Complex-trace analysis

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Problem 9.26a

Given the wavelet , calculate the quadrature function, .

Background

A wave with continuously varying amplitude is of the form


(9.26a)

where we assume that the rate of change of is small compared with and the sampling frequency . If we take fixed, the Hilbert transform (see Sheriff and Geldart, 1995, section 15.2.13) of , written , is


(9.26b)

We note that and differ in phase by . We can combine the signals in equations (9.26a,b) to obtain the complex function :


(9.26c)

(see Sheriff and Geldart, 1995, problem 15.12a for Euler’s formulas and Sheriff and Geldart, 1995, section 15.1.5 for a discussion of complex functions). The quantities and are known as the complex trace and the quadrature trace, respectively. and its mirror image constitute the envelope of both and .


(9.26d)

We think of the complex trace as being traced by the tip of a vector of length rotating in the complex plane as it moves perpendicular to the complex plane in the time direction (Figure 9.26a). The projection of the helical path generated by the tip of this vector onto the real plane is and the projection onto the imaginary plane is . The angle that this vector makes with the real plane is the instantaneous phase and the rotational speed is the instantaneous frequency .


(9.26e)


(9.26f)
Figure 9.26a.  Graphs of (dashed line), (solid line), and (dotted lines).

Because , integral, we generally add (or subtract) multiples of to to make it monotonically increasing (or decreasing). The derivative of the arc tangent in equation (9.26f) usually permits more stable calculation than taking the derivative of itself. However, the finite-difference expression is often sufficiently accurate.

To get we find the Hilbert transform of [see Sheriff and Geldart, 1995, equation (15.176)]:

For a digital function , this becomes [see Sheriff and Geldart, 1995, equation (9.107)]


(9.26g)

where is the quadrature filter. Since or –2 according as is odd or even, the equation reduces to


(9.26h)

Solution

We note that 11 samples constitute 5/4 cycles so, assuming 2-ms sampling, the period is 20(4/5) = 16 ms or = 62 Hz = dominant frequency.

While is causal with elements to , equation (9.26h) shows that is not causal. We use equation (9.26h) to calculate for to 9. Thus,

Continuing the calculation as far as , we obtain for

Figure 9.26a shows and .

Problem 9.26b

What are values of the complex trace () and the amplitude of the envelope .

Solution

Using equations (9.26c,d), we get

Figure 9.26b.  Graph of .

Figure 9.26b is a graph of is in the southeast quadrant, and in the northeast, in the northwest, , , and in the southwest, and in the southeast, and in the northeast. Thus the phase makes a little over one revolution, just as does.

The amplitude is shown dotted in figure 9.26a.

Problem 9.26c

Calculate the instantaneous phase and the instantaneous frequency (adding to the phase multiples of as necessary to obtain a monotonically increasing function. Assume ms.

Solution

We use equations (9.26e,f) to obtain values of and , to 9. To calculate using the derivative operator (problem 9.31) in equation (9.26f) for to 9, we need to . Since , equation (9.26e) shows that both and are , depending on the signs of and ; to determine these signs we use equations (9.26h) to write

Thus, , , , .

We now calculate the values of and , to 9.

Note: Division by 360 gives revolutions/second of the vector , equivalent to hertz.

The average instantaneous frequency is 68 Hz, the standard deviation being 11 Hz.

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