Proakis J (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 12 pptx
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 12 pptx
... relative
difference in SNR of the error probability of 10
−6
is 2 dB.
-7
-6 .5
-6
-5 .5
-5
-4 .5
-4
-3 .5
-3
-2 .5
-2
6
7 8 9 10 11 12 13 14
SNR/bit, dB
log(P(e)
232
... cases.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4 -3 -2 -1 0 1 2 3 45
frequency f
Sv(f)
T=1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-5
-4 -3 -2 -1 0 1 2 3 45
f...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 3 pptx
... f
0
=3.
-1
-0 .8
-0 .6
-0 .4
-0 .2
0
0.2
0.4
0.6
0.8
1
0
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
-1
-0 .8
-0 .6
-0 .4
-0 .2
0
0.2
0.4
0.6
0.8
1
0
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
-1
-0 .8
-0 .6
-0 .4
-0 .2
0
0.2
0.4
0.6
0.8
0
0.2 ... +1)e
j2 πft
df +
1
2j
1
2
0
(−f +1)e
j2 πft
df
=
1
2j
1
j2 πt
fe
j2 πft
+
1
4π
2
t
2
e
j2 πft
...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 1 docx
... then
x
n
=
1
T
T
2
−
T
2
x(t)e
j2 π
n
T
t
dt
=
1
T
T
2
−
T
2
e
j2 π
n
T
t
dt +
1
T
T
4
−
T
4
e
j2 π
n
T
t
dt
=
j
2πn
e
j2 π
n
T
t
T
2
−
T
2
+
j
2πn
e
j2 π
n
T
t
T
4
−
T
4
=
j
2πn
e
j n
− e
j n
+ e
j
n
2
− ... =
n
i=1
α
2
i
1
2
,B=
n
i=1
β
2
i
1
2
8
c)
X(f)=
∞
−∞
x(t)e
j2 πft
dt =
0
−1
(t +1)e
j2 πft
dt +
1
0
(t − 1)e
j2 πft
dt
=
j...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 4 docx
... +
1
T
m
T
m
T
m
2
e
j
e
j2 πnf
m
t
dt
= −
e
j
T
m
j2 πnf
m
e
j2 πnf
m
t
T
m
2
0
−
e
j
T
m
j2 πnf
m
e
j2 πnf
m
t
T
m
T
m
2
=
(−1)
n
− 1
2πn
j( e
j
− e
j
)=
0 n =2l
2
π(2l+1)
sin(1) n =2l +1
Hence,
e
jm(t)
=
∞
l=−∞
2
π(2l ... 1050950
100
2
J
4
(5)
The Fourier Series expansion of e
j sin(2πf
m
t)
is
c
n
= f
m
5
4f
m
1
4f
m
e
j sin(2πf
m
t)
e
j2 πnf
m...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 5 docx
... b3=1.781477937d+00,
+ b4 =-1 .8 2125 5978d+00, b5=1.330274429d+00)
C-
pi=4.*atan(1.)
C-INPUT
PRINT*, ’Enter -x-’
READ*, x
C-
t=1./(1.+p*x)
a=b1*t + b2*t**2. + b3*t**3. + b4*t**4. + b5*t**5.
q=(exp(-x**2./2.)/sqrt(2.*pi))*a
C-OUTPUT
PRINT*, ... Hence,
C
i ,j
= E[(ω
1
+ ω
2
+ ···+ ω
min(i ,j)
)(ω
1
+ ω
2
+ ···+ ω
min(i ,j)
)]
=
min(i ,j)
k=1
min(i ,j)
l=1
E[ω
k
ω
l
]=
min(i ,j)
k=...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 8 docx
... ,y
n
)=
J
j= 1
f
X(t
1
), ,X(t
n
)
(x
1
, ,x
n
)
|J( x
j
1
, ,x
j
n
)|
where J is the number of solutions to the system
y
1
= Q(x
1
),y
2
= Q(x
2
), ···,y
n
= Q(x
n
)
and J( x
j
1
, ,x
j
n
) is the Jacobian ... length 12/ L, the distortion of the vector quantizer is
D =
1
0
12
L
0
[(x, y) − (
1
2
,
12
2L
)]
2
L
12
dxdy
=
L
12
1
0
12
L
0
(x −
1
2
)
2
+(y...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9 pdf
... (see Equation 6.6.17)
-1
-0 .8
-0 .6
-0 .4
-0 .2
0
0.2
0.4
0.6
0.8
1
-1
-0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0.8 1
g(x)
x
D =
1
12 × 4
ν
∞
−infty
[f
X
(x)]
1
3
dx
3
=
1
12 × 4
ν
3
2
3
we have
SQNR ... where
r(t)=
N
i=1
r
i
ψ
i
(t)
s
m
(t)=
N
i=1
s
m,i
ψ
i
(t)
then,
∞
−∞
r(t)s
m
(t)dt =
∞
−∞
N
i=1
r
i
ψ
i
(t)
N
j= 1
s
m ,j
ψ
j
(t)dt...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 10 pot
... (f)
n
k=1
c
k
e
j2 πfkT
c
and therefore, the Fourier transform of the signal matched to s(t)is
H(f )=S
∗
(f)e
j2 πfT
= S
∗
(f)e
j2 πfnT
c
= P
∗
(f)
n
k=1
c
k
e
j2 πfkT
c
e
j2 πfnT
c
= P
∗
(f)
n
i=1
c
n−i+1
e
j2 πf(i−1)T ... and s
j
(t)is
∞
−∞
s
i
(t)s
j
(t)dt =
∞
−∞
n
k=1
c
ik
p(t −kT
c
)
n
l=1
c
jl
p(t −lT
c
)dt
=
n
k=1
n
l=1
c
ik
c
jl
∞
−∞
p(t −kT
c
)p(t −lT
c...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13 pps
... sequences of length 3 and the corresponding
output of the detector.
-1 -1 -1 -4
-1 -1 1 -2
-1 1 -1 0
-1 11 2
1-1 -1 -2
1-1 1 0
1 1-1 2
111 4
As it is observed there are 5 possible output levels b
m
, ... to
d
E
=2
2
+4
2
+2
2
=24
✉
✉
✉
✉✉
✉
✉
✉✉
✉
✉
✉✉
✉
✉
✉
✒
✲✲
❍
❍
❍
❍
❍❥
✟
✟
✟
✟
✟✯
✲
✒
✟
✟
✟
✟
✟✯
❅
❅
❅
❅
❅❘
❍
❍
❍
❍
❍❥
❍
❍
❍
❍
❍❥...
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 14 pot
... figure
♥
❙
❙♦
❄
❅
❅
❅
❅
❅❘
✲✲
✒
✲
D
2
J
DNJ
D
2
J
D
2
J
DNJ
DNJ
D
3
NJ
X
c
X
b
X
a
X
a
Using the flow graph relations we write
X
c
= D
3
NJX
a
+ DNJX
b
X
b
= D
2
JX
c
+ D
2
JX
d
X
d
= DNJX
c
+ DNJX
d
X
a
= D
2
JX
b
Eliminating ... figure
♥
❙
❙♦
❄
❅
❅
❅
❅
❅❘
✲✲
✒
✲
D
2
J
D
2
NJ
D
3
J
DJ
DNJ
DNJ
D
2
NJ
X
c
X
b
X
a
X
a
Using the flow graph relations w...
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