Electronic – First bit of NRZI

codeencodingtelecommunications

I was searching for Non-Return-to-Zero Inverted, but no references I couldn't find any reference about the first bit in NRZI.

I know that, in NRZI, 1 refers to toggle and 0 refers to constant, but I didn't found information about the first bit.

I found examples where the first bit is not specified (unless it's not the first bit ?)

Other examples give a positive voltage (HIGH) when the first bit is 0.

While other examples give a negative voltage (LOW) when the first bit is also 0.

Therefore, is there a rule for determining the voltage of the first bit for a NRZI encoding ?

Best Answer

NRZI codes are polarity insensitive. So in general, no, there is no rule, because all that matters is a change. Many interfaces with differential signals can tolerate a signal that is inverted by accident, perhaps by incorrect wiring.

There is even no general rule whether a change means logical 1 bit or logical 0 bit, those are the two variants of NRZI, called NRZM and NRZS.

In some cases the start voltage would never be seen, for example in a situation where the transmitter has already been transmitting while receiver has been turned off or connected later.

There could exist an interface that defines the polarity somehow, for example there could be a kind of parity bit that could be set in a way to force a certain level or transition after the parity bit.

Related Solutions

Problems with LCD initialization: 4-bit mode

are you setting the RS pin to the correct value?
You seem to think that you can use insert_value( x ) to write 8 bits to the LCD, but this must be done in two steps, like (assuming the 4 bits written by wrire4 are connected to the D7..D4 of your LCD):
```
write4( b >> 4 );
write4( b );
```
You must follow the 4-bit initalization as is shown in the initialization sequence as show in the datasheet EXACTLY, and note that each row in that initialization can imply two write4 commands.

Your insert_value command uses 2 bits of the value you pass to it for the RS and RW bits. Personally I prefer to keep RW on W all the time, and I pass RS as a separate argument. In your version it is very difficult to see what data you are passing to the LCD. But it is clear that you are writing 3 nibbles, where the datasheet specifies 4. This is my sequence:

      write4( 0x03 );
      timing::wait(  15 * timing::ms );
      write4( 0x03 );
      timing::wait(  100 * timing::us );
      write4( 0x03 );
      write4( 0x02 );     // 4 bit mode

A general note on debugging: when you have more than one problem, you can no longer reason "I changed this to to that and it did not work, hence that was not the good idea". It will only work when you have solved ALL problems.

My first STM32 code – please criticize me

On the STM32s, all peripherals are by default disabled and don't get any clock signals. This saves a bunch of power (and forces you to think about interactions between pins as you code). If you want to use GPIOE, you need to tell the STM32 to enable its clock.

1UL << 21 shifts an unsigned long value of 1 21 bits to the left. 1 << 21 shifts some number valued at 1, but it's some default type, which may or may not be unsigned long, and you don't necessarily know how wide your system defaults are. It will vary by platform and compiler. In fact, UL can vary by platform and compiler as well. It is best to use the library typedefs, which are sure things.

(uint32_t)1 << 21 and (uint64_t)1 << 21 are unambiguous.

Best Answer

Related Solutions

Problems with LCD initialization: 4-bit mode

My first STM32 code – please criticize me

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