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\line{\hskip 3 em\hfill \bigbf LAB 5 ANSWERS\hfill}
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\line{\hskip 3 em\hfill\bigbf FOURIER TRANSFORMS  \hfill}
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\vskip 1 em\item{A1)} It has amplitude only at the zero or DC frequency and zero amplitude
everywhere else. A boxcar is supposed to transform to a sync function, so what
is wrong here? The answer is that a continuous Fourier transform will transform
to a sync, but the discrete transform gives just a spike at zero frequency. This
is an important point - the discrete transform is not simply the continuous
transform sampled at discrete points. This is particularly true at the long
period end of any discrete spectrum.

\vskip 1 em\item{A6)} If you look at your tabulated spectrum, you will see that the frequencies
in the discrete transform are: $0,\pm 0.125,\pm 0.250,\pm 0.375, and\, +.5\,
cy/sec$, so the lowest frequency we have available with this transform is $\pm 
0.125\, cy/sec$, and this is not low enough to adequately describe this
function.
The transform compensates by trying to describe the sine wave as a sum of higher
frequency waves, so the spectrum looks broad-band, but it just does not work.
The signal we transform cannot have a wavelength, or a component with a
wavelength, longer than the span of the data. A component with such a wavelength
should be removed as part of the drift. 

\vskip 1 em\item{A7)} This is the same signal we looked at in the prepackaged exercise, so no
need to say anymore.

\vskip 1 em\item{A8)} If the sampling interval is 1 meter, the Nyquist frequency is .5 cy/meter.
This is the highest frequency in the transform. The input wave had a frequency
of 1/1.75 = .57 cy/meter, or higher than the Nyquist. As a result, instead of
getting a nice spike at the frequency of the wave, we do not have that
frequency at our disposal, and the signal is \bf aliased \rm into lower
frequencies in fact to  0.43 cy/meter. This is not one of the Fourier
frequencies, so the spectrum is no a spike but is spread out. This is somewhat like what happened at the
other end of the
frequency scale when we fed the routine a signal with frequency lower than it
could handle. Always remember that  there is an upper, and a lower, limit to the
frequency scale in a discrete Fourier transform. The highest frequency is the
Nyquist, or 1/($2\times$ the sampling interval), the lowest is 1/(data span).

\vskip 1 em\item{A9)} If the samples are a second apart, and we have 8 of them, then the Nyquist
is 0.5 and the frequencies in the transform are $0.0, \pm 0.125,\pm 0.250, \pm 
0.375, and\, 0.5$, so the frequency of the wave 0.25 just happens to be one of
the Fourier frequencies. If the same wave is sampled 8 times at 1.2 seconds ,
the Nyquist is .416 sec, the frequency increments are 0.104, and the Fourier
frequencies are $0.0, \pm 0.104, \pm,0.204, \pm 312, and\, 0.416 $ so the
frequency of the wave is not one of the Fourier frequencies, and the transform
cannot describe the wave with just one spike. It ``leaks'' the signal to the
nearest frequencies. Hence the term frequency leakage.  So a pure sinusoid does
not always appear in the transform as a spike at a single frequency. 

\bye
                              
