Transmission loss (TL) in duct acoustics describes the acoustic performances of a muffler-like system. It is frequently used in the industry areas such as muffler manufacturers and NVH (noise, vibration and harshness) department of automobile manufacturers, and in academic studies. Generally the higher transmission loss of a system it has, the better it will perform in terms of noise cancellation.
Transmission loss (TL) in duct acoustics is defined as the difference between the power incident on a duct acoustic device (muffler) and that transmitted downstream into an anechoic termination. Transmission loss is independent of the source,[1] if only plane waves are incident at the inlet of the device. Transmission loss does not involve the radiation impedance inasmuch as it represents the difference between incident acoustic energy and that transmitted into an anechoic environment. Being made independent of the terminations, TL finds favor with researchers who are sometimes interested in finding the acoustic transmission behavior of an element or a set of elements in isolation of the terminations. But measurement of the incident wave in a standing wave acoustic field requires uses of impedance tube technology, may be quite laborious, unless one makes use of the two-microphone method with modern instrumentation.[2]
By definition the plane wave TL on an acoustic component with negligible mean flow may be described as:
TL=LWi-LWo=10log10\left\vert{Sipi+vi+\over2}{2\overSopov0}\right\vert=10log10\left\vert{Si
2 | |
p | |
i+ |
\overSo
2 | |
p | |
o |
}\right\vert
LWi
LWo
Si,So
pi+
po
vi+
vo
pi+
pi-
po=po+
po-=0
And in most muffler applications, Si and So, the area of the exhaust pipe and tail pipe, are generally made equal, thus we have:
TL=20log10\left\vert{pi+\overpo}\right\vert
Also, since the transmitted sound power cannot possibly exceed the incident sound power (or
\left\vertpi+\right\vert
\left\vertpo\right\vert
If the system contains non-negligible mean flow and have duct sizes which support wave modes of orders higher than the plane wave mode at the frequencies of interest, transmission loss calculations are modified accordingly.
Transmission matrix description
The low-frequency approximation implies that each subsystem is an acoustic two-port (or four-pole system) with two (and only two) unknown parameters, the complex amplitudes of two interfering waves travelling in opposite directions. Such a system can be described by its transmission matrix (or four-pole matrix), as follows[3]
\begin{bmatrix} \hatpi\\ \hatqi \end{bmatrix} = \begin{bmatrix} A&B\\ C&D \end{bmatrix} \begin{bmatrix} \hatpo\\ \hatqo \end{bmatrix}
\hatpi
\hatpo
\hatqi
\hatqo
TL=10log10\left({{1\over4}\left\vert{A+B{S\over\rhoc}+C{\rhoc\overS}+D}\right\vert2}\right)
S
\rhoc
Considering we have the most simplest reactive silencer with only one expansion chamber (length l and cross-sectional area S2), with inlet and outlet both having cross-sectional area S1). As we know the transmission matrix of a tube (in this case, the expansion chamber) is
\begin{bmatrix} A&B\\ C&D \end{bmatrix} = \begin{bmatrix} {\coskl}&{j{\rhoc\overS2}\sinkl}\\ {j{S2\over\rhoc}\sinkl}&{\coskl}\end{bmatrix}
\begin{align}TL&=10log10\left({{1\over4}\left\vert{{\coskl}+{j{S1\overS2}\sinkl}+{j{S2\overS1}\sinkl}+{\coskl}}\right\vert2}\right)\\ &=10log10\left({{\cos2kl}+{1\over4}\left(h+{1\overh}\right)2{\sin2kl}}\right)\\ &=10log10\left({1+{1\over4}\left(h-{1\overh}\right)2{\sin2kl}}\right), \end{align}
h
l
k=\omega/c
c
l
As a simple example, consider a one chamber silencer with h=S1/S2=1/3, at around 400 °C the sound speed is about 520 m/s, with l=0.5 m, one easily calculate the TL result shown on the plot on the right. Note that the TL equals zero when frequency is a multiple of
c\over{2l}
{c\over{4l}}+{n*{c\over{2l}}}
Also note that the above calculation is only valid for low-frequency range because at low-frequency range the sound wave can be treated as a plane wave. The TL calculation will start losing its accuracy when the frequency goes above the cutoff frequency, which can be calculated as
fc=1.84{c\over{\piD}}