what an oscilloscope recommended for a woman passionate about electronics?

[CharlotteSwiss]

I was trying to understand the channel's DC and AC coupling: DC coupling lets pass dc and ac signal; AC coupling blocks the DC part of a signal.
For example, here are the two coupling modes on the siglent compensation signal: with DC coupling we have a square wave 0- 3v, while with AC coupling we have a square wave + 1.5V / -1.5V.






Two questions to understand:
1) to measure the real signal, DC coupling must be used. With AC coupling the signal is filtered (however in the circuit it is not filtered). So what use can AC coupling have?
2) in the DC coupling square wave compensation signal of the photo above, is the DC part of the signal represented by the low horizontal signal of 0V? Or is there no part of DC in this signal? (if there was a part of DC, would the square wave not start higher than the 0V?
thamks 

[tautech]

An example where you need AC coupling might be for an audio signal with a DC offset.

Here below is 1 KHz sinewave with a 1V positive offset but the sinewave is just 5mV in amplitude.
At first glance it's just a DC voltage yet if we increase sensitivity to better inspect it the waveform steps off the display so AC coupling is engaged to remove the DC component then sensitivity increased to see the signal.

[tggzzz]

I was trying to understand the channel's DC and AC coupling: DC coupling lets pass dc and ac signal; AC coupling blocks the DC part of a signal.
For example, here are the two coupling modes on the siglent compensation signal: with DC coupling we have a square wave 0- 3v, while with AC coupling we have a square wave + 1.5V / -1.5V.

That is correct for the square waves you are using. Now, if possible, repeat this with the same peak-to-peak amplitude but a 20% and 80 % duty cycle.

Two questions to understand:
1) to measure the real signal, DC coupling must be used. With AC coupling the signal is filtered (however in the circuit it is not filtered). So what use can AC coupling have?

Not all signals have DC components. The classic example is audio, which is from approx 30Hz to 20kHz.

It can be much easier to create amplifiers that don't have to respond to DC, since that means you can avoid offset and saturation issues. A classic example is accurately measuring low level DC signals, say 1uV, when practical amplifiers have much higher offset voltages. If you amplify the signal enough to be able to measure it, the amplifier output would be saturated against a supply rail. The solution is to "chop" the input signal, i.e. convert it to AC, amplify that without amplifying the DC offset, and measure the AC signal.

For simple measurement with a scope, consider being interested in the ripple on a power supply rail. You might be interested in seeing <100mV noise superimposed on a 10V rail, and that can't be seen when DC coupled since it would be at best 1% of the display's height. AC coupling "removes" the 10V, allowing the 100mV to fill the display.

2) in the DC coupling square wave compensation signal of the photo above, is the DC part of the signal represented by the low horizontal signal of 0V? Or is there no part of DC in this signal? (if there was a part of DC, would the square wave not start higher than the 0V?
thamks 

The DC part is the average (mean) value. Try changing the duty cycle to 20%/80%, and see what happens.

That raises the question of how long do you have to average. A handwaving answer is that if you average for 1s then AC components above 1Hz will be removed, ditto 0.1s and 10Hz.

In practice the removal isn't that sharp, but often if you average for 10 times the longest AC signal period you will be OK. That's equivalent to the cutoff frequency being 1/10th that of your lowest signal frequency.

[CharlotteSwiss]

An example where you need AC coupling might be for an audio signal with a DC offset.

Here below is 1 KHz sinewave with a 1V positive offset but the sinewave is just 5mV in amplitude.
At first glance it's just a DC voltage yet if we increase sensitivity to better inspect it the waveform steps off the display so AC coupling is engaged to remove the DC component then sensitivity increased to see the signal.

So the signal as in your example, is composed of: 1vDC + 5mV AC.
The coupling to be used at the beginning is DC, and we will use AC coupling for convenience in seeing the AC wave enlarged (if we left in DC, we had to move upwards with the position knob)
I hope I understand, thanks Tauthec 

[CharlotteSwiss]

That is correct for the square waves you are using. Now, if possible, repeat this with the same peak-to-peak amplitude but a 20% and 80 % duty cycle.

----

Not all signals have DC components. The classic example is audio, which is from approx 30Hz to 20kHz.
It can be much easier to create amplifiers that don't have to respond to DC, since that means you can avoid offset and saturation issues. A classic example is accurately measuring low level DC signals, say 1uV, when practical amplifiers have much higher offset voltages. If you amplify the signal enough to be able to measure it, the amplifier output would be saturated against a supply rail. The solution is to "chop" the input signal, i.e. convert it to AC, amplify that without amplifying the DC offset, and measure the AC signal.
For simple measurement with a scope, consider being interested in the ripple on a power supply rail. You might be interested in seeing <100mV noise superimposed on a 10V rail, and that can't be seen when DC coupled since it would be at best 1% of the display's height. AC coupling "removes" the 10V, allowing the 100mV to fill the display.

-----

The DC part is the average (mean) value. Try changing the duty cycle to 20%/80%, and see what happens.
That raises the question of how long do you have to average. A handwaving answer is that if you average for 1s then AC components above 1Hz will be removed, ditto 0.1s and 10Hz.
In practice the removal isn't that sharp, but often if you average for 10 times the longest AC signal period you will be OK. That's equivalent to the cutoff frequency being 1/10th that of your lowest signal frequency.

I have looked for the duty cycle setting in the oscilloscope but cannot find it 

----

ok, I understand the example of 10V and 100mV: in AC coupling it is possible to see the beautiful noise wave enlarged to analyze it (since we cut 10V Dc)

-----

So in my example with the square wave in DC coupling, the amplitude is 3v, so can we say we have a DC of 1.5v? (for duty as I said above I don't know how to set it)
the second part of reasoning is currently very complicated for me.
Thank you 

[tautech]

An example where you need AC coupling might be for an audio signal with a DC offset.

Here below is 1 KHz sinewave with a 1V positive offset but the sinewave is just 5mV in amplitude.
At first glance it's just a DC voltage yet if we increase sensitivity to better inspect it the waveform steps off the display so AC coupling is engaged to remove the DC component then sensitivity increased to see the signal.

So the signal as in your example, is composed of: 1vDC + 5mV AC.
The coupling to be used at the beginning is DC, and we will use AC coupling for convenience in seeing the AC wave enlarged (if we left in DC, we had to move upwards with the position knob)
I hope I understand, thanks Tauthec 

Nearly.
As we increase sensitivity to see the AC waveform in DC coupling each vertical division becomes a lesser portion of the total amplitude we need to see so the 1V offset goes off the top of the display and with 200x more sensitivity required to see the 5mV sinewave we cannot get the sinewave displayed easily by just using the Vertical Pos control.
So it is easier to engage AC coupling to remove the DC component (1V offset) and examine the 5mV sinewave.

There are limits as to how far Vertical Pos can be adjusted when using DC coupling and they differ depending on the V/div settings.
You can find these limits listed as channel Offset specs in the datasheet.

There are other tricks to solve DC offsets but we save them for later as it is important to fully understand basic principles.

[rstofer]

The DC part is the average (mean) value. Try changing the duty cycle to 20%/80%, and see what happens.
That raises the question of how long do you have to average. A handwaving answer is that if you average for 1s then AC components above 1Hz will be removed, ditto 0.1s and 10Hz.
In practice the removal isn't that sharp, but often if you average for 10 times the longest AC signal period you will be OK. That's equivalent to the cutoff frequency being 1/10th that of your lowest signal frequency.

I have looked for the duty cycle setting in the oscilloscope but cannot find it 

The idea is to change the duty cycle of the waveform.  Given a higher percentage of on-time, the DC value will be higher.  You can build up a 555 timer circuit that will allow you to change the duty cycle over some range.

https://www.electroschematics.com/pulse-generator-with-555/

Some of the experiments that are suggested require more than just the calibration signal and the 555 is fun when it comes to breadboarding these projects.  It's handy to have a bunch laying around - along with resistors, capacitors and diodes, of course.

I tend to buy things like resistors, capacitors, transistors and diodes in high multiples.  For resistors and capacitors, I almost always buy 100 each.  Transistors and diodes I might just buy 25 or so.  That's one way to build up a stash.

[CharlotteSwiss]

Nearly.
As we increase sensitivity to see the AC waveform in DC coupling each vertical division becomes a lesser portion of the total amplitude we need to see so the 1V offset goes off the top of the display and with 200x more sensitivity required to see the 5mV sinewave we cannot get the sinewave displayed easily by just using the Vertical Pos control.
So it is easier to engage AC coupling to remove the DC component (1V offset) and examine the 5mV sinewave.

There are limits as to how far Vertical Pos can be adjusted when using DC coupling and they differ depending on the V/div settings.
You can find these limits listed as channel Offset specs in the datasheet.

There are other tricks to solve DC offsets but we save them for later as it is important to fully understand basic principles.

this could teach me that when i want to zoom in on a wave that has DC offset, the best thing is to set AC coupling, thanks 

[CharlotteSwiss]

The idea is to change the duty cycle of the waveform.  Given a higher percentage of on-time, the DC value will be higher.  You can build up a 555 timer circuit that will allow you to change the duty cycle over some range.

https://www.electroschematics.com/pulse-generator-with-555/

Some of the experiments that are suggested require more than just the calibration signal and the 555 is fun when it comes to breadboarding these projects.  It's handy to have a bunch laying around - along with resistors, capacitors and diodes, of course.

I tend to buy things like resistors, capacitors, transistors and diodes in high multiples.  For resistors and capacitors, I almost always buy 100 each.  Transistors and diodes I might just buy 25 or so.  That's one way to build up a stash.

now I understand, the cyrcle duty is not a setting of the oscilloscope, but of who generates the signal. Cute that circuit with 555: now I have to learn this siglent well, but in the future a 555 or an Arduino could be a new game for me.
I have hundreds of capacitors, led, resistors; silicon diodes also
thanks rstofer 

[rstofer]

The Arduino is a great device for generating waveforms.  Look for the PWM examples.

[tautech]

now I understand, the cycle duty is not a setting of the oscilloscope, but of who generates the signal.

Yep, correct.

We can select from the Measurements menu duty cycle measurements of both positive and negative duty cycle.
Of course measurement of a sine or square wave should return a 50% result and pulsed waveforms are where we might need precise duty cycle measurement for timing purposes.

Eg: a 4V 20% DC coupled pulse with -2V offset of the 0V channel marker.

[ArthurDent]

I haven’t posted in this thread recently because you are getting plenty of good suggestions from all the other posters who are replying to your questions, but I am following all the back and forth with interest.

My advice is when you have a little spare time, start at the first post and quickly read through all the posts. What you should notice is not how little you thought you knew in your first post but how far you’ve advanced in such a short time. You’re open to suggestions but don’t accept information without questioning it until you really understand it, which is good. 

You’re making great progress and now if we can just convince you to buy some more test equipment…. 

[rstofer]

Here are a couple of AD2 waveforms, one at 20% and the other at 80%, AC coupled.  It'a probably not easy to see and calculate but the area above the 0V reference is the same as the area below the reference.  At 20%, the height above the reference is quite high and the depth below the reference is quite small.

[CharlotteSwiss]

The Arduino is a great device for generating waveforms.  Look for the PWM examples.

yes, I have seen that arduino is excellent for generating signals 

Here are a couple of AD2 waveforms, one at 20% and the other at 80%, AC coupled.  It'a probably not easy to see and calculate but the area above the 0V reference is the same as the area below the reference.  At 20%, the height above the reference is quite high and the depth below the reference is quite small.

I also did the 20-80 duty cyrcle test with the software. In practice it seems to me that if we set 20%, we will have 20% of the signal with positive voltage, while 80% of the signal at zero volts; while if set 80% we will have the reverse.

[CharlotteSwiss]

We can select from the Measurements menu duty cycle measurements of both positive and negative duty cycle.
Of course measurement of a sine or square wave should return a 50% result and pulsed waveforms are where we might need precise duty cycle measurement for timing purposes.

ok understood, in fact the square wave of compensation of the single is calibrated at 50% cyrcle, as the high and low signal have the same execution time

[CharlotteSwiss]

I haven’t posted in this thread recently because you are getting plenty of good suggestions from all the other posters who are replying to your questions, but I am following all the back and forth with interest.

My advice is when you have a little spare time, start at the first post and quickly read through all the posts. What you should notice is not how little you thought you knew in your first post but how far you’ve advanced in such a short time. You’re open to suggestions but don’t accept information without questioning it until you really understand it, which is good. 

You’re making great progress and now if we can just convince you to buy some more test equipment…. 

thanks arthur for your clarification; I try to ask in this discussion, only the things I can't understand while studying the oscilloscope manual. For many people my questions seem trivial, but it is not easy to understand such a complex tool the first time
 

[tggzzz]

The Arduino is a great device for generating waveforms.  Look for the PWM examples.

yes, I have seen that arduino is excellent for generating signals 

Here are a couple of AD2 waveforms, one at 20% and the other at 80%, AC coupled.  It'a probably not easy to see and calculate but the area above the 0V reference is the same as the area below the reference.  At 20%, the height above the reference is quite high and the depth below the reference is quite small.

I also did the 20-80 duty cyrcle test with the software. In practice it seems to me that if we set 20%, we will have 20% of the signal with positive voltage, while 80% of the signal at zero volts; while if set 80% we will have the reverse.

I'm not familiar with that simulation software, but it looks like you have the scope set to DC coupling.

I would connect the generator output to both A and B channels, and ensure both channels have the same vertical sensitivity (say 5V/div) and offset (0V). Make channel A DC coupled, and channel B AC coupled. See the how the two displays do/don't change as you change the duty cycle from 20-50-80%. Remember that AC coupling removes the DC component.

Better still, do it on your real scope

[CharlotteSwiss]

I'm not familiar with that simulation software, but it looks like you have the scope set to DC coupling.

I would connect the generator output to both A and B channels, and ensure both channels have the same vertical sensitivity (say 5V/div) and offset (0V). Make channel A DC coupled, and channel B AC coupled. See the how the two displays do/don't change as you change the duty cycle from 20-50-80%. Remember that AC coupling removes the DC component.

Better still, do it on your real scope

for now I don't have a real function generator, I take it as soon as I have learned well with the oscilloscope.
I did the test you recommended: the difference between 20-50-80 on horizontal times I understood it; I don't understand with 20 and 80 the difference of the waves on the vertical scale (at 50 they overlap)
 

[tggzzz]

I'm not familiar with that simulation software, but it looks like you have the scope set to DC coupling.

I would connect the generator output to both A and B channels, and ensure both channels have the same vertical sensitivity (say 5V/div) and offset (0V). Make channel A DC coupled, and channel B AC coupled. See the how the two displays do/don't change as you change the duty cycle from 20-50-80%. Remember that AC coupling removes the DC component.

Better still, do it on your real scope

for now I don't have a real function generator, I take it as soon as I have learned well with the oscilloscope.
I did the test you recommended: the difference between 20-50-80 on horizontal times I understood it; I don't understand with 20 and 80 the difference of the waves on the vertical scale (at 50 they overlap)

Good; that is what I would expect to see.

The DC component is the average/mean value of the waveform, mathematically the integral of the value over one cycle. For an arbitrary waveform you would need to take a very large number of calculations, but for a square wave  the calculations are very simple - that's why it is a good teaching example. 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.

The DC coupling the trace shows the waveform directly. With AC coupling the trace is that waveform minus V_m. That's all the info you need to calculate the traces you see.

An alternative "hand waving" explanation is that the average value of an AC coupled signal has to be zero. If you remember that the mathematical integral is the area under the curve and eyeball the traces for the AC coupling, you can visually see the effect noted ...

It'a probably not easy to see and calculate but the area above the 0V reference is the same as the area below the reference.

[CharlotteSwiss]

I'm not familiar with that simulation software, but it looks like you have the scope set to DC coupling.

I would connect the generator output to both A and B channels, and ensure both channels have the same vertical sensitivity (say 5V/div) and offset (0V). Make channel A DC coupled, and channel B AC coupled. See the how the two displays do/don't change as you change the duty cycle from 20-50-80%. Remember that AC coupling removes the DC component.

Better still, do it on your real scope

for now I don't have a real function generator, I take it as soon as I have learned well with the oscilloscope.
I did the test you recommended: the difference between 20-50-80 on horizontal times I understood it; I don't understand with 20 and 80 the difference of the waves on the vertical scale (at 50 they overlap)

Good; that is what I would expect to see.

The DC component is the average/mean value of the waveform, mathematically the integral of the value over one cycle. For an arbitrary waveform you would need to take a very large number of calculations, but for a square wave  the calculations are very simple - that's why it is a good teaching example. 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.

The DC coupling the trace shows the waveform directly. With AC coupling the trace is that waveform minus V_m. That's all the info you need to calculate the traces you see.

An alternative "hand waving" explanation is that the average value of an AC coupled signal has to be zero. If you remember that the mathematical integral is the area under the curve and eyeball the traces for the AC coupling, you can visually see the effect noted ...

It'a probably not easy to see and calculate but the area above the 0V reference is the same as the area below the reference.

thanks tggzzz, in the late evening after dinner I think about it and I hope to solve the mathematical operations.
(I don't guarantee anything, but it's time to learn what the DC value of a voltage is: I always hear that it is the average value, but I never investigated! Now it's time to do it)
I will also try with cyrcle 0 and 100

 

[rstofer]

Time to play with colored pencils...

Color in the 20% portion of the AC triggered signal in one color and color in the 80% with another color.  Color just the portion above the reference in the 20% portion and the portion below the reference for the 80% portion.

I hope this will make it clear that the area above the reference for the 20% portion is exactly the same as the area below the reference for the 80% portion.

This only makes sense for AC coupled images but that's what we're talking about.  We're finding the average of the area above the reference and the area below the reference and that average is zero.

How to calculate the area:  Multiply the voltage of the segment above the reference line by the width in divisions.  Then multiply the voltage of the segment below the reference line by the width in divisions.  They should be equal.

We could say that the area above the reference line is positive and the area below the line is negative.  When added together, they equal zero.  The 'average' is zero...

The portion above (yellow) is 1.6V * 2 divisions for an area of 3.2.  Similarly, the portion below the reference (green) is 0.4V * 8 divisions for an area of 3.2.  They're identical areas!  Yellow is positive, green is negative; add them together and you have 0V average.

That's a good reason for making the time/division work out to exactly 10 divisions.  Two divisions high, eight divisions low.  This is probably a little more difficult, mathematically, when dealing with 12 divisions on the screen.

ETA:  It would be high for 2.4 divisions and low for 9.6 divisions.  Workable but not quite as clean.

ETA:  You could think about cutting the green strip into 4 equal length pieces and laying them over the yellow area.  They should cover the area exactly.  Who says adults can't play with crayons!

ETA:  It's not a coincidence that the voltages are 1.6V high and 0.4V low.  The input signal had an amplitude of 1.0V or 2.0V peak-to-peak.  Hey!  1.6 + 0.4 = 2V peak to peak, high to low.

ETA:  What you have learned in this exercise will not have even occurred to the vast majority of scope users.  This is like an advanced course in displaying signals and analyzing the results.  And we did it without calculus.  Finding the area of a rectangle is pretty easy: length times width.  Other signals are a little more complicated but the idea is the same.

[CharlotteSwiss]

Time to play with colored pencils...

Color in the 20% portion of the AC triggered signal in one color and color in the 80% with another color.  Color just the portion above the reference in the 20% portion and the portion below the reference for the 80% portion.

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 

[CharlotteSwiss]

Good; that is what I would expect to see.
The DC component is the average/mean value of the waveform, mathematically the integral of the value over one cycle. For an arbitrary waveform you would need to take a very large number of calculations, but for a square wave  the calculations are very simple - that's why it is a good teaching example. 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.
The DC coupling the trace shows the waveform directly. With AC coupling the trace is that waveform minus V_m. That's all the info you need to calculate the traces you see.
An alternative "hand waving" explanation is that the average value of an AC coupled signal has to be zero. If you remember that the mathematical integral is the area under the curve and eyeball the traces for the AC coupling, you can visually see the effect noted ...

I understood only the last part, in AC the sum of the positive wave + the negative wave = 0
I don't understand what Vh-Vi and Ph and Pi are, if I understand those I can calculate the Vm of the waves as required
 

[CharlotteSwiss]

perhaps to better understand what the Vm of a DC voltage is, I have to do something simpler.
I measured a simple sine wave, and colored the positive part and the negative part.
They have the same area, therefore their sum is equal to 0V (measuring one period)
Does this teach me that a perfect sine wave has no DC component?
 

[tautech]

Yes and the same is true for any other symmetrical waveform.