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9600 baudrate in frequency??

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brunokasimin

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9600 baud rate

Hello,

Anybody know how to convert 9600 baudrate in frequency?

Thx in advance
 

baud rate 9600

9600 baud rate means 9600 bits per seconds f = 9600 Hz or t = 1/9600
 
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baud rate frequency

Does it mean, when i measure it with oscilloscope, i will get 9600 Hz??

Thx
 
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baud rate to frequency

Actually not,
9600 baud rate means when you send data ,one bit will be sent in 1/9600 seconds.That bit can be 1 or 0. To measure a signal with an osiloscope ,that signal must be periodic.
 
baudrate 9600

You will not see a 9600Hz in osciloscope.

Only way to do it, is if you send continuosly 0x55 characters, and use no parity on channel.

0x55 = 01010101, remember that in serial communications, it has a start bit (0) and stop bit (1), so in the channel using a osciloscope you will see:

Start 1 0 1 0 1 0 1 0 Stop Start 1 0 1 0 1 0 1 0 Stop and so on...
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 and so on...

Good luck
 

baud rate and frequency

Is it output from uart must be always 1 byte width? Is this 1 byte correspond to the databits 8 in my Hyperterm?
 

9600 baud frequency

I really don't understand what you want to say.
The output in the uart are normally 8 bits long, plus a start bit and a stop bit.

The time between one bit and next bit is the baudrate, so if you have your uart setted in 9600bps, the bits are 1/9600 segs long, or 0.104uSegs.

So, because you always send bytes of 8 bits each, and 2 aditionally bits (start and stop), the transfer rate of data is 8+2=10, or 1/10 from baudrate. So, you will transfer a maximal data rate of 960 bytes by second.
 

9600 baud

penrico said:
You will not see a 9600Hz in osciloscope.

Only way to do it, is if you send continuosly 0x55 characters, and use no parity on channel.

0x55 = 01010101, remember that in serial communications, it has a start bit (0) and stop bit (1), so in the channel using a osciloscope you will see:

Start 1 0 1 0 1 0 1 0 Stop Start 1 0 1 0 1 0 1 0 Stop and so on...
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 and so on...

Good luck
Click to expand...

If you do this, you will see half of 9600 Hz, because you're using 2T period... so you will have 4800 Hz

1 0 1 0 1 0 1 0
T T
 

convert baud rate to frequency

Jetset, you are right!!!

:D
 

baud rates 9600

There is a common misconception with bauds and bits per second. Actually bauds are number of symbols per unit time.

When you transmit at 9600 bauds, it means you are transmitting 9600 symbols per second. A symbol actually can be 1 bit (in RS-232 it is like that), so you would have 9600 bits per second. But it can happen that you need to use 2 bit-wide symbols. Imagine that instead of having 0 and 1, you had 00, 01, 10 and 11.

If you transmit at 1 baud rate, you will send 1 symbol per second, this is 2 bits. So in this case 9600 bauds would be actually 19200 bits per second...
 

Re: baud rates 9600

hello there.
i have a BPSK signal, i know that in this case , baud rate and bit rate are the same
i know the frequency carrier but i cant determine the bitrate/baudrate,
the work is in matlab.
actually i need to specify this rate in Hertz.
Someone can help me?
 

Have you tried any method for timing recovery? IIRC, fft(x^2)

otherwise the carrier and baudrate are not related. eg, the baudrate could be 1kbps @ 1GHz or 1Mbps @ 1GHz.
 

hy.
so i inspected in time the signal in order to view the type of modulation. i saw it's BPSK
for the carrier i used the commands:
r=czt(s)
r=r/(length(r)/2);
rm = abs(r); %magnitude
rf = unwrap(angle(r)); %angle
i plotted only rm, plot(rm), and there are 2 peaks in mirror, it shows 1841Hz for both .
the length of the signal is 12266.
this signal is stored in a .mat file.

i have a script that demodulates bpsk but it requires the bit rate in hertz.
i just need the bits and reconvert 'em back in ascii ...also. i get only the graph.
anyway there are 230 bits , 8 bit character ascii code. start bit 0, 8 bits ascii code, stop bit 1. so 10 bits.
4 groups of 5 characters separated by 0x20 (space)...in total 230.
i attached the signal.
maybe someone can take a look on this.
 

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can we calculate symbol rate / Frequency Like this :

Suppose Baud Rate(B.R) is 9600,8 bits/symbol, 1 start ,1 Stop bit,
then there will be 10 bits for Each symbol

Symbol rate= (9600 Sym / 10 sec); 960 symbol /sec is the rate

:)
 

but how can i determine the baud rate ?
one tip for this problem (determine baud rate /bit rate) was to
determine it by searching in the carrier the perioad of the biggest change ...
 

You can use oscilloscope to make sure is baud rate correct or not.
you should measure the time between start bit falling edge and the stop bit rising edge then divide this time by 10 (if 1 start bit + 8 data + 1 stop) it should be~ 1.0416 ms. (use cursor of oscilloscope if it has)
Also you can measure the width of start or stop bit, they should be 1/9600 s
Please be aware some UARTs system has an error factor. It means the 9600bps may not be exactly 9600bps in reality. This is depending on source clock of UART and the baud rate. In some case the error is acceptable up to 6% in slow baud rate.
 

I know it's already been answered but I'm not getting it.
what will be the baud rate if we have 9600Hz frequency ?
Are frequency and baud rate are inverse of each other?
 

please read ALL topic, before ask again!
baudrate can't be converted to frequenzy so simple!
 

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
 

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...

we can not define baud rate, because we not know "what character is on serial bus", how many stop bits? parity bits are used?
0xAA=10101010
0xCC=11001100
same baud rate on serial port but double frequency! ... for example!
 

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