what an oscilloscope recommended for a woman passionate about electronics?

[CharlotteSwiss]

I think the Gate control on your scope does the same thing as my cursors and the only way you know I applied cursors (Start Time and End Time on the Rigol) is by the fact that the blue trace only exists during integration.

that's right, gate allows me to have a range of the integral. Here in the image the usual square wave Vpp 3v; I have a sec / div of 100us, so the affected voltage rectangle is 3V * 500us: we can therefore say that the integral in question is 1500uV * s
(and in fact the scale of the integral is 500uV * s, as I marked in red it occupies 3 div, and therefore a total of 1500uV * s)
I hope I got us this time
thanks 

[CharlotteSwiss]

I decided I wasn't happy with the previous integration examples so I made another one.  This time I used cursors to limit the area over which I would integrate (forming a definite integral) and I presented a 1 kHz square wave with 1Vp-p and 500 mV offset such that the signal varies between 0V and 1V.  The integration period is just 500 us so the area under the rectangle is 1V * 500 us or 500 uV s.  Reading from the right edge of the blue trace, we see it is at 5 divisions of 100 V us or 500 uV s.    The value at the far right on the integral does, in fact, give the area under the square wave over the limited period of time.  I love it when a plan comes together.

I saw that the integral has a different trend on the same signal: in the attached image, in the upper part the square wave signal is av coupling, and I have an integral rises / falls / rises / falls ...
In the lower part is the same signal, but DC coupling, therefore it includes the dc offset, and the integral has a trend that rises/ floor / rises / floor ...
Is there an easy explanation for this diversity?

Then I can move on to the last math .. square (then I think the studio is downhill...)
 

[rstofer]

That's because your square wave is non-symmetric about the 0V axis.  When it should be discharging, the logic low value is actually 0V and this neither adds nor subtracts from the integration.  So, it charges, waits, charges, waits, etcl  This is caused by DC coupling and an asymmetric square wave.

[CharlotteSwiss]

I wasn't happy with the previous discussion on the integral of sin(x).  First, we know it should look like a negative cosine but, over the definite integral period of 0 to 2*pi, the integral is 0.  Take a symmetric sine wave and look at the area above the X axis and below the X axis.  They are equal and opposite sign so they cancel each other out.  At multiples of 2pi, the integral is 0.

So, here is a picture of a 1 kHz 1Vp-p sine wave and a definite integral over 1 period.  Note that the value at the far edge of the integral is 0 - just exactly what we expect.  The high point of the -cos(t) occurs when the sin(t) gets back down to 0V - that is because, up until that time, we were still pouring water in the bucket.  Once the sin(t) turns negative, the water starts getting drained out.

I like this "Start Time" and "Stop Time" on the integration function.  It makes everything match the math.

in practice, the integral in question shows us that when the wave is positive we have a rise in the integral, and then descend when the wave is negative and so on; initially I struggled to understand, now it is clearer
 
thanks rstofer

[CharlotteSwiss]

ok I should have understood the logic of the integral, tomorrow I calmly rearrange the things learned in my open writer, then I move on to math-square.
Now I can finally sleep in my bed after two nights at work. See you tomorrow rstofer, good evening
 

[Mechatrommer]

I just noticed your Malaysia flag.  Back in '88-'89, I was living and working in Singapore building a semiconductor plant.  I took up SCUBA diving and made 13 trips to Malaysia that year - several to the resort on Pulau Tioman but more often just sleeping on the beach at Pulau Aur and Pulau Dayang...

if too much lead accumulated in your head, just in case... https://www.trip.com/blog/guide-to-perfect-islands-semporna-malay/ you can drop me a msg i'm here i can help you to look around. we have lots of tourists visit before covid, but none "oldman scam" i ever heard of ...

[CharlotteSwiss]

I rearranged the ideas on the integral mathematical function of the oscilloscope, I divide into two posts for the usual confusion speech on the attached images! I can say I have two doubts, one in this post and one in the next!
In this first attached image, I have rearranged the three phases of the integral relative to the compensation square wave signal; not, I used the signal in AV coupling for the ramp and the descent, so I could also have the signal negative and therefore the ramp down: I instead used the channel in dc coupling to be able to have the zero volt flat integral
I isolated the three phases using the range gate.
if I understood correctly, when the integral ramp is rising it means that the signal is emitting positive voltage; when the ramp is down, it means that the signal is emitting negative voltage; when the ramp is horizontal, at that moment the signal is at rest = zero volts.
It is obvious that if the signal were AC coupling, the integral part would not be flat as it would tansit only for an instant on zero volts ..
I only have a doubt about the unit of measurement: if we look at the rising ramp, we see that the wave area of ​​the integral measures 3V vertically and 500uS horizontally: so we have 3 divs of 1V * 500uS: I see that the scale is indicated in xxxuV * S. Wouldn't it be more accurate to indicate xxxV * uS? (my doubt would be: u is indicated at the time, not the voltage right?)
 

[CharlotteSwiss]

The only thing I didn't understand, is the integral menu offset setting.
The siglent manual on p. 125 says:

The integrate operator provides an Offset softkey that lets you enter a DC offset correction factor for the input signal. Small DC offset in the integrate function input (or even small oscilloscope calibration errors) can cause the integrate function output to "ramp" up or down. This DC offset correction lets you level the integrate waveform.

In the attached image, where you see the integral of the zero volt stretch of the signal, you can see how the offset setting affects (it has a range of + -200mV): in the image I tried both +200 and -200, originally the signal was quite flat.
I didn't understand what it could be for; perhaps in the case of signals with offset, this setting could be used to improve the integral? For example, if the zero volt part were to ramp slightly, could I use offset to make it flat?
thanks 

[rstofer]

If there is a little DC offset in the incoming square wave, the overall integration, over several cycles, will tend to increase or decrease.  Earlier, you had a square wave that showed perfect integration.  That is, it always returned to the lower limit.  If the signal itself needs a little help to prevent it from continually increasing or decreasing, that's what the Offset setting is for.

[rstofer]

Lacking a good source of sine waves, I don't think you can replicate this final experiment so let's walk through the math because I found it interesting.

We know that the integral of sin(x) over the range 0 to 2 pi is exactly 0 because the upper lobe has the same area as the lower lobe (which is negative) and they cancel.  But what about over the range 0 to pi - just the upper lobe.  We know that the integral of sin(x) over that range is 2 (kind of a strange answer).  Just ordinary 2 - no pi's, no weird functions, just 2.  Symbolab.com will do that definite integral.

Suppose we draw a rectangle surrounding that lobe with a sine wave with an amplitude of 1 (Vpeak if you wish) and extending along the X axis from 0 to pi (.3.14).  The area of that rectangle is 3.14 and fully encompasses the sine wave lobe with an area of 2.  So, the sine wave fills up 0.637 of the rectangle and that is esactly the integral of that lobe: 0.637 units (V ms, or whatever).  2 / 3.14159 -> 0.637.  2 is the area of the lobe, 3.14159 is the area of the rectangle enclosing the lobe.

It took me a while to tumble to this solution but, voila', the scope gets the same answer.  The settings aren't exact but here is a scope image using the integral function over that limited range.  The measurement we're looking for is AY = 0.636 V.  Actually, it took more time to tumble to the math than it did to get the scope display.

To make the numbers neat, I used 2 Vp-p and 500 Hz.  That way the upper lobe is 1 ms wide and 1V high.

I have never used the integration function on a scope.  Dave has a video that shows a splendid use:

[rstofer]

I only have a doubt about the unit of measurement: if we look at the rising ramp, we see that the wave area of ​​the integral measures 3V vertically and 500uS horizontally: so we have 3 divs of 1V * 500uS: I see that the scale is indicated in xxxuV * S. Wouldn't it be more accurate to indicate xxxV * uS? (my doubt would be: u is indicated at the time, not the voltage right?)
 

One way or the other but the unit is V s (volts seconds) and the scope chose to keep the time in terms of seconds and uV and V are the same kind of thing, just the decimal point is moving around.  uV seconds makes more sense to me than Volts microseconds.  Both are equally accurate, one is probably more in compliance with the idea.

[CharlotteSwiss]

If there is a little DC offset in the incoming square wave, the overall integration, over several cycles, will tend to increase or decrease.  Earlier, you had a square wave that showed perfect integration.  That is, it always returned to the lower limit.  If the signal itself needs a little help to prevent it from continually increasing or decreasing, that's what the Offset setting is for.

maybe I figured out what offset is for: looking at my attached image, it is the usual square wave with av coupling, so quite pure from DC; in this case I have the high peaks and low peaks of the integral all at the same level; if I had peaks at different heights, then I could have used the offset to balance the integral peaks

[CharlotteSwiss]

Lacking a good source of sine waves, I don't think you can replicate this final experiment so let's walk through the math because I found it interesting.

We know that the integral of sin(x) over the range 0 to 2 pi is exactly 0 because the upper lobe has the same area as the lower lobe (which is negative) and they cancel.  But what about over the range 0 to pi - just the upper lobe.  We know that the integral of sin(x) over that range is 2 (kind of a strange answer).  Just ordinary 2 - no pi's, no weird functions, just 2.  Symbolab.com will do that definite integral.

Suppose we draw a rectangle surrounding that lobe with a sine wave with an amplitude of 1 (Vpeak if you wish) and extending along the X axis from 0 to pi (.3.14).  The area of that rectangle is 3.14 and fully encompasses the sine wave lobe with an area of 2.  So, the sine wave fills up 0.637 of the rectangle and that is esactly the integral of that lobe: 0.637 units (V ms, or whatever).  2 / 3.14159 -> 0.637.  2 is the area of the lobe, 3.14159 is the area of the rectangle enclosing the lobe.

It took me a while to tumble to this solution but, voila', the scope gets the same answer.  The settings aren't exact but here is a scope image using the integral function over that limited range.  The measurement we're looking for is AY = 0.636 V.  Actually, it took more time to tumble to the math than it did to get the scope display.

I follow part of the speech but not all: ok the distance on the X axis for the half period is 3.14, ok the height Vp is 1V
The rectangle containing the wave will be: 3.14 * 1 = 3.14
I don't understand the passage 2 / 3,13 (2 ok should be the area of the positive wave only? .. but why do this division?)
  the result (0.636 what does it represent?)

edit:
now I try to do the same thing with my oscilloscope, and see if I also have that 0.636 in the measurements ..

i can't do the same exercise, the online sine wave has no Vp1v. However, I understood what the measure ay in your example is: it is the height div of the integral

final edit:
I understood what you meant by 0.636: it is the part represented by the lobe of the sinusoid inside the rectangle.
Let's say that it is enough for me to know that the value of the integral is indicated by the oscilloscope with V * s / div (in fact your 0.636 are represented by 3div of 200uV * s plus a small piece)

[CharlotteSwiss]

I only have a doubt about the unit of measurement: if we look at the rising ramp, we see that the wave area of ​​the integral measures 3V vertically and 500uS horizontally: so we have 3 divs of 1V * 500uS: I see that the scale is indicated in xxxuV * S. Wouldn't it be more accurate to indicate xxxV * uS? (my doubt would be: u is indicated at the time, not the voltage right?)
 

One way or the other but the unit is V s (volts seconds) and the scope chose to keep the time in terms of seconds and uV and V are the same kind of thing, just the decimal point is moving around.  uV seconds makes more sense to me than Volts microseconds.  Both are equally accurate, one is probably more in compliance with the idea.

ok, this point is now clear; however, I must always keep in mind that the unit of measurement of the integral is volts * seconds, that is, it represents the voltage strength that that piece of signal is capable of proving where it hits
more or less give, that's the idea
 

[CharlotteSwiss]

come on I can close with the integral, I understand in full lines, I just have to write down the things learned; or I pass to the last function math: square
Note: the paragraph math was a real ordeal for my poor little head, I could have given up, but I resisted
 

[rstofer]

i can't do the same exercise, the online sine wave has no Vp1v. However, I understood what the measure ay in your example is: it is the height div of the integral

Can you do 2 Vp-p?  That's what I did: 2 Vp-p, 500 Hz.

final edit:
I understood what you meant by 0.636: it is the part represented by the lobe of the sinusoid inside the rectangle.
Let's say that it is enough for me to know that the value of the integral is indicated by the oscilloscope with V * s / div (in fact your 0.636 are represented by 3div of 200uV * s plus a small piece)

You got it and you're now in a very small percentage of scope users who have EVER messed with this stuff.  Most often we just look at squiggly lines!

[rstofer]

come on I can close with the integral, I understand in full lines, I just have to write down the things learned; or I pass to the last function math: square
Note: the paragraph math was a real ordeal for my poor little head, I could have given up, but I resisted
 

I have learned a lot in the last few weeks!

I have NEVER messed with this stuff, it didn't exist (for me) until I bought a DSO and, even then, all I did was squiggly lines and protocol decode.  Messing around with the FFT, diff() and integral functions has been a real eye-opener.

Maybe this thread will provide more opportunities to experiment:
https://www.eevblog.com/forum/testgear/what-you-do-with-math-channels-on-oscilloscope/

[CharlotteSwiss]

i can't do the same exercise, the online sine wave has no Vp1v. However, I understood what the measure ay in your example is: it is the height div of the integral

Can you do 2 Vp-p?  That's what I did: 2 Vp-p, 500 Hz.

final edit:
I understood what you meant by 0.636: it is the part represented by the lobe of the sinusoid inside the rectangle.
Let's say that it is enough for me to know that the value of the integral is indicated by the oscilloscope with V * s / div (in fact your 0.636 are represented by 3div of 200uV * s plus a small piece)

You got it and you're now in a very small percentage of scope users who have EVER messed with this stuff.  Most often we just look at squiggly lines!

i tried again with sine wave online, before i had 1Khz, now i did as you suggested 500Hz. (see attached image)
Raising the audio output volume of the pc almost to the limit, ok I got about 2Vpp (unstable, but about)
Ok by doing so in fact the value of the integrated as you can see is 600uV * S (not 636, but as mentioned also the signal was unstable in amplitude ..and the freq is 501, 500 is not select)
In this way, without bothering the calculator, using the oscilloscope I was able to know that that sine wave is capable of a force of 600uV * s where it affects

on the second consideration I was just thinking about it: how many other users would waste so much time in fully understanding the tool? maybe not too many; but I am quite a fussy and stubborn woman; the study of this new siglent could not be different
thanks rstofer 

[CharlotteSwiss]

come on I can close with the integral, I understand in full lines, I just have to write down the things learned; or I pass to the last function math: square
Note: the paragraph math was a real ordeal for my poor little head, I could have given up, but I resisted
 

I have learned a lot in the last few weeks!

I have NEVER messed with this stuff, it didn't exist (for me) until I bought a DSO and, even then, all I did was squiggly lines and protocol decode.  Messing around with the FFT, diff() and integral functions has been a real eye-opener.

Maybe this thread will provide more opportunities to experiment:
https://www.eevblog.com/forum/testgear/what-you-do-with-math-channels-on-oscilloscope/

come on I'm glad the lesson was useful for you too; after all there is always something to learn / experience; helps to stay young ;-)
 

[rstofer]

Nice integration!  If covers exactly the interval of interest  If you were to integrate over 2 pi, you would expect the integral to look like -cos(x).  The first half stays the same but the second half (between pi and 2 pi) comes back down and, at 2 pi, the integral is 0 and this is as expected.

This stuff is close to magic but I doubt that many people play with it.  Sure, the pro's probably use it from time to time but I'll bet there aren't many hobbyists that spend time trying to correlate math and a scope trace.  Math is hard!

 It's been interesting!

[CharlotteSwiss]

Nice integration!  If covers exactly the interval of interest  If you were to integrate over 2 pi, you would expect the integral to look like -cos(x).  The first half stays the same but the second half (between pi and 2 pi) comes back down and, at 2 pi, the integral is 0 and this is as expected.

This stuff is close to magic but I doubt that many people play with it.  Sure, the pro's probably use it from time to time but I'll bet there aren't many hobbyists that spend time trying to correlate math and a scope trace.  Math is hard!

 It's been interesting!

now I take a ride with my wolf dog; when I come back in 1 hour I try to do the integral of 2pi (one period), so I really see what happens.

Meanwhile, in parallel to the integral, I am freeing neurons to understand just the basics of the sinusoid; I used to see it as a wave, but now I want to understand it more; curiosity has come to me since you introduced me to Desmos.com.
By now I should have understood that, in a Cartesian graph, the sinusoid is represented by sin (x); the width of the half wave is pi, the width of the period is 2pi.
Always in a Cartesian plane, sin (x) passes through the zero point and rises ...
Maybe I just need to write these rules for now and stop here, it might be enough ..

[rstofer]

At Desmos, plot sin(x)

Then increase the amplitude by plotting 2 * sin(x) + 3.  The +3 is just an offset to move the waveform away from the baseline
Note that the 2 out front is the amplitude (peak) of the wave.  The waveform is 2 * the amplitude from peak to peak

Third, plot sin(2*x).  This doubles the frequency and I shifted it up 6 units.

Finally, plot sin(x+ pi/4) + 8  again the 8 is a shift and unimportant.  But notice that the waveform is now leading sin(x) by 45 degrees (pi / 4)

There's a lot to learn about trig functions at Desmos.  You can plot sin(x) and its integral -cos(x) and possibly shift the -cos(x) plot up a couple of units.  Or, just leave it at the same baseline.

[CharlotteSwiss]

always on the integral ...

I made these 3 measurements of three different time spaces of the sine wave, looking at the attached image:
  -above a half-period (this measurement is the same as the one already performed before): we can say that the value of the integrated is about 600uV * s.

- in the center 3/4 of a period: in this case the integral always has the ramp of value 600, but from which the half descent of 300 must be removed, for a total integrated value of 300uV * s

- at the bottom the whole period: here the ramp is added to the descent, and being equal in value, it is obvious that the integral is zero

I hope I have said the right ... 

[rstofer]

Perfect!  Until we come up with a real application, I'm willing to move on from the integral.  It's nice that the math and the scope agree.

[CharlotteSwiss]

At Desmos, plot sin(x)

Then increase the amplitude by plotting 2 * sin(x) + 3.  The +3 is just an offset to move the waveform away from the baseline
Note that the 2 out front is the amplitude (peak) of the wave.  The waveform is 2 * the amplitude from peak to peak

Third, plot sin(2*x).  This doubles the frequency and I shifted it up 6 units.

Finally, plot sin(x+ pi/4) + 8  again the 8 is a shift and unimportant.  But notice that the waveform is now leading sin(x) by 45 degrees (pi / 4)

There's a lot to learn about trig functions at Desmos.  You can plot sin(x) and its integral -cos(x) and possibly shift the -cos(x) plot up a couple of units.  Or, just leave it at the same baseline.

ok, now I have learned the instructions to insert to obtain the desired wave: in my example I have the sine wave par excellence sin (x), and another with double frequency (offset to separate it ..)
As we know sin (x) has in the Cartesian a Vp of 1 and a T of 2pi
So we can say that a sine wave with the same waveform as sin (x), must have a division value between Vp and T of 0.159 (1 / 2pi)
Does this reasoning make sense?