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In Section 4.2.2.4, we showed what happens when two copies of the same sound arrive at a listener at different times. For each of the frequencies in the sound, the copy of the frequency coming from speaker B is in a different phase relative to the copy coming from speaker A (Figure 4.27). In the case of frequencies that are offset by exactly one half of a cycle, the two copies of the sound are completely out-of-phase, and those frequencies are lost for the listener in that location. This is an example of comb filtering caused by delay.
To generalize this mathematically, let’s assume that loudspeaker B is d feet farther away from a listener than loudspeaker A. The speed of sound is c. Then the delay t, in seconds, is
[equation caption=”Equation 4.20 Delay t for offset d between two loudspeakers”]
$$!t=\frac{d\: ft}{c\:ft/s}$$
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Assume for simplicity that the speed of sound is 1000 ft/s. Thus, for an offset of 20 ft, you get a delay of 0.020 s.
$$!t=\frac{20\: ft}{1000\:ft/s}$$
$$!t=0.02s=20ms$$
What if you want to know the frequencies of the sound waves that will be combed out by a delay of t? The fundamental frequency to be combed, $$f_{0}$$, is the one that is delayed by half of the period, since this delay will offset the phase of the wave by 180°. We know that the period is the inverse of the frequency, which gives us
$$!t=\frac{1}{2\ast f_{0}}$$
$$!t=\frac{1}{2\ast t}$$
Additionally, all integer multiples of $$f_{0}$$ will also be combed out, since they also will be 180° offset from the other copy of the sound. Thus, we can this formula for the frequencies combed out by delay t.
[equation caption=”Equation 4.21 Comb filtering”]
Given a delay of t seconds between two identical copies of a sound,
then the frequencies $$f_{i}$$ that will be combed out are
$$!f_{i}=\frac{i+1}{2t}for\:all\:integers\:i\geq0$$
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For a 20 foot separation in distance, which creates a delay of 0.02 s, the combed frequencies are 25 Hz, 50 Hz, 75 Hz, and so forth.
In Section 2, we made the point that comb filtering in the air can be handled by increasing the delay between the two sound sources. A 40 foot distance between two identical sound sources results in a 0.04 s delay, which then combs out 12.5 Hz, 25 Hz, 37.5 Hz, 50 Hz, and so forth. The larger delay, the lower the frequency at which combing begins, and the closer the combed frequencies are to one another. You can see this in Figure 4.41. In the first graph, a delay of 0.5682 ms combs out integer multiples of 880 Hz. In the second graph, a delay of 2.2727 ms combs out integer multiples of 220 Hz.
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If the delay is long enough, frequencies that are combed out are within the same critical band as frequencies that are amplified. Recall that all frequencies in a critical band are perceived as the same frequency. If one frequency is combed out and another is amplified within the same critical band, the resulting perceived amplitude of the frequency in that band is the same as would be heard without comb filtering. Thus, a long enough delay mitigates effect of comb filtering. The exercise associated with this section has you verify this point.
Room mode operates by the same principle as comb filtering. Picture a sound being sent from the center of a room. If the speed of sound in the room is 1000 ft/s and the room has parallel walls that are 10 feet apart, how long will it take the sound to travel from the center of the room, bounce off one of the walls, and come back to the center? Since the sound is traveling 5 + 5 =10 feet, we get a delay of $$t=\frac{10ft}{1000\frac{ft}{s}}=0.01s$$. This implies that a sound wave of frequency $$f_{0}=\frac{1}{2\ast 0.01}=50$$ Hz sound wave will be combed out in the center of the room. The center of the room is a node with regard to a frequency of 50 Hz.
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For the second harmonic, 100 Hz, the nodes are 2.5 feet from the wall. The time it takes for sound to move from a point 2.5 feet from the wall and bounce back to that same point is 2.5 + 2.5 = 5 feet, yielding a delay of $$t=\frac{5ft}{1000\frac{ft}{s}}=0.005s$$. This is half the period of the 100 Hz wave, meaning a frequency of 100 Hz will be combed out at those points. However, in the center of the room, we still have a delay of $$t=\frac{10ft}{1000\frac{ft}{s}}=0.001s$$, which is the full period of the 100 Hz wave, meaning the 100 Hz wave gets amplified at the center of the room.
The other harmonic frequencies can be explained similarly.