Exercise C1.13: DTFT Computations using Two-Sided Sequences In this exercise we consider the DTFT of two-sided sequences (including aut ^0 covariance sequences), and in doing so illustrate some basic properties of autocos variance sequences. (a) We first (2024)

Exercise C1.13: DTFT Computations using Two-Sided SequencesIn this exercise we consider the DTFT of two-sided sequences (including aut ${ }^0$ covariance sequences), and in doing so illustrate some basic properties of autocos variance sequences.(a) We first consider how to use the DTFT to determine $\phi(\omega)$ from $r(k)$ on computer. We are given an ACS:$$r(k)= \begin{cases}\frac{M-|k|}{M}, & |k| \leq M \\ 0, & \text { otherwise }\end{cases}$$Generate $r(k)$ for $M=10$. Now, in MatLAB form a vector $\mathbf{x}$ of length $L=2$ : as:$$\mathrm{x}=[r(0), r(1), \ldots, r(M), 0 \ldots, 0, r(-M), \ldots, r(-1)]$$Verify that $\mathrm{x} f=\mathrm{ft}(\mathrm{x})$ gives $\phi\left(\omega_k\right)$ for $\omega_k=2 \pi k / L$. (Note that the elemen of $x f$ should be nonnegative and real.). Explain why this particular choice $\mathrm{x}$ is needed, citing appropriate circular shift and zero padding properties the DTFT.Note that $\mathrm{xf}$ often contains a very small imaginary part due to comput roundoff error; replacing $x f$ by real ( $x f$ ) truncates this imaginary compone: and leads to more expected results when plotting.A word of caution - do not truncate the imaginary part unless you are su it is negligible; the command $\mathrm{zf}=\mathrm{real}(\mathrm{fft}(\mathrm{z})$ ) when$$\mathrm{z}=[r(-M), \ldots, r(-1), r(0), r(1), \ldots, r(M), 0 \ldots, 0]$$gives erroneous "spectral" values; try it and explain why it doesn't work.(b) Alternatively, since we can readily derive the analytical expression for $\phi(\omega$ we can instead work backwards. Form a vector$$\mathrm{yf}=[\phi(0), \phi(2 \pi / L), \phi(4 \pi / L), \ldots, \phi((L-1) 2 \pi / L)]$$and find $y=i f f t(y f)$. Verify that $y$ closely approximates the ACS.(c) Consider the ACS $r(k)$ in Exercise C1.12; let $a_1=-0.9$ and $b_1=0$, and s $\sigma^2=1$. Form a vector $\mathrm{x}$ as above, with $M=10$, and find $\mathrm{xf}$. Why is? not a good approximation of $\phi\left(\omega_k\right)$ in this case? Repeat the experiment fo $M=127$ and $L=256$; is the approximation better for this case? Why?We can again work backwards from the analytical expression for $\phi(\omega)$. For a vector$$\mathrm{yf}=[\phi(0), \phi(2 \pi / L), \phi(4 \pi / L), \ldots, \phi((L-1) 2 \pi / L)]$$and find $y=$ ifft $(y f)$. Verify that $y$ closely approximates the ACS for large (say, $L=256$ ), but poorly approximates the ACS for small $L$ (say, $L=20$ By citing properties of inverse DTFTs of infinite, two-sided sequences, expla how the elements of $\mathrm{y}$ relate to the $\operatorname{ACS} r(k)$, and why the approximation poor for small $L$. Based on this explanation, give a "rule of thumb" on $t$ l choice of $L$ for good approximations of the ACS.(d) We have seen above that the $\mathrm{fft}$ command results in spectral estimates from 0 to $2 \pi$ instead of the more common range of $-\pi$ to $\pi$. The Matlab command fftshift can be used to exchange the first and second halves of the $f$ ft output to correspond to a frequency range of $-\pi$ to $\pi$. Similarly, fftshift can be used on the output of the ifft operation to "center" the zero lag of an ACS, Experiment with $\mathrm{fftshift}$ to achieve both of these results. What frequency vector w is needed so that the command plot(w, fftshift $(f f t(x))$ ) gives the spectral values at the proper frequencies? Similarly, what time vector $t$ is needed to get a proper plot of the ACS with stem(t,fftshift(ifft $(x f))$ )? Do the results depend on whether the vectors are even or odd in length?

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