Estimating the spectra of three very long real-valued data sequences x1[n], x2[n

Question

Estimating the spectra of three very long real-valued data sequences x1[n], x2[n], and x3[n], each consisting of the sum of two sinusoidal components. However, we only have a 256-point segment of each sequence available for analysis. Let x bar_1 [n], x bar_2 [n], and x bar_3[n] denote the 256-point segments of x1[n], x2[n], and x3[n], respectively. We have some information about the nature of the spectra of the infinitely long sequences, as indicated in Eqs. (P10.22-1) through (P10.22-3). Two different spectral analysis procedures are being considered for use, one using a 256-point rectangular window and the other a 256-point Hamming window. These procedures are described below. In the descriptions, the signal RN[n] denotes the N-point rectangular window and HN[n] denotes the N-point Hamming window. The operator DFT2048t- indicates taking the 2048-point DFT of its argument after zero-padding the end of the input sequence. This will give a good interpolation of the DTFT from the frequency samples of the DFT. X_1 (e^j omega) almostequalto delta (omega + 17 pi/64) + delta (omega + pi/4) + delta (omega - pi/4) + delta (omega - 17 pi/64) (P10.22-1) X_2 (e^j omega) almostequalto 0.017 delta (omega + 11 pi/32) + delta (omega + pi/4) + delta (omega - pi/4) + 0.017 delta (omega - 11 pi/32) X_3 (e^j omega) almostequalto 0.01 delta (omega + 257 pi/1024) + delta (omega + pi/4) + delta (omega - 257 pi/1024) Based on Eqs. (P10.22-1) through (P10.22-3), indicate which of the spectral analysis procedures described below would allow you to conclude responsibly whether the anticipated frequency components were present. A good justification at a minimum will include a quantitative consideration of both resolution and side-lobe behavior of the estimators. Note that it is possible that both or neither of the algorithms will work for any given data sequence. Table 7.2 may be useful in deciding which algorithm (s) to use with which sequence. Spectral Analysis Algorithms Algorithm 1: Use the entire data segment with a rectangular window. v [n] = R_256 [n] x bar [n] |V (e^j omega)|_omega = 2 pi k/2048 = |DFT_2048 {v [n]}|. Algorithm 2: Use the entire data segment with a Hamming window. v [n] = H_256 [n] x bar [n] |V (e^j omega)|_omega = 2 pi k/2048 = |DFT_2048 {v[n]}|.

Explanation / Answer

x1(n) and x2(n) are the finite duration sequences for the length N with DFTs X1(K) and X2(k) If X3

(k)=X1(k)X2(k) then the sequence x3(n) can be obtained by circular convolution defined as

N-1

x3(n)=™ [1(m)x2((n-m))N

m=0

Let the sequence x (n) has a length L. If we want to find the N-point DFT(N>L) of the sequence x(n), we have to add (N-L) zeros to the sequence x(n). This is known as zero padding.

uses are:

Better display of frequency spectrum and the DFT can be used in linear filtering

And periodicity , linearity , time interval , circular time shift are the following properties of DFT

If x(k) is N-point DFT of a finite duration sequence x(n),

then x(n+N) = x(n) for all n. x(k+N) = x(k) for all k.

If x1(k) and x2(k) are N-point DFTs of finite duration sequences x1(n) and x2(n), then DFT [a x1(n) + b x2(n)] = a x1(k) + b x2(k), a, b are constants

Taking the DFT, and converting to polar notation, results in the 129 point frequency spectrum in (d). Unfortunately, this also looks like a noisy mess. This is because there is not enough information in the original 256 points to obtain a well behaved curve. Using a longer DFT does nothing to help this problem. For example, if a 2048 point DFT is used, the frequency spectrum becomes 1025 samples long. Even though the original 2048 points contain more information, the greater number of samples in the spectrum dilutes the information by the same factor. Longer DFTs provide better frequency resolution, but the same noise level.

The answer is to use more of the original signal in a way that doesn't increase the number of points in the frequency spectrum. This can be done by breaking the input signal into many 256 point segments. Each of these segments is multiplied by the Hamming window, run through a 256 point DFT, and converted to polar notation. The resulting frequency spectra are then averaged to form a single 129 point frequency spectrum.

The second factor limiting resolution is more subtle. Imagine a signal created by adding two sine waves with only a slight difference in their frequencies. Over a short segment of this signal, say a few periods, the waveform will look like a single sine wave. The closer the frequencies, the longer the segment must be to conclude that more than one frequency is present. In other words, the length of the signal limits the frequency resolution. This is distinct from the first factor, because the length of the input signal does not have to be the same as the length of the DFT. For example, a 256 point signal could be padded with zeros to make it 2048 points long. Taking a 2048 point DFT produces a frequency spectrum with 1025 samples. The added zeros don't change the shape of the spectrum, they only provide more samples in the frequency domain. In spite of this very close sampling, the ability to separate closely spaced peaks would be only slightly better than using a 256 point DFT. When the DFT is the same length as the input signal, the resolution is limited about equally by these two factors. We will come back to this issue shortly.

cos 0n for several different values of 0 . As 0 increases from zero toward (parts a-d), the sequence oscillates more rapidly. As 0 increases from to 2 (parts d-a), the oscillations become slower.

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