dibit value phase shift
00 00
01 900
11 1800
10 2700
The last three phases are out by a factor of ten. They should be:
dibit value phase shift
00 0
01 90
11 180
10 270
The phase shift is in degrees.
Pete
Thanks for noticing the irregularities. This happened due to not looking properly in the formatting:
dibit value phase shift
00 00
01 900
11 1800
10 2700
Paul_KD7HB:
Absolutely different, but have become interchangeable with the demise of telephone modems.Baud refers to the sine wave of the tone used to send data via modems. The old 300 baud and 600 baud modems encoded one bit per single sine wave. There the bit rate and baud rate were the same.
More generally, baud refers to the Symbol rate. Simple modulation schemes such as baseband NRZ, PSK, and FSK encode one bit per symbol. Thus, with these schemes the baud rate is equal to the symbol rate. More sophisticated modulation format such as QPSK, nQAM, and PAM pack multiple bits per symbol.
The advantage of these schemes is that they increase the bit rate (for the same symbol rate) without increasing the required channel bandwidth. The disadvantage (besides complexity) is that they require a higher SNR to achieve the same Bit Error Rate performance. Or, equivalently, they have a poorer BER at the same SNR as the simpler techniques. This can be overcome with the use of Forward Error Correction coding.
brice3010:
Are bitrate and baud rate the same, as I read differently?
AFAIK, baud is state transitions per second. But for serial connections, one state transfer is one bit.
gfvalvo:
1 Start Bit + 8 Data Bits + 1 Stop Bit = 10 bits per data byte transmitted.
2400 bits / second
How many bytes you sending?
This answer came closest to my initial question: how long does it take to transmit a specific number of bytes at 2400baud.
Transmitting this message: 100,109.89,11.00,5,10,B and one startmarker, and one endmarker transmitted as a string (each character = 1 byte) at 1200 baud takes +/-225ms. See attachment (blue = TX, yellow = RX).
Edit: This is using HC-12 radio's set at 1200 baud for the serial link.
Transmitting this message: 100,109.89,11.00,5,10,B and one startmarker, and one endmarker transmitted as a string (each character = 1 byte) at 1200 baud takes +/-225ms.
How can anything take negative time?
25 characters should take just less than 209 ms.
TolpuddleSartre:
25 characters should take just less than 209 ms.
It is about 208.3 ms ((23 + 2)*(1/1200)*10) which is indeed less than 209 ms.
TolpuddleSartre:
How can anything take negative time?25 characters should take just less than 209 ms.
..negative time..?
TolpuddleSartre:
How can anything take negative time?25 characters should take just less than 209 ms.
But the scope printout shows more like 225ms..
brice3010:
But the scope printout shows more like 225ms..
Whatever is transmitting the bytes might be leaving a small gap between each byte, or it might be configured to transmit two stop bits instead of one.
christop:
Whatever is transmitting the bytes might be leaving a small gap between each byte, or it might be configured to transmit two stop bits instead of one.
HC12 datasheet says "8 databits, no parity, one stopbit"
HC-12 v2.3A.pdf (279 KB)
brice3010:
..negative time..?
You wrote "+/-225".
-225 is negative.
TolpuddleSartre:
You wrote "+/-225".
-225 is negative.
This is funny, sorry: I meant "more or less 225", so I wrote "+/- 225"
brice3010:
Are bitrate and baud rate the same, as I read differently?
Bitrate and baud are often the same, and in most practical purposes. A bit is either a 1 or 0. In a communication, such as TTL or CMOS logic, there are also 2 states. In RS-232, 2 states for valid data, other states considered invalid. However, if one were to create a communication system in which 10 states existed, say tones, for example, data could be sent in decimals and baud would not equal bit rate.
