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Modulation Techniques, CRC, Spread Spectrum, WLAN, TCP/IP, Cellular Communications

   

Added on  2022-11-14

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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 two level amplitude shift keying or BASK (Binary Amplitude Shift Keying)or ON-
OFF keying only amplitude changes frequency and phase are constant.
In BASK the amplitude of the carrier signal changes according to the input binary
data.
b. Two level Frequency Shift Keying
In two level frequency shift keying or BFSK the frequency of the sinusoidal
carrier is changes according to the input discrete values.
In BFSK only two frequency levels are present High frequency for binary 1and
low frequency for binary 0.
c. Two level Phase Shift Keying
Two level frequency shift keying or BPSK uses two signal phases(0 degree & 180 degree)
BPSK is used for high bit rate

d. Differential Phase shift keying
 The input sequence of binary bits are modified such that the next bit depends upon the
previous bit.
 In DPSK two operation are perform differential encoding and phase shift keying.
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 degree)
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 levels, L=4bits
So, it is 16 level FSK
out=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 out
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)

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