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Linear Frequency Chirps

We presently apply the algorithm suggested in section 4 to several linear chirp examples where we treat $\alpha $ as unknown initially. In each of these cases, figures 12 through 17, the first subplot shows the magnitude FFT of a signal characterized by some $a$ value noted above the plot. We have also marked the half height point with a dashed line. Without verifying that the corresponding $k_{hh}$ range suggests a sufficiently large $\alpha $, we estimate an $\alpha $ value by inverting the magnitude expression as described above (and use it to plot our estimate of the instantaneous frequency function in the third subplot). We do not do this verification in order to allow the illustration of ``failed attempts'' by the algorithm when $\alpha $ (or $a$) is too small.

In the second subplot, we see the second order difference of the phase, again with the half height region marked. We also show - with dashed vertical lines - bounds for an often more limited region in cases corresponding to $b<1$ in the phase inversion technique. A horizontal dashed line represents the average phase concavity in this valid range, which we use to backsolve for $\alpha $ via inversion of the phase expression.

In the third subplot, we show the estimated frequency characteristic obtained by backsolving for $\alpha $ using each of the two techniques, along with the original. The phase model is shown with a dashed line and the magnitude model with a dotted line. Each estimated $a$ value is also shown above the plot, next to the actual $a$ value for comparison. We see greater accuracy from the magnitude model in all cases. For these examples and those in the next subsections, we have used a Hann window (as required), $F=8000$, $N = 201$, and $K = 5\cdot N$.

Figure 12: $a=5$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig1.eps}}

Figure 13: $a=2$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig2.eps}}

Figure 14: $a=1$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig3.eps}}

Figure 15: $a=.67$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig4.eps}}

Figure 16: $a=.33$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig5.eps}}

Figure 17: $a=.167$
\resizebox{5.5in}{3.2in}{\includegraphics{backsolvefig6.eps}}


next up previous contents
Next: Other Monotonic Nonstationary Signals Up: Examples of Estimating by Previous: Examples of Estimating by   Contents
Aaron S. Master 2002-10-17