9600 baudrate in frequency??

If we use QAM-16 then is it possible to find baud rate ? If yes then how?

sunjava1 said:

If we use QAM-16 then is it possible to find baud rate ? If yes then how?

Click to expand...

This is "Quadrature amplitude modulation" and have carrier frequency which not determine baud rate!

Taken From Google E-book:

In ASK the baud rate is the same as the bandwidth

For PSK the baud rate is the same as the bandwidth

QAM is a
combination of ASK and PSK


If above is true then,
9600 HZ = 96000 Bauds (with ASK)
and
9600 HZ = 96000 Bauds (with PSK)

Then can it be for QAM too? i.e. bandwidth = baud rate, as its the combination of PSK and ASK

ASK/FSK/PSK are all 2-level signals. there are two patterns that can be tramsitted, and thus 1b per baud.

for QAM, there are possibly more. eg, QPSK or "4QAM" has 4 patterns, or 2b of data that can be transmitted per symbol.
for 16QAM, 4b can be transmitted per symbol at maximum.

When ECC and other protocol is added, you can end up with things like 2.8b/symbol. This would mean 28 data bits an 12 other bits per 10 symbols. The commentary about ECC is a separate issue, and also applies to ASK/PSK/FSK if these systems make use these features. the opposite also applies to compression. if the source is compressed before transmission, it might be possible to transmit 5.2 data bits in the 4b symbol. of course, the source data would need to be compressible for that aspect to work.

Didn't understand what you say, Can you tell me just relation between bandwidth and baud rate for QAM-16 ?

In telecommunications and electronics, baud (/ˈbɔːd/, unit symbol "Bd") is synonymous to symbols per second or pulses per second. It is the unit of symbol rate, also known as baud rate or modulation rate; the number of distinct symbol changes (signaling events) made to the transmission medium per second in a digitally modulated signal or a line code. The baud rate is related to but should not be confused with gross bit rate expressed in bit/s.

The symbol duration time, also known as unit interval, can be directly measured as the time between transitions by looking into an eye diagram of an oscilloscope. The symbol duration time Ts can be calculated as:
Ts=1/fs
where fs is the symbol rate.

A simple example: A baud rate of 1 kBd = 1,000 Bd is synonymous to a symbol rate of 1,000 symbols per second. In case of a modem, this corresponds to 1,000 tones per second, and in case of a line code, this corresponds to 1,000 pulses per second. The symbol duration time is 1/1,000 second = 1 millisecond.
The baud unit is named after Émile Baudot, the inventor of the Baudot code for telegraphy, and is represented as SI units. That is, the first letter of its symbol is uppercase (Bd), but when the unit is spelled out, it should be written in lowercase (baud) except when it begins a sentence.

The symbol rate is related to but should not be confused with gross bit rate expressed in bit/s. The term baud rate has sometimes incorrectly been used to mean bit rate, since these rates are the same in old modems as well as in the simplest digital communication links using only one bit per symbol, such that binary "0" is represented by one symbol, and binary "1" by another symbol. In more advanced modems and data transmission techniques, a symbol may have more than two states, so it may represent more than one bit (a bit (binary digit) always represents one of exactly two states).

If N bits are conveyed per symbol, and the gross bit rate is R, inclusive of channel coding overhead, the symbol rate fs can be calculated as:
fs=R/N

In that case M=2^N different symbols are used. In a modem, these may be sinewave tones with unique combinations of amplitude, phase and/or frequency. For example, in a 64QAM modem, M=64, and so the bit rate is N=6 times the baud rate. In a line code, these may be M different voltage levels.

The ratio might not even be an integer; in 4B3T coding, the bit rate is 4/3 the baud rate. (A typical basic rate interface with a 160 kbit/s raw data rate operates at 120 kbaud.) On the other hand, Manchester coding has a bit rate equal to 1/2 the baud rate.

By taking information per pulse N in bit/pulse to be the base-2-logarithm of the number of distinct messages M that could be sent, Hartley[1] constructed a measure of the gross bitrate R as: R= fs * log2(M).

---------- Post added at 05:37 ---------- Previous post was at 05:20 ----------

sunjava1 said:

Don't want to convert baud rate into frequency, I'm trying to say,

Suppose the frequency is: 12000HZ, how do we find the baud rate ?

and I've read all the posts above, and didn't understand, so I decided to register and ask

Click to expand...

Ok I'll do some simple calculation using this equations to solve this issue
Fs=R/N; right [R is gross bit rate & N is number of bits conveyed per smbole]

here you did not provide complete info so i am assuming some parameters you can cahnge it accordingly .
Suppose one symbol contain 8data-bits in addition of two bits(start stop) it will be 10bit per symbol
so in this way the gross bit l rate will be 120Kbps .

:shock:

halee awan said:

In telecommunications and electronics, baud (/ˈbɔːd/, unit symbol "Bd") is........
.......................a measure of the gross bitrate R as: R= fs * log2(M).

Click to expand...

Well said, but this is a published resource and didn't see anywhere mentioned that this wikipedia's article on baud.
Baud - Wikipedia, the free encyclopedia

yep exactly this is Baud - Wikipedia, the free encyclopedia, then?
But its correct

halee awan said:

yep exactly this is Baud - Wikipedia, the free encyclopedia, then?
But its correct

Click to expand...

Since it is published from wikipedia I also suppose it's correct, then?

alexxx said:

Since it is published from wikipedia I also suppose it's correct, then?

Click to expand...

Then it supposed to be closed ..

Still Didn't get it, is frequency and bandwidth are equal ? I don't think so,

Suppose we have given a bandwidth of 5000HZ then how do we find the baud rate and bit rate using QAM16 encoding ?

Bit rate will be
Bit rate= Baud rate * 4

how do we find the baud rate from bandwidth?
is it 5000 bauds / seconds ?

In my view .......Actually baud rate is no of characters per second to find out the maximum frequency one can send 10101010 binary sequence i.e ascii values of 170 to get maximum frequency .....

Maybe it is useful to look at the definition and the different methods of serial transmission where the baud rate expression is used.

As rightly said baud mean only information bits transmitted per second. It does not say anything about the method of transmitting nor the protocol used.

If baud is used for asynchronous serial communication it means that the time slot for transmission one bit is defined by the Baud rate. With 9600 baud it means the time slot is 10.42 microsecond.

The protocol for asynchronous transmission is that a start bit is the first high to low transition and it stays low for the time of one bit. The receiver samples usually the transmission line in the middle of the agreed bit time (10.42 us in case of 9600 Baud). Since there is no clock transmitted the receiver has to generate the sampling points based on the defined baud rate, the number of bits, odd/even parity bit and the number of stop bits(transmitted at the end). A stop bit is always a high level on the transmission line and will be followed by a low start bit on the next transmission. If for example the number of data bits are defined with 8, no parity and one stop bit a total of 10 bits is the agreed protocol.

The data information transmitted could be anything from all low(0) to a low for the start bit and the rest all high(1). In this case the data bits are all low or all high, the transmission line carries just one pulse(length one stop bit or followed by 8 low data bits). If the sender would do every 1ms an all low(0) or all high(1) transmission the frequency would be 1kHz. If the information send is anything different the frequency is actual changing with the information pattern transmitted.

Here are some diagrams which might be useful to get a clearer picture: **broken link removed**

Can You just solve this :

Suppose we have given a bandwidth of 5000HZ then how do we find the baud rate and bit rate using QAM16 encoding ?

is it 5000 bauds / seconds ?

Maybe it is useful to look at the definition and the different methods of serial transmission where the baud rate expression is used.

As rightly said baud mean only information bits transmitted per second. It does not say anything about the method of transmitting nor the protocol used.

If baud is used for asynchronous serial communication it means that the time slot for transmission one bit is defined by the Baud rate. With 9600 baud it means the time slot is 10.42 microsecond.

The protocol for asynchronous transmission is that a start bit is the first high to low transition and it stays low for the time of one bit. The receiver samples usually the transmission line in the middle of the agreed bit time (10.42 us in case of 9600 Baud). Since there is no clock transmitted the receiver has to generate the sampling points based on the defined baud rate, the number of bits, odd/even parity bit and the number of stop bits(transmitted at the end). A stop bit is always a high level on the transmission line and will be followed by a low start bit on the next transmission. If for example the number of data bits are defined with 8, no parity and one stop bit a total of 10 bits is the agreed protocol.

The data information transmitted could be anything from all low(0) to a low for the start bit and the rest all high(1). In this case the data bits are all low or all high, the transmission line carries just one pulse(length one stop bit or followed by 8 low data bits). If the sender would do every 1ms an all low(0) or all high(1) transmission the frequency would be 1kHz. If the information send is anything different the frequency is actual changing with the information pattern transmitted.

Here are some diagrams which might be useful to get a clearer picture: **broken link removed**

hi,

Baud Rate is the total number of bits i.e 1's and 0's transmitted per second.

And Frequency is number of complete (sinusoidal or square or triangular ) cycles per second.

Therefore, if only the sequence of 0's & 1's are transmitted then the frequency is 9600/2=4800 Hz

Re: baud rate to frequency

got it...................