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Page 1: Basic Questions and Answers Digital Networks

A noiseless 4-Khz channel is sampled every millisecond. What is the maximum data rate? The maximum data rate can be calculated by:

maximumdata rate=2∗H∗log2 (V )bits /sec

Where H is the frequency and V is the amount of levels (2 for binary, for example). We need to know the amount of levels to calculate the maximum data rate in this case.

If a binary signal is sent over a 3-Khz channel whose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate?

The maximum data rate can be calculated by:

maximumdata rate=H∗log 2 (1+S / N ) bits /sec

Where S/N is the signal to noise ratio and is often stated in decibels. So for this case:

maximumdata rate=3000∗log2 (1+20 ) bits /sec

Resulting in approximately 13,177 bps maximum achievable data rate.

Sketch the Manchester encoding for the bit stream: 0001110101

Sketch the differential Manchester encoding for the bit stream of the previous problem. Assume the line is initially in the low state.

What is the difference between a passive star and an active repeater in a fiber optic network?

A passive star simply receives a signal while an active repeater actually repeats the signal, effectively multiplying the range of the signal.

Page 2: Basic Questions and Answers Digital Networks

How long does it take to transmit an 8 inch by 10 inch image by facsimile over an ISDN B channel? The facsimile digitizes the image into 300 pixels per inch and assigns 4 bits per pixel.

The image has a total of 8x10 or 80 square inches. Then each square inch has 300 pixels so we have 19,200 pixels for the image. We then have 4 bits per pixel so in total we have 76,800 bits for the image.

ISDN B can transmit at a maximum of 64kbps so it would take 76.8/64 = 1.2 seconds to transmit the image.

Same as last problem but with a RGB image with 8 bits per pixel.

It would take twice the time, so 2.4 seconds.

How many lines can a PBX handle using time division switching if the RAM access time is 50 nsec?

We know that the time needed to process a frame is 2nT microsec and one frame period is 125 microsec. Therefore n=(125 microsec/(2*T)) and since T = 50 nanosec then:

n= 1252∗0.05

=1250 lines

A terminal multiplexer has six 1200-bps terminals and n 300-bps terminal connected to it. The outgoing line is 9600 bps. What is the maximum value of n?

We can establish the equation:

6∗1200+n∗300=9600

From which we solve that the maximum number for n is 8. A maximum of eight 300 bps lines can be connected to the same TDM as the six 1200 lines.


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