what an oscilloscope recommended for a woman passionate about electronics?

[rstofer]

I liked coloring with crayons at school
But I tried with the same signal in AC mode to make 20 cyrcles and 80 cyrcles, but I don't get different rectangles like froma, but the same. I think the zero volt on the display is the red dot on the left. I must have got lost somewhere 

Your waveforms are correct but, for this square wave experiment, I only want one waveform on the screen.  So, take a 1 kHz squarewave and set the timebase to 100 us/div.  100 us/div * 10 div = 1000 us or 1 ms.  A 1 kHz signal will have a total period of 1 ms so it all works out.

The dotted line in the middle of the screen is the 0V reference.  When you do a 20% waveform, you get a narrow but tall block above the 0V reference and a wide but short block below the 0V reference,  What you have is correct,

[CharlotteSwiss]

I liked coloring with crayons at school
But I tried with the same signal in AC mode to make 20 cyrcles and 80 cyrcles, but I don't get different rectangles like froma, but the same. I think the zero volt on the display is the red dot on the left. I must have got lost somewhere 

Your waveforms are correct but, for this square wave experiment, I only want one waveform on the screen.  So, take a 1 kHz squarewave and set the timebase to 100 us/div.  100 us/div * 10 div = 1000 us or 1 ms.  A 1 kHz signal will have a total period of 1 ms so it all works out.

The dotted line in the middle of the screen is the 0V reference.  When you do a 20% waveform, you get a narrow but tall block above the 0V reference and a wide but short block below the 0V reference,  What you have is correct,

ok, I tried a 1khz square wave.
50% cyrcle test: the area of the positive signal is equal to the air of the negative signal (amplitude Vp-p = 4v)
20% cyrcle test: the area of the positive signal is equal to the air of the negative signal (amplitude Vp-p = 3.2 + 0.8 = 4v)
Now I have to understand what all this can teach me 
thanks

[CharlotteSwiss]

I made some progress to understand the average value of a periodic signal. 

I used the siglent square wave compensation signal measured in DC coupling.



Area of the period (marked in red): 3v x 500us (0.005s) = 0.0015
Vm: 0.0015 / T (0.001) = 1.5Vdc
So I really tried to measure the Vm of the signal with my multimeter set in DC, and it showed me exactly 1.5vdc.
Great!

For a UNIPOLAR periodic square wave signal I understand!
Maybe I can't understand when the signal has both positive and negative parts, I have difficulty calculating the periodic area of the signal on paper ...

[tggzzz]

For a UNIPOLAR periodic square wave signal I understand!
Maybe I can't understand when the signal has both positive and negative parts, I have difficulty calculating the periodic area of the signal on paper ...

Good.

Go back to the maths, and plug in the values. Consider that the mean (i.e. average) y_mean of two values y₁ and y₂ is simply y_mean = (y₁ + y₂) / 2.

If  y₁=3 and  y₂=0, then y_mean = 1.5
If  y₁=-3 and  y₂=0, then y_mean = -1.5
If  y₁=1.5 and  y₂=-1.5, then y_mean = 0

[CharlotteSwiss]

Go back to the maths, and plug in the values. Consider that the mean (i.e. average) y_mean of two values y₁ and y₂ is simply y_mean = (y₁ + y₂) / 2.

If  y₁=3 and  y₂=0, then y_mean = 1.5
If  y₁=-3 and  y₂=0, then y_mean = -1.5
If  y₁=1.5 and  y₂=-1.5, then y_mean = 0

so to find the average value of the DC voltage in a bipolar square wave like this:



is it enough to make Vm = 3v - 1v / 2 = 1vdc?
Or just do: (yellow area - orange area - white area) / T period

Are they there now?
thanks   

[tggzzz]

Go back to the maths, and plug in the values. Consider that the mean (i.e. average) y_mean of two values y₁ and y₂ is simply y_mean = (y₁ + y₂) / 2.

If  y₁=3 and  y₂=0, then y_mean = 1.5
If  y₁=-3 and  y₂=0, then y_mean = -1.5
If  y₁=1.5 and  y₂=-1.5, then y_mean = 0

so to find the average value of the DC voltage in a bipolar square wave like this:

(Attachment Link)

is it enough to make Vm = 3v - 1v / 2 = 1vdc?
Or just do: (yellow area - orange area - white area) / T period

Are they there now?
thanks   

I previously gave the equation: for a square wave with voltages V_h and V_l for proportions of the time P_h and P_l, the mean value V_m = V_h*P_h + V_l*P_l. I'll leave it to you to calculate V_m for the 20%, 50% and 80% duty cycles It is also worth doing a "sanity check" with 100% and 0% duty cycle.

For the trace you show, V_h=3, V_l=-1, P_h=0.25 and P_l=0.75. In other words the yellow and white areas bounded by the green trace and the 0V. The orange area isn't an area to be included, since it isn't bounded by a y value.

[CharlotteSwiss]

I previously gave the equation: for a square wave with voltages V_h and V_l for proportions of the time P_h and P_l, the mean value V_m = V_h*P_h + V_l*P_l. I'll leave it to you to calculate V_m for the 20%, 50% and 80% duty cycles It is also worth doing a "sanity check" with 100% and 0% duty cycle.

For the trace you show, V_h=3, V_l=-1, P_h=0.25 and P_l=0.75. In other words the yellow and white areas bounded by the green trace and the 0V. The orange area isn't an area to be included, since it isn't bounded by a y value.

yes, but I didn't understand those acronyms Ph, Pi etc, now I understand 
Ok I'm calculating the Vm of the AC 20% (I start with this because I have doubts that the calculation is right)



Vm: (positive yellow area) + (negative yellow area) / T period
       (3,2v x 0,0002s) + (-0,8v x 0,0008s) / 0,001s
       (0,00064) + (-0,00064) / 0,001s  = 0

So even with duty cyrcle 20% we have a Vm = zero volts as with duty cyrcle 50%? and the same thing will also be with 80% ...
Or am I wrong?

slowly I will understand too... 

[tggzzz]

I previously gave the equation: for a square wave with voltages V_h and V_l for proportions of the time P_h and P_l, the mean value V_m = V_h*P_h + V_l*P_l. I'll leave it to you to calculate V_m for the 20%, 50% and 80% duty cycles It is also worth doing a "sanity check" with 100% and 0% duty cycle.

For the trace you show, V_h=3, V_l=-1, P_h=0.25 and P_l=0.75. In other words the yellow and white areas bounded by the green trace and the 0V. The orange area isn't an area to be included, since it isn't bounded by a y value.

yes, but I didn't understand those acronyms Ph, Pi etc, now I understand 
Ok I'm calculating the Vm of the AC 20% (I start with this because I have doubts that the calculation is right)

(Attachment Link)

Vm: (positive yellow area) + (negative yellow area) / T period
       (3,2v x 0,0002s) + (-0,8v x 0,0008s) / 0,001s
       (0,00064) + (-0,00064) / 0,001s  = 0

So even with duty cyrcle 20% we have a Vm = zero volts as with duty cyrcle 50%? and the same thing will also be with 80% ...
Or am I wrong?

slowly I will understand too... 

Yes, exactly. Or with any other duty cycle or with any other waveform whatsoever. Over the course of one cycle with an AC coupled signal, the area above the 0V line will equal the area below the 0V line. That is equivalent to saying the mean value is zero.

Now, if the waveform isn't as simple as a rectangular wave, you might have to increase the number of rectangles of area V_h*P_h, and add all those together. But you will recognise those are precisely the operations performed in calculus when calculating the integral.

[CharlotteSwiss]

Yes, exactly. Or with any other duty cycle or with any other waveform whatsoever. Over the course of one cycle with an AC coupled signal, the area above the 0V line will equal the area below the 0V line. That is equivalent to saying the mean value is zero.

Now, if the waveform isn't as simple as a rectangular wave, you might have to increase the number of rectangles of area V_h*P_h, and add all those together. But you will recognise those are precisely the operations performed in calculus when calculating the integral.

I finally begin to understand something 
Moreover, we said that the AC coupling blocks the DC component, so it is reasonable to expect an average value = 0.
Now I have understood better also the duty cyrcle, that is the working time of the signal: intervening on its percentage, it is possible to vary the working time of the load.
But I saw that in the software it allows you to set 1-99% (not 0 and 100), but so by eye, with 0 or 100% you have a linear DC signal I think ...
Note: I had never been interested in duty cyrcle, now I realized that my multimeter has the duty function and works well: I measured the siglent compensation signal, and it makes me 50%
thanks 

[rstofer]

So even with duty cyrcle 20% we have a Vm = zero volts as with duty cyrcle 50%? and the same thing will also be with 80% ...
Or am I wrong?

You are correct and, lacking a DC component, it will apply to all other waveforms.  It's the removal of the DC component that makes this work.  Try it with a sawtooth.  It's harder to compute the areas and it may be more difficult to see but the areas above and below the 0V reference are equal.

I generated a 1kHz triangle with a 1V DC offset.  It's not intuitive where the 0V reference is but the entire waveform is at or above 0V.  That's what is fed to the AC coupled scope.

On the scope there are two iterations but notice that the triangles have equal area above and below the 0V reference.  The DC offset has been removed.

I suppose there is a signal processing reason why we talk about this stuff but it also comes up in audio amplifiers where we use rather large capacitors (like 100 ufd) to remove the DC bias that results from most amplifier stages.  In the old days, we used transformers to isolate the high DC voltages used in vacuum tube circuits.  Same idea except the transformer also performs impedance matching between the high impedance of the vacuum tube circuit and the low impedance of the speaker.

DC won't go through a transformer either.

[tggzzz]

Good

If you ever get a signal generator, then it is worth setting up the scope as shown in your simulation and then simply varying the duty cycle. You will see how that changes the trace on the scope, and will begin to have a visceral feel for what's happening. Such playing around twiddling controls can be enlightening - and that's especially true when debugging real circuits.

If your simulation is real-time, you might be able to do that in the simulator.

Your next task will be to understand the limits of the averaging done by the AC coupling. You will find the scope has a "low frequency cutoff" in the specification; if the waveform frequency is above that cutoff, the averaging will be good. If it is lower (i.e. the period is longer) then the averaging will be imperfect and the imperfection will be visible as a "distorted" trace. (Mathematically it will be the input signal passed through a high-pass filter.)

[CharlotteSwiss]

tomorrow I see your new suggestions better, now it's time to stretch out horizontally that here we are in the middle of the night
 

[tggzzz]

I suppose there is a signal processing reason why we talk about this stuff

Depends on what you mean by "signal processing". Separating the AC component from a DC level is pretty fundamental to many types of theoretical analysis, practical analysis, and practical circuit design.

If you want to consider audio circuits, then have a look at the "bias frequency" used in magnetic tape recorders to overcome the non-linearities in the tape's B-H curve.

but it also comes up in audio amplifiers where we use rather large capacitors (like 100 ufd) to remove the DC bias that results from most amplifier stages. 

In that example, usually it is best to think of it in two ways:

• level shifting, e.g. one amplifier's output is 5Vdc+the AC signal, but the next amplifier's input is centred on -1Vdc. The AC coupling capacitor will have 5V on one terminal and -1V on the other (i.e it will block the 6V difference), but the AC only signal will pass through the capacitor (i.e. not be blocked)
• avoiding amplifying a DC input so much that it would saturate amplifier's output. Consider an input with a 10mV offset voltage. If that was amplified 1000 times, the amplifier output would be sitting at 10V. That would be challenging in an amplifier with +-5V power supply

In the old days, we used transformers to isolate the high DC voltages used in vacuum tube circuits.  Same idea except the transformer also performs impedance matching between the high impedance of the vacuum tube circuit and the low impedance of the speaker.

True, but that is entirely different to AC coupling. The important transformer function is to change the impedance; that it also blocks the DC component is unimportant.

[tggzzz]

I suppose there is a signal processing reason why we talk about this stuff

Depends on what you mean by "signal processing". Separating the AC component from a DC level is pretty fundamental to many types of theoretical analysis, practical analysis, and practical circuit design.

If you want to consider audio circuits, then have a look at the "bias frequency" used in magnetic tape recorders to overcome the non-linearities in the tape's B-H curve.

but it also comes up in audio amplifiers where we use rather large capacitors (like 100 ufd) to remove the DC bias that results from most amplifier stages. 

In that example, usually it is best to think of it in two ways:

• level shifting, e.g. one amplifier's output is 5Vdc+the AC signal, but the next amplifier's input is best when centred on -1Vdc. The AC coupling capacitor will have 5V on one terminal and -1V on the other (i.e it will block the 6V difference), but the AC only signal will pass through the capacitor (i.e. not be blocked)
• avoiding amplifying a DC input so much that it would saturate amplifier's output. Consider an input with a 10mV offset voltage. If that was amplified 1000 times, the amplifier output would be sitting at 10V. That would be challenging in an amplifier with +-5V power supply

In the old days, we used transformers to isolate the high DC voltages used in vacuum tube circuits.  Same idea except the transformer also performs impedance matching between the high impedance of the vacuum tube circuit and the low impedance of the speaker.

True, but that is entirely different to AC coupling. The important transformer function is to change the impedance; that it also blocks the DC component is unimportant.

[rstofer]

If you want to consider audio circuits, then have a look at the "bias frequency" used in magnetic tape recorders to overcome the non-linearities in the tape's B-H curve.

We do the same thing with a 'dither frequency' on hydraulic servo valves to improve linearity and to prevent 'sticking'.

[rstofer]

Note: I had never been interested in duty cyrcle, now I realized that my multimeter has the duty function and works well: I measured the siglent compensation signal, and it makes me 50%

That feature will come in handy when you start playing with PWM or RC servos (model airplane type servos).  Just remember that you have it.

[tautech]

Charlotte
Now you are getting some good core understanding of scope use we need advance you further to using the power of a modern DSO. 

Measurements.
At one time we lined up waveform to screen graticules and did mental calcs based on s/div and V/div settings to obtain measurements and for those of us used to working like that many of us still do.
However with the modern DSO we have a powerful Measurement suite where we can select just a few we need or engage all but the display becomes quite busy when all are used.

Probe Comp signal:
Here I have selected 4 that you might commonly use and assigned them all to a particular channel (ch4 SDS1104X-E) although you can select and have measurements displayed from more than 1 channel.
The multifunction control is used to scroll through the list and a press selects the measurement required and a brief description of what each does.




Further on these X-E's we can select Statistics to gather further info about each measurement.
Take note of the Mean measurement value. 

[rstofer]

I suppose there is a signal processing reason why we talk about this stuff

Depends on what you mean by "signal processing". Separating the AC component from a DC level is pretty fundamental to many types of theoretical analysis, practical analysis, and practical circuit design.

We might run into the DC component issue when we get this thread up to FFT.  Sometimes there is a DC spike, sometimes there isn't.

[tggzzz]

If you want to consider audio circuits, then have a look at the "bias frequency" used in magnetic tape recorders to overcome the non-linearities in the tape's B-H curve.

We do the same thing with a 'dither frequency' on hydraulic servo valves to improve linearity and to prevent 'sticking'.

That is very different.

With B-H curve, the bias frequency is much higher than the tape's frequency response (80kHz springs to mind), and the amplitude is much larger than the audio signal amplitude. See, for example, http://hccc.org.uk/acbias.html

It looks like "your" dither frequency is just within the transducer's response bandwidth, and of a small amplitude. Or perhaps just of an amplitude that will cause a small transducer response, but I would have thought that might waste power.

[CharlotteSwiss]

You are correct and, lacking a DC component, it will apply to all other waveforms.  It's the removal of the DC component that makes this work.  Try it with a sawtooth.  It's harder to compute the areas and it may be more difficult to see but the areas above and below the 0V reference are equal.

I generated a 1kHz triangle with a 1V DC offset.  It's not intuitive where the 0V reference is but the entire waveform is at or above 0V.  That's what is fed to the AC coupled scope.

On the scope there are two iterations but notice that the triangles have equal area above and below the 0V reference.  The DC offset has been removed.

ok I understand, it is clear that the AC coupling has blocked the DC offset of the input signal. As it is evident that the positive and negative areas are identical, making an average value of zero.
If I measured that signal with the multimeter in dc mode, I would have a measurement of 0v
thanks 

Note: I had never been interested in duty cyrcle, now I realized that my multimeter has the duty function and works well: I measured the siglent compensation signal, and it makes me 50%

That feature will come in handy when you start playing with PWM or RC servos (model airplane type servos).  Just remember that you have it.

In fact, reading about how duty cyrcle affects the workload, I also got involved in pwm, now I understand something more about this function. I will keep this in mind, moreover that Uni-T multimeter has many useful resources

[CharlotteSwiss]

Good

If you ever get a signal generator, then it is worth setting up the scope as shown in your simulation and then simply varying the duty cycle. You will see how that changes the trace on the scope, and will begin to have a visceral feel for what's happening. Such playing around twiddling controls can be enlightening - and that's especially true when debugging real circuits.

If your simulation is real-time, you might be able to do that in the simulator.

Your next task will be to understand the limits of the averaging done by the AC coupling. You will find the scope has a "low frequency cutoff" in the specification; if the waveform frequency is above that cutoff, the averaging will be good. If it is lower (i.e. the period is longer) then the averaging will be imperfect and the imperfection will be visible as a "distorted" trace. (Mathematically it will be the input signal passed through a high-pass filter.)

yes of course, when I take a signal generator I will already have some basic notions and I will be able to experience what I have already learned.
I looked at the specs, I see a "low frequency response" =< 2Hz, maybe you mean this limit?
 

[tggzzz]

Good

If you ever get a signal generator, then it is worth setting up the scope as shown in your simulation and then simply varying the duty cycle. You will see how that changes the trace on the scope, and will begin to have a visceral feel for what's happening. Such playing around twiddling controls can be enlightening - and that's especially true when debugging real circuits.

If your simulation is real-time, you might be able to do that in the simulator.

Your next task will be to understand the limits of the averaging done by the AC coupling. You will find the scope has a "low frequency cutoff" in the specification; if the waveform frequency is above that cutoff, the averaging will be good. If it is lower (i.e. the period is longer) then the averaging will be imperfect and the imperfection will be visible as a "distorted" trace. (Mathematically it will be the input signal passed through a high-pass filter.)

yes of course, when I take a signal generator I will already have some basic notions and I will be able to experience what I have already learned.
I looked at the specs, I see a "low frequency response" =< 2Hz, maybe you mean this limit?
 

Quite possibly, but I'd have to see the surrounding context to be sure.

[CharlotteSwiss]

Charlotte
Now you are getting some good core understanding of scope use we need advance you further to using the power of a modern DSO. 

Measurements.
At one time we lined up waveform to screen graticules and did mental calcs based on s/div and V/div settings to obtain measurements and for those of us used to working like that many of us still do.
However with the modern DSO we have a powerful Measurement suite where we can select just a few we need or engage all but the display becomes quite busy when all are used.

Probe Comp signal:
Here I have selected 4 that you might commonly use and assigned them all to a particular channel (ch4 SDS1104X-E) although you can select and have measurements displayed from more than 1 channel.
The multifunction control is used to scroll through the list and a press selects the measurement required and a brief description of what each does.

Further on these X-E's we can select Statistics to gather further info about each measurement.
Take note of the Mean measurement value. 

I am happy that DSO do many mathematical calculations: this, however, also has a sad dark side, that is, machines replace our brains. If we look at modern children, they no longer make any mental calculations, they use the cellphone calculator ...
I suppose even retired electronic technicians are having a hard time seeing how oscilloscopes make life easier today

I had already seen these mathematical functions in the overview of the manual in the first pages, for now I have written on my text document only a few things, when I get to the paragraph that deepens this theme I study it better ..
Little by little I get there, and what I don't understand I ask you experts. You are very kind and helpful tutit, thank you
 

[CharlotteSwiss]

Quite possibly, but I'd have to see the surrounding context to be sure.

it is reported in the vertical system specifications: it says "low frequency response" AC-3db / = <2Hz (at the bnc input)

[tautech]

You are very kind and helpful tutit, thank you

tutit 

tautech is shortened from my company name being the district Taupaki where I live in NZ and Technologies but if you insist call me tutit or even better just plain Rob.
When I have time I drop in here and offer screenshots from very similar DSO's so it's easier for you to understand.
I have been using and selling these X-E's for some years and know them very well so it's easy for me to provide good examples as your thread develops.

You must know this is already a well followed thread as you are not the only novice working hard to understand what you see on a scope display and why.
You like many others before you have come a long way from opening the box, pressing the power button and the Autoset and threads like this really help other novices too in their understanding of circuits and scopes.

Please carry on.