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Sir Tanika Mukopadhyay taught us Probability at Homi Bhabha National Institute. He gave us assignments so that we can practice what we learned in form of problems. Here is solution to those problems. Its main emphasis is on following points: Rectangular, Interval, Appendix, Densites, Negativ, Autocorrelation, Instance, Orthogonal, Fourier, Transform
Typology: Exercises
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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
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
0
e−j^2 πf tdt =^1 T
1 − e−j^2 πf t j 2 πf
=
ej^2 πf T /^2 − e−j^2 πf T /^2 j 2 πf e−j^2 πf T /^2
=
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
4 −^ cos(2πf^ ) and through the second system,
SZ (f ) = |G(f )|^2 SY (f ) =
5 4 −^ cos(2πf^ )
1 2 cos(2πf^ )
(b) Note that
SW,Y (f ) = H(f )SX (f ) =
1 − 12 e−j^2 πf
Taking the inverse Fourier transform,
RW,Y (k) =
u(k)
Considering both systems,
SW,Z (f ) = H(f )G(f )SX (f ) =
(1 − 12 e−j^2 πf^ )(1 − 14 e−j^2 πf^ )
=
1 − 12 e−j^2 πf^
1 − 14 e−j^2 πf
Taking the inverse Fourier transform,
RW,Z (k) = N 0
)k u(k) −
)k u(k)
(c) We can find SZ (f ), RZ (k), and then E(Z n^2 ) = RZ (0). From part (a),
SZ (f ) =
4 −^ cos(2πf^ )
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
)|k| − 64 105
)|k|
(c) First, SY (f ) = |H(f )|^2 σ W^2
= σ
(^2) W 4
4 −^ cos(2πf^ )
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)