Question

Can someone please explain to me the solution to this problem! I don't understand the solution, I just need a detail explanation step by step so I can understand this problem with all the subparts! Thanks!

3.5 Find the power spectrum for each of the following wide-sense stationary random processes that have the given autocorrelation sequences (a) rx(k) 26(k)j8(k -1)-j(k+1) (b) T(k)(k)2(0.5) (c) T(k)26(k)+cos(Tk/4) (d) rx(k)=' 10 k k< 10 ; otherwise Solutioin (a) This autocorrelation sequence is finite in length, and the powcr spcctrum is simply Note that, as required, P(e) is real and non-negative (b) With a real, using the DTFT pair we have - COS w 5 - 4 cos w (c) Since the DTFT of a complex exponential is an impulse, it follows that the power spectrum of ra(k) 26(k) cos(Tk/4) is (d) Observe that ra(k) is a triangle that is symmetric about k0 and extends from k-9 to k = 9. Therefore, Tz(k) may be written as the convolution of two pulses r(kp(k) p(-k) where 10skss9 0 else Since the DTFT of p(k) is 9 sinw/2 then the power spectrum is 2 sin- 5w

Answer 1

Given Auto correlation process v_x(k) and its power spectrum [also called power spectral density] p_x(e^j omega). They both form Fourier transform pair. i.e. Fourier transform v_x(k) = p_x(e^6jw) v_x [k] p_x(e^j omega) = > F.T pair therefore to find power spectrum we need to find fourier transform of ACE v_x(k) v_x(k) = 2 delta(k) + j delta(k - 1) - j delta(k + 1) F.T of delta(k) 1 By time shifting property delta(k + k_0) doubleheadarrow 1 middot e^plusminus j omega middot k_0 Here k_0 = 1 then the final answer is F.T [v_x(k)] = p_x(e^j omega) = 2.1 + 2.1 middot j e^-j omega - j middot 1 middot e^j omega p_x(e^j omega) = 2 + j e^-jw - j e^j omega and sin x = e^jx - e^-jx/2j & cos x = e^jx + e^-jx/2 Bu using above p_x(e^j omega) can be written as because [-j = 1/j j = -1/j] p_x(e^j omega) = 2 + e^-jw/-j + e^j omega/j \r\n so p_x(e^j omega)