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# Unit v rpq1

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### Unit v rpq1

1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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