Generalization of the Redfield-Kunz treatment of quadrature phase time data - technical

Generalization of the Redfield-Kunz Treatment of Quadrature Phase Time Data

CLASICAL, DISCRETE, QUADRATURE PHASE Fourier transformation stipulates that the analog signals x(t) and y(t) to be Fourier transformed are sampled and digitized simultaneously and that all time increments are equal. Digitization is often done by two-channel waveform recorders or equivalent instruments. Many of these sample the two input signals x(t) and y(t) alternately and not simultaneously. These instruments also do ot store the data in separate sets [x.sub.i.] and [y.sub.i.], i=0, ..., N - 1 with N = 2.sup.2 for some integer n, but rather in a single consecutive set [x.sub.0., y.sub.0., x.sub.1., y.sub.1., ..., x.sub.N-1., y.sub.N-1.]. Redfield and Kunz have indicated an elegant way to handle the computation of the Fourier transform or spectrum in this situation. The data set [x.sub.0, ..., y.sub.N-1] is multipied, element by element, with +1, -1, -1, +1, +1, -1, -1, ... The discrete cosine (sine) Fourier transform of the resulting set [x.sub.0., -y.sub.0., -x.sub.1., y.sub.1., x.sub.2., ...] gives, without any further data manipulation, the quadrature phase absorption (dispersion) spectrum with the center or zero frequency automatically in the middle of the respective partial spectrum.

Hewlett-Packard offers a two-channel waveform recorder (the HP 5180A) which is very well-suited for digitization of quadrature phase time signals of many kinds, for which the Fourier transform, or spectrum, is ultimately to be computed. The data obtained in pulsed nuclear magnetic resonance (NMR) spectroscopy is just one example. The problem is that the HP 5180A samples input signals x(t) and y(t) neither simultaneously nor alternately. Wahtever digitization rate is chosen, y(t) is always sampled 100 ns later than x(t). This problem could be cured with a delay line and appropriate compensation of the losses in the RF part of the quadrature phase receiver, but it is preferable to seek the solution on the computational side. It is the purpose of this communication to indicate the appropriate steps.

To explain the procedure we apply these steps to a test signal x(t) = cos(2[pi]ft/T), y(t) = sin(2[pi]ft/T) with f being an integer and T = N[delta]t the total measuring time, where [delta]t is the sampling interval. T is thus an integer multiple of the period of the test signal and we can clearly formulate what we want to see in the resulting spectrum: the absoption part should have a line with some normalized amplitude at the position [upsilon].sub.k = k[delta][upsilon] with k = +f and [delta][upsilon] = 1/T, and zeros everywhere else. All entries of the dispersion part should be zero. The signal is sampled as shown in Fig. 1, where the 100-ns difference in the sampling times for x(t) and y(t) is generalized to an interval dt.

The stored data consists of the set [s.sub.q] with s.sub.q = x.sub.21 [right arrow] cos(2[pi]fl/N) for q = 2l even = y.sub.21+1 [right arrow] sin(2[pi]fl/N) for q = 21 + 1 odd where l = 0, ..., N - 1. The arrows indicate the result in the case of our test signal. The Redfield-Kunz-treatment leaves us with the set [r.sub.q.] with r.sub.q = r.sub.21 = (-1).sup.21 x.sub.21 [right arrow] (-1).sup.1cos(2[pi]fl/N), q = 21 = r.sub.21+1 = -(-1).sup.1 y.sub.21+1 [right arrow] -(-1).sup.1sin(2[pi]fl/N), q = 21 + 1.

The Fourier transform or complex spectrum g([upsilon].sub.k) of the set [r.sub.q.] consists of the set of N complex numbers

We now insert our test signal into equation 1 and consider first the Redfield-Kunz case [epsilon] = dt/[delta]t = 2/2.

Expressing i by e.sub.i[pi]/2 and (-1).sup.l by e.sup.i[pi]l leaves us with

Since we get g.sub.k = N/2 [(1 + e.sup.-i[pi](k-f-N/2)/N)[delta].sub.k,N/2+f + (1 - e.sup.-i[pi](k+f-N/2)/N)[delta].sub.k,N/2-f].

Note that both exponentials are equal to one whenever the Kronecker symbols are nonzero. Therefore, g.sub.k = N[delta].sub.k,N/2+f.

This is the Redfield-Kunz result: a test signal of the type we have chosen gives a single line at the desired position, i.e., f steps to the right of the center in the absoption spectrum, whereas the dispersion spectrum (imaginary parts of the g.sub.k) is zero throughout the entire range of k. If we had chosen x(t) = sin(2[pi]ft/T) and y(t) = cos(2[pi]ft/T) as test signals we would have obtained a single nonzero entry in the dispersion part of the complex spectrum.

We now turn to the general or HP case where [epsilon] = dt/[delta]t [is not =] 1/2. With the identical sequence of steps in the treatment of the data, i.e., Redfield-Kunz multiplication and Fourier transformation as above, we obtain which are the modified equations 2 and 2a. Instead of equation 3 we get g.sub.k = N/2 [(1 + e.sup.-[pi](k-2f[epsilon]-N/2)/N)[delta].sub.k,N/2+f + (1 - e.sub.-i[pi](k+2f[epsilon]-N/2)/N)[delta].sub.k,N/2-f].

Applying the action of the Kronecker symbols to the exponentials gives g.sub.k = N/2 [(1 + e.sup.-i[pi](k-N/2)(1-2[epsilon])/N)[delta].sub.k,N/2+f + (1 - e.sup.-i[pi](k-N/2)(1-2[epsilon])/N)[delta].sub.k,N/2-f].

The problem with the HP waveform recorders is now evident: for [epsilon] [is not =] 1/2 the exponentials are not equal to one, which means that our test signal produces nonzero entries in the dispersion part of the spectrum as well as a "ghost line" at k = N/2 - f in the absorption part. We also recognize that the ghost lines are small in the central region of the spectrum where k [nearly equals] N/2 whereas they become large near the edges, whre k [nearly equals] 0 and k [nearly equals] N - 1. An example for [epsilon] = 0.25 and f = 3N/8 is shown in Fig. 2a.

To "exorcise" the ghosts we begin by defining a set (h.sub.k) by h.sub.k * g.sub.N-k

This expression for h.sub.k follows immediately from equation 2a'. Again introducing Kronecker symbols gives h.sub.k = N/2 [(1 - e.sup.i[pi](k-N/2)(1-2[epsilon]/N)[delta].sub.k,N/2+f + (1 + e.sup.i[pi](k-N/2)(1-2[epsilon])[delta].sub.k,N/2-f].

By comparing equations 5 and 6 we see immediately that g.sub.k + h.sup.*.sub.k = g.sub.k + g.sup.*.sub.N-k = N([delta].sub.k,N/2+f + [delta].sub.k,N/2-f.) and g.sub.k.-h.sup.*.sub.k.=g.sub.k.-g.sup.*.sub.N-k.=N([delta].sub.k,N/ 2+f.-[delta].sub.k,N/2-f.) e.sup.-i[pi](k-N/2)(1-2[epsilon])/N from which we obtain our final result which replaces equation 4, valid for the Redfield-Kunz-case [epsilon] = 1/2 in the case of the use of an HP 5180A or equivalent waveform recorder. The digitization scheme of these waveform recorders can thus be taken care of by this simple prescription: treat the data exactly as in the Redfield-Kunz-case where the x and y samples are taken alternately. This gives the set [g.sub.k]. Then use these g.sub.k to compute the set [g.sub.k] according to equation 7. The real (imaginary) part of this set represents the desired discrete absoption (dispersion) spectrum.

COPYRIGHT 1988 Hewlett Packard Company
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