Phase shift keying (PSK.) method is used to modulate digital signals at 9600 bps using 16 levels. The line signals speed (i.e. modulation rate) will be
A. | 2400 bands |
B. | 1200 bands |
C. | 4800 bands |
D. | 9600 bands |
Option: A Explanation :
9600 corresponds to bits per second and the level is 16. For a 16-PSK, given the baud rate N, the bit rate would be 4N. Here bps is provided which is 9600.
In order to calculate the baud rate we have to divide bps/4, because the signal level is 16.So the answer would be 9600/4=2400 baud. Option is (A).
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Television channels are 6 MHz. How many bits can be sent if four level digital signals are used (assume a noiseless channel)?
A. | 24 Mbps |
B. | 12 Mbps |
C. | 6 Mbps |
D. | 36 Mbps |
Option: A Explanation : Click on Discuss to view users comments. |
What is the maximum burst length on an 155.52 Mbps ATM ABR connection whose PCR value 200,000 and whose L value is 25 msec ?
A. | 9 cells |
B. | 12 cells |
C. | 15 cells |
D. | 18 cells |
Option: B Explanation :
Given : T = 5 μ sec Click on Discuss to view users comments. |
Poll/select line discipline requires what to identify the packet recipient ?
A. | Timer |
B. | Buffer |
C. | Address |
D. | Dedicated line |
Option: C Explanation : In Poll / select method, every device on a link has an address that can be used for identification Click on Discuss to view users comments. |
In sliding window flow control, if window size is 63, what is the range of sequence numbers ?
A. | 0 to 63 |
B. | 0 to 64 |
C. | 1 to 63 |
D. | 1 to 64 |
Option: A Explanation : In sliding window flow control the window size is 63, the range of sequence numbers are 0 to 63 { 0.....63}. Click on Discuss to view users comments. |