General: Aspect Ratio Clarification

[edDV]

If an SVCD was encoded at 480 x 480 as MPEG 2 with a dar 4:3, since my current analog tv does not deal really with pixels, why would the beam(?) or waveform kind of incorrectly show the picture wrong on my TV with some of the picture cut off the sides (the left and right)? Is possible because the DVD player tried to stretch out the 480 x 480 video to 720 x 480 before sending it to the DAC?

It could be.

480x480 is not a standard DVD resolution so the DVD player may have been making assumptions. Maybe the DVD player made the assumption this 480x480 video was 352x480 so scaled it with PAR = 0.9090 x2 = 1.81818 causing the left and right edges to go off screen.

When you feed non-strandard signals into a DVD player, you never know what it will do.

Experiment instead with standard resolutions.

[jagabo]

Analog video is a continuous waveform. It has a bandwidth -- a limit on how high a frequency sine wave can pass through it. As the frequency of a sinusoidal signal rises and approaches the bandwidth of the device the output amplitude of the signal drops. There is no exact frequency at which the signal can no longer pass through the device. But often the "resolution" is stated as that frequency at which the output amplitude is 1/2 the input amplitude. Look at the oscilliscope graph in this post and read the accompanying text (it's about the horizontal color resolution of VHS -- VHS color resolution is much lower than the the luma resolution):

https://forum.videohelp.com/threads/319420-Who-uses-a-DVD-recorder-as-a-line-TBC-and-wh...=1#post1981589

Several posts earlier you can see frames from the actual video capture that graph was made from. The second image in this post:

https://forum.videohelp.com/threads/319420-Who-uses-a-DVD-recorder-as-a-line-TBC-and-wh...=1#post1980652

(The input amplitude of the sinusoidal wave was constant across the frequency sweep.)

Video (analog TV) resolution is stated over a portion of a width equal to the height of the frame. So a stated "480 lines of resolution" covers a 3:3 portion of the frame, making the resolution 640 lines over the full width of the 4:3 frame. But again, you need to stop thinking of pixels. This is an arbitrarily determined number (using the half amplitude technique in the above linked post) and will vary from device to device.

You should look at this issue the other way around: how is an analog waveform digitized? It is sampled with an ADC at some frequency. To convert that digital data back to an analog waveform it is sent to a DAC at the same frequency it was sampled. If you sample at 1 MHz you playback a 1 MHz, if you sample at 1 KHz you playback at 1 KHz. If you don't sample at high enough a rate you will not capture all the details of the analog signal.

The length (in time) of one scanline of analog NTSC video is 63.5 microseconds. The viewable image occupies 52.6 microseconds within that scanline. See the diagram at the top of page 2 of this PDF file:

http://www.intersil.com/data/tb/tb368.pdf

That 52.6 microseconds of waveform can be digitally sampled with an ADC as many or as few times as you want. If you sample it too few times you will not catch all the detail contained in the signal. If you sample too often you will end up with much more data than you need to reconstruct the limited (by the device's bandwidth) amount of detail in the signal. To playback the signal you send the digital data back out to the DAC at the same speed it was sampled by the ADC.

For ITU rec.601 video, samples are collected every ~75 nanoseconds. Over the 52.6 microseconds of the image portion of the NTSC scanline that corresponds to ~703 pixels (0.075 * 703 ~= 52.6). 720 pixels are collected to have a little leeway on each side of the image and to have a mod16 frame size. To reconstruct the analog waveform you would send those digital samples to the DAC at the rate of one every 75 nanoseconds.

An SVCD would sample the analog waveform less often (every ~110 ns), and play those samples back at the slower rate. VCD would sample at an even slower rate (every ~150 ns).

The reason I believe a DVD player would play a 480x480 SVCD frame by programming the DAC to output 480 samples directly (rather than digitally resizing the frame to 720x480 and sending it out the same way it does a 720x480 DVD) is because DACs are very programmable. Consider the DAC on your computer's graphics card. It can be programmed to output anything from 640x350 at 60 Hz to maybe 2048x1536 at 85 Hz (or more). So it's much easier for the player to program the DAC (we're probably talking about changing the values of a few registers in the DAC before starting playback) than to resize each and every frame of the video.

[edDV]

NTSC Analog and DVD ADR:Analog TV:

I found an article on the internet that described an analog TV as having a vertical and horizontal resolution. The Vertical resolution would be the number of horizontal lines scanned from left to right and then counting down from top to bottom. Thus, this would be the height of the display which never changes--it is fixed. That makes sense. And then it described the horizontal resolution as counting the number of vertical lines, thus getting the displays Width, which changes dependent on the video's signal. The thing that confused me in the article is that it said DVD had a horizontal resolution as having 480 lines, VHS as having 240 lines and broadcast having 330 lines. My question is: if the Vertical resolution is fixed at 480(or 486) and the Horizontal resolution of DVD is 480, how does that 480= 720 x 480 or even 640 x 480. The vertical resolution is fixed in an analog signal, so need to explain that. what I am not getting is the Horizontal Resolution. What would make sense to me would be Width 480(dvd horizontal resolution) x 480(fixed analog Vertical resolution)= 480 x 480. what am I missing?

Analog video and terminology are different from digital. Analog NTSC video is vertically sampled into ~480 horizontal lines (formerly 486). It is actually scanned 240 even lines then 240 odd lines for interlace 480 but for now just think 480 progressively scanned lines for now. So using the definition of sampled video, analog NTSC is vertically sampled. The horizontal lines are not sampled. The horizontal resolution depends on analog bandwidth.

An analog NTSC TV station was allowed 4.2 MHz analog bandwidth for transmission (think monochrome only). If you point a TV camera at a chart that shows alternating horizontal white and black at increasing frequency, you would reach a point where black and white resolve to gray. That point is defined as maximum "horizontal lines of resolution". It is the point where alternating black and white dots can no longer be resolved on a high quality monitor with the human eye. Further, these alternating black and white lines don't cover the full horizontal width, but a circle whose diameter is approimately equal to vertical height. That way you can compare horizontal to vertical resolution without taking into account 4:3 aspect ratio.

This was the basis of the EIA resolution chart (c)1956.



The wedges in the chart increase resolution as they narrow. They are calibrated in "lines of resolution". The numbers are only accurate when the framing arrows on the four sides touch the edge of the original camera viewfinder frame.

The chart itself is a high resolution photo image placed in front of a high quality camera. The high quality camera feeds the 4.2 MHz NTSC transmitter. At the other end, a high quality NTSC tuner feeds a high quality monitor. The only part of the equipment that will cause the alternating white/black dots on the line to turn gray, is the 4.2 MHz transmission channel. At a 4.2 MHz the dots will go to gray at ~330 on the horizontal lines of resolution scale. On the vertical axis, interlace sampling limits resolution by the so called Kell Factor. The Kell factor can be argued but it is generally taken to be 0.7 for still video. So, assuming 0.7, expected vertical lines of resolution will be 480 x 0.7 = 336. So in theory, an ideal 4.2 MHz channel will transmit interlace video with equivalent ~330-336 horizontal and vertical resolution, both measured in "lines of resolution".

Now replace the 4.2 MHz TV transmitter with a perfect VHS VCR. An ideal VHS VCR passes 3MHz monochrome video on playback. This gives about 240 "horizontal lines of resolution". Vertical 336 remains the same if you ignore VHS noise.

None of this relates exactly to digital display pixels, but an approximation can be drawn.
Minimum digital sample rate should be twice the highest frequency captured. 4.2 MHz (330 horizontal lines of resolution) should be sampled at or above 8.4 Ms/s. That sample rate creates a digital frame ~440 pixels wide. Since the vertical is already sampled, the vertical height is 480.

So in a perfect world 440x480 is the minimum NTSC sample rate. But the world is not perfect so some additional sample rate (width) is required. The satellite digital channels determined this should be 480x480. Cable went higher at 524x480 for SD digital channels. Broadcasters wanted to produce programs at 704-720 for 6.75 MHz bandpass (~530 maximum horizontal lines of resolution) or 720x480.

To make a DVD, you must choose 720x480, 704x480, 352x480 or 352x240. Other resolutions are not supported.

I may have made an error above in my quick analysis.

[Krelmaneck]

If there is one thing I should get from any of these posts on this board about a composite signal transforming to a wavefrom on my analog TV is that I really should not think in terms of Pixels; however, even though NTSC has the capability of showing the perfect 640 x 480 picture, it is still better to think in terms of Line Length when thinking about analog signals. I might even kill the term aspect ratio when thinking about how this all works because DAR is just something I might want to think about when encoding, but really has no bearing on the technical sides of things.


Now please bare with me. So I am going to kill everything I have learned. But I still can still think in terms of horizontal samples, specifically the bt.470 (analog) or bt.601 which are non-square samples. Yes, I am trying not not even think of the word pixel. I do not know the sampling rate of bt.470, but I do know the sampling rate for bt.601 which is 13.5 MHZ. Question clarifier: Is that 13,500 samples per 1 video frame (active and inactive picture portions) OR is that just the number of samples it is per second and contain a few pictures? I think it is one frame, but I just want to make sure of that. Also; what is the sampling rate of bt.407?


Last thing I understand is that each line line as it get process through the waveform contains the amplitude (brightness, Voltage in (IRE) and Color information. And one more clarifier: the scan line length of each scanline you said is in nanoseconds. So how can one figure out how many digital width height= one scanline in nanoseconds. So if I had a mp4 that I turned into a 704 x 480 DVD MPG-2 that got scaled up to 720 x 480 but the DVD player that got fed through the component video signal--how does one figure out how many 720 digital pixels equal the number of Nanseconds (iUS) in one NTSC scanline, including the blanking and active picture? Or is there no such relationship--a picture is just a picture and the beam just scans it into one line. And if there IS a relationship, I have the ENCODING QUESTION: if I got a video from an internet source (legal), and I do not know the sampling rate, do I just guess the sampling rate--hit or miss? Back to how a video is played( DVD), when the player feeds the info through the component signal to its waveform, how many 720 digital pixels equal one NTSC scaneline in nanseconds?


As you can see I have based my digital measurements on Pixels per W/H, so I decided to measure analog tv by Nanoseconds per scanline in a given sampling rate (wither 13.5 or the other one). I think that is a reasonable thing. But even as edDV did point out originally, NTSC still has the capability of showing 640 x 480, but just for curiosity, how many nanoseconds in one NTSC scanline equals 640 digital square pixels? Or is that not a possible calculation?

[edDV]

If there is one thing I should get from any of these posts on this board about a composite signal transforming to a wavefrom on my analog TV is that I really should not think in terms of Pixels; however, even though NTSC has the capability of showing the perfect 640 x 480 picture, it is still better to think in terms of Line Length when thinking about analog signals. I might even kill the term aspect ratio when thinking about how this all works because DAR is just something I might want to think about when encoding, but really has no bearing on the technical sides of things.

Actually, 16:9 analog wide format also exists in at least two forms.

Now please bare with me. So I am going to kill everything I have learned. But I still can still think in terms of horizontal samples, specifically the bt.470 (analog) or bt.601 which are non-square samples. Yes, I am trying not not even think of the word pixel. I do not know the sampling rate of bt.470, but I do know the sampling rate for bt.601 which is 13.5 MHZ. Question clarifier: Is that 13,500 samples per 1 video frame (active and inactive picture portions) OR is that just the number of samples it is per second and contain a few pictures? I think it is one frame, but I just want to make sure of that. Also; what is the sampling rate of bt.407?

13.5 MHz is the sample rate across each line. It works out to 704 luminance pixels in the visible space between blanking.

bt.470 is analog NTSC. There is no horizontal sampling. It is analog. You could say analog NTSC video is vertically sampled into 525 lines (486 or 480 lines visible).

[edDV]

Last thing I understand is that each line line as it get process through the waveform contains the amplitude (brightness, Voltage in (IRE) and Color information. And one more clarifier: the scan line length of each scanline you said is in nanoseconds. So how can one figure out how many digital width height= one scanline in nanoseconds. So if I had a mp4 that I turned into a 704 x 480 DVD MPG-2 that got scaled up to 720 x 480 but the DVD player that got fed through the component video signal--how does one figure out how many 720 digital pixels equal the number of Nanseconds (iUS) in one NTSC scanline, including the blanking and active picture? Or is there no such relationship--a picture is just a picture and the beam just scans it into one line. And if there IS a relationship, I have the ENCODING QUESTION: if I got a video from an internet source (legal), and I do not know the sampling rate, do I just guess the sampling rate--hit or miss? Back to how a video is played( DVD), when the player feeds the info through the component signal to its waveform, how many 720 digital pixels equal one NTSC scaneline in nanseconds?

This represents one line of NTSC. The "video" area is 52.6 microseconds wide.This part gets sampled to 704 pixels @13.5 Ms/s.



When you sample NTSC 720 wide (8 extra pixels left and right) you capture some blanking (black side stripes) as seen here. This was just captured 720x480 from live TV.



Above is 720x480 square pixel (a bit wide). Click on the picture to see full size.

Below is the same picture with center 704x480 h squeezed to 640x480 square pixel.

[edDV]

As you can see I have based my digital measurements on Pixels per W/H, so I decided to measure analog tv by Nanoseconds per scanline in a given sampling rate (wither 13.5 or the other one). I think that is a reasonable thing. But even as edDV did point out originally, NTSC still has the capability of showing 640 x 480, but just for curiosity, how many nanoseconds in one NTSC scanline equals 640 digital square pixels? Or is that not a possible calculation?

For DVD sampled 704x480 or 720x480, one pixel would be

52.6 usec / 704 = 0.0747 usec or 74.7 nanoseconds wide

If you sampled the picture 640x480 (not DVD compliant), one pixel would be

52.6 usec / 640 = 0.082 usec or 82.2 nanoseconds wide

When the picture is converted back to analog, it no longer has pixels.

When you look at the waveform for a 720x480 capture, you can see the "blacker than black" blanking showing lower left and right.

[edDV]

I suggest you read one of these books
http://www.google.com/search?tbs=bks:1&tbo=p&q=video&num=10

The first one is quite excellent
http://books.google.com/books?id=VRailj6TKqUC&printsec=frontcover&dq=video&hl=en&ei=nf...page&q&f=false

This one is more basic
http://books.google.com/books?id=KGbFAYGt5qsC&printsec=frontcover&dq=video&hl=en&ei=uQ...page&q&f=false

[Alex_ander]

how does one figure out how many 720 digital pixels equal the number of Nanseconds (iUS) in one NTSC scanline, including the blanking and active picture?

720/13.5MHz=53.333(3)mkS

For NTSC analog line full scan period is 63.555 microseconds, of which active part is 52.65 mkS (52.65x13,5=~711 samples) and analog blanking interval is 10.9 mkS. So of 720 samples per line 9 samples appear in blanking interval after sampling analog signal. The area limited by 711 and full number of analog lines 485 corresponds to 4:3 (or 16:9 by extention) analog image. In digital storage version of that video only 480 lines are left, so both 711(visible) or 720 represent an image 'wider' than 4:3. While the 704x480 part of the same stored image has proportions very close to 4:3 when displayed in centimeters (711/485~=704/480). That is why the numbers 704/480 are widely used as target for resizing videos with standard DAR's.

[Krelmaneck]

how does one figure out how many 720 digital pixels equal the number of Nanseconds (iUS) in one NTSC scanline, including the blanking and active picture?

720/13.5MHz=53.333(3)mkS

For NTSC analog line full scan period is 63.555 microseconds, of which active part is 52.65 mkS (52.65x13,5=~711 samples) and analog blanking interval is 10.9 mkS. So of 720 samples per line 9 samples appear in blanking interval after sampling analog signal. The area limited by 711 and full number of analog lines 485 corresponds to 4:3 (or 16:9 by extention) analog image. In digital storage version of that video only 480 lines are left, so both 711(visible) or 720 represent an image 'wider' than 4:3. While the 704x480 part of the same stored image has proportions very close to 4:3 when displayed in centimeters (711/485~=704/480). That is why the numbers 704/480 are widely used as target for resizing videos with standard DAR's.

But edDV has stated that once the picture is converted to analog, there are no pixels. But I just wanted to make sure if a pixel would be the same thing as a sample


Also, if there are 710.85 samples per scanline, is that a fixed number of samples for NTSC? Or could you also have a different pixels per scanline if you used the SMPTE 244m pixles?

[Krelmaneck]

And another question:

When dealing with a SIZE of a 4:3 VIDEO, how does one measure the video in Centimeters? I have measured my physical screen with a ruler and it came to apx. 4:3. But how can one measure a playing video by centimeters? I know it sounds like a dumb question, but once I get this question answered, I think I will have a basic understanding. I do not need a whole course, but I think I got a basic understanding, more or less.

[edDV]

...

But edDV has stated that once the picture is converted to analog, there are no pixels. But I just wanted to make sure if a pixel would be the same thing as a sample


Also, if there are 710.85 samples per scanline, is that a fixed number of samples for NTSC? Or could you also have a different pixels per scanline if you used the SMPTE 244m pixles?

Samples that occur during the active video portion of a line form pixels. Samples that occur in the horizontal or vertical blanking area are not part of the active picture area. Analog video can be sampled at any rate but some rates are more optimal.

SMPTE 244m is the 14.31818 MHz (4xfsc) sample rate described above where fsc equals the 3.579545 (exactly 315/88) MHz NTSC subcarrier. It was optimized for composite video capture. The D2 and D3 digital recording formats used this sample rate.

ITU Rec 601 is a component Y,Cb,Cr standard where luminance is sampled at the 13.5 MHz rate. With component video there is no subcarrier. The 13.5 MHz sample rate worked equally well for NTSC and PAL frame rates and raster size. This allowed dual standard equipment to be built. The D1 based component formats DV, DVD, DVB and ATSC use the 13.5 MHz sample rate.

http://en.wikipedia.org/wiki/NTSC
http://broadcastengineering.com/mag/broadcasting_understanding_composite_digital/

[edDV]

And another question:

When dealing with a SIZE of a 4:3 VIDEO, how does one measure the video in Centimeters? I have measured my physical screen with a ruler and it came to apx. 4:3. But how can one measure a playing video by centimeters? I know it sounds like a dumb question, but once I get this question answered, I think I will have a basic understanding. I do not need a whole course, but I think I got a basic understanding, more or less.

Analog video has no spacial size, just an aspect ratio and horizontal lines. You can have any screen size you want to build from tiny camcorder viewfinders up to Jumbotrons.

[Alex_ander]

But edDV has stated that once the picture is converted to analog, there are no pixels. But I just wanted to make sure if a pixel would be the same thing as a sample

I did'n use the word 'pixel' (picture element) which is related to how a picture is displayed on screen in case it is composed of discrete elements. Neither BT.601 nor H.262 (mpeg2 doc) use word 'pixel'.
Sample is a general term in signal processing for representing an analog function (2-dimensional here) by a finite quantity of numbers. This is only done for storage or wireless transmission (both original picture and human perception of it from screen are analog). By sampling theorem* (aka Kotelnikov or Nyquist theorem) any function with limited spectrum can be completely restored from point samples taken by intervals less than 1/(double bandwidth). In case of video signal, samples taken with intervals defined in BT.601 (1/13.5 MHz for both PAL/NTSC timings) and encoded to mpeg2 by H.262, can be either restored back to analog signal (e.g. analog output of a DVD player) or digitally transformed for some other standard interface. DAR is defined independently by AR flag of mpeg2 and any type of display (analog, discrete square/non-square screen elements) must show e.g. a 4:3 image with the same ratio in centimeters.

* also explained in this useful article by a M$'s display developing technician:
"A pixel is not a little square"
http://alvyray.com/memos/6_pixel.pdf

Also, if there are 710.85 samples per scanline, is that a fixed number of samples for NTSC? Or could you also have a different pixels per scanline if you used the SMPTE 244m pixles?

The number of samples would be different due to different sample frequency. But that standard is not related to DVD.

[jagabo]

When dealing with a SIZE of a 4:3 VIDEO, how does one measure the video in Centimeters? I have measured my physical screen with a ruler and it came to apx. 4:3. But how can one measure a playing video by centimeters?

The analog signal has no size, only timing characteristics. It takes on a size when it is instantiated. Ie, when it is displayed on a TV. Be careful, CRTs are notoriously inaccurate with aspect ratios. And you can't see the entire picture because they overscan.

[Krelmaneck]

When dealing with a SIZE of a 4:3 VIDEO, how does one measure the video in Centimeters? I have measured my physical screen with a ruler and it came to apx. 4:3. But how can one measure a playing video by centimeters?

The analog signal has no size, only timing characteristics. It takes on a size when it is instantiated. Ie, when it is displayed on a TV. Be careful, CRTs are notoriously inaccurate with aspect ratios. And you can't see the entire picture because they overscan.

That is what I kind of meant. I wanted to ask how do you measure the video playing on your tv screen in centimeters? I did not know about the overscan, but it was talked about earlier by you or someone else. Is there a way to decrease overscan for CRT for example in an avisynth script, or does overscan just happen in analog pictures because it leaves alot of room for blanking since a video could seem unstrudy or it could have extra stuff such as subtitles or other "junkm" thus leaving no way to cut out overscan since it is a built-in feature of a player or analog tv screen.

[Krelmaneck]

But edDV has stated that once the picture is converted to analog, there are no pixels. But I just wanted to make sure if a pixel would be the same thing as a sample

I did'n use the word 'pixel' (picture element) which is related to how a picture is displayed on screen in case it is composed of discrete elements. Neither BT.601 nor H.262 (mpeg2 doc) use word 'pixel'.
Sample is a general term in signal processing for representing an analog function (2-dimensional here) by a finite quantity of numbers. This is only done for storage or wireless transmission (both original picture and human perception of it from screen are analog). By sampling theorem* (aka Kotelnikov or Nyquist theorem) any function with limited spectrum can be completely restored from point samples taken by intervals less than 1/(double bandwidth). In case of video signal, samples taken with intervals defined in BT.601 (1/13.5 MHz for both PAL/NTSC timings) and encoded to mpeg2 by H.262, can be either restored back to analog signal (e.g. analog output of a DVD player) or digitally transformed for some other standard interface. DAR is defined independently by AR flag of mpeg2 and any type of display (analog, discrete square/non-square screen elements) must show e.g. a 4:3 image with the same ratio in centimeters.

* also explained in this useful article by a M$'s display developing technician:
"A pixel is not a little square"
http://alvyray.com/memos/6_pixel.pdf

Also, if there are 710.85 samples per scanline, is that a fixed number of samples for NTSC? Or could you also have a different pixels per scanline if you used the SMPTE 244m pixles?

The number of samples would be different due to different sample frequency. But that standard is not related to DVD.

Ok. that makes sense. So would it be accurate to say that the samples are one of the black or white dots(or a varied version of a black or white dot such as the color grey) ? And if so, does each little dot have to have a certain spacing? Or can one picture element be 3 inches a way and another reference dot can be 1 inch away from another dot, or does the dots of a precise accurate timing? I think if I remember correctly, the reference points can be almost stuck together which would indicate a picture was super sharp.


Could 1 (one) scanline look like this Example, and also can each scanline have two rows of reference dots like this?

. . .... . . . . . . . ... . . ..... . ..... ..
..... ... . . . . . . . . . . . . . . . . . .... .... . .... ... =710.85 (I am not going to draw 711 dots)



Or would it have to be exact like this example?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . = 710.85 points (i cant draw white dots)



I think the answer is YES, it could look like EITHER example. I just want to make sure.


Now, here is some confusion, let us say there 720 x 480 sampling area with a rate of 13.5 mhz, now, if 13,500 samples are sampled per scanline, how can only 710.85 be the active picture? What happened to the other 12,789 picture elements--do they go bye bye?


Let's refer to Posting #67. Now let's look at Jimmy's shirt, and then look at the waveform. I can see that the waveform, left some white picture elements where his shirt was as well as with some of his desk. And the rest of the black kind of references the darker parts of the picture, roughly, right? Now, the parts of the bottom and top are for the blanking, right? Ok now here is the question: what does the numbers 0-100 represent in this chart? Is that mhz or khz? Is that power?


And we know that analog signals or the display do not have pixels or square pixels for that fact. But with the analog signal, will the analog tv still use the reference dots in the waveform? Just double checking. I need to ask stupid questions to make sure.

[jagabo]

That is what I kind of meant. I wanted to ask how do you measure the video playing on your tv screen in centimeters?

You get out a ruler and measure it.

I did not know about the overscan, but it was talked about earlier by you or someone else. Is there a way to decrease overscan for CRT for example in an avisynth script, or does overscan just happen in analog pictures because it leaves alot of room for blanking since a video could seem unstrudy or it could have extra stuff such as subtitles or other "junkm" thus leaving no way to cut out overscan since it is a built-in feature of a player or analog tv screen.

Overscan is a feature of the TV. There were usually no user controls to adjust overscan on CRT based monitors. There were usually service controls inside the TV to adjust overscan, centering, size (width and height), barrel and pincushion distortion, etc.

[jagabo]

Let's refer to Posting #67. Now let's look at Jimmy's shirt, and then look at the waveform. I can see that the waveform, left some white picture elements where his shirt was as well as with some of his desk. And the rest of the black kind of references the darker parts of the picture, roughly, right? Now, the parts of the bottom and top are for the blanking, right?

Your interpretation is completely wrong. Start with a single scan line. In the waveform graph the height of the line indicates the intensity of each pixel along the scan line (in an analog video signal it's the voltage of the signal at that point along the scan line). Here's one scanline (it's actually several lines thick so you can see it) at the top of the image and the graph of it below:



As you can see, the bottom of the graph represents dark, the top bright. With a full frame the waveform graph is simply an overlay of the graphs from all scanlines, ie, the sum of all 480 graphs in a 720x480 frame. The graph has no representation of the horizontal or vertical blanking.

Here's graphical image on one line of NTSC video including the horzontal blanking interval and all timing pulses with the one scanline graph overlaid:

Ok now here is the question: what does the numbers 0-100 represent in this chart? Is that mhz or khz? Is that power?

The numbers are IRE levels -- ie the brightness (relative voltage in the analog signal). 0 IRE is full darkness, 100 IRE is full brightness. There is room at the top and bottom for occasional overshoot. 0 IRE corresponds to digital luma=16. 100 IRE corresponds to digital luma=235. In an 8 bit YUV image Luma can go as low as 0 and as high as 255.

[Krelmaneck]

Let's refer to Posting #67. Now let's look at Jimmy's shirt, and then look at the waveform. I can see that the waveform, left some white picture elements where his shirt was as well as with some of his desk. And the rest of the black kind of references the darker parts of the picture, roughly, right? Now, the parts of the bottom and top are for the blanking, right?

Your interpretation is completely wrong. Start with a single scan line. In the waveform graph the height of the line indicates the intensity of each pixel along the scan line (in an analog video signal it's the voltage of the signal at that point along the scan line). Here's one scanline (it's actually several lines thick so you can see it) at the top of the image and the graph of it below:


[Attachment 3464 - Click to enlarge]

As you can see, the bottom of the graph represents dark, the top bright. With a full frame the waveform graph is simply an overlay of the graphs from all scanlines, ie, the sum of all 480 graphs in a 720x480 frame. The graph has no representation of the horizontal or vertical blanking.

Here's graphical image on one line of NTSC video including the horzontal blanking interval and all timing pulses with the one scanline graph overlaid:


[Attachment 3465 - Click to enlarge]

Ok now here is the question: what does the numbers 0-100 represent in this chart? Is that mhz or khz? Is that power?

The numbers are IRE levels -- ie the brightness (relative voltage in the analog signal). 0 IRE is full darkness, 100 IRE is full brightness. There is room at the top and bottom for occasional overshoot. 0 IRE corresponds to digital luma=16. 100 IRE corresponds to digital luma=235. In an 8 bit YUV image Luma can go as low as 0 and as high as 255.

So if we had a DVD video playing through an analog signal, the higher the voltage of each graph, the more intense it is? And the lower the intensity on the graph represents less intense or darker images? The 480 graphs are still there in the tv's scanlines when it is a DVD video playing, right?

[edDV]

* also explained in this useful article by a M$'s display developing technician:
"A pixel is not a little square"
http://alvyray.com/memos/6_pixel.pdf

This article illustrates the parallel worlds of computer graphics and broadcast engineering as it still existed in 1995. Keep in mind that the broadcast engineers worked out the currently used sampling standards ten to twenty years before (e.g. 4xfsc sampling and CCIR-601).

From the article

So the correct terminology for this case is that the monitor has a
“non-square pixel spacing ratio”, not that it has “non-square pixels”. Most modern
computer displays, if correctly adjusted, have square PSR—ie, PSR = 1.

Every point he makes in this article assumes a PSR = 1. Digital television mostly works in PSR's other than one. That is the concept that is foreign to computer graphics scientists and engineers. Everything else in this article applies equally to video.

[edDV]

Ok. that makes sense. So would it be accurate to say that the samples are one of the black or white dots(or a varied version of a black or white dot such as the color grey) ? And if so, does each little dot have to have a certain spacing? Or can one picture element be 3 inches a way and another reference dot can be 1 inch away from another dot, or does the dots of a precise accurate timing? I think if I remember correctly, the reference points can be almost stuck together which would indicate a picture was super sharp.

Could 1 (one) scanline look like this Example, and also can each scanline have two rows of reference dots like this?

. . .... . . . . . . . ... . . ..... . ..... ..
..... ... . . . . . . . . . . . . . . . . . .... .... . .... ... =710.85 (I am not going to draw 711 dots)

Or would it have to be exact like this example?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . = 710.85 points (i cant draw white dots)

I think the answer is YES, it could look like EITHER example. I just want to make sure.

Absolutely not!

Video sampling has equally spaced samples with 8 bits (256 levels)* of luminance or more per sample.


* Black is defined as level 16, white is defined as level 235 for 8 bit video.

[edDV]

Now, here is some confusion, let us say there 720 x 480 sampling area with a rate of 13.5 mhz, now, if 13,500 samples are sampled per scanline, how can only 710.85 be the active picture? What happened to the other 12,789 picture elements--do they go bye bye?

There aren't 13,500,000 samples per scanline. There are 13,500,000 samples per second. For NTSC there are 29.97 frames of video per second.

Let's refer to Posting #67. Now let's look at Jimmy's shirt, and then look at the waveform. I can see that the waveform, left some white picture elements where his shirt was as well as with some of his desk. And the rest of the black kind of references the darker parts of the picture, roughly, right? Now, the parts of the bottom and top are for the blanking, right? Ok now here is the question: what does the numbers 0-100 represent in this chart? Is that mhz or khz? Is that power?

Also look at the spike for the whitish cup.

100 = white = digital level 235 (levels up to 255 are allowed)

Zero = black = digital level 16

-7.5 = blanking = digital level 0
(look at the left and right excursions into blankng for 720 sampling)

And we know that analog signals or the display do not have pixels or square pixels for that fact. But with the analog signal, will the analog tv still use the reference dots in the waveform? Just double checking. I need to ask stupid questions to make sure.

The analog waveform after D/A would look similar. Analog NTSC is measured on a 1 volt peak to peak scale.

[Krelmaneck]

Wow. This is mind-boggling.

It seems like a lot can happen in just a few nanoseconds in a waveform.

The thing that I can only really comprehend for now is that as the waveform goes from peak to peak, it kind of gathers sample information about the brightness or non-brightness of the picture information, in which the voltage which is measured in IRE which can be converted to digital luminance. I also know that the spacing between reference dots are equal and the same in an 8 bit YUV setting.

I know that is kind of like a 1st grade level understanding. So I need to ask more questions about the most recent illustration by edDV's waveform.

edDV stated that the peaks are measured from peak to peak by voltage. what exactly are they measured for in each u part of the form(or the bottom of the hill)? I know the waves are looking for black and white levels, but what else are they being measured for? Also, in each part of the PEAK of the hill, is it always the same. For example, can you have a waveform whose peaks are really low in part of the scanline or are the waves equal? In other words, lets just say that there is movie in a dark alley, and the first scanline is black in most of the scanline, thus giving it a O IRE. Is it possible for that waveform to almost be a straight line as it goes across the scanline. Can the height of each peak be different? Does each peak to peak always the same in length in nanoseconds? Also,, it looks like it has 40 IRE for blanking. When it comes to blanking what does the amplitude of 40 IRE mean since there is no IRE=Luma factor in blanking?




The reference dots kind of remember me of a DIRTY NINTENDO game that needs to be clean or almost broken. Of course, you cannot see the waveforms, but you will see something that looks like reference dots if the cartridge is not clean.

[Alex_ander]

And we know that analog signals or the display do not have pixels or square pixels for that fact. But with the analog signal, will the analog tv still use the reference dots in the waveform?

No, it will not/could not (only standard analog synch pulses), an analog TV uses the signal restored to analog from digital samples stored for each frame (the reconstruction is similar to linking dots on some graph with a smooth curve). Actually 3 different signals are restored: one Y (the one sampled with 13.5 MHz) - from luminance values and two color-difference signals R-Y & B-Y, both originally sampled with twice lower frequency 6.75MHz (=lower number of samples per line). Those signals can be converted to composite (1 signal, with color components modulating a sub-carrier), component S-video (2 signals), or RGB (3 signals) for sending from DVD player to TV or monitor. If it's a digital monitor, it applies its own internal procedures to video for getting correct proportions.

[edDV]

Wow. This is mind-boggling.

It seems like a lot can happen in just a few nanoseconds in a waveform.

The thing that I can only really comprehend for now is that as the waveform goes from peak to peak, it kind of gathers sample information about the brightness or non-brightness of the picture information, in which the voltage which is measured in IRE which can be converted to digital luminance. I also know that the spacing between reference dots are equal and the same in an 8 bit YUV setting.
I know that is kind of like a 1st grade level understanding. So I need to ask more questions about the most recent illustration by edDV's waveform.

edDV stated that the peaks are measured from peak to peak by voltage. what exactly are they measured for in each u part of the form(or the bottom of the hill)? I know the waves are looking for black and white levels, but what else are they being measured for?

Note that the RS-170A diagram is for the analog waveform. It doesn't apply to digital. An analog line starts with blanking which is the 0 IRE reference. Next comes the sync pulse which extends down to -40 IRE. The -20 IRE point on the sync pulse starts the CRT horizontal scan line (off screen at this point). The burst is the color subcarrier reference for the line. The active picture starts at the end of the blanking pulse. For analog NTSC, black is at 7.5 IRE and nominal white is at 100 IRE.

Now, if this analog line is digitized, the first sample of a 704 sampled line would start approximately at that 4 IRE mark on the trailing edge of the blanking pulse and the 704th sample would occur just before the next blanking pulse. This would be repeated 240 lines per field to make a 704x480 frame.

Also, in each part of the PEAK of the hill, is it always the same. For example, can you have a waveform whose peaks are really low in part of the scanline or are the waves equal? In other words, lets just say that there is movie in a dark alley, and the first scanline is black in most of the scanline, thus giving it a O IRE.

If the picture is dark, the line would be a constant 7.5 IRE. If there is a bright light bulb in the center of the picture, the line would go from black to white (100 IRE) for the width of the light bulb then return to black. If the light bulb was replaced with a lit multitone object, the line would vary over the 7.5 to 100 IRE range to represent the gray tones of the object.

Is it possible for that waveform to almost be a straight line as it goes across the scanline. Can the height of each peak be different? Does each peak to peak always the same in length in nanoseconds?

Yes, yes and no. A line of active video rises and falls with the object's gray tones. There's a limit to how sharp an edge transition can be. In broadcast NTSC the maximum sharpness is limited by the 4.2 MHz luminance bandpass.

[Krelmaneck]

Wow. This is mind-boggling.

It seems like a lot can happen in just a few nanoseconds in a waveform.

The thing that I can only really comprehend for now is that as the waveform goes from peak to peak, it kind of gathers sample information about the brightness or non-brightness of the picture information, in which the voltage which is measured in IRE which can be converted to digital luminance. I also know that the spacing between reference dots are equal and the same in an 8 bit YUV setting.
I know that is kind of like a 1st grade level understanding. So I need to ask more questions about the most recent illustration by edDV's waveform.

edDV stated that the peaks are measured from peak to peak by voltage. what exactly are they measured for in each u part of the form(or the bottom of the hill)? I know the waves are looking for black and white levels, but what else are they being measured for?

Note that the RS-170A diagram is for the analog waveform. It doesn't apply to digital. An analog line starts with blanking which is the 0 IRE reference. Next comes the sync pulse which extends down to -40 IRE. The -20 IRE point on the sync pulse starts the CRT horizontal scan line (off screen at this point). The burst is the color subcarrier reference for the line. The active picture starts at the end of the blanking pulse. For analog NTSC, black is at 7.5 IRE and nominal white is at 100 IRE.

Now, if this analog line is digitized, the first sample of a 704 sampled line would start approximately at that 4 IRE mark on the trailing edge of the blanking pulse and the 704th sample would occur just before the next blanking pulse. This would be repeated 240 lines per field to make a 704x480 frame.

Also, in each part of the PEAK of the hill, is it always the same. For example, can you have a waveform whose peaks are really low in part of the scanline or are the waves equal? In other words, lets just say that there is movie in a dark alley, and the first scanline is black in most of the scanline, thus giving it a O IRE.

If the picture is dark, the line would be a constant 7.5 IRE. If there is a bright light bulb in the center of the picture, the line would go from black to white (100 IRE) for the width of the light bulb then return to black. If the light bulb was replaced with a lit multitone object, the line would vary over the 7.5 to 100 IRE range to represent the gray tones of the object.

Is it possible for that waveform to almost be a straight line as it goes across the scanline. Can the height of each peak be different? Does each peak to peak always the same in length in nanoseconds?

Yes, yes and no. A line of active video rises and falls with the object's gray tones. There's a limit to how sharp an edge transition can be. In broadcast NTSC the maximum sharpness is limited by the 4.2 MHz luminance bandpass.

It is very clear to me now that each wavefrom in a scanline is measured by its gray tones just as you mentioned in the last sentence. And taking info from another previous post, I guess you could also call this the Y signal? Now, I kind of read some conflicting information. One person stated that each wavefrom tells about the intensity of a PIXEL. I might have read that wrong. And from what I just read, there are NO pixels in an analog signal (it only gets pixels when it is digitized)--there are just sych pulses in an analog wavefom (Posting 85 by Alex_ander). And I think you wrote there are still a few reference dots (not square) in an analog signal with each reference point (or dot) equally spaced out (just like in Jagabo's diagram in posting 79) And I remember that posting. So what is right?

[edDV]

It is very clear to me now that each wavefrom in a scanline is measured by its gray tones just as you mentioned in the last sentence. And taking info from another previous post, I guess you could also call this the Y signal? Now, I kind of read some conflicting information. One person stated that each wavefrom tells about the intensity of a PIXEL. I might have read that wrong. And from what I just read, there are NO pixels in an analog signal (it only gets pixels when it is digitized)--there are just sych pulses in an analog wavefom (Posting 85 by Alex_ander).

The problem is this discussion keeps jumping between a digital line (samples and pixels) and an analog line ... a varying gray tone waveform between H sync pulses.

And for simplicity we have only been talking about Y luminance (gray levels). Color gets much more complicated.

And I think you wrote there are still a few reference dots (not square) in an analog signal with each reference point (or dot) equally spaced out (just like in Jagabo's diagram in posting 79) And I remember that posting. So what is right?

There are no "dots" in analog. The dots in Jagabo's display are just for measurement reference, similar to lines on graph paper.

For NTSC analog video, the sync references are three. TV sets lock in this order.

1. Vertical Sync Pulse - marks the start of a new field (vertical interval). The pulses differ for even or odd fields.

2. Horizontal Sync Pulse - marks the start of a scan line. Active video starts on the trailing edge of Blanking.

3. 3.58 MHz* "Color Burst" on the "back porch" of blanking - this gives a frequency and phase reference for color decoding.

*3.579545 MHz ... another "magic frequency"

[jagabo]

It is very clear to me now that each wavefrom in a scanline is measured by its gray tones just as you mentioned in the last sentence. And taking info from another previous post, I guess you could also call this the Y signal?

Yes, analog video consists of luma (Y) and two color signals (U and V). It was done this way to keep the color signal compatible with black and white TVs (which only see the luma channel).

Now, I kind of read some conflicting information. One person stated that each wavefrom tells about the intensity of a PIXEL. I might have read that wrong. And from what I just read, there are NO pixels in an analog signal (it only gets pixels when it is digitized)--there are just sych pulses in an analog wavefom (Posting 85 by Alex_ander). And I think you wrote there are still a few reference dots (not square) in an analog signal with each reference point (or dot) equally spaced out (just like in Jagabo's diagram in posting 79) And I remember that posting. So what is right?

If you use an oscilloscope (an analog device) to view the analog waveform (no pixels) you will get continuous lines. Since I was showing you a waveform generated by a computer the video and the waveform were digital (pixels). Ie, digital pixels are an approximation of the analog waveform.