Consider a UART with the following setup: 8 data bits . EVEN parity 2 stop bits

Question

Consider a UART with the following setup: 8 data bits . EVEN parity 2 stop bits LSB first a) If the receiver gets the following frame: 111100111011001111, has there been any data corruption in this communication? Explain and show your work to a reasonable extent b) If the baudrate is 9600 baud/sec, approximately how long does it take to transfer one character using this UART? c) If one end wishes to send 'G' to another, what will be the full data frame including start, data, parity and stop bits?

Explanation / Answer

a.

111100111011001111

when the line is idle then the signal looks like 1111...., after a run of 1's if you see 0 it means data is transmitting so from there we need to count.

so the actual frmane is : 0 01110110 0 1111

first bit is start bit i.e., 0

we are considering parity so data filed is 8 bit that are 01110110

next is parity field here we are considering even parity is that bit is set to 0

next bit stop which is set to 1

To detect error we need to look into data part 01110110 and check parity.

Here we are using even parity means if we have even number of 0's in data then parity is 0, but here in our data we have five 1's so by this we can conclude that the data is correpted.

b.

one bit takes 1/9600 sec = 0.00014 sec

(includes 1 start bit, no parity, 1 stop bit)

8+1+1 =10 bits which takes 10*1/9600 = 1.04 msec

c.

Start Data Parity Stop ---> frame format

L 1000111 L H                       L=low, H= high

-> start bit is low when sending data

-> in parity we represent even parity as LOW and odd parity as HIGH

-> stop bit set to high when sending data.

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