Analog scope in X-Y mode

[atferrari]

Analog oscilloscope in X-Y mode.

I am not sure how to quantify this based on the scope bandwith, 5 MHz:

I locate the beam inside a grid of 200x200 values by applying any of 200 possible steps (0 to +3V range) to each axis.

Beam initially located at (50,50). My questions:

a) How long it will take to get the beam at (150,150)?
b) How long to reach 0,150?
c) How long to reach 150,0?
d) How long for the three cases above if I start at (0,0)?

I do not have my analog scope with me right now but spent some time wondering how could I actually measure this by myself. (I am affraid it is way too abvious).

 

[cowboybob]

Re-thinking this...

 

[ronsimpson]

scope is at 50,50 and you want to move it to 150,150. So the movement will be +100, +100.
questions? I do not know what type of answer you are looking for.

1) "200 steps" It will take 100 steps. If you can make a X step at the same time as you make a Y step then it will take 100 steps in time.

2) "5 mhz" So you scope will move to with in 80% of the way on 70nS. It will be very close by 200nS. (assuming the X and Y channels are both 5mhz) If your question is about scope speed, then it will not matter if the scope is moving 3 volts or 1 volt or 0.3 volts it will take the same amount of time.

 

[cowboybob]

Analog oscilloscope in X-Y mode.

I am not sure how to quantify this based on the scope bandwith, 5 MHz:

I locate the beam inside a grid of 200x200 values by applying any of 200 possible steps (0 to +3V range) to each axis.

Beam initially located at (50,50). My questions:

a) How long it will take to get the beam at (150,150)?
b) How long to reach 0,150?
c) How long to reach 150,0?
d) How long for the three cases above if I start at (0,0)?

I do not have my analog scope with me right now but spent some time wondering how could I actually measure this by myself. (I am affraid it is way too abvious).

Bandwidth is only an issue if either the X or Y signal frequency exceeds 5Mhz.

Does either axis in your grid arrangement have a known time division?

 

[ronsimpson]

Does either axis have a known time division?

In XY mode there is no time.

Bandwidth is only an issue if either the X or Y signal frequency exceeds 5Mhz.

I don't understand the question so I don't know if the scope bandwidth is the question.

 

[atferrari]

Sorry Ron, as a non native I believed that with "how long" I was asking "how much time".

How many units of time after applying the new pair of voltages the beam will be at destination.

Voltages are applied both at the same time.

 

[Nigel Goodwin]

I do not know what type of answer you are looking for.

I know what you mean - I don't even understand the question, never mind what kind of answer he's looking for.

He really needs to post some context for the question to give us a clue.

But essentially X-Y mode is 'comparing' the frequencies of the two signals.

cowboybob is quite correct that bandwidth makes little difference, unless he's comparing RF frequencies - and I can't say I've ever had occasion to use lissajous with RF?.

 

[Nigel Goodwin]

Sorry Ron, as a non native I believed that with "how long" I was asking "how much time".

How many units of time after applying the new pair of voltages the beam will be at destination.

As fast as the voltages change (as near as makes no odds - and FAR, FAR faster than you can see).

 

[MrAl]

Hello,

To get the position from one place to anther you have to know the rise time. The vertical rise time is a different spec than the bandwidth, and you would get this when you buy the scope as part of the specs.

I am not sure however if there would be a separate spec for the horizontal time, usually the sweep is linear and goes all the way across one division in the time it takes depending on the horizontal sweep setting. It might not be as good as the vertical because it doesnt usually have to go as fast, but i cant say for sure until checking the manufacturers data.

So this would be a two RC time constant problem, and if you know how to do the vertical then you know how to do the horizontal, and that will give you the new position time.

Vertical rise times are typically 10ns, 20ns, or faster. For say 10ns that would mean that the vertical trace can move from 0 to 90 percent of the measured voltage in 10ns if it is a square wavefront. So if you wanted to you could do the math:
0.9*V=V*(1-e^(-t/RC))

and solve for RC:
RC=t/2.3

and then you can go back to the original equation:
Vnew=Vstep*(1-e^(-t/RC))

where Vstep is the step voltage you apply to the vertical input, and Vnew is the actual voltage, and of course these voltages translate to physical position through the graticule setting and an actual distance measurement of the graticule height. You could then solve for the time.
You would do the same thing for the horizontal, and that would lead to your new x,y position, and again solve for the time.
Since one time will probably be longer than the other, you could probably just use one RC equivalent to get the time, as long as it was the slowest RC (probably the horizontal), if the two distances were always to be the same. If different, then you'd have to just use the slowest time which could be either.

A slightly better estimate for RC is:
RC=t/2.2

because the rise time is actually from 10 percent to 90 percent, so we have to find the time between those two levels not from 0 to 90.

An interesting question is of course, why do you want to know this information? Do you intend to make the scope display pictures or characters or something like that?

 

[ronsimpson]

X-Y on my scope; One channel moves the beam up/down while the other channel moves the beam left/right.
So if both channels are +3V the beam is in the upper right corner.
Here is a picture using XY where volts and current are displayed.

Another picture using XYZ.
**broken link removed**

 

[atferrari]

Here we go.

No Lissajous, no frequency comparisons.

In a test made long time ago, drawing those symbols ("L" letters if you like), I got some on the leftmost position, notoriously distorted.

Symbols are drawn as in normal text lines from left to right, from top to bottom, one "character" at time (built spot after spot).

Knowing that, you will see that the distorted ones are preceded by a a character on the righmost position in the previous line. Otherwise, no distortion.

I asumed that the beam was arriving late for the first spot in the 1st char of the new line (left bottom spot in the character). In other words, code seemed to be faster than the beam. I still feel it is true.

I recall having solved this, brute force style, applying a 19-NOPs delay (16MHz PLL clock 18F micro) somewhere in the code. I believed that, in that way I was actually "waiting for the beam". Sole thing I can say, the symbols came OK after that.

Since I decided to discard the previous one before reworking anything, I want to know what to look at.

Gentlemen...?

 

[Nigel Goodwin]

Well that's rather a different story to what you've posted previously

Presumably you have a Z input on your scope then?, and you're using that to turn the beam off when you don't want an area to light up? - because just X/Y can only draw a continuous line.

So how are you getting characters all over the screen?, without the earlier ones fading? - scopes are normally fairly short persistence.

I don't think your problem is the speed of moving the beam, as it's extremely fast, but more a question of you're doing so much moving in order to create a display like this - however, the Z input could be pretty slow, as it's probably not really made for fast signals.

 

[Tony Stewart]

This is a question about settling time to within 1% of target and not about rise time of 1/(1-e) or 63% of the rated slew rate for a given bandwidth. Often this settling time for a 1st order filter is 3x the rated rise time t(99%) = 3 x t (63%) of target


If you are drawing vectors with long retrace from last character (end of row) to first character on next row, them you must allow more blank time for this retrace in order for the 1st "L" in the row to be drawn vertical and straight.

1) It looks like you have to add more NOPs for the wait time for vector draw after large retrace vectors from far right to far left to the 1st character position.

2) It also looks like some bright dots have excessive dwell time before blanking.

3) the blanking bandwidth looks adequate as the dots are "fairly crisp"

4) there is some circular distortion perhaps maybe it is in the photograph but if in the display ...
it means it will need magnetic pincushion adjustment in the corners to flatten the outer rows and straighten the outside columns.

 

[Nigel Goodwin]

it means it will need magnetic pincushion adjustment in the corners to flatten the outer rows and straighten the outside columns.

Makes me think of Trinitron

 

[atferrari]

Well that's rather a different story to what you've posted previously

No story but, basically I posted a multiple-points question, Nigel. The handling of voltages is as described.

Presumably you have a Z input on your scope then?, and you're using that to turn the beam off when you don't want an area to light up? - because just X/Y can only draw a continuous line.

So how are you getting characters all over the screen?, without the earlier ones fading? - scopes are normally fairly short persistence.

Yes Z input and nothing else. No fancy A gate or B gate outputs like some Teks.

I get them all over simply by building, dot by dot, one char after the other and once the screen is done, repeating the whole business.

I don't think your problem is the speed of moving the beam, as it's extremely fast, but more a question of you're doing so much moving in order to create a display like this - however, the Z input could be pretty slow, as it's probably not really made for fast signals.

Yes after so many saying that, I realized that I've put the focus (no pun intended) on how fast and not on when it settles. I think I learnt something.

Gracias.

 

[atferrari]

This is a question about settling time to within 1% of target and not about rise time of 1/(1-e) or 63% of the rated slew rate for a given bandwidth. Often this settling time for a 1st order filter is 3x the rated rise time t(99%) = 3 x t (63%) of target.

OK. It seems that that is the core of my problem.

If you are drawing vectors with long retrace from last character (end of row) to first character on next row, them you must allow more blank time for this retrace in order for the 1st "L" in the row to be drawn vertical and straight.

1) It looks like you have to add more NOPs for the wait time for vector draw after large retrace vectors from far right to far left to the 1st character position.

So, my solution that reads " wait for it to settle", was right.

2) It also looks like some bright dots have excessive dwell time before blanking.

Pure bad code, Tony, waiting too long for shifting to next char. It was corrected after this picture.

3) the blanking bandwidth looks adequate as the dots are "fairly crisp"

Found the Z input quite responsive for this.

4) there is some circular distortion perhaps maybe it is in the photograph but if in the display ...
it means it will need magnetic pincushion adjustment in the corners to flatten the outer rows and straighten the outside columns.

It is the camera. Actual image had not distortion and symbols came even crispier.

OK. Back to the code part of this.

Gracias for your help.

 

[ronsimpson]

I know video. The bright spot is because the beam is at that position for a long time.
It appears that after you draw a character you go away and leave the the Z on.

Look at the top row. "LLLLLLLLLLLLLLLL" That looks OK but the right dot is probably on 2x too long. My thought is that the beam is on for the time it would take to draw the black dot between characters.

Now look at the last L in in any line. The last dot is TOO BRIGHT. Because you left the Z on during the black time of the next empty characters. Any time there is a L ahead of a black space, (including the right edge) the last dot is too much time.

Because there is only "L"s I don't know enough. If this problem is only on the last dot then it on one problem.
BUT
If the problem shows with a "8" or "H" and it is the last dot in a line then it is another problem.
Either way it is a timing problem in your code.

Please try "T ", "L ","H ", "M ", or "Z " or "X ", "F ". This will give more information.

 

[KeepItSimpleStupid]

FWIW: http://www.tuxgraphics.org/electronics/scopeclock/oscilloscope_clock.shtml

 

[cowboybob]

In XY mode there is no time.

a) How long it will take to get the beam at (150,150)?

My bold emphasis.

Ron, just took the OP's question verbatim. I wasn't entirely sure if he actually meant XY mode (which atferrari did, but...). And, as it turns out, it really was a timing issue, just not as I first interpreted the question (not even close )...

 

[dr pepper]

I have messed with vector graphics, fairly recent, but it was with a tv crt repurposed.
Bright spot was a problem for me too, you need a good blanking amplifier that will switch off the beam quickly, a 'scope ought to be able to do this as you can easily display video on most 'scopes so the z input must have a b/w of around 4mc's.
I got around the issue by using another o/p on the micro for blanking the beam as well as the dac's for x/y/z, works well.
To get some ideas on blanking or even xy vectoring google amplifone monitor, they were a xy monitor popular for video games back in the 80's.