Download E4702 HW#1 solutions

E4702 HW#1 solutions
by Anmo Kim ([email protected])
office hour: 2-3PM TUE&THR (1247 Mudd)
1. (P1.15) A running integrator is defined by
Z
t
y(t) =
x(τ )dτ
(1)
t−T
where x(t) is the input, y(t) is the output, and T is the integration period. Both x(t) and y(t) are sample
functions of stationary processes X(t) and Y (t), respectively. Show that the power spectral density of the
integrator output is related to that of the integrator input by
SY (f ) = T 2 sinc2 (f T )SX (f )
(2)
(Solution)
The relation between SY (f ) and SX (f ) can be described by (1.58) in Haykins.
SY (f ) = |H(f )|2 SX (f )
Thus, we need to show that |H(f )|2 = T 2 sinc2 (f T ).
The impulse response h(t) can be directly obtained from (1) by giving the impulse function, δ(t), as an input.
h(t) = {
1,
0,
0≤t≤T
otherwise
By the Fourier transform,
Z
H(f ) =
T
1 · e−j2πf t dt
0
¡ −j2πf T
¢
1
e
−1
−j2πf
1
=
· [sin 2πf T + j(cos 2πf T − 1)]
2πf
=
Then,
£
¤
1
· (sin 2πf T )2 + (cos 2πf T − 1)2
(2πf )2
1
· [2 − 2 cos 2πf T ]
=
(2πf )2
|H(f )|2 =
=
4 sin2 πf T
(2πf )2
= T 2 sinc2 (f T )
Thus,
SY (f ) = T 2 sinc2 (f T )SX (f )
1
2. (P1.16) A zero-mean stationary process X(t) is applied to a linear filter whose impulse response is defined by
a truncated exponential:
ae−at , 0 ≤ t ≤ T
h(t) = {
0,
otherwise
Show that the power spectral density of the filter output Y (t) is
SY (f ) =
a2
a2
(1 − 2 exp(−aT ) cos(2πf T ) + exp(−2aT ))SX (f )
+ 4π 2 f 2
where SX (f ) is the power spectral density of the filter input.
(Solution)
Since X(t) is a stationary process, then as in (P1.15), SY (f ) can be described by the following equation.
SY (f ) = |H(f )|2 SX (f )
The Fourier transform of the given h(t) is
Z
∞
H(f ) =
h(t)e−j2πf t dt
−∞
Z T
ae−(a+j2πf )t dt
h
i
a
=−
· e−(a+j2πf )T − 1
a + j2πf
=
0
Then,
∗
|H(f )|2 = H(f )H(f )
½
h
i¾ ½
h
i¾
a
a
−(a+j2πf )T
−(a−j2πf )T
= −
· e
−1 · −
· e
−1
a + j2πf
a − j2πf
£
¤
a2
= 2
· 1 − 2e−aT cos(2πf T ) + e2aT
a + 4π 2 f 2
Thus,
SY (f ) =
a2
a2
(1 − 2 exp(−aT ) cos(2πf T ) + exp(−2aT ))SX (f )
+ 4π 2 f 2
3. (P1.17) The output of an oscillator is described by
X(t) = A cos(2πF t − Θ)
where A is a constant, and F and Θ are independent random variables. The probability density function of
Θ is defined by
1
, 0 ≤ θ ≤ 2π
fΘ (θ) = { 2π
0, otherwise
Find the power spectral density of X(t) in terms of the probability density function of the frequency F . What
happens to this power spectral density when the frequency assumes a constant value?
(Solution)
Because Θ and F are independent, the joint probability density function becomes:
fΘ,F (θ, x) = fΘ (θ) · fF (x)
2
Then, autocorrelation function of X(t) becomes:
RX (t, u) = E[X(t)X(u)]
Z Z
=
fΘ,F (θ, x)X(t)X(u)dθdx
Z
x θ
∞ Z 2π
=
−∞
1
=
2π
2
=
A
2π
Z
0
∞
1
fF (x)X(t)X(u)dθdx
2π
Z 2π
fF (x)
A2 cos(2πxt − θ) · cos(2πxu − θ)dθdx
−∞
∞
0
Z
fF (x)
−∞
2 Z ∞
=
Let τ = t − u,
A
2
1
2
Z
2π
[cos(2πx(t + u) − 2θ) + cos(2πx(t − u))] dθdx
0
fF (x) cos(2πx(t − u))dx
−∞
A2
RX (τ ) =
2
Z
∞
fF (x) cos(2πxτ )dx
(3)
−∞
And RX (τ ) can be represented as a Fourier transform pair of power spectral density, SX (x):
Z ∞
RX (τ ) =
SX (x) exp(j2πxτ )dx
(4)
−∞
Since RX (τ ) is a real and even function,
Z
∞
RX (τ ) =
SX (x) cos(2πxτ )dx
(5)
−∞
From the similarity between (3) and (5), we can get SX (x) without calculating the Fourier transform of
RX (τ ) directly.
SX (x) =
A2
fF (x)
2
If the frequency x is a constant value, x0 , fF (x) = 12 δ(x − x0 ) + 21 δ(x + x0 ). Thus, SX (x) becomes:
SX (x) =
A2
A2
δ(x − x0 ) +
δ(x + x0 )
4
4
3