How to control ultra low frequency and improve the low frequency uniformity

1. The direction of speaker 1 and speaker 2 are the same, and the front-rear distance is 1/4 wavelength.
2. The two speakers are the ideal ones, the signal conversion delay time is the same, and the same ideal cosine wave is sent in the A direction.
3. The sounding point distance from point A to speaker 2 is SA.
4. The sounding point distance from point B to speaker 1 is SB.
The key step now is to have a time difference between Speaker 1 and Speaker 2. Setting a system time difference is speaker 1/4 delay 1/4T.
Front: Let's analyze, the speaker 1 first sounds, after 1/4T, it reaches the position of the speaker 2. Because the speaker 2 has a delay of 1/4T, so the speaker 2 just starts to sound, then their sound waves are synchronized and the phase is very close. For coherent waves, theoretically we can get a sound pressure gain of +6dB.
Rear: The rear analysis is a bit difficult. Let us understand this. I don't know if I can understand it. Let the reference point be at point B, the speaker 2 will be later than the sound wave of the speaker 1 (distance 1/4 wavelength + delay 1/4T), so that they are 180 degrees apart. This is a phase that cancels each other out.
The characteristics of this method:
1. It is very important to choose the system frequency. Because of the relationship between 1/4T and speaker distance, this amount is very important. If the adjustment is not good, it is counterproductive.
2. Since it is ultra low frequency, it is generally below 100HZ. If the set frequency is Fk, the effective frequency that can be controlled by this method is: 0.707*Fk to 1.414*FK. The control effect of the center frequency is most obvious.
Improve low frequency uniformity:
The second speaker is 1/4 wavelength away, and the speaker 2 is delayed by 1/4T. Set the sensitivity of the speaker to 100db, so good. The speaker distance is 1.34 meters and the delay is 4ms.
Let's assume that the listening point is 1 meter. We are calculating the sound pressure.
Sound pressure level at 1 meter:
Speaker 2: 100DB*1W-20LG1=100DB, speaker 1:100DB*1W-20*LG(1+1.34)m=92.6dB. How to calculate the sound pressure level superposition of coherent wave? Let us first calculate the sound level to the sound Press, then count to the sound level. Sound pressure of speaker 2 at 1 meter: 10^(100DB/20)*P0, the sound pressure of speaker 1 at this point is: 10^(92.6DB/20)*P0, the superposition of coherent waves is the sound pressure vector sum, And because of the phase one, their sum is: 142658 * P0, the sound pressure level is: 20LG142658 = 103db.
Therefore, at a meter in front of the speaker 2, the sound pressure level is 103DB. After superimposition, the sound level is increased by 3DB (this is if there is no sound blocking)
Let us look at the sound pressure level at 16 meters:
Speaker 2: 100DB*1W-20LG16m = 76DB, speaker 1: 100DB*1W-20*LG (16+1.34) m = 75.2dB. First calculate the sound level to the sound pressure, speaker 2 at 16 meters sound pressure: 10^(76DB/20)*P0, the sound pressure of the speaker 1 at this point is: 10^(75.2DB/20)*P0, the superposition of the coherent waves is the sound pressure vector sum, and because of the phase one, their And: 12064 * P0, the sound pressure level is: 20LG12064 = 81.6db.
Therefore, at the front of the speaker 2, the sound pressure level is: 81.6db. The two speakers are superimposed, which is 5.6DB higher than the sound pressure level of a single speaker.

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