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Unformatted text preview: ndependent, then P (Z = 1jX = 1; Y = 1) should equal P (Z = 1). P (Z = 1jX = 1; Y = 1)
is easily seen to be 1, but P (Z = 1) = 1=2. Since the two are not equal, X , Y , and Z are not independent. Problem 2
a) SW (f ) = N =2. Therefore,
0 E jW (t)j2] = RW ( )j =0 = N0 (0) = 1
2 b) Since H (f ) is a stable LTI system and the input fW (t)g is a WSS process, the output fY (t)g is also WSS.
Furthermore, c)
Taking the inverse Fourier transform, SY (f ) = jH (f )j2 SW (f ) = jH (f )j2 N0
2
SY (f ) = jH (f )j2 N0 = N0 ( 2f )
2
2
W
RY ( ) = N0 W sinc(2W )
1 Then, E jY (t)j2 ] = RY ( )j =0 = N0 W sinc(0) = N0 W d) Sampling the output fY (t)g at rate 1=T = 2W samples per second means that we are looking at samples 1=2W
seconds apart. The time di erence between any two samples of this discretetime process fY (nT )g = fY n]g is
k=2W where k is an integer. Therefore,
RY n] (k) = RY (k=2W ) = N0 W sinc(k) = N0 W k]
The process is white because its powerspectral density is at. Furthermore, it is Gaussian since the...
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This document was uploaded on 01/24/2014.
 Spring '09

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