How do you determine the data rate capacity of a channel allocation?
 

I'm not sure where else to ask this question, but I guess since this
group deals with antennas then maybe you guys are familiar with this.
This is regarding HDTV broadcast:

"Because of the 6 MHz. channel bandwidth allocated, each channel will
only support a data rate of 19.2 Mb/sec."

How is 19.2Mb/sec calculated?

Thanks!

How do you determine the data rate capacity of a channel allocation?
 

On Thu, 08 Nov 2007 00:40:50 +0000, MRW wrote:

I'm not sure where else to ask this question, but I guess since this group
deals with antennas then maybe you guys are familiar with this. This is
regarding HDTV broadcast:

"Because of the 6 MHz. channel bandwidth allocated, each channel will only
support a data rate of 19.2 Mb/sec."

How is 19.2Mb/sec calculated?

Thanks!


I would guess that with the 8vsb modulation schema that was selected, they
took the maximum cymbal rate that would fit into a 6Mhz channel, then
selected a FEC rate that would sustain a reliability link budget. This
dictated the bitrate of 19392659bps

http://www.broadcast.net/~sbe1/8vsb/8vsb.htm
http://en.wikipedia.org/wiki/8VSB#Bi...ission_systems

Other modulation schemes exists that would fit into a 6Mhz channel such as
256QAM but theory dictates that the higher the bitrate, the more energy
per bit that is required at the receive end to make a usable signal.

How do you determine the data rate capacity of a channel allocation?
 

On Thu, 08 Nov 2007 00:40:50 -0000, MRW wrote:

How is 19.2Mb/sec calculated?

It's called digital compression.

73's
Richard Clark, KB7QHC

How do you determine the data rate capacity of a channel allocation?
 

"Richard Clark" wrote in message
...
On Thu, 08 Nov 2007 00:40:50 -0000, MRW wrote:

How is 19.2Mb/sec calculated?

It's called digital compression.

73's
Richard Clark, KB7QHC


http://www.maxim-ic.com/appnotes.cfm/appnote_number/750

has an interesting chart that shows the analog bandwidth requirements for
various resolutions, including the 1080i and 720p HD formats, which are both
shown as 26 MHz. This is misleading, however, since digitizing the signal
results in data rates in the range of 1 Gb/s to 1.6 Gb/sec before
compression, according to one source.

The compression algorithm can be "tweaked" to put the original program into
almost any bandwidth. (The US Navy's "TV-Direct-to-Sailors" puts three TV
programs into a 3.6 Mbps data stream.
http://www.sia.org/2007DoDSatcomWork...y/DoD/Navy.ppt I've seen it and
it's OK.) Most compression algorithms are termed "lossy" because they lose
some of the original data. The resulting deficiencies, called "artifacts"
may be visible, depending on the amount of compression, the observer and the
program material.

There are several modulation methods which will all put about the same data
rate into a 6 MHz-wide channel.

How do you determine the data rate capacity of a channel allocation?
 

MRW wrote:
I'm not sure where else to ask this question, but I guess since this
group deals with antennas then maybe you guys are familiar with this.
This is regarding HDTV broadcast:

"Because of the 6 MHz. channel bandwidth allocated, each channel will
only support a data rate of 19.2 Mb/sec."

How is 19.2Mb/sec calculated?

Thanks!


The channel capacity (idealized) can be computed with Shannon's equation:

C = B * log2(1+ S/N)

From this we can see that you can get an arbitrary number of bits per
second through any bandwidth, if you have a high enough SNR and smart
enough channel coding. (viz, 56kbps telephone modems through a 3kHz
telephone wire channel)

there's a tradeoff, of course..

You can only radiate so much power (regulator and practical limits)
You can only point that power in so tight a beam (broadcast has to be
semi-omnidirectional, at least in a horizontal plane)
There's a practical limit to antenna size both at transmit and receive side.
You're ultimately limited by kTB noise at the receiver (most consumers
would balk at liquid helium cooled receivers)

So, what they did was pick a transmitter EIRP, a notional consumer
receive antenna gain and receiver noise figure, a typical radius to some
"service contour", and do the link budget..(with most of the numbers
chosen to match the existing analog services)

from that, they figure that they get enough SNR to support 19.2 Mbps
(with a substantial margin to allow for fading)

In this case, we can calculate backwards:
2^(19.2/6)-1 = S/N

S/N = 8.2 or about 9dB


The goal of the designer, then, is to figure out a coding and modulation
strategy that works well with the impairments typical of the path
(multipath probably is the most annoying one) and which allows
inexpensive receivers to be made, potentially at the expense of a more
complex transmitter.

The 4 popular strategies a
some form of n-ary PSK (8PSK)
QAM
COFDM
8VSB

all have pros and cons (some deal with multipath better, some are more
efficient for transmitting with better peak/average ratio, some deal
with doppler better).. For instance, on a satellite link from GEO, you
don't have to agonize about doppler,and you want to run saturated
amplifiers(for efficiency), so the PSK modulations are popular.


Once you know the bit rate, then it's a matter of coding/compressing the
raw video (which for HD is several Gbps) to fit in the 19Mbps channel.
Again, you want to push for simple decoders (which are going to be made
and sold in millions) at the expense of complex encoders (which are
going to be made and sold in dozens or hundreds).

Most of the compression standards allow for varying sophistication in
the encoding process, all with the same decoder. For instance, you can
do no frame-to-frame compression, in which case you'll have to lose a
lot of detail in the frames to get it to fit (this is the cheap
encoder). Or, you can use a fancier encoder that uses frame to frame
compression (i.e. figure out the background once and send it once, then
just send changes). Same decoder, but image quality is MUCH better.

Anyone who has "digital TV" from satellite or cable co has seen the
difference in encoder quality. Things encoded at the broadcast network
headend (where they've got more money) tend to look better than things
encoded in the cable TV local headend (like local commercials).

Likewise, one can tradeoff speed of encoding.. you can do a good
encoding job in much slower than real time cheaper than doing it in real
time. So a pre-recorded program could be encoded/compressed to a high
quality, without spending a huge amount of money.

How do you determine the data rate capacity of a channelallocation?
 

On Nov 9, 3:18 pm, Jim Lux wrote:
MRW wrote:
I'm not sure where else to ask this question, but I guess since this
group deals with antennas then maybe you guys are familiar with this.
This is regarding HDTV broadcast:

"Because of the 6 MHz. channel bandwidth allocated, each channel will
only support a data rate of 19.2 Mb/sec."

How is 19.2Mb/sec calculated?

Thanks!

The channel capacity (idealized) can be computed with Shannon's equation:

C = B * log2(1+ S/N)

From this we can see that you can get an arbitrary number of bits per
second through any bandwidth, if you have a high enough SNR and smart
enough channel coding. (viz, 56kbps telephone modems through a 3kHz
telephone wire channel)

there's a tradeoff, of course..

You can only radiate so much power (regulator and practical limits)
You can only point that power in so tight a beam (broadcast has to be
semi-omnidirectional, at least in a horizontal plane)
There's a practical limit to antenna size both at transmit and receive side.
You're ultimately limited by kTB noise at the receiver (most consumers
would balk at liquid helium cooled receivers)

So, what they did was pick a transmitter EIRP, a notional consumer
receive antenna gain and receiver noise figure, a typical radius to some
"service contour", and do the link budget..(with most of the numbers
chosen to match the existing analog services)

from that, they figure that they get enough SNR to support 19.2 Mbps
(with a substantial margin to allow for fading)

In this case, we can calculate backwards:
2^(19.2/6)-1 = S/N

S/N = 8.2 or about 9dB

The goal of the designer, then, is to figure out a coding and modulation
strategy that works well with the impairments typical of the path
(multipath probably is the most annoying one) and which allows
inexpensive receivers to be made, potentially at the expense of a more
complex transmitter.

The 4 popular strategies a
some form of n-ary PSK (8PSK)
QAM
COFDM
8VSB

all have pros and cons (some deal with multipath better, some are more
efficient for transmitting with better peak/average ratio, some deal
with doppler better).. For instance, on a satellite link from GEO, you
don't have to agonize about doppler,and you want to run saturated
amplifiers(for efficiency), so the PSK modulations are popular.

Once you know the bit rate, then it's a matter of coding/compressing the
raw video (which for HD is several Gbps) to fit in the 19Mbps channel.
Again, you want to push for simple decoders (which are going to be made
and sold in millions) at the expense of complex encoders (which are
going to be made and sold in dozens or hundreds).

Most of the compression standards allow for varying sophistication in
the encoding process, all with the same decoder. For instance, you can
do no frame-to-frame compression, in which case you'll have to lose a
lot of detail in the frames to get it to fit (this is the cheap
encoder). Or, you can use a fancier encoder that uses frame to frame
compression (i.e. figure out the background once and send it once, then
just send changes). Same decoder, but image quality is MUCH better.

Anyone who has "digital TV" from satellite or cable co has seen the
difference in encoder quality. Things encoded at the broadcast network
headend (where they've got more money) tend to look better than things
encoded in the cable TV local headend (like local commercials).

Likewise, one can tradeoff speed of encoding.. you can do a good
encoding job in much slower than real time cheaper than doing it in real
time. So a pre-recorded program could be encoded/compressed to a high
quality, without spending a huge amount of money.


Thank you very much! This has to be one of the most useful posts I've
read.