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

16.       Calculate the power of a signal with dB_m = −30.

Solution

Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dB_m and is calculated as dB_m = 10 log₁₀ P_m , where P_m is the power in milliwatts.

dB_m = 10 log₁₀ P_m => -30 = 10 log₁₀ P_m

                            => log₁₀ P_m = -3

                            => P_m = 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 μW; what are the values of SNR and SNR_dB ?

Solution

The values of SNR and SNR_dB can be calculated as follows:

                SNR = 10,000 μW / 1 μW = 10,000

                SNR_dB = 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 SNR_dB.

The values of SNR and SNR_dB for a noiseless channel are

SNR = signal power/0 = ∞

SNR_dB = 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

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