M-Ary Signalling MCQ || M-Ary Signalling Questions and Answers

Answer.2. M²

Explanation

The distance of each signaling point from the origin is given by √E_s 

Where E_s = E_blog₂M.

Power in M-array communication is given by E_s

As Es = Eblog2M

To increase power by log₂M,

Es = E_b(log₂M)²

Es = E_blog₂M²

Es = 2E_blog₂M

Hence to increase the rate of communication by log2M in M-array communication,

the power requirement increased by M².

Answer.3. Shannon limit

Explanation

The Shannon capacity theorem defines the maximum amount of information, or data capacity, which can be sent over any channel or medium (wireless, coax, twisted pair, fiber, etc.). What this says is that the higher the signal-to-noise (SNR) ratio and the more the channel bandwidth, the higher the possible data rate.

Eb/N0 curve has a waterfall shape. Shannon limit gives the threshold value below which reliable communication cannot be maintained.

Answer.2. 2.0 km

Explanation

Bandwidth = 500 mbps

Frame size = 10,000 bits

Signal speed = 2,00,000 km/sec

Transmission delay = 2 × propagation delay

frame size/bandwidth = 2 × (length of the cable /signal speed)

Length of the cable =  (20000 × 10000)/(2 × 500 × 10⁶)

Length of the cable =  2 Km

Answer.1. QPSK

Explanation

Probability of error (P_e) = $Q\sqrt {\frac{{{E_b}}}{{{N_0}}}}$

P_e is minimum when energy (E_b) is maximum.

Now,

E_b ∝ d² (distance between the symbol)

i.e., P_e is minimum when the distance is maximum.

Analysis:

BER (Bit Error Ratio) Is a Signal Quality Quantitative Measurement of Digital Communication Systems. Out of the given scheme, QPSK has the maximum distance, and hence P_e is minimum in QPSK.  For M-ary as M increases, the distance between symbols decreases.

Answer.2. Improved, degraded

Explanation

In M-ary signaling as k increases, the curve moves towards the degraded error performance. It produces improved error performance in the case of orthogonal signaling and degraded error performance in the case of multiple-phase signaling.

Answer.2. 3000 bps

Explanation

The bandwidth for a full-duplex binary FSK transmission is given by:

B.W. = Baud rate + f₂ – f₁

Calculation:

Because the transmission is full-duplex only 5000 (5 kHz) is allocated for each direction.

Given f₂ – f₁ 2k – 2000

Band rate = BW – (f₂ – f₁) – 5000 – 2000

Band rate = 3000 bps

Answer.3. B = B_m log₂ M

Explanation

M-array PSK:

For m-array PSK minimum bandwidth to pass digitally modulated carrier is given as:

$B.W. = \frac{{{R_b}}}{{{{\log }_2}M}}$ ——– 1

BPSK:

For BPSK M = 2 so,

$B = \frac{{{R_b}}}{{{{\log }_2}2}} = {R_b}$ ——- 2

From Equation 1 and 2;

B = Bm log₂ M

Answer.1. 4-ary system

Explanation

An M-ary transmission is a type of digital modulation where instead of transmitting one bit at a time, two or more bits are transmitted simultaneously.

QPSK is a 4-ary system, that is, it has 4 different states each represented by a set of 2 binary digits, i.e., 00, 01, 10, 11. Likewise, an 8-ary system has 8 different states each represented by a set of 3 binary digits, i.e., 000, 001, ….., 111, a 16-ary system has 16 different states represented by a set of 4 binary digits, i.e., 0000, 0001, ….., 1111, and so on.

The minimum energy noise vector for 4-ary system is smaller than 2-ary system. So 4-ary system is more vulnerable to noise.

Answer.3. 16

Explanation

For M-PSK bandwidth of signal given by B $ = \frac{{{R_b} \times \left( {1 + \propto } \right)}}{{logM}}$

Given bit rate  = 200kbps, excess bandwidth = 100

⇒ 100 KHz $ = \frac{{200\: \times \:{{10}^3} \times \:2}}{{logM}}$

⇒ Log M = 4

M = 16

Answer.4. 4

Explanation

M-array PSK:

For m-array PSK minimum bandwidth to pass digitally modulated carrier is given as:

$B.W. = \frac{{{R_b}}}{{{{\log }_2}M}}$

M-array Bandwidth of FSK is given as;

$BW = \frac{{{2^{N + 1}}{R_b}}}{N}$

Where N = log₂ M

R_b = Bit wave

Bandwidth efficiency = Bit wave (Rb)/Bandwidth

Calculation:-

(A) 8-array PSK:–

Bandwidth $ = \frac{{2{R_b}}}{{{{\log }_2}8}} = \frac{2}{3}{R_b}$

Bandwidth efficiency $ = \frac{{2{R_b}}}{{\frac{2}{3}{R_b}}} = 1.5$

(B) 8-array FSK:-

N = log₂ 8 = 3

Bandwidth = $\begin{array}{l} = \frac{{{2^{N + 1}}{R_b}}}{N}\\ \\ = \frac{{{2^4}{R_b}}}{3} = \frac{{16}}{3}{R_b} \end{array}$

(C) 16-array FSK:-

N = log₂ 16 = 4

Bandwidth $= \frac{{{2^5}{R_b}}}{4} = \frac{{32}}{4}{R_b} = 8\:{R_b}$

Bandwidth efficiency $= \frac{{{R_b}}}{{8\:{R_b}}} = 0.125$

(D) 16-array PSK:-

Bandwidth $= \frac{{2{R_b}}}{{{{\log }_2}16}} = \frac{2}{4}{R_b} = \frac{{{R_b}}}{2}$

Bandwidth efficiency $= \frac{{{R_b}}}{{\frac{{{R_b}}}{2}}} = 2$

So, the decreasing order of bandwidth efficiency is:-

D > A > E > B > C