CurtPalme.com Home Theater Forum

[AnalogRocks]

How much bandwidth does 1080P require?

[AnalogRocks]

How much bandwidth does 1080P require?

[hailtoby]

It depends on the the refresh rate. I think you take the screen resolution in pixels and multiply by refresh and its in Hz.

1920x1080=2,073,600

2.07 MHz x 60 = 124.4 MHz


That kinda seems right.

[AnalogRocks]

[AnalogRocks]

[AnalogRocks]

Interesting tidbit. Still not sure how the calculation goes though. Maybe I'm to tired or to dumb today to get it.

http://eepn.com/locator/products/ArticleID/34510/34510.html


Calculating Video Bandwidth for VGA Systems
March 2007 Issue
Published Date: March 05, 2007 Fred Zlotnick,
Maxim Integrated Products

In 1987 IBM introduced the VGA video standard using RGB analog signals, which specifies a 640x480-pixel format with 60 Hz refresh rate and the capability of producing sixteen colors. The analog RGB standard quickly became known as the VGA standard, and most computers today still retain the 15-pin, D-Sub blue connector for VGA monitors. The VGA standard is now maintained by the Video Electronics Standards Association (VESA), which defines the signals and how to measure and test them.

Today the VGA family includes an alphabet soup of names, ranging from the original VGA (Video Graphics Array) up through UXGA (Ultra Extended Graphics Array). Industry experts
expect that this standard will remain a part of the PC world until 2015 or later.

Some systems must route the VGA signal through an amplifier or switch, yet it is not obvious how a designer calculates the bandwidth required for such a switch or amplifier. Making that calculation requires some mathematics and an understanding of CRT monitors. Since the CRT was once the most common display device by far, a basic video bandwidth calculation still depends on the physics of the old-style CRT monitors.

Even though CRT monitors are rapidly being displaced by LCD monitors, they are part of the standard, and one must therefore understand them to do the bandwidth calculations. Within a CRT, the electron beam is magnetically deflected across the screen from top to bottom, thereby tracing a "raster" pattern (see Figure 1). Information is displayed during the active part of the display (the trace). At the end of a trace the beam must return to the left side to begin another trace, and during that re-trace it must be invisible, which is accomplished by "blanking" it to the black level. The time taken for re-trace must be taken into account. Similarly, a vertical re-trace to the start position is made when the beam reaches the bottom of the CRT, and that time must also be accounted for in the bandwidth calculation.

Example: 1280x1024 pixels; 60Hz refresh

For a 60Hz refresh rate, the time required to display one frame is 16.6 ms, and we assume the vertical retrace time is 1.6 ms. That leaves 15 ms for the active scan time. If a frame includes 1024 active lines, then each line is completed in 14.6 ms. We assume the horizontal-retrace time is 1.8 ms, leaving 12.8 ms for the active trace.

We need to display 1280 pixels in 12.8 ms, so the duration of each pixel is 10 ns. To calculate the worst-case bandwidth needed, we assume an alternate on, off, on, off display of any color vs. white. If, for example, we apply a square wave to the green signal, with 10 ns per pixel, the result is a square-wave with a 20 ns period, which is 50 MHz (see Figure 2).

This waveform can display alternate G, BK, G, BK (512 alternating green/black pixels) across the screen. What happens, however, if the waveform is a sine wave instead of a square wave?

Figure 3 shows a sine wave superimposed on a square wave. If the sine wave alone is displayed on a CRT monitor, it is clear that the average value of the pixel would be less, and that it would not become as intense as quickly. The pixel would be a little "fuzzy," achieving full brightness only in the middle of its 10ns interval. In place of a nice crisp dot, the sine wave signal would produce edges somewhat less defined, or "smeared."

A perfect signal passing though a 50 MHz filter might produce a usable display, but the image would not appear as "crisp" as that produced by a device with a wider-bandwidth. Because most of today's monitors are based on the LCD instead of the CRT, it is no longer obvious how much bandwidth is necessary. We know, however, that the system must be able to pass the fundamental frequency as shown in the Table.

The Table below lists the combination of horizontal and vertical pixels, refresh rate, and fundamental video frequency corresponding to each standard of resolution. It is assumed that 66% of the signal is active, and the video bandwidth equals the fundamental frequency. Note that increasing the refresh rate (from 60 Hz to 80 Hz, for example) has the same effect on bandwidth as does an increase in the pixel rate. For today's LCD monitors, the refresh rate for optimum viewing is rarely higher than 60 Hz. Thus, a reasonable rule of thumb for determining the video bandwidth of any amplifier or switch in the system is 2x to 3x the fundamental frequency.

This table is useful when selecting components to meet one or more resolutions. For example, if the target resolution is a VGA switch (1280 X 1024 at 60Hz), we see that the bandwidth (fundamental) is 60 MHz. As an example, the MAX4885 or MAX4887 VGA switch chips offer a bandwidth of greater than 400 MHz. The fundamental and 3rd harmonic(180 MHz) pass almost unattenuated. Either of these switches would work quite well, inserting less than 0.5 dB loss into the system, with no need for video buffers. If a video buffer were to be used, then the designer should select a buffer with a bandwidth of greater than 200 MHz to match the switch, such as the MAX4219.


Fred Zlotnick is a Senior Corporate Applications Engineer at Maxim Integrated Products Inc., Sunnyvale, CA

[AnalogRocks]

Interesting tidbit. Still not sure how the calculation goes though. Maybe I'm to tired or to dumb today to get it.

http://eepn.com/locator/products/ArticleID/34510/34510.html


Calculating Video Bandwidth for VGA Systems
March 2007 Issue
Published Date: March 05, 2007 Fred Zlotnick,
Maxim Integrated Products

In 1987 IBM introduced the VGA video standard using RGB analog signals, which specifies a 640x480-pixel format with 60 Hz refresh rate and the capability of producing sixteen colors. The analog RGB standard quickly became known as the VGA standard, and most computers today still retain the 15-pin, D-Sub blue connector for VGA monitors. The VGA standard is now maintained by the Video Electronics Standards Association (VESA), which defines the signals and how to measure and test them.

Today the VGA family includes an alphabet soup of names, ranging from the original VGA (Video Graphics Array) up through UXGA (Ultra Extended Graphics Array). Industry experts
expect that this standard will remain a part of the PC world until 2015 or later.

Some systems must route the VGA signal through an amplifier or switch, yet it is not obvious how a designer calculates the bandwidth required for such a switch or amplifier. Making that calculation requires some mathematics and an understanding of CRT monitors. Since the CRT was once the most common display device by far, a basic video bandwidth calculation still depends on the physics of the old-style CRT monitors.

Even though CRT monitors are rapidly being displaced by LCD monitors, they are part of the standard, and one must therefore understand them to do the bandwidth calculations. Within a CRT, the electron beam is magnetically deflected across the screen from top to bottom, thereby tracing a "raster" pattern (see Figure 1). Information is displayed during the active part of the display (the trace). At the end of a trace the beam must return to the left side to begin another trace, and during that re-trace it must be invisible, which is accomplished by "blanking" it to the black level. The time taken for re-trace must be taken into account. Similarly, a vertical re-trace to the start position is made when the beam reaches the bottom of the CRT, and that time must also be accounted for in the bandwidth calculation.

Example: 1280x1024 pixels; 60Hz refresh

For a 60Hz refresh rate, the time required to display one frame is 16.6 ms, and we assume the vertical retrace time is 1.6 ms. That leaves 15 ms for the active scan time. If a frame includes 1024 active lines, then each line is completed in 14.6 ms. We assume the horizontal-retrace time is 1.8 ms, leaving 12.8 ms for the active trace.

We need to display 1280 pixels in 12.8 ms, so the duration of each pixel is 10 ns. To calculate the worst-case bandwidth needed, we assume an alternate on, off, on, off display of any color vs. white. If, for example, we apply a square wave to the green signal, with 10 ns per pixel, the result is a square-wave with a 20 ns period, which is 50 MHz (see Figure 2).

This waveform can display alternate G, BK, G, BK (512 alternating green/black pixels) across the screen. What happens, however, if the waveform is a sine wave instead of a square wave?

Figure 3 shows a sine wave superimposed on a square wave. If the sine wave alone is displayed on a CRT monitor, it is clear that the average value of the pixel would be less, and that it would not become as intense as quickly. The pixel would be a little "fuzzy," achieving full brightness only in the middle of its 10ns interval. In place of a nice crisp dot, the sine wave signal would produce edges somewhat less defined, or "smeared."

A perfect signal passing though a 50 MHz filter might produce a usable display, but the image would not appear as "crisp" as that produced by a device with a wider-bandwidth. Because most of today's monitors are based on the LCD instead of the CRT, it is no longer obvious how much bandwidth is necessary. We know, however, that the system must be able to pass the fundamental frequency as shown in the Table.

The Table below lists the combination of horizontal and vertical pixels, refresh rate, and fundamental video frequency corresponding to each standard of resolution. It is assumed that 66% of the signal is active, and the video bandwidth equals the fundamental frequency. Note that increasing the refresh rate (from 60 Hz to 80 Hz, for example) has the same effect on bandwidth as does an increase in the pixel rate. For today's LCD monitors, the refresh rate for optimum viewing is rarely higher than 60 Hz. Thus, a reasonable rule of thumb for determining the video bandwidth of any amplifier or switch in the system is 2x to 3x the fundamental frequency.

This table is useful when selecting components to meet one or more resolutions. For example, if the target resolution is a VGA switch (1280 X 1024 at 60Hz), we see that the bandwidth (fundamental) is 60 MHz. As an example, the MAX4885 or MAX4887 VGA switch chips offer a bandwidth of greater than 400 MHz. The fundamental and 3rd harmonic(180 MHz) pass almost unattenuated. Either of these switches would work quite well, inserting less than 0.5 dB loss into the system, with no need for video buffers. If a video buffer were to be used, then the designer should select a buffer with a bandwidth of greater than 200 MHz to match the switch, such as the MAX4219.


Fred Zlotnick is a Senior Corporate Applications Engineer at Maxim Integrated Products Inc., Sunnyvale, CA

[Nashou66]

124 is with no retrace added this formula takes that into account.

http://www.csgnetwork.com/videosignalcalc.html?horizontal=1920&vertical=1080&refresh=48&tpixels=2073600&horscannr=138.2&horscanwer=141.2&bandwidth=223.9

Nashou

[hailtoby]

Does anyone know what the actual formula is? Now Im interested in how its calculated with retrace.

[hailtoby]

I just looked at all the projector specifications and now I am confused.

I know the vertical and horizontal scanrates also come into play, but I thought low bandwidth is what limited most projectors resolution. Not even taking retrace into acount, only a handful of projectors have a high enough bandwidth to display 1080p60. Once you factor in retrace, it looks like only the BR909 can handle it, and just barely.

What am I missing?

[CRT_Ben]

You're not missing anything...that's why people like Mike Parker and Greg Eismann can sell bandwidth modded board sets for $1k+

[hailtoby]

That makes sense.

[Nashou66]

I think the bandwidth on "paper" is there for the ones you listed, but getting the electronics to work in an optimized environment and under their laboratory specs is difficult to do at the factory. And That is what Mike and Greg attempt to get done. To allow the bandwidth to make its way through the video chain to the CRT's. Soem chips are rated at 800Mhz for example but that spec was done under lab environment with al the proper external parts needed to make that claim. When the Electronics manufacturer design the equipment they are constrained by cost, board real-estate, and time in designing the product to get it into production. So they are more concerned with getting it as close as they can. So I think it's fair to say that those two can get them to handle higher bandwidths than spec'd in the brochures .

Athanasios

[Gary M.]

http://myhometheater.homestead.com/bandwidthcalculator.html

that is the correct formula which is from Extron, true 1920x1080p at 60hz requires 186.6 Mhz

-Gary

[kschmit2]

The Extron formula only applies to equipment in the middle of a chain (switching, amplifiers, splitters), not to the last device in the chain (display).
Unfortunately the calculator on csgnetwork is also based on the Extron formula.

[Nashou66]

[WTS]

Well I think you'd be surprised to know that MPs ICs don't have that high of BW, I don't recall off hand what it is.

[Nashou66]

Yes he mentioned that the higher flat gain bandwidth is more important. At least 3 times the required bandwith is all that is needed for proper head room, so a 800 MHz chip at a gain of 2 is enough i would think. I use the EL5166 op amp but I think the OPA695 will do better after looking at the flat gain bandwidth of 300, even the original CLC449 has the same spec at 200 which is not bad. I have seen a close up pic of his VNB's and notice he doesn not change the feedback or gain resistors to alter the gain of the chip, so i wonder if he is still using the CLC449? Hmmm I have been searching for chips that wont use any new feedback resistor values and only the EL6166 is close I think the OPA695 might work also with out a resistor change. Its the decoupling caps that make the difference i would think. Once again I go Off topic...

Athanasios

[PaulB]

[AnalogRocks]

Hey now Nash. No reverse engineering.

Driving trians backwards can be hazardous.

[AnalogRocks]

Hey now Nash. No reverse engineering.

Driving trians backwards can be hazardous.

[Nashou66]

[AnalogRocks]

[AnalogRocks]

[Nashou66]