "polarity" not phase.

Phase implies a fraction of a cycle offset, and for a sine wave opposite polarity looks like 180' of phase shift, but it isn't.

JR

PS: It sure is easier to say in phase and you won't confuse many people, just trying to be accurate.

Click to expand...

Yes you are right. And all because so many manufacturer print "phase bottom" in their manuals or on front panels. Polarity switch would be better.  I do it too :'(

Click to expand...

I have had to deal with this on consoles myself and polarity takes up more of the often tight panel real estate.  ;D ;D

I can't even get the industry to spell "bus" correctly so I try to be relaxed about this small stuff...

JR

JohnRoberts said:

Not to be overly pedantic but the proper terminology is same "polarity" not phase.

Click to expand...

For me "inverting" is synonym for "out of phase" and "non-inverting" is synonym for "in phase". If you google " inverting opamp"and "out of phase", you will find lot of results from edu sites where they say "inverting=180 out of phase". IMO, that's the reason why all correlation scales are marked from +1 to -1. Just for simplifying math (cos(0)=+1, cos(180)=-1). 

JohnRoberts said:

.... and for a sine wave opposite polarity looks like 180' of phase shift, but it isn't.

Click to expand...

You are wrong here, IMHO.
from:
sin(x + y) = sin(x)cos + cos(x)sin
and y=180 we get:
sin(x+180)=sin(x)cos(180)+cos(x)sin(180)
for cos(180)=-1 and sin(180)=0
the final result is:
sin(x+180)=-sin(x)

I can only lead you to the water I can not make your drink...

Real world signals are not continuous steady state sine waves.

You are welcome to believe what you want.

JR

moamps said:

JohnRoberts said:

Not to be overly pedantic but the proper terminology is same "polarity" not phase.

Click to expand...

For me "inverting" is synonym for "out of phase" and "non-inverting" is synonym for "in phase". If you google " inverting opamp"and "out of phase", you will find lot of results from edu sites where they say "inverting=180 out of phase". IMO, that's the reason why all correlation scales are marked from +1 to -1. Just for simplifying math (cos(0)=+1, cos(180)=-1). 

JohnRoberts said:

.... and for a sine wave opposite polarity looks like 180' of phase shift, but it isn't.

Click to expand...

You are wrong here, IMHO.
from:
sin(x + y) = sin(x)cos + cos(x)sin
and y=180 we get:
sin(x+180)=sin(x)cos(180)+cos(x)sin(180)
for cos(180)=-1 and sin(180)=0
the final result is:
sin(x+180)=-sin(x)

Click to expand...

Try with a sawtooth wave, inverting polarity gives a ramp up or down, shifting 180º phase is always ramp up (or down, depending on what you started) but half cycle later. When you speak of phase shift you do the phase shift on the time domain, so you delay 1/2 cycle, not in the frequency domain, where you will do as you said with all the components of the Fourier transform where you will get what you want.

JS

JohnRoberts said:

Real world signals are not continuous steady state sine waves.

Click to expand...

I didn't said that they are.
I just wanted to help some people here, but that obviously isn't welcome.

joaquins said:

Try with a sawtooth wave, inverting polarity gives a ramp up or down, shifting 180º phase is always ramp up (or down, depending on what you started) but half cycle later.

Click to expand...

 
Sorry, but you lost me here. Are you talking about DC shifted sawtooth? Never mind.

Oh, it's welcome, we never said it isn't... But JR was saying that's not the same and he had a reason. I explained you with a n example, no DC shifted, here is a pic with what I'mm saying. First plot is original, second 180º phase shift, 3rd reversed polarity.


Your post was in fact useful, we learn from mistakes, what you said is from where the confusion comes, what you said it's true, if you add 180º to a sin you get the same as reversing it, but that's for a sin signal. If you get a more complex signal, as the sawtooth for example it isn't anymore, you could take the fourier transform which has all the sin separately and add to each 180º and finally get the reversed polarity, but not to the signal in the time domain, imagine you have an unknown periodic function, and the only thing you know about it is that the period is T:

f(t)=f(t+T)

You can't say only knowing this that half period later will be reverse polarity, or the same function without that half period multiplied by -1:

f(t+T/2)=-f(t) (this is what you can't be sure if you don't know the function)

JS