There are a number of reasons why an understanding of sound and speech production, hearing and perception are relevant to speech and language therapy:

• The acoustic form of language is part of the speech chain that links speaker to hearer. If you didn't study acoustics, your understanding would stop at articulation and start again at phonetic transcription.
• Speech is highly variable and this influences its effectiveness as a means of communication. Effective communication means proper sound generation, transmission, analysis and decoding. To understand, say, how one vowel sound is 'clear' while another is 'muffled' you need to appreciate how the sound energy is generated and shaped by the vocal tract, how it can be affected by its transmission to the listener, how the listener is able to tell which vowel has been articulated.
• Science is built on quantitative analysis; on measurements not just opinions. How can we measure speech without instruments to capture and represent the acoustic signal. How can you do science without the knowledge about how such instruments operate and what it is possible and practical to measure?
• Clinical work is not about saying speech is 'good' or 'bad' but by understanding what about the speech is different to normal, what the possible causes for a disorder might be, and what the consequences are for normal communication. In acoustics we provide the concepts which explain in what ways, for example, disorders of the vocal folds, disorders of hearing, or background noise affect the ability to communicate.
• Clinical work and research work require you to write reports on your activities in a scientific (i.e. non-subjective) manner. In Acoustics we get you to do simple experimental work and then try to describe what you have done, what you found out and why you did it.
• Clinical work will require you to use machines such as tape recorders, computers or even an electro-laryngograph. An understanding of how these machines work and are used will enhance your effectiveness.
• RCSLT specifies that you have to do it.

Basically a decibel is a measure of power that can be applied in a number of fields and to a number of physical phenomena. It is used in Acoustics, as we have seen, to measure the intensity of a sound; it is also used in radio telecommunications to measure the intensity of a radio signal, or used in telecommunications to measure the power in a signal communicated over a telephone line.

What makes a decibel different from other units is that it always expresses the power in the signal as a ratio to some standard or reference power. In other words, rather than being like length and having units such as metres, the decibel describes a power as being 'ten-times' or 'one-hundredth' of the power of some reference power. Thus when we say that the gain in an amplifier is 20 decibels, we mean that the amplifier has changed the power in the signal by 20 decibel units, or that the output of the amplifier is 20 decibels of power greater than the input to the amplifier. This is more informative than to say that the output of the amplifier is 20 watts of power, since we don't know how many watts of power were put in!

The other useful aspect of decibels is that they express these ratios to a standard power using a logarithmic scale rather than a linear scale. This may seem more awkward than useful to you, but in fact it arises because many physical processes work multiplicatively on power rather than as addition. If you imagine a sound signal going through a wall, clearly it loses power (it has more intensity on one side than on the other) but the question is: does the wall remove a fixed amount of power (say 20watts) or does it change the power by a constant fraction (say reduce by a factor of 2)? Well, the latter is the right answer. We can express the effect of the wall on the sound as a fixed ratio of powers, but we can't express it as a fixed number of watts. Thus, if we put 10 watts in we might get 5 watts out, and if we put 20 watts in we might get 10watts out. Using decibels we might say that the wall has attenuated the signal by 3 decibels, and this would be true regardless of the input power.

Finally, some mathematics. If we have a power P and some reference power Pref, then the ratio expressed in decibels is just

That is, we divide the measured power by the reference power, take the logarithm base 10, then multiply by 10.

Sometimes we do not have measurements of the power in the signal, but we do have measurements of the amplitude of the signal. Fortunately we can use the fact that the power in the signal is proportional to the square of the amplitude of the signal (don't worry about this, it is because power is current times voltage, and that both the voltage in the signal and the current the signal can generate are proportional to amplitude). Thus we can substitute the ratio of powers in the decibel formula with a squared ratio of amplitudes, which because of the effect of the log reduces to:

Where A is the measured amplitude and Aref is the reference amplitude.

Finally, some useful identities: