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copied!<p>A couple of programming changes, first of all:</p> <pre><code>xAxis += pieSteps; if (xAxis >= fullSinWave) xAxis -= fullSinWave; //wrap x back into 0-2pi period </code></pre> <p>will help reduce numeric error.</p> <pre><code>in_buf[i].r = dataStream[i]; in_buf[i].i = 0; </code></pre> <p>will set the input buffer to <code>sin(x)</code>, previously you had it set to <code>sin(x) + j*sin(x)</code>, where <code>j = sqrt(-1)</code>.</p> <p>Moving <code>wantedHz = 1000;</code> out of the loop looks better.</p> <p>And a more fundamental issue: you set <code>wantedHz = 1000</code>. With a sample rate of 44.1 kHz this corresponds to <code>44100 points/sec * (1/1000) sec/cycle = 44.1 points/cycle</code>. With a buffer of 512 points, you will get 11.6 cycles of the sine wave in the buffer. Non-integer cycles lead to <a href="http://en.wikipedia.org/wiki/Spectral_leakage" rel="nofollow">leakage</a>. </p> <p>Before getting into this, though, try setting <code>wantedHz = 12*44100.0/512</code> to give exactly 12 cycles in the buffer. You should see two spikes in the transform: one at index 12, and one at index 511-12.</p> <p>The reason you'll see two spikes is that the transform of <code>sin(w_0*x)</code> is <code>j*{-delta(w-w_0) - delta(w+w_0)}</code>. That is, you get an impulse function at w_0 and -w_0 in the imaginary part of the transform. The reason they are at the places they are is that the transform goes from 0 to 2*pi.</p> <p>After you do this, go back to <code>wantedH = 1000</code>, giving you a non-integer number of cycles in the buffer. You should see a wide tent-shaped result, centered around bins 11 and 511-11. You should multiply <code>dataStream</code> by a window function (<a href="http://en.wikipedia.org/wiki/Hann_function" rel="nofollow">Hann</a> is good) to reduce the impact of this effect.</p>

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