# Computer Networks problems and solutions for UGC NET/GATE exam - Set 4

16.       Calculate the power of a signal with dBm = −30.
Solution
Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts.
dBm = 10 log10 Pm => -30 = 10 log10 Pm
=> log10 Pm = -3
=> Pm = 10-3 mW
17.       The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as dB = 10log10P2/P1 = -1.5
log10P2/P1 = -0.15
P2/P1 = 10-0.15 = 0.71
P2 = 0.71P1 = 0.7 x 2 = 1.4 mW
18.       The power of a signal is 10 mW and the power of the noise is 1 Î¼W; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as follows:
SNR = 10,000 Î¼W / 1 Î¼W = 10,000
SNRdB = 10log1010,000 = 10log10104
= 10 x 4 = 40
Note:
In communication systems, noise is an error or undesired random disturbance of a useful information signal.
There are different types of noises.
Thermal - random noise of electrons in the wire creates an extra signal
Induced - from motors and appliances, devices act are transmitter antenna and medium as receiving antenna.
Crosstalk - same as above but between two wires.
Impulse - Spikes that result from power lines, lightning, etc.
Signal to Noise Ratio (SNR): To measure the quality of a system the SNR is often used. It is defined as the ratio of signal power to the noise power. It is usually given in dB and referred to as SNRdB.

The values of SNR and SNRdB for a noiseless channel are
SNR = signal power/0 = ∞
SNRdB = 10log10∞ = ∞
We can never achieve this ratio in real life; it is an ideal.

19.       Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as
Solution
C = 2 B log22n (C= capacity in bps, B = bandwidth in Hz)
= 2 x 3000 x log22 = 6000 bps
Note:
Nyquist Theorem: Nyquist gives the upper bound for the bit rate of a transmission system by calculating the bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth of the system (assuming 2 symbols/per cycle and first harmonic).
Nyquist theorem states that for a noiseless channel:
C = 2 B log22n
C = capacity in bps
B = bandwidth in Hz
2n = the number of signal levels
20.       Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as
Solution
C = 2 B log22n = 2 x 3000 x log24 = 12000 bps