1. UNIT –V
LINEAR SYSTEMS WITH RANDOM INPUTS
PART – A
1. Define a system. When is it called a linear system?
A system is a functional relationship between the input
( )tx and the output
( )ty . The functional relationship is written as ( ) ( )[ ]txfty = .
If ( ) ( )[ ] ( )[ ] ( )[ ]tXfatXfatXatXaf 22112211 ±=± , then f is called a
linear system.
2. Write a note on linear system.
If ( ) ( )[ ] ( )[ ] ( )[ ]tXfatXfatXatXaf 22112211 ±=± , then f is called
a linear system.
If ( ) ( )[ ]htXfhtY +=+ where ( ) ( )[ ]tXftY = , then f is called a
time – invariant
system or ( )tX and ( )tY are said to form a time invariant system.
If the output ( )1tY at a given time 1tt = depends only on ( )1tX
and not on any other
past or future values of ( )tX , then the system f is called a
memoryless system.
If the value of the output ( )tY at 1tt = depends only on the past
values of the input
( ) 1, tttX ≤ . (ie) ( ) ( )[ ]11 ; tttXftY ≤= , then the system is called a
casual system.
3. State the properties of linear system.
The properties of linear system are
(i) If a system is such that its input ( )tX and its output ( )tY
are related by a convolution integral, then the system is a
linear time invariant system.
(ii) If the input to a time-invariant, stable linear system is a
WSS process, the output will also be a WSS process.
(iii) The power spectral densities of the input and output
processes in the system are connected by the relation
( ) ( ) ( )ωωω XXYY SHS 2
= , where ( )ωH is the Fourier transform
of unit impulse response function ( )th .
140

2. 4. Define system weighting function.
If the output ( )tY of a system is expressed as the convolution
of the input ( )tX
and a function ( )th (ie) ( ) ( ) ( )∫
∞
∞−
−= ,duutXuhtY then ( )th is called
the system
weighting function.
5. What is unit impulse response of a system? Why is it called so?
If a system is of the form ( ) ( ) ( )∫
∞
∞−
−= ,duutXuhtY then the
system weighting
function ( )th is also called unit impulse response of the system.
It is called so because the response (output) ( )tY will be ( )th ,
when the input
( )tX = the unit impulse function ( )tδ .
6. If the input of a linear system is a Gaussian random process,
comment about the
output random process.
If the input of a linear system is a Gaussian random process,
then the output will
also be a Gaussian random process.
7. If the input ( )tX of the system ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY is the unit
impulse function,
prove that ( ) ( )thtY = .
Given ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY
Put ( ) ( )ttX δ=
Therefore ( ) ( )ututX −=− δ
( ) ( ) ( )∫
∞
∞−
−= duutuhtY δ
( ) ( )∫
∞
∞−
−= duuuth δ [ by the property of convolution ]
( ) ( )thth =−= 0 .
8. If a system is defined as ( ) ( )∫
∞
−
−=
0
1
dueutX
T
tY T
u
, find its unit
impulse function.
141

3. Given ( ) ( )∫
∞
−
−=
0
1
dueutX
T
tY T
u
( )∫
∞
−
−=
0
1
duutXe
T
T
u
The unit impulse function ( )





≥
=
−
elsewhere
te
Tth
T
t
,0
0,
1
.
9. If ( ){ }tX and ( ){ }tY in the system ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY are WSS
processes, how are
their autocorrelation functions related?
( ) ( ) ( )τττ hRR XYYY
*
=
( ) ( ) ( )τττ −= hRR XXXY
*
, where * denotes convolution.
10. If the input and output of the system ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY are
WSS processes, how
are their power spectral densities related?
( ) ( ) ( ) 2
ωωω HSS XXTT = , where ( )ωH is the Fourier
transform of ( ).th
11. Define the power transfer function or system function of the
system.
The power transfer function or system function of the
system is the Fourier
transform of the unit impulse function of the system.
12. If the system function of a convolution type of linear system is
given by
( )





>
≤
=
ctfor
ctfor
cth
0
2
1
. Find the relation between power spectral density
functions of
the input and output processes.
142

4. ( ) ( )[ ] ( ) ∫∫ −
−
∞
∞−
−
===
c
c
titi
dte
c
dteththFH ωω
ω
2
1
( ) dtt
c
idtt
c
dttit
c
c
c
c
c
c
c
∫∫∫ −−−
−=−= ωωωω sin
2
1
cos
2
1
sincos
2
1
( )0
2
1
cos2
2
1
0
c
idtt
c
c
−= ∫ ω [since the first and second
integrand are even
and odd functions]
[ ]
ω
ω
ω
ωω
ω
c
c
c
c
t
c
c
sin
0sin
1sin1
0
=−=





=
( ) ( ) ( ) ( )ω
ω
ω
ωωω XXXXYY S
c
c
SHS 22
2
2 sin
== .
13. What is thermal noise? By what type of random processes is it
represented?
Thermal noise is the noise because of the random motion
of free electrons in
conducting media such as a resistor. It is represented by
Gaussian random
processes.
14. Why is thermal noise called white noise?
( )ωNNS is a constant for all values of ω (ie) ( )ωNNS
contains all frequencies in
equal amount. White noise is called so in analogy to white light
which consists of
all colour.
15. Define white noise.
Let ( )tX be a sample function of a WSS noise process,
then ( ){ }TttX ∈, is
called the white noise if the power density spectrum of
( ){ }TttX ∈, is constant at
all frequencies . (ie) ( )
2
0N
SNN =ω where 0N is a real positive
constant.
143

5. 16. If the power spectral density of white noise is
2
0N
, find its
autocorrelation function.
( ) ( ) ( ) ττδττδτδ τωτω
de
N
de
NN
F ii
∫∫
∞
∞−
−
∞
∞−
−
==





222
000
( )
22
000 N
e
N
==
Therefore ( ) ( )τδτ
22
001 NN
FRNN =





= −
.
17. If the input to a linear time invariant system is white noise
( ){ }tN , what is power
spectral density function of the output?
If the input to a linear time invariant system is white
noise ( ){ }tN , then the
power spectral density of the output ( )ωYYS is given by
( ) ( ) ( ) ( ) 202
2
ωωωω H
N
HSS XXYY == .
where ( ){ }tY is the output process and ( )ωH is the power
transfer function.
18. Find the average power or the mean square value of the white
noise ( ){ }tN ?
Average power = ( )[ ] ( )02
NNRtNE =
( ) ( )
( )∫∫
∞
∞−
∞
∞−
== ωωωω dSdeS NNNN
0
∞→= ∫
∞
∞−
ωd
N
2
0
19. A wide sense stationary noise process ( )tN has an autocorrelation
function
( ) ∞<<∞−=
−
ττ
τ
,
3
ePRNN with P as a constant. Find its power
density spectrum.
( ) ( ) ( ) τωτωττττω ττωττω
diePdeePdeRS ii
NNNN sincos
33
−=== ∫∫∫
∞
∞−
−
∞
∞−
−−
∞
∞−
−








−= ∫∫
∞
∞−
−
∞
∞−
−
τωτττω ττ
deideP sincos
33
144

6. ( )0cos2
0
3
ideP −= ∫
∞
−
τωττ
[since the first and
second integrand are
even and odd
functions]
( )
( )
∞
−∞
−






+−
+−
== ∫
0
22
3
0
3
sincos3
3
2cos2 ωτωωτ
ω
τωτ
τ
τ e
PdeP
( ) 22
9
6
03
9
1
02
ωω +
=





+−
+
−=
P
P .
20. Define band limited white noise.
Noise having a non-zero and constant spectral density over
a finite frequency
band and zero elsewhere is called band limited white noise.
If ( ){ }tN is a band limited white noise, then
( )





≤
=
elsewhere
N
S
B
NN
,0
,
2
0
ωω
ω .
21. State a few properties of band limited white noise.
Properties of band limited white noise are
(i) ( )[ ] B
N
tNE ω
π2
02
=
(ii) ( ) 





=
τω
τω
π
ω
τ
B
BB
NN
N
R
sin
2
0
(iii) ( )tN and 





+
B
k
tN
ω
π
are independent, where k is
a non-zero integer.
22. Find the autocorrelation function of the band-limited white noise.
( ) ( ) ( ) ττωτω
π
ω
π
ωω
π
τ
ω
ω
ω
ω
ωττω
di
N
de
N
deSR
B
B
B
B
ii
NNNN sincos
22
1
22
1
2
1 00
+=== ∫∫∫ −−
∞
∞−
145

7. ( )








+=








+= ∫∫∫ −−
0cos2
4
sincos
4 0
00
id
N
did
N BB
B
B
B
ωω
ω
ω
ω
ωωτ
π
ωωτωωτ
π
146

8. [since the first and second integrand are even and odd functions]
[ ]
B
BB
B
NNN B
ωτ
ωτ
π
ω
ωτ
τπτ
ωτ
π
ω
sin
2
0sin
2
sin
2
00
0
0
=−=





= .
23. Find the average power of the band – limited white noise.
Average power ( )[ ] ( )02
NNRtNE ==






=
→
B
BBN
ωτ
ωτ
π
ω
τ
sin
lim
2 0
0
( ) 





==
→
1
sin
limsin1
2 0
0
θ
θ
π
ω
θ
ce
N B
π
ω
2
0 BN
= .
147

9. PART – B
1. Show that if ( ){ }tX is a WSS process then the output ( ){ }tY is a WSS
process.
Solution: ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY
( )[ ] ( ) ( )[ ]∫
∞
∞−
−= duutXEuhtYE
( )∫
∞
∞−
= duuhxµ [ since ( ){ }tX is a WSS process ]
= a finite constant independent of t
( ) ( ) ( )[ ]2121 , tYtYEttRYY =
( ) ( ) ( ) ( )








−−= ∫ ∫
∞
∞−
∞
∞−
21221121 duduutXutXuhuhE
( ) ( ) ( ) ( )[ ]∫ ∫
∞
∞−
∞
∞−
−−= 21221121 duduutXutXEuhuh
( ) ( ) ( )∫ ∫
∞
∞−
∞
∞−
−−= 21221121 , duduututRuhuh XX
( ) ( ) ( ) ( )[ ]∫ ∫
∞
∞−
∞
∞−
−−−= 21212121 duduuuttRuhuh XX [ since ( ){ }tX is
a WSS ]
Since the R.H.S of the above equation is a function of ( )21 tt − ,
so will be the L.H.S.
Therefore, ( )21 ,ttRYY will be a function of ( )21 tt − .
Therefore, ( ){ }tY is a WSS process.
2. For a linear system with random input ( )tx , the impulse response
( )th and output ( )ty , obtain the cross correlation function ( )τXYR
and the output autocorrelation function ( )τYYR .
Solution: (1) Let ( ) ( ) ( )∫
∞
∞−
−= ααα dhtXtY
Therefore, ( ) ( ) ( ) ( ) ( )∫
∞
∞−
−+=+ αααττ dhtXtXtYtX
( ) ( )[ ] ( ) ( )[ ] ( )∫
∞
∞−
−+=+ αααττ dhtXtXEtYtXE
( ) ( ) ( )∫
∞
∞−
+= αααττ dhRR XXXY
Putting βα −= when ∞=∞−= βα ,
βα dd −= when ∞−=∞= βα ,
148

10. ( ) ( ) ( )( )∫
∞−
∞
−−−= βββττ dhRR XXXY
( ) ( ) ( )∫
∞−
∞
−−−= βββττ dhRR XXXY
( ) ( ) ( )∫
∞
∞−
−−= βββττ dhRR XXXY
( ) ( ) ( )τττ −= hRR XXXY *
(2) ( ) ( ) ( ) ( ) ( )∫
∞
∞−
−−=− αατατ dhtYtXtYtY
( ) ( )[ ] ( ) ( )[ ] ( )∫
∞
∞−
−−=− αατατ dhtYtXEtYtYE
( ) ( ) ( ) αααττ dhRR XYYY −=
Assuming that ( ){ }tX and ( ){ }tY are jointly WSS
Therefore, ( ) ( ) ( )τττ hRR XYYY *= .
3. For a linear system with random input ( )tX , the impulse response
( )th and output ( )tY , obtain the power spectrum ( )ωYYS and cross
power spectrum ( )ωXYS .
Solution: By the previous problem, we have
( ) ( ) ( )τττ −= hRR XXXY * ----------------(1)
( ) ( ) ( )τττ hRR XYYY *= -----------------(2)
Taking Fourier transform of equations (1) and (2) , we have
( )[ ] ( )[ ] ( )[ ]τττ −= hFRFRF XXXY .
( ) ( ) ( )ωωω *.HSS XXXY = --------------(3)
where ( )ωH is the Fourier transform of unit impulse response
function ( )th
and ( )ω*H is the conjugate of ( )ωH .
and ( )[ ] ( )[ ] ( )[ ]τττ hFRFRF XYYY .=
( ) ( ) ( )ωωω HSS XYYY .= ------------------(4)
Inserting (3) in(4), we have
( ) ( ) ( ) ( )ωωωω HHSS XXYY *.=
( ) ( ) 2
ωω HS XX=
Therefore, ( ) ( ) ( ) 2
ωωω HSS XXYY = .
149

11. 4. Show that ( ) ( ) ( ) 2
ωωω HSS XXYY = where ( )ωXXS and ( )ωYYS are the
power spectral density functions of the input ( )tX and the output
( )tY and ( )ωH is the system transfer function.
Solution: ( ) ( ) ( )[ ]ττ −= tYtYERYY
( ) ( ) ( ) ( )








−−−= ∫ ∫
∞
∞−
∞
∞−
dvvtXvhduutXuhE τ.
( ) ( ) ( ) ( )








−−−= ∫ ∫
∞
∞−
∞
∞−
dvduvhuhvtXutXE τ
( ) ( )[ ] ( ) ( )∫ ∫
∞
∞−
∞
∞−
−−−= dvduvhuhvtXutXE τ
If ( ){ }tX is WSS, then
( ) ( )( ) ( ) ( )∫ ∫
∞
∞−
∞
∞−
−−= dvduvhuhvuRR XXYY ττ
The above integral is a two fold convolution of autocorrelation with
impulse
response ( )th
Therefore, ( ) ( ) ( ) ( )ththRR XXYY −= **ττ
Taking Fourier transform on both sides, we have
( )[ ] ( )[ ] ( )[ ] ( )[ ]thFthFRFRF XXYY −= ..ττ
( ) ( ) ( )[ ] ( )[ ]thFthFSS XXYY −= ..ωω --------------------------
(1)
( ) ( )[ ]thFH =ω
( ) ( )∫
∞
∞−
−
= dtethH ti ω
ω
---------------------------------------(2)
( ) ( )∫
∞
∞−
= dtethH ti ω
ω*
Putting st −= when ∞=∞−= st ,
dsdt −= when ∞−=∞= st ,
( ) ( ) ( )∫
∞−
∞
−
−−= dseshH si ω
ω*
( ) ( )∫
∞−
∞
−
−−= dseshH si ω
ω*
( ) ( )∫
∞
∞−
−
−= dseshH si ω
ω*
( )[ ]thF −= ---------------------------------(3)
Inserting (2) and (3) in (1), we have
( ) ( ) ( ) ( )ωωωω *.. HHSS XXYY =
( ) ( ) ( ) 2
. ωωω HSS XXYY = .
150

12. 5. A system has an impulse response function ( ) ( ),tUeth tβ−
= find the
power spectral
density of the output ( )tY corresponding to the input ( )tX .
Solution: Given ( ) ( )tUeth tβ−
=
0, ≥= −
te tβ
where ( )



<
>
=
0,0
0,1
t
t
tU
( ) ( )[ ] ∫
∞
−−
==
0
dteethFH tit ωβ
ω
( )
∫
∞
+−
=
0
dte ti ωβ
( )
( )
∞+−






+−
=
0
ωβ
ωβ
i
e tii
( )
ωβωβ ii +
=−
+
−=
1
10
1
( )
ωβ
ω
i
H
−
=
1
*
( ) ( ) ( )ωωω *
2
HHH =
22
11
.
1
ωβωβωβ +
=
−+
=
ii
Therefore, ( ) ( ) ( ) 2
. ωωω HSS XXYY =
( ) ( ) 22
1
.
ωβ
ωω
+
= XXYY SS .
6. ( )tX is the input voltage to a circuit and ( )tY is the output voltage.
( ){ }tX is a stationary
random process with 0=xµ and ( ) τα
τ −
= eRXX . Find ( )ωµ YYy S, and
( )τYYR if the
power transfer function is ( )
WLiR
R
H
+
=ω .
Solution: ( ) ( ) ( )∫
∞
∞−
−= duutXuhtY
( )[ ] ( ) ( )








−= ∫
∞
∞−
duutXuhEtYE
( )[ ] ( ) ( )[ ]∫
∞
∞−
−= duutXEuhtYE
( ) ( )∫
∞
∞−
= duuh 0 [ since 0=xµ is given ]
( )[ ] 0=tYE
151

13. Therefore, ( )[ ] 0== tYEyµ .
( ) ( ) ( ) 2
. ωωω HSS XXYY = where ( ) ( )[ ]τω XXXX RFS =
( ) ( )∫
∞
∞−
−
= ττω τω
deRS i
XXXX
( )∫∫
∞
∞−
−
∞
∞−
−−
−== ττωτωτ τατωτα
diedee i
sincos.
∫∫
∞
∞−
−
∞
∞−
−
−= ττωττω τατα
deide sincos
( )0cos2
0
ide −= ∫
∞
−
ττωτα
[since the first and
second integrand
are even and
odd functions]
∫
∞
−
=
0
cos2 ττωτα
de
( )
( )
∞
−






+−
+−
=
0
22
sincos2 τωωτωα
ωα
τα
e
( ) 2222
2
0
1
02
ωα
α
α
ωα +
=





+−
+
−=
( ) 22
2
ωα
α
ω
+
=XXS
( )
2
22
2
LWiR
R
SYY
++
=
ωα
α
ω
( ) 222
2
22
.
2
WLR
R
SYY
++
=
ωα
α
ω
( )
( )( )22222
2
2
WLR
R
SYY
++
=
ωα
α
ω
( ) ( )[ ]ωτ YYYY SFR 1−
=
7. A wide sense stationary random process ( )tX with autocorrelation
function
( ) τα
τ −
= eARXX where A and α are real positive constants is
applied to the
152

14. input of an linearly time invariant system with impulse response
( ) 0, >= −
teth tb
where b is a real positive constant. Find the
spectral density of
the output ( )tY of the system.
Solution: ( ) ( )∫
∞
∞−
−
= ττω τω
deRS i
XXXX
( ) ∫
∞
∞−
−−
= τω τωτα
deeAS i
XX
( )∫
∞
∞−
−
−= ττωτωτα
dieA sincos








−= ∫ ∫
∞
∞−
∞
∞−
−−
ττωττω τατα
deideA sincos
( )





−= ∫
∞
−
0
0cos2 ideA ττωτα
∫
∞
−
=
0
cos2 ττωτα
deA
( )
( )
∞
−






−−
+−
=
0
22
sincos2 τωωτωα
ωα
τα
e
A
( )





+−
+
−= 0
1
02 22
α
ωα
A
( ) 22
2
ωα
α
ω
+
=
A
S XX
( ) ( )[ ]thFH =ω
( ) ( )∫
∞
∞−
−
= dtethH ti ω
ω
( ) ∫
∞
−−
=
0
dteeH titb ω
ω
( )
∫
∞
+−
=
0
dte tib ω
( )
( )
∞+−






+−
=
0
ω
ω
ib
e tib
( )10
1
−
+
−=
ωib
ωib +
=
1
( ) 22
2 11
ωω
ω
+
=
+
=
bib
H .
( ) ( ) ( ) 2
. ωωω HSS XXYY =
153

15. 2222
1
.
2
ωωα
α
++
=
b
A
( )
( )( )2222
2
b
A
SYY
++
=
ωαω
α
ω .
8. Establish the spectral representation theorem.
Solution: Let ( )tX be a signal and define
( )
( )




<<−
=
otherwise
T
t
T
tX
tXT
,0
22
,
Represent ( )tXT in a Fourier series ( complex form )
( ) ∑
∞
∞−
= tni
nT ectX 0ω
where ( )∫
−
−
==
2
2
0
2
,
1 0
T
T
tni
Tn
T
wheredtetX
T
c
π
ωω
( )∫
−
−
=
2
2
0
1
T
T
tni
n dtetX
T
c ω
( )0
1
ωniX
T
= [ taking the envelope ( )ωiX of nc ]
Therefore, ( ) ( )∑
∞
∞−
= tni
T eniX
T
tX 0
0
1 ω
ω
( ) ( )∑
∞
∞−






= tni
T eniXtX 0
0
0
2
1 ω
ω
ω
π
( ) ( ) 00 .
2
1 0
ωω
π
ω
∑
∞
∞−
= tni
T eniXtX
As ∞→T , ( ) ( )tXtXT →
Therefore, ( ) ( )∫
∞
∞−
= ωω
π
ω
deiXtX ti
2
1
-----------------------(1)
and equivalently, ( by Fourier integral theorem )
( ) ( )∫
∞
∞−
−
= dtetXiX ti ω
ω -------------------------(2)
( )ωiX is usually called as spectrum of the aperiodic signal ( )tx
154

16. 9. Show that in an input – output system the energy of a signal is equal
to the energy
of the spectrum.
Solution: Energy of a signal ( )∫
∞
∞−
= dttX
2
( ) ( )∫
∞
∞−
= dttXtX *
( ) ( )∫ ∫
∞
∞−
∞
∞−
−








= dtdeiXtX ti
ωω
π
ω
*
2
1
[ by equation (1) of problem 8 above ]
( ) ( )∫ ∫
∞
∞−
∞
∞−
−








= dttXdeiX ti
ωω
π
ω
*
2
1
( ) ( )∫ ∫
∞
∞−
∞
∞−
−








= ωω
π
ω
diXdtetX ti
*
2
1
[ by changing
the
order of integration ]
( ) ( )∫
∞
∞−
= ωωω
π
diXiX *
2
1
[ by equation (2) of problem 8 above ]
( )∫
∞
∞−
= ωω
π
diX
2
2
1
= Energy of the spectrum.
10. If ( ){ }tX is a band limited process such that ( ) 0=ωXXS , when σω> ,
prove
that ( ) ( )[ ] ( )002 22
XXXXXX RRR τστ ≤− .
Solution: ( ) ( )∫
∞
∞−
= ωω
π
τ ωτ
deSR i
XXXX
2
1
( ) ( ) ( )∫
∞
∞−
+= ωωτωτω
π
τ diSR XXXX sincos.
2
1
( ) ( ) ωωτω
π
ωωτω
π
dSdS XXXX sin
2
1
cos
2
1
∫∫
∞
∞−
∞
∞−
+=
( ) ( )0
2
1
cos
2
1
π
ωωτω
π
+= ∫
∞
∞−
dS XX
155

17. [ since ( )ωXXS is an even function and therefore the first integrand is
even and the second is an odd function ].
( ) ( ) ( )
∫
∞
∞−
= ωω
π
deSR XXXX
0
2
1
0
( ) ωω
π
dS XX∫
∞
∞−
=
2
1
( ) ( ) ( ) ( )∫∫
∞
∞−
∞
∞−
−=− ωωτω
π
ωω
π
τ dSdSRR XXXXXXXX cos
2
1
2
1
0
( ) ( ) ( ) ( )∫
∞
∞−
−=− ωωτω
π
τ dSRR XXXXXX cos1
2
1
0
( ) ( ) ( ) ( )∫−
−=−
σ
σ
ωωτω
π
τ dSRR XXXXXX cos1
2
1
0 [since ( ){ }tX is band
limited ]
( ) ( ) ( )∫−






=−
σ
σ
ω
ωτ
ω
π
τ dSRR XXXXXX
2
sin2
2
1
0 2
-----------------------(1)
From trigonometry, 0sin ≤θ
22
sin θθ ≤
22
sin2
22
2 ωτωτ
≤





---------------------------
(2)
Inserting (2) in (1) , we have
( ) ( ) ( )∫−






≤−
σ
σ
ω
ωτ
ω
π
τ dSRR XXXXXX
22
1
0
22
( )∫−
≤
σ
σ
ωω
π
τσ
dS XX
4
22
( )∫
∞
∞−
≤ ωω
π
τσ
dS XX
4
22
( )∫
∞
∞−
≤ ωω
π
τσ
dS XX
2
1
2
22
( )0
2
22
XXR
τσ
≤
( ) ( ) ( )0
2
0
22
XXXXXX RRR
τσ
τ ≤−
( ) ( )[ ] ( )002 22
XXXXXX RRR τστ ≤−
156

18. 11. If ( ) ( ) ( )tNtAtY ++= θω0cos , where A is a constant, θ is a random
variable
with a uniform distribution in ( )ππ,− and ( ){ }tN is a band limited
Gaussian
white noise with a power spectral density ( )




<−
=
elsewhere
for
N
S B
NN
,0
,
2
0
0
ωωω
ω .
Find the power spectral density of ( ){ }tY . Assume that ( )tN and θ
are
independent.
Solution: Given ( ) ( ) ( )tNtAtY ++= θω0cos
Since θ is uniformly distributed in ( )ππ,− , the density function
is given by
( ) πθπ
π
θ <<−= ,
2
1
f
( ) ( ) ( )[ ]2121 , tYtYEttRYY =
( ) ( )( ) ( ) ( )( )[ ]220110 coscos tNtAtNtAE ++++= θωθω
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]211202102010
2
coscoscoscos tNtNtNtAtNtAttAE +++++++= θωθωθωθω
( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ 11202102010
2
coscoscoscos NtNEtNEtAEtNEtAEttEA +++++++= θωθωθωθω
( since ( )tN and θ are independent )
( ) ( )[ ] ( )[ ]( ) ( )[ ]( ) ( )2120102010
2
,0cos0coscoscos ttRtAEtAEttEA NN+++++++= θωθωθωθω
( ) ( ) ( ) ( ) ( )212010
2
21 ,coscos, ttRdfttAttR NNYY +++= ∫−
θθθωθω θ
π
π
( ) ( ) ( ) ( )212010
2
21 ,
2
1
coscos2
2
, ttRdtt
A
ttR NNYY +++= ∫−
θ
π
θωθω
π
π
( ) ( ) ( )[ ] ( )2120102010
2
21 ,coscos
4
, ttRdtttt
A
ttR NNYY +−−+++++= ∫−
π
π
θθωθωθωθω
π
( ) ( )( ) ( )( ) ( )21210210
2
21 ,cos2cos
4
, ttRdttdtt
A
ttR NNYY +








−+++= ∫∫ −−
π
π
π
π
θωθθω
π
( )
( )( )
( )( ) ( )21210
210
2
21 ,cos
2
2sin
4
, ttRdtt
ttA
ttR NNYY +








−+




 ++
= ∫−−
π
π
π
π
θω
θω
π
157

19. ( ) ( )( ) ( )( )( ) ( )( )( ) ( )21210210210
2
21 ,cossinsin
2
1
4
, ttRtttttt
A
ttR NNYY +



−++−+= −
π
πθωωω
π
( ) ( )( )[ ] ( )21210
2
21 ,cos2
4
, ttRtt
A
ttR NNYY +−= ωπ
π
( ) ( ) ( )21210
2
21 ,cos
2
, ttRtt
A
ttR NNYY +−= ω
( ) ( )ττωτ NNYY R
A
R += 0
2
cos
2
Taking Fourier transform on both sides , we have
( )[ ] [ ] ( )[ ]ττωτ NNYY RFF
A
RF += 0
2
cos
2
( ) ( ) ( )[ ] ( )ωωωδωωδπω NNYY S
A
S +++−= 00
2
2
where [ ] ( ) ( )[ ]pppF ++−= ωδωδπτcos .
158