1.0) = 0.440 -&gt; 0.575
  J2(1.0) = 0.115 -&gt; 0.150
  J3(1.0) = 0.019 -&gt; 0.025
  J4(1.0) = 0.002 -&gt; 0.003

(* Jn values normalized to match
the peak values given above)
</pre>
</td><td>
<pre>
  J0(2.0) = 0.224 -&gt; 0.388 (*)
  J1(2.0) = 0.577 -&gt; 1.0
  J2(2.0) = 0.353 -&gt; 0.611
  J3(2.0) = 0.129 -&gt; 0.223
  J4(2.0) = 0.034 -&gt; 0.058
  J5(2.0) = 0.007 -&gt; 0.012
  J6(2.0) = 0.001 -&gt; 0.002
</pre>
<small>(A larger FFT reduces the mismatch)</small>
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