Chapter 9

Frequency analysis and spectral estimation

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Problem 9:1

Show that the Fourier transform of {x(t),t=0,1,…n−1}\{ x(t), t = 0, 1, \ldots n-1\},

X(f)=∑t=0n−1x(t)e−i2πft, \mathscr {X}(f) = \sum _{t=0}^{n-1} x(t) e^{-i2\pi f t},

included in the periodogram, R^x(f)=1n∣X(f)∣2\widehat{R}_x(f) = \frac{1}{n} \left|\mathscr {X}(f)\right|^2, is periodic with period one.

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Problem 9:2

Show that the mean square error of the spectrum estimate is equal to the sum of the variance and the squared bias,

E[(R^x(f)−Rx(f))2]=Var⁡[R^x(f)]+Bx2(f), \mathbb {E}\left[(\widehat{R}_x(f) - R_x(f))^2\right] = \operatorname {Var}\left[\widehat{R}_x(f)\right] + B_x^2(f),

where the variance defined as

Var⁡[R^x(f)]=E[(R^x(f))2]−E[R^x(f)]2, \operatorname {Var}\left[\widehat{R}_x(f)\right] = \mathbb {E}\left[(\widehat{R}_x(f))^2\right] - \mathbb {E}\left[\widehat{R}_x(f)\right]^2,

and the squared bias

Bx2(f)=(E[R^x(f)]−Rx(f))2. B_x^2(f) = (\mathbb {E}\left[\widehat{R}_x(f)\right] - R_x(f))^2.
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Problem 9:3

Figure 9.11(a) shows the periodogram and Figure 9.11(b) shows the modified periodogram using a Hanning window of the same data sequence with n=64n = 64 samples. We know that the sequence consists of a number of sinusoids. By studying the figures, what can you conclude about the sequence of data?

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Problem 9:4
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(a)

Determine the condition of the data window function w(t)w(t), for an unbiased spectral estimate of a white noise process, x(t)x(t), t=0…n−1t = 0 \ldots n-1, with variance σ2\sigma^2.

(b)

If we also consider the variance, how is the ``best'' spectrum estimate (lowest variance) for a white noise process achieved?

Problem 9:5

Figure 9.12(a) shows the modified periodogram and Figure 9.12(b) the Welch method of 10 realizations of an AR(2)-process with a peak at f=0.2f = 0.2. The number of data samples is n=64n = 64 and the number of averages in Welch method is K=8K = 8. Which of the two methods gives the best estimate? Motivate your answer.

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