# Bits and multiplexing data

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#1
OK so there's this question about binary data and multiplexed data.
So each signal is read in 2 microseconds. The signal is:
101101011001000101010010

Now the signals are multiplexed every 6 microseconds, which means it is 3-bit data. I need to work out the binary data in each of these signals.

http://www.contentextra.com/bacconli...chapter11.html

The whole link is not coming: so just click on problem number 9.

Now my question is: does it always have to be alternate bits? Can't consecutive bits be in one signal? How does it work?
0
8 years ago
#2
OK so there's this question about binary data and multiplexed data.
So each signal is read in 2 microseconds. The signal is:
101101011001000101010010

Now the signals are multiplexed every 6 microseconds, which means it is 3-bit data. I need to work out the binary data in each of these signals.

http://www.contentextra.com/bacconli...chapter11.html

The whole link is not coming: so just click on problem number 9.

Now my question is: does it always have to be alternate bits? Can't consecutive bits be in one signal? How does it work?
There is information missing from your statement above to make a full answer.

At face value, this is a Time Division Multiplexing problem but without a diagram or more information I am making a calculated assumption:

There appears to be two input channels both comprising 3-bit data.

These can be presented to the multiplexer as either parallel or serial input data streams but most likely parallel in this case.

The mutiplexer has a serial data stream output with a clock periodof 2uS.

The parallel input data will be presented as two 3-bit parallel words and latched first then held in a buffer, first one channel and then the other ready for muxing.

Once latched, each parallel buffer is read bit by bit by the mux and presented to the output in 3 x 2us bits of serial data.

The output thus alternates between parallel input streams which can be read and processed within the 6uS window.

To work out the binary data you need to know which way around the parallel streams are encoded. i.e. MSB or LSB first in each 3-bit serial word.
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#3
(Original post by uberteknik)
There is information missing from your statement above to make a full answer.

At face value, this is a Time Division Multiplexing problem but without a diagram or more information I am making a calculated assumption:

There appears to be two input channels both comprising 3-bit data.

These can be presented to the multiplexer as either parallel or serial input data streams but most likely parallel in this case.

The mutiplexer has a serial data stream output with a clock periodof 2uS.

The parallel input data will be presented as two 3-bit parallel words and latched first then held in a buffer, first one channel and then the other ready for muxing.

Once latched, each parallel buffer is read bit by bit by the mux and presented to the output in 3 x 2us bits of serial data.

The output thus alternates between parallel input streams which can be read and processed within the 6uS window.

To work out the binary data you need to know which way around the parallel streams are encoded. i.e. MSB or LSB first in each 3-bit serial word.

The numbers are 0, 10, 20, 30, 40, in case the diagram cannot be seen clearly. And there are 10 divisions each.

This shows a digital signal. The time scale is in microseconds.
This signal is read at a rate of 500 kHz. (from which I got the binary data)
If the signal contains two multiplexed signals each 6 s long, work out the binary data in eachof the signals.

0
8 years ago
#4

The numbers are 0, 10, 20, 30, 40, in case the diagram cannot be seen clearly. And there are 10 divisions each.

This shows a digital signal. The time scale is in microseconds.
This signal is read at a rate of 500 kHz. (from which I got the binary data)
If the signal contains two multiplexed signals each 6 s long, work out the binary data in eachof the signals.

Yes, that's Time Division Multiplexing:

As you have correctly identified, each 6uS signal must comprise 3 bit words (signal time / sampling period) = 6x10-6/ 2x10-6 = 3 samples per signal.

In this example of TDMuxing, each signal is allocated a recurring 6uS window and the question states there are only two signals. So the words must alternate. If there were 6 channels say, then they would be muxed as recurring 6 x 6uS words: 3-bit channel1, 3-bit channel 2, 3-bit channel 3 etc.

In reality, there are no hard and fast rules unless the data stream must comply with a defined international specification so that diferent manfacturers' equipment can communicate with each other as well as decode the data stream.

For this question you still have to make assumptions, For instance the serial stream is transmitted MSB to LSB in order to reconstruct the data.

Given that the raw serial data stream is

(put spaces in as you go to mimimise reading error):

101 101 011 001 000 101 010 010

Then reconstruct the channel data by alternately breaking out.

i.e. the data stream is structured c1w1, c2w1, c1w2, c2w2, c1w3, c2w3 etc. where c = channel, w = 3-bit word.

Channel 1: 101 011 000 010

Channel 2: 101 001 101 010
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#5
(Original post by uberteknik)
Yes, that's Time Division Multiplexing:

As you have correctly identified, each 6uS signal must comprise 3 bit words (signal time / sampling period) = 6x10-6/ 2x10-6 = 3 samples per signal.

In this example of TDMuxing, each signal is allocated a recurring 6uS window and the question states there are only two signals. So the words must alternate. If there were 6 channels say, then they would be muxed as recurring 6 x 6uS words: 3-bit channel1, 3-bit channel 2, 3-bit channel 3 etc.

In reality, there are no hard and fast rules unless the data stream must comply with a defined international specification so that diferent manfacturers' equipment can communicate with each other as well as decode the data stream.

For this question you still have to make assumptions, For instance the serial stream is transmitted MSB to LSB in order to reconstruct the data.

Given that the raw serial data stream is

(put spaces in as you go to mimimise reading error):

101 101 011 001 000 101 010 010

Then reconstruct the channel data by alternately breaking out.

i.e. the data stream is structured c1w1, c2w1, c1w2, c2w2, c1w3, c2w3 etc. where c = channel, w = 3-bit word.

Channel 1: 101 011 000 010

Channel 2: 101 001 101 010
Ok, that was really good. I'm sorry if I'mm asking too many questions but why does it have to be c1w1, c2w1, c1w2, c2w2, c1w3, c2w3 etc. where c = channel, w = 3-bit word.
Why can't it be c1w1, c1w2, c1w3, c1w4, c2w1, c2w2 etc.
0
8 years ago
#6
Ok, that was really good. I'm sorry if I'mm asking too many questions but why does it have to be c1w1, c2w1, c1w2, c2w2, c1w3, c2w3 etc. where c = channel, w = 3-bit word.
Why can't it be c1w1, c1w2, c1w3, c1w4, c2w1, c2w2 etc.
There are no hard rules for how channes are organised in the serial data stream - that's up to the designer to decide.

The multiplexer requires an overhead of time to convert the channels to serial data. The channel data needs time to stabilise before encoding. Alternating means one channel can be processed while a different channel is transmitted. In other words its a processing time saver.

Additionally, alternating channels are good practice to reduce the impact of data loss or corruoption and it leads on to bit error rates, signal to noise ratios, data corruption, checksums etc.

In a real application, each channel and data word will be wrapped between synchronisation bits at the start of the word, then the channel data and finally some kind of checksum like odd or even parity or a more advanced technique like hamming at the end of each word.

When a serial word is transmitted, the longer the word length, the higher the probability that it will end up being corrupted by noise or drop-outs.

Keeping the word length small and alternating between channels makes the whole tranmission more robust because any random corruption is likely to affect a small part of the whole message which together with the checksum can be more easily detected and corrected.

e.g if the serial stream comprised data for frames of a video, where channels represented the row or column data of consecutive video frames passing from say a video processor to a remote display, then stringing the whole frame information together would mean a whole frame has a higher probability of corruption rather than say just one row or column of pixels in any individual frame.

i.e. the difference between blanking a whole frame vs blanking one row or column within a whole frame.
0
#7
(Original post by uberteknik)
There are no hard rules for how channes are organised in the serial data stream - that's up to the designer to decide.

The multiplexer requires an overhead of time to convert the channels to serial data. The channel data needs time to stabilise before encoding. Alternating means one channel can be processed while a different channel is transmitted. In other words its a processing time saver.

Additionally, alternating channels are good practice to reduce the impact of data loss or corruoption and it leads on to bit error rates, signal to noise ratios, data corruption, checksums etc.

In a real application, each channel and data word will be wrapped between synchronisation bits at the start of the word, then the channel data and finally some kind of checksum like odd or even parity or a more advanced technique like hamming at the end of each word.

When a serial word is transmitted, the longer the word length, the higher the probability that it will end up being corrupted by noise or drop-outs.

Keeping the word length small and alternating between channels makes the whole tranmission more robust because any random corruption is likely to affect a small part of the whole message which together with the checksum can be more easily detected and corrected.

e.g if the serial stream comprised data for frames of a video, where channels represented the row or column data of consecutive video frames passing from say a video processor to a remote display, then stringing the whole frame information together would mean a whole frame has a higher probability of corruption rather than say just one row or column of pixels in any individual frame.

i.e. the difference between blanking a whole frame vs blanking one row or column within a whole frame.
Oh, ok, I got some of that. Thanks!
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