# A (super short) introduction to acoustic physical modeling

**Posted:**October 14th, 2011 |

**Author:**Gina Collecchia | Comments Off on A (super short) introduction to acoustic physical modeling

What does it take to understand the acoustics of a physical space, like a room or a musical instrument? You don’t have to be a musician to desire—or even require—good sound. The acoustics of an amphitheater or simple rectangular room dictate the the signal-to-noise ratios of sonic signals like speech and music. When sound reflects off of the walls, ceiling, floor, and other objects in a room, it loses energy (volume) each time it does so. In highly reverberant rooms such as churches (the Hagia Sophia in Turkey has something like a 10-second reverberation time!), the reflections are numerous because they don’t lose as much energy with each bounce, so signals originating from the space will retain less information because previous signals are still bouncing around. So, the greater the reverberation, the less signal and greater noise.

But, not all signals are equal. Acoustic spaces (and musical instruments) are ideally designed to give preferences to a certain frequency range, which for speech is about 1000-5000 Hz. We can test these “frequency preferences” by measuring how the room behaves to specific frequencies. This measurement is called a **frequency response**. One way to do this is by playing a **sine sweep**—a sine wave starting from 0 Hz and constantly increasing (linearly or logarithmically) until ending at 25000 Hz (or whatever desired range)—and recording it in the room. This method requires no transformation to a frequency domain because time and frequency are tied! The recorded amplitude is the **amplitude response** of the room. From this, it should be clear that a room and a musical instrument alike are both types of **filters**.

However, speakers have frequency responses, too, so unless that response is known (and even if it is), the sine sweep can pose problems and return inaccuracies. Fortunately, there is another, more clever way to do this: by an impulse function, such as one of those mentioned in the post below. Refresher: an impulse function \delta(t-a) is defined as zero everywhere that *t* does not equal *a* (*t* is time, *a* is some instant of time), and non-zero where *t=a* (infinity in the continuous case and 1 in the discrete case). An impulse function is not differentiable, but it is integrable (in the continuous case), and its integral is equal to 1.

Now, what does an impulse sound like? This two-fold jump in amplitude from 0 to maximum value to 0 again is colloquially called a **tone burst**: it’s like dropping the needle on a record player, or sampling audio at non-zero endpoints and causing a speaker to emit a ‘click’—the sound of it trying to move infinitely fast (because velocity is the derivative of position! So it all comes down to differentiability!) from one position to another. This is also known as **clipping**, an undesirable effect in virtually every aspect of life save hygiene (okay, there may exist others). What does clipping have to do with the frequency responses of acoustic spaces? Well, what is the frequency response of an impulse?

This is quickly calculated by taking the Fourier transform of \delta(t-a), which is equal to exp(-ia\omega), where \omega is frequency. When a=0, which is the case for a single impulse, and since time is relative (don’t sue; yeah I’ve seen the article about the neutrinos), it’s okay to call a=0 in most situations. Thus, F{\delta(t-0)}=exp(-i0\omega)=1 for all \omega. This is called a **flat (frequency) response**, the desired response for speakers, and when we apply it to a room and record the results, the Fourier transform of the results shows the frequency response of the room!

The sound of an impulse is that of **white noise**, which has equal energy spread to all frequencies, much like white light contains the entire visible spectrum. The Fourier transform of \delta(t-a) is exp(-ia\omega) because of the **Shifting theorem**, which states that F{x(t-a)}=F{x(t)}*exp(-ia\omega). Finally, the Fourier transform of a finite-time-domain signal is infinite, and the Fourier transform of an infinite-time-domain signal is finite. This is a result of **Gibb’s Phenomenon**, the subject of my next post!

Happy DSP’ing!