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

1. To increase the rate of communication by log2M in M-array communication, the power requirement increased by

1. 2M
2. M2
3. MM
4. 2M

Explanation

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

Where Es = Eblog2M.

Power in M-array communication is given by Es

As Es = Eblog2M

To increase power by log2M,

Es = Eb(log2M)2

Es = Eblog2M2

Es = 2Eblog2M

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

the power requirement increased by M2.

2. The limit which represents the threshold Eb/N0 value below which reliable communication cannot be maintained is called as

1. Probability limit
2. Error limit
3. Shannon limit
4. Communication 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.

3. The length of cable required for transmitting data at the rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits and for signal speed 2,00,000 km/s is

1. 2.5 km
2. 2.0 km
3. 1.5 km
4. 1.0 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

4. Which modulation scheme will have minimal BER?

1. QPSK
2. 8-PSK
3. 16-PSK
4. 16-QAM

Explanation

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

Pe is minimum when energy (Eb) is maximum.

Now,

Eb ∝ d2 (distance between the symbol)

i.e., Pe 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 Pe is minimum in QPSK.  For M-ary as M increases, the distance between symbols decreases.

5. M-ary signaling produces _______ error performance with orthogonal signaling and _______ error performance with multiple phase signaling.

3. Improved, improved

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.

6. A full-duplex binary FSK transmission is made through a channel of bandwidth 10 kHz. In each direction of transmission, the two carriers used for the two states are separated by 2 kHz. The maximum baud rate for this transmission is:

1. 2000 bps
2. 3000 bps
3. 5000 bps
4. 10000 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

7. What is the relation between bandwidth B of BPSK signal and the bandwidth Bm of the M-ary PSK signal, for a given data rate?

1. Bm = MB
2. Bm = B log2 M
3. B = Bm log2 M
4. B = M⋅Bm

Answer.3. B = Bm 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

8. Which M-ary is more vulnerable to noise?

1. 4-ary system
2. 2-ary system
3. Binary system
4. None of the mentioned

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.

9. An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with a bandwidth of 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value of M is

1. 10
2. 14
3. 16
4. 20

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

10. Arrange the bandwidth efficiency of the following M-ary modulation schemes in descending order

(A) 8 – ary PSK

(B) 8 – ary FSK

(C) 16 – ary FSK

(D) 16 – ary PSK

(E) 4 – ary FSK

Choose the correct answer from the options given below:

(1) (D), (C), (A). (B), (E)

(2) (D), (E), (A). (C), (B)

(3) (C), (D), (B). (A), (E)

(4) (D), (A), (E). (B), (C)

1. 1
2. 2
3. 3
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

Rb = 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

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