SHIFTOUT MSB/LSBFIRST clarification

[iw2fvo]

good day to all in this forum,
I am using PicBasic pro to load an 8 bit shift register and I wrote a very simple program like this:

INCLUDE "MODEDEFS.BAS"
DT var byte
DT=128 ' decimal
SHIFTOUT H_DATA,H_CLK,msbfirst ,[DT\8]

Then I do suppose that the MSB ( in this case "1" ) is transmitted first and then "0000000" is transmitted after it.
By looking to the output stream with a scope, I find that the bit at logic level = high ( dec 128) is shown to the left side of the scope and not at the right side of the scope display.
I am just a little bit confuesd about that: any clarification please ?
Thanks in advance.
bye

[peterdeco1]

I haven't done this in a long time but here is a snippet of code I used to send by SPI to a device:

Low PORTA.3 : Pause 10 'CHIP SELECT
ShiftOut PORTC.5,PORTC.3,0,[9254\16] 'playconfig0
ShiftOut PORTC.5,PORTC.3,0,[114\8] '1st play byte
High PORTA.3 : Pause 10 'deselect - end of word

The 0 after portc.3 means send code LSB. A 1 would send it MSB.

[iw2fvo]

thanks a lot,
Is your SPI device set to receive MSBfirst or LSBfirst please ?
Bye

[richard]

time on a scope flows from left to right
9052

[iw2fvo]

I am so sorry Richard:
if the time flows from left to right, then the first bit sent should be on the right and not to the left of the scope !
Where am I wrong please ?
Bye

[richard]

which second was recorded first in time sequence ?

9053

[richard]

slow your timbase right down , what direction does the dot travel

the dot having writ moves on

[iw2fvo]

the first bit sent out by the pic should be on the right ( 9 seconds old )
The last bit sent by the pic should be on the left ( 0 seconds old )

I am sorry for not understanding...
thanks

[richard]

last try
9054

[iw2fvo]

thanks again,
I am sorry for not understand...
If I slow down the scope time base the dot goes from left to right.
So the first bit transmitted by the PIC is always to the right of the scope and the last transmitted by the PIC is shown to the left.
Am I right ?
Then the 9 seconds old bit is the first transmitted by the PIC.
The 0 secs old bit is the last transmitted by the PIC.
Is it right ?
Thanks

[iw2fvo]

sorry: I did not tell that I am talking about the PIC output .
Thanks

[richard]

completely wrong , the first bit is on the left

[iw2fvo]

could you please check in the real situation:

connect the PIC DATA output to scope ch_1
connect the PIC clock output to scope ch_2 and set the trigger to this channel.
Tranmit bin 00000011 ( dec 3 ) \8 selecting MSB FIRST. and insert a small delay in the main loop just for better scope triggering. My scope is a old style !

What do you see on the scope ? 00000011 or 11000000 ?
Or better: do you see "11" to the right side or to the left side of the scope ?

Thanks for helping
regards

[peterdeco1]

You will send 00000011.

[iw2fvo]

Hi Peter,
could you please clarify better ?
( my English is not a first language )
Thanks

[iw2fvo]

iT IS MY FAULT:
Richard explanation is correct as usual !
Thanks for all.
All the best
regards,
Bye

[Ioannis]

In the first post you said the number was 128 (10000000 in binary) so the MSB is indeed a '1'.

But on the last post you saw a number 3 (00000011 in binary) so the MSB is a '0'.

Ioannis

[iw2fvo]

thanks for the response,
I had on my mund a shift register that will receive the serial stream from the PIC and I did associated the scope display to the shift register...
I was in error when I thought that the first bit on the left hand side of the scope is the last transmitted by the PIC: the first bit bit on the left of the scope is the first transmitted and the last bit to the right is the last transmitted.
Again, it was my fault .
Thanks for the assistance.
regards,Bye