Click here to purchase the entire book in PDF format. Specular Reflections and Comb Filters Let's stand in front of a perfectly reflective wall in an otherwise anechoic environment. We'll put an omnidirectional loudspeaker and a microphone somewhere relatively near each other and near the wall. Let's say, for the sake of argument that the loudspeaker and the microphone are 34.4 cm away from each other and the total distance from the loudspeaker to the wall and back to the microphone (the path of the reflected sound) is 68.8 cm. This is shown in Figure 3.22 Figure 3.22: An omnidirectional loudspeaker, an omnidirectional microphone and a wall in space. Now we'll connect a sine wave generator to the loudspeaker and look at the output of the microphone on an oscilloscope. We'll start by putting out a 1 Hz wave from the loudspeaker. The wavefront will radiate away from the loudspeaker in all directions (including the directions towards the microphone and towards the wall). One millisecond after the wavefront was emitted by the loudspeaker, it will reach the microphone. One millisecond later, the reflection off the wall will arrive at the microphone. This means that there is a difference of 1 ms between the time of arrival of the direct sound and of the early reflection off the wall. In real life, we would also know that the reflection off the wall will be one half the amplitude of the direct sound because the wavefront had to travel twice as far to get to the microphone. (Don't forget that the wall is perfectly reflective, so there is no loss on the reflection itself.) However, we're going to pretend for the first half of this section that sound does not decay with distance. We'll come back to this issue and correct for it later. Question: What is the phase difference between the direct sound and the reflection off the wall? Well, we know that the difference in the time of arrival is 1 ms. We can therefore calculate this as a phase difference using the frequency. The loudspeaker is putting out 1 Hz, which is 360$^\circ$ per second (or 2 $\pi$ radians per second). There is a 0.001 s difference in the time of arrival, so the phase difference is 0.001 * 360$^\circ$ = 0.36$^\circ$. This is a very small difference. In fact, it's so small that we can assume that the difference is 0. Therefore, the wave arriving as the direct sound is in phase with the wave arriving as the reflection. Therefore, these two waves add constructively at the microphone to make a sine wave that is twice as loud as either one of them individually (we're pretending that sound doesn't decay with distance, remember?). Now, we'll increase the frequency of the sine wave generator. As we get higher and higher in frequency, we are saying that we have more and more waves per second coming from the loudspeaker. This means that we have more degrees of phase per second coming from the loudspeaker. For example, if the frequency is 100 Hz, then we have 360$^\circ$ * 100 Hz = 36000$^\circ$ per second. the time difference between the arrival of the direct and reflected sounds is still 1 ms, however, since we are now at 100 Hz, that means that there is a 36000$^\circ$ * 0.001 = 36$^\circ$ difference in phase between the two. The higher in frequency we go, the larger the phase difference. This is because the difference in the time of arrival does not change. Let's move all the way up to a 500 Hz sine wave. Now the phase difference at the microphone between the direct and the reflected sound is 180$^\circ$ (you can do the math yourself...). This means that we will get destructive interference between the two, and there will be no output from the microphone. Essentially, the reflection is ``cancelling'' the direct sound. Keep going up in frequency to 1 kHz. At this frequency, the phase difference is 360$^\circ$, so we get back to having perfect constructive interference and two times the output again, just like we did at 1 Hz. At 1.5 kHz, we'll get no output, at 2 kHz we get a boost and so on and so on. Notice that these frequencies are just multiples of the first cancellation frequency at 500 Hz. If we graph this response of the output of the microphone vs. frequency and we draw the frequency scale linearly, it will look like Figure . (Interestingly, you might notice that the shape of this plot is a cosine curve with a DC offset.) COMB FILTER SHOWING LIN FREQUENCY AND LINEAR AMPLITUDE If you squint your eyes just right, the shape of this curve looks like a comb. And since, the effect at the microphone is as if we had done something to the original sound (we did do something to the original sound...) it sounds like it's been filtered. Therefore, we call this effect a comb filter. If we draw the same response on a dB scale, it will look like Figure which still looks like a comb. COMB FILTER SHOWING LIN FREQUENCY AND dB AMPLITUDE If we draw the same response on a logarithmic frequency scale, it will look like Figure which still looks a little less like a comb. COMB FILTER SHOWING LOG FREQUENCY AND dB AMPLITUDE What happens if we change the difference in the time of arrival of the direct and the reflected sound? We still wind up with a comb filter, however, the frequencies of its boosts and cuts will change. The larger the difference in the time of arrival, the lower the frequency will shift in the filter's response. If the distance changes in time, you will perceive it as a kind of ``swishing'' sound. If you want to hear what this sounds like, put your hand in front of your face, palm towards you. Make a ``ssshhhhh'' sound and move your hand closer and farther from your face. You'll hear a weird effect on the sound you're making. This is a comb filter. The direct sound from your mouth to your ear is interfering with the reflection off your hand. As you change the distance to your hand, you change the difference in the time of arrival of the two sounds, so you hear the frequency response of the comb filter shifting as well. The Real World Up to now in this section, we have been pretending that sound doesn't decay with distance. In real life, however, it does. So, what effect does this have on the response of the comb filter? It's actually pretty simple. The basic frequency response shape won't change much. However, since the reflection is now not quite as loud as the direct sound (because it had to travel farther), the constructive interference won't give you two times the output, and the destructive interference won't completely cancel. To get an idea of the result, take a look at Figure which shows the case of the example we started with. In this case, we have a difference in the time of arrival of 1 ms, and the reflection is 6 dB lower than the direct sound (because it had to travel 68.8 cm instead of 34.4 cm - half the distance). COMB FILTER SHOWING LOG FREQUENCY AND dB AMPLITUDE WHERE REFLECTION IS 6 DB LOWER