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 log_{10}P_{m}, where P_{m}is the power in milliwatts.
dB

_{m}= 10 log_{10}P_{m}=> -30 = 10 log_{10}P_{m}
=>
log

_{10}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 = 10log

_{10}P_{2}/P_{1}= -1.5
log

_{10}P_{2}/P_{1}= -0.15
P

_{2}/P_{1}= 10^{-0.15}= 0.71
P

_{2}= 0.71P_{1}= 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 SNR

_{dB}?__Solution__

The values of SNR and SNR

_{dB}can be calculated as follows:
SNR
= 10,000 Î¼W / 1 Î¼W = 10,000

SNR

_{dB}= 10log_{10}10,000 = 10log_{10}10^{4}
=
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}= 10log_{10}∞ = ∞
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 log

_{2}2^{n}(C= capacity in bps, B = bandwidth in Hz)
= 2 x 3000 x log

_{2}2 = 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 log

_{2}2^{n}
C = capacity in bps

B = bandwidth in Hz

2

^{n}= 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 log

_{2}2^{n}= 2 x 3000 x log_{2}4 = 12000 bps
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