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EE 5375/7375 Random Processes October 23, 2003
Homework #7 Solutions
Problem 1. textbook problem 7. Let p(x) be the rectangular function shown in Fig. P7.2. Is RX (τ ) = p(τ /T ) a valid autocorrelation function?
RX (τ ) is a rectangle with height A in the interval (−T, T ). According to Appendix B in the textbook, its Fourier transform is A 2 T sin(2 2 πf Tπf T ). Since this is negative sometimes, but power spectral densities must always be non-negative, p(τ /T ) cannot be a valid autocorrelation function.
Problem 2. textbook problem 7. Let Z(t) = X(t) + Y (t). Under what conditions does SZ (f ) = SX (f ) + SY (f )?
We know (from notes and Example 7.4 in the textbook) that
RZ (τ ) = RX (τ ) + RY,X (τ ) + RX,Y (τ ) + RY (τ )
and SZ (f ) = SX (f ) + SY,X (f ) + SX,Y (f ) + SY (f )
It is clear that SZ (f ) = SX (f ) + SY (f ) only if RY,X (τ ) + RX,Y (τ ) = 0. This would be true, for instance, if RY,X (τ ) = RX,Y (τ ) = 0 for all τ ; then X and Y are said to be orthogonal to each other.
Problem 3. textbook problem 7. Show that (a) RX,Y (τ ) = RY,X (−τ ) (b) SX,Y (f ) = S∗ Y,X (f ).
(a) RX,Y (τ ) = E(X(t + τ )Y (t)) = E(Y (t)X(t + τ )) = RY,X (−τ ) (b) The spectral density is the Fourier transform of the autocorrrelation function:
SX,Y (f ) =
−∞
RX,Y (τ )e−j^2 πf τ^ dτ
=
−∞
RY,X (−τ )e−j^2 πf τ^ dτ
=
−∞
RY,X (τ )ej^2 πf τ^ dτ
= S Y,X∗ (f )
The last step follows from observing that the usual Fourier transform uses e−j^2 πf τ^ = cos(2πf τ )−j sin(2πf τ ) in the integral, but the second to last step has ej^2 πf τ^ = cos(2πf τ ) + j sin(2πf τ ). Therefore the result will be the complex conjugate of the usual Fourier transform.
Problem 4. textbook problem 7. Let Y (t) = X(t) − X(t − d). (a) Find RX,Y (τ ) and SX,Y (f ). (b) Find RY (τ ) and SY (f ).
(a) We find RX,Y (τ ) = E(X(t + τ )(X(t) − X(t − d))) = E(X(t + τ )X(t)) − E(X(t + τ )X(t − d)) = RX (τ ) − RX (τ + d)
Taking the Fourier transform, SX,Y (f ) = SX (f ) − SX (f )ej^2 πf d
1
where we have used the property that a time shift of T in the time domain corresponds to an additional exponential factor in the frequency domain, for example, a Fourier transform pair is
R(τ − T ) ⇔ S(f )e−j^2 πf T
(b) Similarly, we find
RY (τ ) = E((X(t + τ ) − X(t + τ − d))(X(t) − X(t − d))) = 2RX (τ ) − RX (τ + d) − RX (τ − d)
Taking the Fourier transform,
SY (f ) = 2SX (f ) − SX (f )ej^2 πf d^ − SX (f )e−j^2 πf d = 2SX (f )(1 − cos(2πf d))
Problem 5. textbook problem 7. Let X(t) and Y (t) be independent wide-sense stationary random processes, and define Z(t) = X(t)Y (t). (a) Show that Z(t) is wide-sense stationary. (b) Find RZ (τ ) and SZ (f ).
(a) To be WSS, Z(t) must have a constant mean and an autocorrelation function that depends on only on the lag. The mean is E(Z(t)) = E(X(t)Y (t)) = E(X(t))E(Y (t)) = μX μY
which is constant because μX and μY are constants. The autocorrelation is
RZ (t, t + τ ) = E(X(t)Y (t)X(t + τ )Y (t + τ )) = E(X(t)X(t + τ ))E(Y (t)Y (t + τ )) = RX (τ )RY (τ )
which depends only on the lag τ. Hence Z(t) is WSS. (b) From part (a), we found the autocorrelation function RZ (τ ) = RX (τ )RY (τ ). As listed in Appendix B in the textbook, taking the Fourier transform of a product corresponds to the convolution of the individual transforms: SZ (f ) = SX (f ) ∗ SY (f ) where ∗ is the convolution operator.
Problem 6. textbook problem 7. Let Dn = Xn − Xn−d, where d is an integer constant and Xn is a zero-mean, WSS random process. (a) Find RD (k) and SD (f ) in terms of RX (k) and SX (f ). (b) Find E(D^2 n).
(a) It is straightforward to find
RD (k) = E(DnDn+k) = E((Xn − Xn−d)(Xn+k − Xn+k−d)) = 2RX (k) − RX (k + d) − RX (k − d)
Taking the Fourier transform,
SD (f ) = 2SX (k) − SX (k + d) − SX (k − d) = 2SX (f )(1 − cos(2πf d))
(b) E(D^2 n) = RD (0) = 2RX (0) − 2 RX (d)
Problem 7. textbook problem 7.
2
Let Y (t) be a short-term integration of X(t):
Y (t) =
T
∫ (^) t
t−T
X(t′) dt′
Find SY (f ) in terms of SX (f ).
The impulse response can be written
h(t) =^1 T
∫ (^) t
t−T
δ(s) ds =^1 T
∫ (^) t
−∞
δ(s) ds − 1 T
∫ (^) t−T
−∞
δ(s) ds
=^1 T (u(t) − u(t − T ))
H(f ) =^1 T
∫ T
0
e−j^2 πf tdt =^1 T
1 − e−j^2 πf t j 2 πf
=
T
ej^2 πf T /^2 − e−j^2 πf T /^2 j 2 πf e−j^2 πf T /^2
=
T
sin(πf T ) πf e−jπf T
Now,
SY (f ) = |H(f )|^2 SX (f ) = 1 T 2
sin^2 (πf T ) π^2 f 2 SX (f )
Problem 10. textbook problem 7. The input into a filter is 0-mean white noise with noise power density N 0 /2. The filter has transfer function
H(f ) = 1 1 + j 2 πf
(a) Find SY,X (f ) and RY,X (τ ). (b) Find SY (f ) and RY (τ ). (c) What is the average power of the output?
(a) SY,X (f ) = H(f )SX (f ) = (^) 1+j^12 πfN 20. Taking the inverse Fourier transform, RY,X (τ ) = N 20 e−τ^ for τ > 0 (see Appendix B in the textbook). (b) SY (f ) = |H(f )|^2 SX (f ) = N 20 1+4^1 π (^2) f 2. Taking the inverse Fourier transform, RY (τ ) = N 40 e−|τ^ |^ (see Appendix B in the textbook). (c) The average power of the output is RY (0) = N 40.
Problem 11. textbook problem 7. Let Y (t) be the output of a linear system with impulse response h(t) and input X(t). Find RY,X (τ ) when the input is white noise. Explain how this result can be used to estimate the impulse response of a linear system.
We know that SY,X (f ) = H(f )SX (f ). For white noise input, SX (f ) = N 20 and SY,X (f ) = H(f ) N 20. Taking the inverse Fourier transform, RY,X (τ ) = N 20 h(t). The result is the impulse response h(t) scaled by a factor N 0
Problem 12. textbook problem 7. A zero-mean white noise sequence is input into a cascade of two systems (as shown in Fig. P7.6). System 1 has impulse response hn = (1/2)nu(n) and system 2 has impulse response gn = (1/4)nu(n) where u(n) = 1 for n ≥ 0, and 0 elsewhere. (a) Find SY (f ) and SZ (f ). (b) Find RW,Y (k) and RW,Z (k); find SW,Y (f ) and SW,Z (f ). (c) Find E(Z n^2 ).
4
(a) First we find
H(f ) =
∑^ ∞
k=
(1/2)ne−j^2 πf n^ =
1 − 12 e−j^2 πf
G(f ) =
∑^ ∞
k=
(1/4)ne−j^2 πf n^ = 1 1 − 14 e−j^2 πf
Now, through the first system,
SY (f ) = |H(f )|^2 N^0 2
= 5 N^0 /^2
4 −^ cos(2πf^ ) and through the second system,
SZ (f ) = |G(f )|^2 SY (f ) =
( N^0 /^2
5 4 −^ cos(2πf^ )
16 −^
1 2 cos(2πf^ )
(b) Note that
SW,Y (f ) = H(f )SX (f ) =
N 0 / 2
1 − 12 e−j^2 πf
Taking the inverse Fourier transform,
RW,Y (k) =
N 0
u(k)
Considering both systems,
SW,Z (f ) = H(f )G(f )SX (f ) =
N 0 / 2
(1 − 12 e−j^2 πf^ )(1 − 14 e−j^2 πf^ )
=
N 0
1 − 12 e−j^2 πf^
N 0 / 2
1 − 14 e−j^2 πf
Taking the inverse Fourier transform,
RW,Z (k) = N 0
)k u(k) −
N 0
)k u(k)
(c) We can find SZ (f ), RZ (k), and then E(Z n^2 ) = RZ (0). From part (a),
SZ (f ) =
N 0 / 2
4 −^ cos(2πf^ )
16 −^
1 2 cos(2πf^ )
8 7 N^0 5 4 −^ cos(2πf^ )^
4 7 N^0 17 16 −^ cos(2πf^ )
We will make use of the Fourier transform pair
|α|k^ ⇔ 1 − α^2 1 + α^2 − 2 α cos(2πf )
to find the inverse Fourier transform of SZ (f ). After rearrangement of SZ (f ) into the appropriate form,
RZ (k) =^32 21
N 0
)|k| − 64 105
N 0
)|k|
(c) First, SY (f ) = |H(f )|^2 σ W^2
= σ
(^2) W 4
4 −^ cos(2πf^ )
16 −^
1 2 cos(2πf^ )
3 4 σ X^2 5 4 −^ cos(2πf^ )^
15 16 σ (^2) X 17 16 −^
1 2 cos(2πf^ ) Taking the inverse Fourier transform,
RY (k) =
)|k| −
)|k|
Problem 14. Matlab (optional for EE 5375) First generate 1024 i.i.d. normal random samples by typing:
y = randn(1024,1); Next, set vectors a and b by typing: a = [1.0 -1.2 +0.4]; b = [1.0 -.8 -0.1]; An ARMA process will be generated by a filter with the normal process used as input: x = filter(b,a,y); The process can be viewed by typing: plot(x) Compute the periodogram: X = fft(x,1024); P=X.*conj(X)/1024; The periodogram can be viewed by plotting: plot(P)
hn gn
Fig. P7.6 for Problem 12
Wn
Yn
Zn
Fig. P7.2 for Problem 1
A
x
p(x)
