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Car Audio Discussion
General Car Audio
Time Alignment vs. Phase Shift
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<blockquote data-quote="Sarthos" data-source="post: 7030310" data-attributes="member: 610335"><p>Let's see if this works. Here's a hypothetical graph for a note played at 1000 Hz, with the front left speaker 2 feet from the driver's ear, and the front right speaker 6 feet from the driver's ear.</p><p></p><p><a href="http://graph-plotter.cours-de-math.eu/graph.php?a0=2&amp;a1=sin%281000%2A%28x-0.001776%29%29&amp;a2=sin%281000%2A%28x-0.005329%29%29&amp;a3=sin%281000%2A%28x-0.001776%29%29%2Bsin%281000%2A%28x-0.005329%29%29&amp;a4=1&amp;a5=3&amp;a6=7&amp;a7=1&amp;a8=1&amp;a9=1&amp;b0=500&amp;b1=500&amp;b2=0&amp;b3=0.01&amp;b4=-2&amp;b5=2&amp;b6=10&amp;b7=10&amp;b8=5&amp;b9=5&amp;c0=3&amp;c1=0&amp;c2=1&amp;c3=1&amp;c4=1&amp;c5=1&amp;c6=1&amp;c7=0&amp;c8=0&amp;c9=0&amp;d0=1&amp;d1=20&amp;d2=20&amp;d3=0&amp;d4=&amp;d5=&amp;d6=&amp;d7=&amp;d8=&amp;d9=&amp;e0=&amp;e1=&amp;e2=&amp;e3=&amp;e4=14&amp;e5=14&amp;e6=13&amp;e7=12&amp;e8=0&amp;e9=0&amp;f0=0&amp;f1=1&amp;f2=1&amp;f3=0&amp;f4=0&amp;f5=&amp;f6=&amp;f7=&amp;f8=&amp;f9=&amp;g0=&amp;g1=1&amp;g2=&amp;g3=0&amp;g4=0&amp;g5=0&amp;zalt=" target="_blank">http://graph-plotter.cours-de-math.eu/graph.php?a0=2&amp;a1=sin%281000%2A%28x-0.001776%29%29&amp;a2=sin%281000%2A%28x-0.005329%29%29&amp;a3=sin%281000%2A%28x-0.001776%29%29%2Bsin%281000%2A%28x-0.005329%29%29&amp;a4=1&amp;a5=3&amp;a6=7&amp;a7=1&amp;a8=1&amp;a9=1&amp;b0=500&amp;b1=500&amp;b2=0&amp;b3=0.01&amp;b4=-2&amp;b5=2&amp;b6=10&amp;b7=10&amp;b8=5&amp;b9=5&amp;c0=3&amp;c1=0&amp;c2=1&amp;c3=1&amp;c4=1&amp;c5=1&amp;c6=1&amp;c7=0&amp;c8=0&amp;c9=0&amp;d0=1&amp;d1=20&amp;d2=20&amp;d3=0&amp;d4=&amp;d5=&amp;d6=&amp;d7=&amp;d8=&amp;d9=&amp;e0=&amp;e1=&amp;e2=&amp;e3=&amp;e4=14&amp;e5=14&amp;e6=13&amp;e7=12&amp;e8=0&amp;e9=0&amp;f0=0&amp;f1=1&amp;f2=1&amp;f3=0&amp;f4=0&amp;f5=&amp;f6=&amp;f7=&amp;f8=&amp;f9=&amp;g0=&amp;g1=1&amp;g2=&amp;g3=0&amp;g4=0&amp;g5=0&amp;zalt=</a></p><p></p><p>The blue line is that of the driver's side speaker</p><p></p><p>The red line is that of the passenger's side speaker</p><p></p><p>The green line is that of their overlap</p><p></p><p>As you can see, the two waves mostly offset each other. The result is that the two speakers aren't as loud combined as either individual speaker</p><p></p><p>Now, if the driver's side speaker is delayed</p><p></p><p><a href="http://graph-plotter.cours-de-math.eu/graph.php?a0=2&amp;a1=sin%281000%2A%28x-0.005329%29%29&amp;a2=sin%281000%2A%28x-0.005329%29%29&amp;a3=2%2Asin%281000%2A%28x-0.005329%29%29&amp;a4=1&amp;a5=3&amp;a6=7&amp;a7=1&amp;a8=1&amp;a9=1&amp;b0=500&amp;b1=500&amp;b2=0&amp;b3=0.01&amp;b4=-2&amp;b5=2&amp;b6=10&amp;b7=10&amp;b8=5&amp;b9=5&amp;c0=3&amp;c1=0&amp;c2=1&amp;c3=1&amp;c4=1&amp;c5=1&amp;c6=1&amp;c7=0&amp;c8=0&amp;c9=0&amp;d0=1&amp;d1=20&amp;d2=20&amp;d3=0&amp;d4=&amp;d5=&amp;d6=&amp;d7=&amp;d8=&amp;d9=&amp;e0=&amp;e1=&amp;e2=&amp;e3=&amp;e4=14&amp;e5=14&amp;e6=13&amp;e7=12&amp;e8=0&amp;e9=0&amp;f0=0&amp;f1=1&amp;f2=1&amp;f3=0&amp;f4=0&amp;f5=&amp;f6=&amp;f7=&amp;f8=&amp;f9=&amp;g0=&amp;g1=1&amp;g2=&amp;g3=0&amp;g4=0&amp;g5=0&amp;zalt=" target="_blank">http://graph-plotter.cours-de-math.eu/graph.php?a0=2&amp;a1=sin%281000%2A%28x-0.005329%29%29&amp;a2=sin%281000%2A%28x-0.005329%29%29&amp;a3=2%2Asin%281000%2A%28x-0.005329%29%29&amp;a4=1&amp;a5=3&amp;a6=7&amp;a7=1&amp;a8=1&amp;a9=1&amp;b0=500&amp;b1=500&amp;b2=0&amp;b3=0.01&amp;b4=-2&amp;b5=2&amp;b6=10&amp;b7=10&amp;b8=5&amp;b9=5&amp;c0=3&amp;c1=0&amp;c2=1&amp;c3=1&amp;c4=1&amp;c5=1&amp;c6=1&amp;c7=0&amp;c8=0&amp;c9=0&amp;d0=1&amp;d1=20&amp;d2=20&amp;d3=0&amp;d4=&amp;d5=&amp;d6=&amp;d7=&amp;d8=&amp;d9=&amp;e0=&amp;e1=&amp;e2=&amp;e3=&amp;e4=14&amp;e5=14&amp;e6=13&amp;e7=12&amp;e8=0&amp;e9=0&amp;f0=0&amp;f1=1&amp;f2=1&amp;f3=0&amp;f4=0&amp;f5=&amp;f6=&amp;f7=&amp;f8=&amp;f9=&amp;g0=&amp;g1=1&amp;g2=&amp;g3=0&amp;g4=0&amp;g5=0&amp;zalt=</a></p><p></p><p>In this graph the two speakers have perfectly matched waves, which are indistinguishable. As you can see from their resultant, it is far larger than the original green wave before time alignment was added, in fact it is double the volume of either individual speaker. That is perfect overlap.</p><p></p><p>High frequencies suffer the greatest offset. To figure out how far offset a wave is, multiply the frequency of the note by the distance difference in the distance between the speakers and divide it by 1126. That is the offset of the sine wave. If the number happens to equal 3.14159... you (theoretically) have perfect cancellation. In this example 1000 Hz will suffer a lot since 1000*(6-2)/1126=4.44, which isn't total cancellation, but it's still bad. In this example of a speaker 6 feet away and one 2 feet away the speaker will have complete cancellation at 884 Hz, as well as any multiple of 884 * (2n+1) so 884, 2652, 4420, 6188 Hz, etc. will all be cancelled nearly completely. This is devastating to sound quality.</p><p></p><p>I could continue with calculations on this, but instead I'll just state the fact that this is a problem more with high frequencies than with low frequencies. a note at 1000 Hz has a wavelength of just over a foot, but a frequency of 40 Hz has a wavelength of over 28 feet, so the distance between different subs won't make a big difference. But with high notes that have frequencies of a foot or less are easily thrown off by the difference in distance between the speakers.</p></blockquote><p></p>
[QUOTE="Sarthos, post: 7030310, member: 610335"] Let's see if this works. Here's a hypothetical graph for a note played at 1000 Hz, with the front left speaker 2 feet from the driver's ear, and the front right speaker 6 feet from the driver's ear. [URL="http://graph-plotter.cours-de-math.eu/graph.php?a0=2&a1=sin%281000%2A%28x-0.001776%29%29&a2=sin%281000%2A%28x-0.005329%29%29&a3=sin%281000%2A%28x-0.001776%29%29%2Bsin%281000%2A%28x-0.005329%29%29&a4=1&a5=3&a6=7&a7=1&a8=1&a9=1&b0=500&b1=500&b2=0&b3=0.01&b4=-2&b5=2&b6=10&b7=10&b8=5&b9=5&c0=3&c1=0&c2=1&c3=1&c4=1&c5=1&c6=1&c7=0&c8=0&c9=0&d0=1&d1=20&d2=20&d3=0&d4=&d5=&d6=&d7=&d8=&d9=&e0=&e1=&e2=&e3=&e4=14&e5=14&e6=13&e7=12&e8=0&e9=0&f0=0&f1=1&f2=1&f3=0&f4=0&f5=&f6=&f7=&f8=&f9=&g0=&g1=1&g2=&g3=0&g4=0&g5=0&zalt="]http://graph-plotter.cours-de-math.eu/graph.php?a0=2&a1=sin%281000%2A%28x-0.001776%29%29&a2=sin%281000%2A%28x-0.005329%29%29&a3=sin%281000%2A%28x-0.001776%29%29%2Bsin%281000%2A%28x-0.005329%29%29&a4=1&a5=3&a6=7&a7=1&a8=1&a9=1&b0=500&b1=500&b2=0&b3=0.01&b4=-2&b5=2&b6=10&b7=10&b8=5&b9=5&c0=3&c1=0&c2=1&c3=1&c4=1&c5=1&c6=1&c7=0&c8=0&c9=0&d0=1&d1=20&d2=20&d3=0&d4=&d5=&d6=&d7=&d8=&d9=&e0=&e1=&e2=&e3=&e4=14&e5=14&e6=13&e7=12&e8=0&e9=0&f0=0&f1=1&f2=1&f3=0&f4=0&f5=&f6=&f7=&f8=&f9=&g0=&g1=1&g2=&g3=0&g4=0&g5=0&zalt=[/URL] The blue line is that of the driver's side speaker The red line is that of the passenger's side speaker The green line is that of their overlap As you can see, the two waves mostly offset each other. The result is that the two speakers aren't as loud combined as either individual speaker Now, if the driver's side speaker is delayed [URL="http://graph-plotter.cours-de-math.eu/graph.php?a0=2&a1=sin%281000%2A%28x-0.005329%29%29&a2=sin%281000%2A%28x-0.005329%29%29&a3=2%2Asin%281000%2A%28x-0.005329%29%29&a4=1&a5=3&a6=7&a7=1&a8=1&a9=1&b0=500&b1=500&b2=0&b3=0.01&b4=-2&b5=2&b6=10&b7=10&b8=5&b9=5&c0=3&c1=0&c2=1&c3=1&c4=1&c5=1&c6=1&c7=0&c8=0&c9=0&d0=1&d1=20&d2=20&d3=0&d4=&d5=&d6=&d7=&d8=&d9=&e0=&e1=&e2=&e3=&e4=14&e5=14&e6=13&e7=12&e8=0&e9=0&f0=0&f1=1&f2=1&f3=0&f4=0&f5=&f6=&f7=&f8=&f9=&g0=&g1=1&g2=&g3=0&g4=0&g5=0&zalt="]http://graph-plotter.cours-de-math.eu/graph.php?a0=2&a1=sin%281000%2A%28x-0.005329%29%29&a2=sin%281000%2A%28x-0.005329%29%29&a3=2%2Asin%281000%2A%28x-0.005329%29%29&a4=1&a5=3&a6=7&a7=1&a8=1&a9=1&b0=500&b1=500&b2=0&b3=0.01&b4=-2&b5=2&b6=10&b7=10&b8=5&b9=5&c0=3&c1=0&c2=1&c3=1&c4=1&c5=1&c6=1&c7=0&c8=0&c9=0&d0=1&d1=20&d2=20&d3=0&d4=&d5=&d6=&d7=&d8=&d9=&e0=&e1=&e2=&e3=&e4=14&e5=14&e6=13&e7=12&e8=0&e9=0&f0=0&f1=1&f2=1&f3=0&f4=0&f5=&f6=&f7=&f8=&f9=&g0=&g1=1&g2=&g3=0&g4=0&g5=0&zalt=[/URL] In this graph the two speakers have perfectly matched waves, which are indistinguishable. As you can see from their resultant, it is far larger than the original green wave before time alignment was added, in fact it is double the volume of either individual speaker. That is perfect overlap. High frequencies suffer the greatest offset. To figure out how far offset a wave is, multiply the frequency of the note by the distance difference in the distance between the speakers and divide it by 1126. That is the offset of the sine wave. If the number happens to equal 3.14159... you (theoretically) have perfect cancellation. In this example 1000 Hz will suffer a lot since 1000*(6-2)/1126=4.44, which isn't total cancellation, but it's still bad. In this example of a speaker 6 feet away and one 2 feet away the speaker will have complete cancellation at 884 Hz, as well as any multiple of 884 * (2n+1) so 884, 2652, 4420, 6188 Hz, etc. will all be cancelled nearly completely. This is devastating to sound quality. I could continue with calculations on this, but instead I'll just state the fact that this is a problem more with high frequencies than with low frequencies. a note at 1000 Hz has a wavelength of just over a foot, but a frequency of 40 Hz has a wavelength of over 28 feet, so the distance between different subs won't make a big difference. But with high notes that have frequencies of a foot or less are easily thrown off by the difference in distance between the speakers. [/QUOTE]
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Time Alignment vs. Phase Shift
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