Solution to EE 504 Test #3 F03
1.
Below are some functions which are candidates for being valid autocorrelation
functions. For each one indicate by circling the correct answer whether it is possible or
impossible that the function could be the autocorrelation function of a real random variable
and, if it is impossible, state why in the space provided.
(a)
R X (τ ) = 10 sinc(20τ )
Possible
Reason: Even function, Maximum at zero, Fourier transform is non-negative
(b)
R X (τ ) = 100(u(τ − 4 ) − u(τ + 4 ))
Impossible
Reason: Value at τ = 0 is negative. Fourier transform is not non-negative everywhere.
(c)
R X (τ ) = 100(u(τ − 2) − u(τ + 6))
Impossible
Reason: Not an even function.
(d)
R X (τ ) = 20 cos( 400πτ )
Possible
Reason: Even function, Maximum at zero, Fourier transform is non-negative
(e)
R X (τ ) = 5e −3τ sin(25πt)
Impossible
Reason: Not an even function
(f)
R X (τ ) = 10 tri(τ ) − 6 tri(2τ )
Impossible
Reason: Maximum not at zero.
(g)
R X (τ ) = 3 tri(τ ) + 4 cos(10πt) − 2
Impossible
Reason: Fourier transform is not non-negative everywhere. Also, using the rule for
finding the expected value of X would produce an imaginary number.
2.
The autocorrelation of a random process is R X (τ ) = 3 tri(τ ) + 4 cos(10πt) + 2.
(a)
What is the expected value, E ( X ) ?
(
)
E ( X ) = lim ± 3 tri(τ ) + 2 = ± 2 = ±1.414
τ →∞
(b)
What is the mean-squared value, E ( X 2 ) ?
E ( X 2 ) = R X (0) = 3 + 4 + 2 = 9
3.
A finite-time integrator has an impulse response,
 t − 2
h( t) = −10[u( t) − u( t − 4 )] = −10 rect 
 .
 4 
Noise, X, with a double-sided PSD of G X ( f ) = 20 + 4δ ( f ) and a positive expected value
excites (is the input signal to) this finite-time integrator.
(a)
The expected value, E (Y ) , of the response (output signal), Y.
R X (τ ) = 20δ (τ ) + 4
∞
E (Y ) = E ( X ) ∫ h( t) dt
E ( X ) = lim
∞
τ →∞
(
−∞
)
R X (τ ) = 2
∫ h(t)dt = −40
−∞
Therefore
E (Y ) = 2(−40) = −80
Alternate Solution:
(
E (Y ) = lim ± R Y (τ )
τ →∞
)
R Y (τ ) = F −1 (GY ( f ))
GY ( f ) = G X ( f ) H( f )
2
H( f ) = −40 sinc( 4 f )e − j 4 πf ⇒ H( f ) = 1600 sinc 2 ( 4 f )
2
[
]
GY ( f ) = [20 + 4δ ( f )]1600 sinc 2 ( 4 f ) = 32000 sinc 2 ( 4 f ) + 6400δ ( f )
R Y (τ ) =
32000  t 
tri  + 6400
 4
4
 32000  t 

E (Y ) = lim  ±
tri  + 6400  = ±80
τ →∞
 4
4


This gets the expected value of Y to within a sign but does not resolve the sign like the first
method does.
(b)
The mean-squared value, E (Y 2 ) , of the response, Y.
E (Y 2 ) = R Y (0)
R Y (τ ) = R X (τ ) ∗ h(τ ) ∗ h(−τ )



 τ − 2   
 −τ − 2   
h(τ ) ∗ h(−τ ) = −10 rect 
   ∗ −10 rect 

 4   
 4   



 τ + 2
 τ − 2
h(τ ) ∗ h(−τ ) = 100 rect 

 ∗ rect 
 4 
 4 

τ
h(τ ) ∗ h(−τ ) = 100 4 tri  
 4 

τ
R Y (τ ) = [20δ (τ ) + 4 ] ∗ 400 tri 
 4

τ
τ
R Y (τ ) = 400 20δ (τ ) ∗ tri  + 4 ∗ tri  


 4 
4



τ
R Y (τ ) = 400 20 tri  + 16 
 4


R Y (0) = 400[20 tri(0) + 16] = 400 × 36 = 14400
Alternate Solution:
E (Y
∞
2
∞
) = ∫ G ( f )df = ∫ G ( f ) H( f )
Y
X
−∞
2
df
−∞
H( f ) = −40 sinc( 4 f )e − j 4 πf ⇒ H( f ) = 1600 sinc 2 ( 4 f )
2
E (Y
∞
2
∞
) = ∫ [20 + 4δ ( f )]1600 sinc (4 f )df = 1600 ∫ [20 + 4δ ( f )]sinc (4 f )df
2
2
−∞
−∞
∞
∞

2
E (Y ) = 1600  ∫ 20 sinc ( 4 f ) df + ∫ 4δ ( f ) sinc 2 ( 4 f ) df 
−∞
−∞

2
E (Y
∞
2
∞
) = 32000 ∫ sinc (4 f )df + 6400 ∫ δ ( f ) sinc (4 f )df = 8000 + 6400 = 14400
2
144244
3
−∞
1 t
1
= tri   =
4  4  t =0 4
2
−∞
4.
A circuit consists of a resistor, R, at an absolute temperature, T, and a capacitor, C,
in parallel. (The double-sided PSD of the short-circuit Johnson noise current of a resistor
2kT
and the double-sided PSD of the open-circuit Johnson noise voltage is 2kTR where
is
R
J
k is Boltzmann’s constant, 1.38 × 10 −23 .)
K
(a)
Find a general formula for the mean-squared voltage across the capacitor (and the
resistor) in terms of k, R, T and C. This formula should contain no integral signs, Fourier
transforms or convolution operators.
You may use the integration formula,
dx
1
bx

−1 
∫ a2 + (bx )2 = ab tan  a  , if needed during the process of finding the formula.
Using a current source model for the Johnson noise of the resistor, the transfer function
from the noise current to the capacitor voltage is the impedance of the parallel R-C circuit.
The mean-squared noise voltage is
E (V
∞
2
) = ∫ G ( f )df
ov
−∞
2
G ov ( f ) = G i ( f ) H i ( f ) = G i ( f ) H*i ( f ) H i ( f )
2 kT
Gi( f ) =
R
R
R
j 2πfC
Hi( f ) =
=
1
j 2πfRC + 1
R+
j 2πfC


R
R
2 kT 
G ov ( f ) =



R  j 2πfRC + 1  − j 2πfRC + 1
R2
2 kT
2 kTR
=
G ov ( f ) =
2
R (2πfRC ) + 1 (2πRC ) 2
∞
2 kTR
E (V ) =
(2πRC) 2 −∞∫
1
 1 
f 2 +

 2πRC 
2
df
2
2
 1 
f +

 2πRC 
∞
kT
2πRCkT
kT
E (V 2 ) = 2 2 2πRC tan −1 (2πRCf ) −∞ =
2
2 π =
C
2π RC
2π RC
[
2
]
Alternate Solution:
If we use a voltage source model for the resistor noise,
G ov ( f ) = G v ( f ) H v ( f ) = G v ( f ) H*v ( f ) H v ( f )
2
G v ( f ) = 2 kTR
1
1
j 2πfC
Hv ( f ) =
=
1
j 2πfRC + 1
R+
j 2πfC



1
1
G ov ( f ) = 2 kTR


 j 2πfRC + 1  − j 2πfRC + 1
G ov ( f ) = 2 kTR
1
(2πfRC)
2
+1
=
2 kTR
(2πRC) 2
1
 1 
f +

 2πRC 
2
2
From this step on, the solution is the same as the previous one.
(b)
You should find that the mean-squared voltage is the independent of resistance, R.
By sketching the PSD of the noise voltage across the capacitor for the two cases,
R = 1 , C = 1 , T = 300 and R = 2 , C = 1 , T = 300 , explain how that can happen even
though the PSD of the resistor Johnson noise current (or voltage) noise is a function of R.
-20
1.8
PSD of Noise Voltage Across Capacitor for Two Different Resistances
x 10
1.6
1.4
1.2
R=1
0.8
R=2
G(f)
1
0.6
0.4
0.2
0
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
Frequency, f (Hz)
0.4
0.6
0.8
1
The two PSD’s have different peak amplitudes and different shapes but their areas are the
same.