Answer:
The extra bits that the transmitter sent - 500 kbps
Solution:
As per the question:
The given data rate is 100 Mbps
The clock of the digital transmitter is 0.5% faster than the receiver clock.
Now,
The extra bits send are given by:
[tex]\frac{5}{1000}\times 100\times 10^{3}} = 500 kbps[/tex]
