1. Convert the binary data “011010” into analog wavefor

1. Convert the binary data “011010” into analog waveforms using following modulation
techniques:
a. Two level Amplitude Shift Keying
b. Two level Frequency Shift Keying
c. Two level Phase Shift Keying
d. Differential Phase shift keying
e. Four level Amplitude Shift Keying
f. Four level Phase Shift Keying
g. Eight level Amplitude Shift Keying
Ans: a. Two level Amplitude Shift Keying
 In this modulation technique only amplitude changes frequency and phase are
constant. It is also called as On-OFF keying.
 The amplitude of the analog carrier signal varies in accordance with the bit
stream(Modulating Signal).
b. Two level Frequency Shift Keying
 In two level frequency shift keying the frequency of the sinusoidal carrieris
shifted between two discrete values.
 These frequencies are fH and fL.(fH for binary 1 and fL for binary 0)
c. Two level Phase Shift Keying
 Two level frequency shift keying is also called as Binary Phase Shift Keying. It uses two
opposite signal phases(0 degree & 180 degree)

d. Differential Phase shift keying
 The input sequence of binary bits are modified such that the next bit depends upon the
previous bit.
 Therefore, in the receiver, the previous received bits are used to detect the present bit.
e. Four level Amplitude Shift Keying
 M=2^m where M=4 levels, then m=2 bit, So These four levels are 00 ,01, 10, 11
 Given Binary input is 011010 ,so divide this data into 2 bit like 01 10 10 because
m=2.

f. Four level Phase Shift Keying
 In Four level Phase Shift Keying 4 Phase shift are there(0, 90, 180, 270)
 Binary input signal divide into two bits (symbol).
Symbols Phases(degree)
00 0
01 90
10 180
11 270
h. Eight level Amplitude Shift Keying
 M=2^m, Here M=8 so m=3bits
 So in that example 8 Amplitude layers are there for 3 bit input symbol. Here input is
011010 divide this into 3 bit symbols like 011 010.

2. With fc = 500 kHz, fd = 25 kHz, and M = 16 (L = 4 bits), compute the frequency assignments for
each of the sixteen possible 4-bit data combinations.
Ans: Given fc=500kHz, fd=25kHz, M=16, L=4
V(FSK)=cos(wc*t+M*fd*t)
=cos(2*pi*fc*t +M*fd*t)
=cos(2*pi*500*t+M*25*t)
=cos(1000 *pi*t+25*M*t)
M L V(FSK)
1 0000 cos(1000*pi*t+25*t)
2 0001 cos(1000*pi*t+50*t)
3 0010 cos(1000*pi*t+75*t)
4 0011 cos(1000*pi*t+100*t)
5 0100 cos(1000*pi*t+125*t)
6 0101 cos(1000*pi*t+150*t)
7 0110 cos(1000*pi*t+175*t)
8 0111 cos(1000*pi*t+200*t)
9 1000 cos(1000*pi*t+225*t)
10 1001 cos(1000*pi*t+250*t)
11 1010 cos(1000*pi*t+275*t)
12 1011 cos(1000*pi*t+300*t)
13 1100 cos(1000*pi*t+325*t)
14 1101 cos(1000*pi*t+350*t)
15 1110 cos(1000*pi*t+375*t)
16 1111 cos(1000*pi*t+400*t)