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)

This question was previously asked in

UGC NET Paper 2: Electronic Science Nov 2020 Official Paper

Option 4 : 4

Official Paper 1: Held on 24 Sep 2020 Shift 1

12478

50 Questions
100 Marks
60 Mins

__Concept__:-

M-array Bandwidth of PSK is given as;

\(BW = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

M-array Bandwidth of FSK is given as;

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

Where N = log_{2} M

R_{b} = Bit wave

Bandwidth efficiency \(=\frac{{{\rm{Bit\;wave\;}}\left( {{{\rm{R}}_{\rm{b}}}} \right)}}{{{\rm{Bandwidth}}}}\)

__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 = log_{2} 8 = 3

Bandwidth \(=\frac{{{2^{N + 1}}{R_b}}}{N}\)

\(= \frac{{{2^4}{R_b}}}{3} = \frac{{16}}{3}{R_b}\)

Bandwidth efficiency \(=\frac{{{R_b}}}{{\frac{{16}}{3}{R_b}}}\)

\(= \frac{3}{{16}} = 0.1875\)

(C) __16-array FSK__:-

N = log_{2} 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 \(=\frac{{{R_b}}}{{\frac{{{R_b}}}{2}}} = 2\)

(E) __4-array FSK__:-

N = log_{2} 4 = 2

Bandwidth \(=\frac{{{2^3}{R_b}}}{2} = 4\;{R_b}\)

Bandwidth efficiency \(=\frac{{{R_b}}}{{4\;{R_b}}} = 0.25\)

So, the decreasing order of bandwidth efficiency is:-

D > A > E > B > C

Option (4) is correct ans.