hi,


Well, take a look at next lines

mov th1,#0fdh 
mov tl1,#0fdh ;TH1=TL1 = 256 - (freq / (32 * 12 * BAUD)) 
mov pcon,#00h 
mov scon,#50h ;8-bit UART mode1 (Serial Mode 1 Set by Timer 1) 

Let me calculate:
256-(11059200/32/12/9600) = 256-3 = 0xFD
that`s correct for 9600.
Now the question: how do you obtain 19200? The code in your first message:

mov th1,#0fah ;baud rate (fd=19200,fa=9600) 
mov tl1,#0fah ;TH1=TL1 = 256 - (freq / (32 * 12 * BAUD)) 
mov pcon,#00h 
mov scon,#50h ;8-bit UART mode1 (Serial Mode 1 Set by Timer 1)

So you have used 0xFA (250).
Now try calculate:
BAUD = 11059200/32/12/(256-250) = 4800.
You have configured UART to 4800, not 19200.

My suggestion to obtain 19200:

mov th1,#0fdh 
mov tl1,#0fdh ;TH1=TL1 = 256 - (freq / (32 * 12 * BAUD)) 
mov pcon,#80h 
mov scon,#50h ;8-bit UART mode1 (Serial Mode 1 Set by Timer 1) 

It is the same as for 9600 but PCON is loaded with 0x80 and so it sets bit SMOD which doubles baud rate.

Regards,
Oleg