# Computer Networks problems and solutions for UGC NET/GATE exam - Set 3

11.       What is the bit rate for high-definition TV (HDTV)?
Solution
HDTV uses digital signals to broadcast high quality video signals.
The HDTV screen is normally a ratio of 16 : 9.
There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second.
Twenty-four bits represents one color pixel.
1920 x 1080 x 30 x 24 = 1,492,992,000 bps = 1.5 Gbps
Note:
The TV stations reduce this rate to 20 to 40 Mbps through compression.
12.       What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?
Solution
The answer depends on the accuracy desired.
a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.

b. A better solution is to use the first and the third
harmonics with  B = 3 × 500 kHz = 1.5 MHz.

c. Still a better solution is to use the first, third, and fifth
harmonics with B = 5 × 500 kHz = 2.5 MHz.
Note:
A digital signal is a composite analog signal with an infinite bandwidth.

Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.

If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission.

In baseband transmission, the required bandwidth is proportional to the bit rate;
if we need to send bits faster, we need more bandwidth.

Bandwidth requirements

13.       We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?
Solution
The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.
14.       Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as
Solution
10log10P2/P1 = 10log10 0.5P1/P1 = 10log10 0.5
= 10(-0.3) = -3 dB
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
Note:
Attenuation means loss of energy -> weaker signal
When a signal travels through a medium it loses energy overcoming the resistance of the medium.
Amplifiers are used to compensate for this loss of energy by amplifying the signal.
To show the loss or gain of energy the unit “decibel” is used.
dB = 10log10P2/P1
P1 - input signal
P2 - output signal
15.       A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1 . In this case, the amplification (gain of power) can be calculated as
Solution
10log10P2/P1 = 10log10 10P1/P1
= 10log10 10 = 10(1) = 10 dB