Question

MTN operating on a frequency of 2.3MHz aim at achieving 15Mbps of data throughput with a successful receiver interpreting an incoming signal to noise (S/N)dB = 12, 18, 25. Calculate the maximum bite rate of each S/N without using calculator explain the tradeoff factor, compare your results with the calculator results, which of this is enough for 4G which aim at achieving 14 Mbps and why?

Answer 1

Explanation :

S/N or Signal to Noise Ratio :

Signal-to-noise ratio (SNR or S/N) is a measure used in science and engineering that compares the level of a desired signal to the level of background noise. SNR is defined as the ratio of signal power to the noise power, often expressed in decibels. A ratio higher than 1:1 (greater than 0 dB) indicates more signal than noise.

SNR = P-signal / P-noise where P is average power.

When the signal to noise ratio, signal and noise are all in decibels :

SNR db = P-signal db – P-noise db .

The Shannon–Hartley theorem :

In information theory, the Shannon–Hartley theorem tells the maximum rate at which information can be transmitted over a communications channel of a specified bandwidth in the presence of noise. It is an application of the noisy-channel coding theorem to the archetypal case of a continuous-time analog communications channel subject to Gaussian noise. The theorem establishes Shannon's channel capacity for such a communication link, a bound on the maximum amount of error-free information per time unit that can be transmitted with a specified bandwidth in the presence of the noise interference, assuming that the signal power is bounded, and that the Gaussian noise process is characterized by a known power or power spectral density.

The Shannon–Hartley theorem states the channel capacity C , meaning the theoretical tightest upper bound on the information rate of data that can be communicated at an arbitrarily low error rate using an average received signal power S through an analog communication channel subject to additive white Gaussian noise (AWGN) of power N is given by:

C = B log base2 ( 1 + S /N )  where :

• C is the channel capacity in bits per second, a theoretical upper bound on the net bit rate (information rate, sometimes denoted I, excluding error-correction codes;
• B is the bandwidth of the channel in hertz (passband bandwidth in case of a bandpass signal);
• S is the average received signal power over the bandwidth (in case of a carrier-modulated passband transmission, often denoted C), measured in watts (or volts squared);
• N is the average power of the noise and interference over the bandwidth, measured in watts (or volts squared); and
• S / N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the noise and interference at the receiver (expressed as a linear power ratio, not as logarithmic decibels).

In the light of the above discussions, the solution to the given set of questions are as follows:

The Solution :

As per the given data, here,

C = channel capacity = 15 Mbps = 15 * 10^6 bps ;

B = bandwidth = 2.3 MHz = 2.3 * 10^6 Hz ;

S / N in db =

• a. 12 ;
• b. 18 ;
• c. 25 .

Again, as per formula as discussed above, C = B log base2 ( 1 + S /N ).

Calculation of maximum bit rate for each S/N without using calculator :

Without using calculator, we know that log base 2 of a positive integer is a positive number >1 and in the given data all the S/N values are positive integers, viz 12, 18 and 25. From this it can be concluded that the value of the log part of the formula is a posituve integer >1. Using this understanding, applying the formula to the 3 given S/N values yield :

a. For S/N = 12, Ca = B * Va where Va is some positive integer depending on S/N =12;

b. For S/N = 18, Cb = B * Vb where Vb is some positive integer depending on S/N =18;

c. For S/N = 25, Cc = B * Vc where Vc is some positive integer depending on S/N =25.

The Tradeoff Factor :

As seen in each case above, the trade off factor in all the above is B = bandwidth = 2.3 * 10^6 Hz for the given data.

According to the discussions on the top, B is the bandwidth of the channel in hertz (passband bandwidth in case of a bandpass signal) and this is the trade off factor in determining the value of C in each case.

Calculation of maximum bit rate for each S/N by using calculator :

a. Ca = B * [ log base2 ( 1 + S /N ) ] = 2.3 * 10^6 * [  log base2 (1 + 12 ) ] =  2.3 * 10^6 *  log base2 (13) = 8.5 * 10^6Hz = 8.5 MHz ;

b. Cb = B * [ log base2 ( 1 + S /N ) ] = 2.3 * 10^6 * [  log base2 (1 + 18 ) ] =  2.3 * 10^6 *  log base2 (19) = 9.77* 10^6Hz = 9.77 MHz ;

c. Cc = B * [ log base2 ( 1 + S /N ) ] = 2.3 * 10^6 * [  log base2 (1 + 25 ) ] =  2.3 * 10^6 *  log base2 (26) = 10.81 * 10^6Hz = 10.81 MHz .

Which of the above is enough for achieving 4G at 14 Mbps and reason for same :

Roughly speaking the MHz is a clear indicator of speed in Mbps. Moreoevr, 4G offers maximum real-world download speeds up to around 100Mbps, making it over 20 times faster than 3G. Theoretical maximum 4G speeds are significantly higher at 300Mbps, although such speeds are only achievable in controlled laboratory environments. Therefore, the Case c. above where S/N is 25 denoting a very less noise level, is a close approximation to 4G speed at 14 Mbps.

This concludes the answer to all parts of the question along with the necessary explanations.

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