Communication Systems Engineering Episode 1 Part 9 pptx
Communication Systems Engineering Episode 1 Part 9 pptx
... 0000 010 1 10 10 11 11 Syndrome
10 00 11 01 0 010 011 1 10
010 0 00 01 111 0 10 11 01
110 0 10 01 011 0 0 011 11
Suppose 011 1 is received, S = 10 , co-set leader = 10 00
Decode: C = 011 1 + 10 00 = 11 11
G
HH
T
=
=
=
10 10
010 1
10 10
010 1
10
01
10
01
Eytan ... = 3, R = 1/ 2
000 →→
→→
000000 10 0 →→
→→
10...
Communication Systems Engineering Episode 1 Part 1 pps
... Lecture
4-Feb L1
6-Feb L2
11 -Feb L3
13 -Feb L4
18 -Feb
20-Feb L5
25-Feb L6
27-Feb L7
4-Mar L8
6-Mar L9
11 -Mar L10
13 -Mar L 11
18 -Mar L12
20-Mar L13
25-Mar
27-Mar
1- Apr L14
Eytan Modiano ...
8-Apr L16
10 -Apr L17
15 -Apr L18
17 -Apr L 19
22-Apr
24-Apr L20
29- Apr L 21
1- May L22
6-May L23
8-May L24
13 -May L25
15 -May L26
5 / 19 - 5/23
Topic Reading
Pa...
Communication Systems Engineering Episode 1 Part 2 ppsx
...
M
M
1 1 1
∑
P
i
Log( ) =
∑
P
i
Log( ) +
∑
P
i
Log( ) ≤
0
, equality when P
i
=
i =1
MP
i
i =1
P
i
i =1
M M
M
M
1
That is
,
∑
P
i
Log
(
) ≤
∑
P
i
Log(M) = Log(M)
i =1
P
i
i =1
Eytan ... P
Proof, continued
M
1
Consider the sum
∑
P
i
Log( ), by log inequality:
i =1
MP
i
M M
1 1 1
≤
∑
P
i
( −
1
) =
∑
( − P
i
) =
0
, equality when P
i
=
i =1
MP...
Communication Systems Engineering Episode 1 Part 3 docx
...
Slide 6
jt jf
Rectangle pulse
t
1 ||< 1/ 2
t / /
Π() =
12 | t |= 12
0 otherwise
12
Π
∞
2
π
/
F[(t)]
=
∫
−∞
Π(t)e
− jft
dt
=
∫
12
e
− j 2
π
ft
dt
/
e
−
π
−
e
π
π
f ... the
Fourier transform of the original separated by 1/ Ts intervals
-1/ Ts
-w
w
1/ Ts
2/Ts
Eytan Modiano
Slide 10
Proof, continued
• If 1/ Ts > 2W then the replicas of X(f)...
Communication Systems Engineering Episode 1 Part 4 docx
... 299 ) gives optimal quantizer for Gaussian source
• E.g., N=4,
– D = 0 .11 75, H(x) = 1. 91 1
– Recall: uniform quantizer, D= 0 .11 88, H(x) = 1. 90 4 (slight improvement)
0 .9 816
1. 51
0.4528
-0 .9 816 ...
∆∆
∆
fx
Uniform quantizer example
• N=4, X~N(0 ,1)
x
() =
2
πσ
1
e
− x
2
/ 2
σ
2
,
σ
2
= 1
• From table 6.2,
∆
=0 .99 57, D=0 .11 88, H(Q)= 1. 90 4
– Notic...
Communication Systems Engineering Episode 1 Part 5 ppt
... ≤
⇒≤≤
⇒
⇒
=
<+
−
−
==
∑∑
log( ) log( )
log( ) log( )
11
2
21
11
1
11
n
i
=
<+
=< +
=+
≤< +
==
∑∑
log( ) log( ) ,
,
log( ) ( ) .
,
() ()
11
1
1
11
1
11
pp
Now
npnp
p
HX
Hence
HX n HX
ii
ii
i
k
i
i
i
k
Eytan ...
0 .1
a
5
0 .1
0.3
0.25
0.25
0.2
0.3
0.25
0.45
+
+
+
0.55
0.45
+
1. 0
1
0
0
1
0
1
0
1
Letter Codeword
a
1...
Communication Systems Engineering Episode 1 Part 6 potx
... coding achieves 1 bit difference between adjacent levels
• Example M= 8 (can be generalized)
A
1
000
A
2
0 01
A
3
011
A
4
010
A
5
11 0
A
6
11 1
A
7
10 1
Eytan Modiano
A
8
10 0
Slide ...
Two-dimensional signals
• S
i
= (S
i1
, S
i2
)
• Set of signal points is
called a constellation
S
1
= (1, 1)
S
2
=( -1, -1)
S
3
= (1, -1)
S
4
=( -1, 1)
• 2-D constellat...
Communication Systems Engineering Episode 1 Part 7 pps
... points
10
-1
P
e
10
-5
12
14
antipodal
orthogonal
3dB
E
b
/N
0
(dB)
Eytan Modiano
Slide 25
∈∈
∈
Signal Detection
• After matched filtering we receive r = S
m
+ n
– S
m
∈
{S
1
, ... continued
g(t)
0 => S
1
= g(t)
1 => S
2
= -g(t)
A
Eytan Modiano
S(t)
Y(t)
T
2T
3T
“S
1
(t)”
T
“S
2
(t)”
T
T
“Y
1
(t)”
“Y
2
(t)”
2T
2T
Slide 14
Detection i...
Communication Systems Engineering Episode 1 Part 8 pdf
... attenuation
P
R1
= P
T
/L, P
T2
= P
R1
A, P
R2
= P
T2
/L, …
P
N1
= P
N
, P
N2
= P
N1
A/L + P
N
, …
Let A = L => P
RK
= P
T
/L, P
NK
= KP
N
P
RK
/P
NK
= P
T
/LKP
N
= 1/ K (P
R1
/P
N1
)
Received ... received (E
b
/N
0
)
d
= 10 dB
• PAM modulation
P
b
= Q( E N2
0 b
/ )
• Repeater: Received (E
b
/N
0
)
u/d
= 1/ 2 (E
b
/N
0
)
u
= 10 dB - 3dB = 7dB
– => Pb = 5x10...
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