By Jyrki Kauppinen

ISBN-10: 3527402896

ISBN-13: 9783527402892

This contemporary method of the topic is obviously and logically dependent, and offers readers an figuring out of the essence of Fourier transforms and their purposes. All very important facets are incorporated with recognize to their use with optical spectroscopic facts. in response to well known lectures, the authors give you the mathematical basics and numerical functions that are crucial in sensible use. the most a part of the publication is devoted to purposes of toes in sign processing and spectroscopy, with IR and NIR, NMR and mass spectrometry handled either from a theoretical and sensible perspective. a few facets, linear prediction for instance, are defined right here completely for the 1st time.

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Content material:

**Additional info for Fourier Transforms in Spectroscopy**

**Example text**

29) Similarly, −1 |h(t)|2 = H ( f ) H ( f ). 30) These equations are the autocorrelation theorem of Fourier transforms. 4. It is equal to the inverse Fourier transform of the autocorrelation function: |H ( f )|2 = −1 {h(t) h(t)}. 31) The power spectrum is used when the phase spectrum θ ( f ) can be omitted. This is the case in the examination of noise, shake, seismograms, and so forth. 1: Tex Willer shoots six shots in one second and then loads his revolver for four seconds. These periods of ﬁve seconds repeat identically for the whole duration of one minute of a gunﬁght.

Show that autocorrelation function h (h where (h h)(−t) = [h(t) h(t)]∗ , h)(−t) means the value of h 4. Compute the convolution function 2T (t) = 2T (t) ∗ h at the point −t. 2T (t), where 2T (t) is the boxcar function 1, |t| ≤ T, 0, |t| > T. Plot the result. 5. Prove that {H ( f )G( f )} = {H ( f )} ∗ {G( f )} (the convolution theorem). 6. Show that the power spectrum of a real signal is always symmetric. 7. , determine the transform −1 {h(t) sin(2π f 0 t)}. 8. Let us assume that we know the convolution theorem −1 {h(t) ∗ g(t)} = H ( f )G( f ).

The result is stored back in d. *) var i, istep, j, m, mmax :integer; ct, st, z, theta, wstpr, wstpi, tr, ti:real; begin (* fft *) with d do begin j: = 0; for i: = 0 to n − 1 do begin if i < j then begin z: = re [i]; re [i] : = re [ j]; re [ j] : = z; z: = im [i]; im [i] : = im [ j]; im [ j] : = z end; m: = n div 2; while ( j > m) and (m >= 0) do begin j: = j − m − 1; m: = m div 2 end; j: = j + m + 1 end; mmax: = 1; while mmax < n do begin istep: = 2 ∗ mmax; for m: = 1 to mmax do begin i: = m − 1; theta: = isign ∗ pi ∗ (m − 1)/mmax; ct: = cos(theta); st: = sin(theta); while i < n do begin j: = i + mmax; tr: = re [i]; 57 58 4 Fast Fourier transform ti: = im [i]; wstpr: = ct ∗ re [ j] + st ∗ im [ j]; wstpi: = ct ∗ im [ j] − st ∗ re [ j]; re [i] : = tr + wstpr; im [i] : = ti + wstpi; re [ j] : = tr − wstpr; im [ j] : = ti − wstpi; i: = i + istep end end; mmax: = istep end; for i: = 0 to n − 1 do begin re [i] : = re [i] /sqrt(n); im [i] : = im [i] /sqrt(n) end end end; (* fft *) Problems 1.

### Fourier Transforms in Spectroscopy by Jyrki Kauppinen

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