Set-1 Final by Ali (1-9) (R).p65 - SIA GROUP

Probability Theory and Stochastic Processes (May-2012, Set-1) JNTU-Kakinada
Code No: R21043/R10
II B.Tech. I Semester Supplementary Examinations
May - 2012
PROBABILITY THEORY AND STOCHASTIC PROCESSES
( Electronics and Communications Engineering )
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S.1
Time: 3 Hours Max. Marks: 75
Answer any FIVE Questions
All Questions carry equal marks
- - -
1. (a) A survey of 100 companies shows that 75 of them have installed wireless local area network (WLANS) on their
premises. If three of these companies are chosen at random without replacement, what is the probability that
each of the three has installed WLANS. (Unit-I, Topic No. 1.4)
(b) State and prove the Baye’s theorem. [8+7] (Unit-I, Topic No. 1.5)
⎧ x
⎪
2. (a) The PDF of a random variable X is given by f x
(x) = ⎨2
− x
⎪
⎩ 0
(i) Find the CDF of X
(ii) Find P [0.2 < X < 0.8]
(iii) Find P [0.6 < X < 1.2] (Unit-II, Topic No. 2.2)
0 < x < 1
1 ≤ x ≤ 2
otherwise
(b) Write short notes on “Gaussian distribution”. [8+7] (Unit-II, Topic No. 2.3)
3. (a) The PDF of a random variable X is given by f x
(x) = 4x(9 – x 2 )/ 81 0 ≤ x ≤ 3. Find the mean, variance and third
moment of X. (Unit-III, Topic No. 3.3)
⎧1
⎪
(b) The random variable X has the following PDF f x
(x) = ⎨3
⎪⎩ 0
−1
< x < 2
Otherwise
If we desire Y = 2X + 3, what is the PDF of Y. [7+8] (Unit-III, Topic No. 3.4)
4. (a)
2 2
Two random variables X and Y have zero mean and variance σ x
= 16 and σ y = 36. If their correlation
coefficient is 0.5 determine the following,
(i) The variance of the sum of X and Y
(ii) The variance of the difference of X and Y. (Unit-IV, Topic No. 4.3)
(b) State and prove the ‘Central limit theorem’. [8+7] (Unit-IV, Topic No. 4.4)
⎧ 1
⎪
24
5. (a) A joint probability density function is f xy
(x, y) = ⎨
⎪ 0
⎪
⎩
(b)
0 < x < 6
0 < y < 4
elsewhere
Find out the expected value of the function g(X, Y) = (XY) 2 . (Unit-V, Topic No. 5.1)
Set-1
Solutions
Prove that the moment generating function of the sum of two independent variables is the product of their
moment generating function. [8+7] (Unit-V, Topic No. 5.1)

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012
6. (a) Explain the following,
(i) Stationarity
(ii) Ergodicity
(iii) Statistical independence with respect to random processes. (Unit-VI, Topic No. 6.4)
(b) State and prove the properties of cross correlation function. [8+7] (Unit-VI, Topic No. 6.5)
7. (a) A random process Y(t) has the power spectral density S YY
(ω) =
Find,
(i) The average power of the process
(ii) The auto correlation function. (Unit-VII, Topic No. 7.1)
9
ω 2 + 64
(b) State the properties of power spectral density. [8+7] (Unit-VII, Topic No. 7.1)
8. Write short notes on the following,
(i) Thermal noise (Unit-VIII, Topic No. 8.5)
(ii) Effective noise temperature. [7+8] (Unit-VIII, Topic No. 8.5)
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Probability Theory and Stochastic Processes (May-2012, Set-1) JNTU-Kakinada
S.3
SOLUTIONS TO MAY-2012, SET-1, QP
Q1. (a) A survey of 100 companies shows that
75 of them have installed wireless local
area network (WLANS) on their premises.
If three of these companies are chosen
at random without replacement, what is
the probability that each of the three has
installed WLANS.
Answer :
May-12, Set-1, Q1(a) M[8]
Given that,
In a survey,
Total number of companies = 100
Number of WLAN installed companies = 75
Number of WLAN installed companies chosen at
random without replacement = 3
P(All the three WLAN installed companies) =
Let A be the event of selecting first company that
installed WLAN.
Then, the probability of A is given by,
P(A)=
Number of
75c
=
100
75
= 100
1
c1
∴ P( A)
= 0.75
companies installed WLAN
companies
Total number of
Now, the total number of companies remained = 99
Number of WLAN installed companies remained = 74
Let B be the event of selecting second company that
installed WLAN.
Then, the probability of selecting B after selecting A
is given by,
P(B)=
74
99
c1
c1
74
= = 0.7475 99
∴ P( B)
= 0.7475
Now,
The total number of companies remained = 98
Number of WLAN installed companies remained = 73
Let C be the event of selecting third company that
installed WLAN.
Then, the probability of selecting C after selecting A
and B is given by,
73c1
P(C) =
98
c1
73
= = 0.7449 98
∴ P( C)
= 0. 7449
∴ The probability that all the three companies have
installed WLAN is given by,
P(A ∩ B ∩ C) = P(A) P(B) P(C)
[Q Selection of companies A, B and
C is independent to each other]
= 0.7449 × 0.7475 × 0.75
= 0.4176
∴ P( A ∩ B ∩C)
= 0.4176
(b) State and prove the Baye’s theorem.
Answer :
May-12, Set-1, Q1(b) M[7]
For answer refer Unit-I, Q23.
Q2. (a) The PDF of a random variable X is given
⎧ x
⎪
by f x
(x) = ⎨2
− x
⎪
⎩ 0
(i)
Find the CDF of X
0 < x < 1
1≤
x ≤ 2
otherwise
(ii) Find P[0.2 < X < 0.8]
(iii) Find P[0.6 < X < 1.2]
Answer :
May-12, Set-1, Q2(a) M[8]
The given probability density function of a random
variable ‘X’ is,
⎧ x
⎪
F X
(x) = ⎨ 2 − x
⎪⎩
0
0 < x < 1
1 ≤ x ≤ 2
Otherwise
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S.4 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012
(i)
(i) CDF of X, F X
(x) =
(ii) P(0.2 < X < 0.8) =
(ii) P(0.6 < X < 1.2) =
The cumulative distributive function of ‘X’ is given
by,
∞
F X
(x) = ∫
f X ( x)
dx
−∞
1
=
∫xdx
+
∫
2 −
0
1
⎡
2
x
=
2 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
0
2
1
( x)
dx
+ 2 [] x 2 1 –
⎡
2
x
2 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
⎡1 2 − 0⎤
⎡2
= ⎢ ⎥ + 2 [2 – 1] – ⎢
⎢⎣
2 ⎥⎦
⎢⎣
1 ⎡ 4 −1⎤
= + 2 [1] – ⎢ ⎥⎦ 2 ⎣ 2
2
1
2
−
2
2
1 ⎤
⎥
⎥⎦
(iii)
0.64
− 0.04
=
2
= 0.6/2
= 0.3
∴ P( 0.2 < X < 0.8 = 0.3
The value of P(0.6 < X < 1.2) is obtained as,
1. 2
P(0.6 < X < 1.2) = ∫
0.6
1
f X
( x)
dx
= ∫
f X x)
dx +
∫
0.6
1
1.2
( f ( x)
dx
=
∫xdx
+
∫
2 −
0.6
1
⎡
2
x
=
2 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
0.6
1.2
1
1
X
( x)
dx
⎡
2
1. 2
x
+ 2 [x] 1
–
2 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
1.2
1
(ii)
∴
F X
= 2
1 + 2 + 2
3
1 + 4 + 3
=
2
= 2
8
= 4
( x)
= 4
The value of P(0.2 < X < 0.8) is obtained as,
0.8
P(0.2 < X < 0.8) = ∫
0.2
f X
0
= ∫ . 8
xdx
0. 2
⎡
2
x
=
2 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
( x)
dx
0.8
0.2
⎡
2 2
1 − (0.6) ⎤
⎡
2 2
(1.2) −1
⎤
= ⎢ ⎥ + 2 [1.2– 1] – ⎢ ⎥
⎢⎣
2 ⎥⎦
⎢⎣
2 ⎥⎦
⎡1−
0.36⎤
⎡1.44
−1⎤
= ⎢ ⎥ + 2 (0.2) –
⎣ 2
⎢ ⎥ ⎦ ⎣ 2 ⎦
=
0.64
2
∴ P( 0.6 < X < 1.2) = 0.5
+ 0.4 –
= 0.32 + 0.4 – 0.22
= 0.5
0.44
2
(b) Write short notes on “Gaussian distribution”.
Answer :
May-12, Set-1, Q2(b) M[7]
For answer refer Unit-II, Q20.
Q3. (a) The PDF of a random variable X is given
by f x
(x) = 4x(9 – x 2 )/ 81 0 ≤ x ≤ 3. Find the
mean, variance and third moment of X.
Answer :
May-12, Set-1, Q3(a) M[7]
The given probability density function of a random
variable ‘X’ is,
=
(0.8)
2 −
(0.2)
2
2
f X
(x) =
2
4x(9
− x )
, 0 ≤ x ≤ 3
81
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Probability Theory and Stochastic Processes (May-2012, Set-1) JNTU-Kakinada
Mean, E[X] =
Consider,
2
Variance, σ X
=
∞
Third moment of X, E[X 3 E[X 2 2
]=
] =
∫
x f X ( x)
dx
−∞
The mean of random variable ‘X’ is given by,
3
2
∞
2 4x(9
− x )
=
E[X] = ∫
xf X ( x)
dx
∫
x
dx
81
0
−∞
3
3
2
4 3 5
x x
= ∫ x 4 (9 − )
=
dx
81
∫
( 9x
− x ) dx
81
0
0
= 4 3
⎧
3 3
⎫
4 ⎪ ⎡
4
⎤ ⎡
6
x x ⎤ ⎪
2 2
81
∫
x ( 9 − x ) dx
= ⎨9⎢
⎥ − ⎢ ⎥ ⎬
81 ⎢
0
⎪⎩ ⎣ 4 ⎥⎦
⎢⎣
6
0
⎥⎦
0 ⎪⎭
= 4 3
4
2 4
⎪⎧
⎛
4
⎞ ⎛
6
⎞⎪⎫
81
∫
( 9x
− x ) dx
= ⎨ ⎜
3 − 0
⎟ ⎜
3 − 0
9 − ⎟
81
⎬
0
⎪⎩ ⎝ 4 ⎠ ⎝ 6 ⎠⎪⎭
⎧
3 3
⎫
4 ⎪ ⎡
3
⎤ ⎡
5
4 ⎧ ⎛ 81⎞
⎛ 729 ⎞⎫
x x ⎤ ⎪
= ⎨9⎜
⎟ − ⎜ ⎟⎬
= ⎨9⎢
⎥ − ⎢ ⎥ ⎬
81 ⎩ ⎝ 4 ⎠ ⎝ 6 ⎠⎭
81 ⎢ ⎥ ⎢ ⎥
⎪⎩ ⎣ 3 ⎦0
⎣ 5 ⎦ 0 ⎪⎭
4 81 4 729
= × 9 × – ×
4 ⎪⎧
⎡
3
3 − 0 ⎤ ⎡
5
3 − 0⎤⎪⎫
81 4 81 6
= ⎨9⎢
⎥ − ⎢ ⎥⎬
81 ⎪⎩ ⎢⎣
3 ⎥⎦
⎢⎣
5 ⎥⎦
= 9 – 6
⎪⎭
= 3
4 ⎧ ⎛ 27 ⎞ ⎛ 243⎞⎫
∴ E[X 2 ]= 3
= ⎨9⎜
⎟ − ⎜ ⎟⎬
81 ⎩ ⎝ 3 ⎠ ⎝ 5 ⎠⎭
Then,
2
σ
4 4 243
X
= E[X 2 ] – E[X] 2
= × 3 × 27 – × 81 81 5
= 3 – (1.6) 2
= 0.44
12
= 4 –
5
∴Variance,
σ 2 X = 0.44
20 −12
The third moment of ‘X’ is given by,
=
5
∞
E[X 3 3
]=
8
∫
x f X ( x)
dx
=
−∞
5
3
2
= 1.6
3 4x(9
− x )
=
∫
x
dx
81
∴ Mean, E[
X ] = 1.6
0
The variance of random variable ‘X’ is given by,
3
4 4 2
σ = E[X 2 ] – E[X] 2 = 81
∫
x ( 9 − x ) dx
2
X
0
S.5
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S.6 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012
3
= 4 4 6
81
∫
( 9x
− x ) dx
4
= 81
4
= 81
4
= 81
0
⎧
⎪ ⎡
5
x ⎤
⎨9⎢
⎥
⎢ ⎥ ⎪⎩ ⎣ 5 ⎦
3
0
3
⎡
7 ⎫
x ⎤ ⎪
− ⎢ ⎥ ⎬
⎢⎣
7 ⎥⎦
0 ⎪⎭
⎪⎧
⎛
5
⎞ ⎛
7
⎞⎪⎫
⎨ ⎜
3 − 0
⎟ ⎜
3 − 0
9 − ⎟
⎬
⎪⎩ ⎝ 5 ⎠ ⎝ 7 ⎠⎪⎭
⎧ ⎛ 243 ⎞
⎨9⎜
⎩ ⎝ 5
⎟ −
⎠
⎛ 2187 ⎞⎫
⎜ ⎟⎬
⎝ 7 ⎠⎭
4 243 4 2187
= × 9 × – × 81 5 81 7
=
108 108 –
5 7
⎡ 7 − 5⎤
= 108 ⎢ ⎥
⎣ 35 ⎦
216
= 35
= 6.17
∴ Third moment of X , E[
X 3 ] = 6.17
(b)
The random variable X has the following
PDF f x
(x) =
⎧1
⎪
⎨3
⎪⎩ 0
−1<
x < 2
Otherwise
If we desire Y = 2X + 3, what is the PDF
of Y.
Answer :
May-12, Set-1, Q3(b) M[8]
The given probability density function of random
variable ‘X’ is,
⎧1
⎪
F X
(x) = ⎨ 3
⎪⎩ 0
−1
< x < 2
Otherwise
Transformation function, Y = 2 X + 3.
Probability density function of Y, f Y
(y) =
From the monotonic transformation function, the
probability density function of transformation ‘Y’ is given
by,
dx
f Y
(y)= f X
(x)
dy
From the transformation ‘Y’ we have,
Y = 2X + 3
Y − 3
X =
2
or
y − 3
x =
2
Limits
For x = – 1, y = 2(– 1) + 3
= – 2 + 3
y = 1
For x = 2, y = 2(2) + 3
= 4 + 3
y = 7
... (1)
... (2)
By differentiating the equation (2) with respect to ‘y’
we get,
dx 1 = ... (3)
dy 2
By substituting equations (2) and (3) in equation (1),
we get,
∴
f Y
⎛ y − 3 ⎞
f Y
(y)= f ⎜ ⎟
X ⎝ 2 ⎠
( y)
=
3
1
=
=
1
6
0
1 1 =
2 6
;
;
1
2
1 < y < 7
Otherwise
Q4. (a) Two random variables X and Y have
2
zero mean and variance σ x
= 16 and
2
σ y = 36. If their correlation coefficient
is 0.5, determine the following,
(i) The variance of the sum of X and Y
(ii) The variance of the difference of X
and Y.
May-12, Set-1, Q4(a) M[8]
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Probability Theory and Stochastic Processes (May-2012, Set-1) JNTU-Kakinada
Answer :
Given that,
For two random variables X and Y,
Variance of X,
2
σ X
= 16
S.7
2
Variance of Y, σ Y
= 36
Correlation coefficient, ρ XY
= 0.5
(i) The variance of sum of X and Y,
2
( X + Y )
σ =
(i)
2
( X −Y )
(ii) The variance of difference of X and Y, σ =
The variance of the sum of X and Y is obtained as,
2
( X + Y )
σ = E[(X + Y) 2 ] – (E[X + Y]) 2
= E[X 2 + 2 × Y + Y 2 ] – (E[X] + E[Y]) 2
= E[X 2 ] + 2 E[X Y] + E[Y 2 ] – (E[X] 2 + 2 E[X] E[Y] + E[Y] 2 )
= E[X 2 ] + 2 E[X Y] + E[Y 2 ] – E[X] 2 – 2 E[X] E[Y] – E[Y] 2
= E[X 2 ] – E[X] 2 + E[Y 2 ] – E[Y] 2 + 2(E[X Y] – E[X] E[Y])
2 2
= σ X
+ σ Y
+ 2(E[X Y] – E[X] E[Y]) ... (1)
The correlation coefficient of X and Y is given by,
ρ XY
=
E[
XY]
− E[
X ] E[
Y ]
σ σ
X
Y
E[X Y] – E[X] E[Y] = ρ XY
σ X
σ Y
= 0.5 × 4 × 6
= 12 ... (2)
By substituting equation (2) in equation (1), we get,
2
( X + Y )
σ
2 2
= σ X
+ σ Y
+ 2(E[X Y] – E[X] E[Y])
= 16 + 36 + 2 (12)
= 76
(ii)
2
∴ X +Y )
The variance of sum of X and Y , σ ( = 76
The variance of the difference of X and Y is given by,
2
( X −Y )
σ = E[(X – Y) 2 ] – (E[X – Y]) 2
= E[X 2 – 2 X Y + Y 2 ] – (E[X] – E[Y]) 2
= E[X 2 ] – 2 E[X Y] + E[Y 2 ] – (E[X] 2 – 2 E[X] E[Y] + E[Y] 2 )
= E[X 2 ] – 2 E[X Y] + E[Y 2 ] – E[X] 2 + 2 E[X] E[Y] – E[Y] 2
= (E[X 2 ] – E[X] 2 ) + (E[Y 2 ] – E[Y]) 2 – 2(E[X Y] – E[X] E[Y])
2 2
= σ X
+ σ Y
– 2(E[X Y] – E[X] E[Y])
= 16 + 36 – 2 (12)
= 28
2
X −Y )
∴ The variance of difference of X and Y , σ ( = 28
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S.8 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012
(b) State and prove the ‘Central limit theorem’.
Answer :
May-12, Set-1, Q4(b) M[7]
For answer refer Unit-IV, Q26.
Q5. (a) A joint probability density function is
f XY
(x, y) =
⎧ 1
⎪
24
⎨
⎪ 0
⎪
⎩
0 < x < 6
0 < y < 4
Elsewhere
Find out the expected value of the function
g(X, Y) = (XY) 2
Answer :
May-12, Set-1, Q5(a) M[8]
Given that,
For a two random variables X and Y,
Joint probability density function,
⎧ 1
⎪
24
f XY
(x, y)= ⎨
⎪ 0
⎪
⎩
0 < x < 6
0 < y < 4
Elsewhere
g(X, Y)= (X Y) 2
Expected value of g(X, Y), E[g(X, Y)] =
Then, the expected value of g(X, Y) is obtained as,
∞
∞
E[g(X, Y)] = ∫∫
g(
X , Y ) f XY ( x,
y)
dxdy
−∞ −∞
6 4
= ∫∫
0 0
1
= 24
1
=
24
1
=
24
1
=
24
= 64
∴E[ g(
X , Y)
= 64
2 1
( XY)
dxdy
24
6
∫
0
X
2
dx
⎡
3
X
3 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
6
0
4
∫
0
Y
2
dy
3
⎡Y
3 ⎥ ⎥ ⎤
⎢
⎢⎣
⎦
⎡6 3 − 0⎤
⎡ 4 3 − 0
⎢ ⎥
⎢⎣
3 ⎥⎦
⎥ ⎥ ⎤
⎢
⎢⎣
3 ⎦
216
3
64
3
4
0
(b) Prove that the moment generating
function of the sum of two independent
variables is the product of their moment
generating function.
Answer :
May-12, Set-1, Q5(b) M[7]
Let, A 1
and A 2
are two independent random variables
with moment generating function MA ( ) and M A ( ) .
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1 V
2 V
From the definition of moment generating function,
M A
(V)= E[e VA ]
Then, the moment generating function for the sum of
two independent random variables is given by,
M A + A ( V ) = E[
e ]
1
2
V ( A 2 ) 1 + A
2
= E[ e VA . e ]
1 VA
Since, A 1
and A 2
are independent random variables.
E[A 1
A 2
]= E[A 1
] E[A 2
]
⇒ M A + A ( V )
2
= E[ e
VA 1 ]. E[
e
VA ]
1
∴
1+
2
M A A ( V ) = M
2
A1
( V ). M
A2
( V )
Hence, proved.
Q6. (a) Explain the following,
(i) Stationarity
(ii) Ergodicity
(iii) Statistical independence with
respect to random processes.
Answer :
May-12, Set-1, Q6(a) M[8]
(i) Stationarity
For answer refer Unit-VI, Q5, Topic: Stationary
Random Process.
(ii) Ergodicity
For answer refer Unit-VI, Q13, Topic: Ergodic Random
Process.
(iii) Statistical Independence with Respect to Random
Processes
For answer refer Unit-VI, Q5, Topic: Statistical
Independence.
(b) State and prove the properties of cross
correlation function.
Answer :
May-12, Set-1, Q6(b) M[7]
Cross Correlation
For answer refer Unit-VI, Q20, Topic: Cross Correlation.
Cross Correlation Properties
For answer refer Unit-VI, Q19(ii).

Probability Theory and Stochastic Processes (May-2012, Set-1) JNTU-Kakinada
Q7. (a) A random process Y(t) has the power
spectral density S YY
(ω) =
Find,
(i) The average power of the process
(ii) The auto correlation function.
Answer :
May-12, Set-1, Q7(a) M[8]
The given power spectral density of a random process
Y(t) is,
ω
9
2 +
64
(ii)
S.9
The auto correlation function of random process Y(t)
is given by,
R YY
(τ) = F.T –1 [S YY
(ω)]
⎡ 9 ⎤
= F.T –1 ⎢ 2 ⎥
⎣ω + 64 ⎦
9
= F.T
–1
⎡
⎢ 8 ⎣ ω
2
8
+ 8
2
⎤
⎥
⎦
(i)
9
S YY
(ω) =
ω 2 + 64
(i) Average power of the process, P YY
=
(ii) Auto correlation function, R YY
(τ) =
The average power of the process Y(t) is given by,
9
= sin 8τ 8
⎡ −1
⎡ a ⎤ ⎤
⎢Q F.
T ⎢ ⎥ = sin at u(
t)
2 2
⎥
⎣ ⎣a
+ ω ⎦ ⎦
P YY
=
=
=
=
=
=
1
2π
1
2π
9
2π
9
2π
∞
∫
−∞
S YY ( ω)
dω
∞
∫
−∞
∞
∫
−∞
9
dω
2
ω + 64
ω
2
⎡
⎢ tan
⎣8
1
+ 8
1 − 1
2
dω
⎛ ω ⎞⎤
⎜ ⎟⎥
⎝ 8 ⎠⎦
∞
−∞
9 1 × [tan –1 (∞) – tan –1 (– ∞)]
2π 8
9 1 ×
2π 8
⎡
⎢tan
⎣
−1
⎛ π ⎞ −1⎛
π ⎞⎤
⎜ tan ⎟ + tan ⎜ tan ⎟⎥
⎝ 2 ⎠ ⎝ 2 ⎠⎦
∴
R XX
9
( τ)
= sin8τu(
τ)
8
(b) State the properties of power spectral
density.
Answer :
May-12, Set-1, Q7(b) M[7]
For answer refer Unit-VII, Q1.
Q8. Write short notes on the following,
(i) Thermal noise
(ii) Effective noise temperature.
Answer :
May-12, Set-1, Q8 M[7+8]
(i) Thermal Noise
For answer refer Unit-VIII, Q19, Topic: Thermal Noise.
(ii) Effective Noise Temperature
For answer refer Unit-VIII, Q21(i).
=
9
16π
⎡ π π ⎤
⎢ + ⎥
⎣ 2 2 ⎦
=
= 16
9
9 [π]
16π
∴ Average power, P =
XX
9
16
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