# Why does a smaller acoustic impedance mismatch produce lower transmission?

I'm attempting to solve a relatively trivial problem, but cannot seem to convince myself the answer I'm getting is true.

I have two scenarios.

Scenario 1.

An acoustic wave propagates from Medium W to Medium G.

Medium W    ¦    Medium G


Scenario 2.

An acoustic wave propagates from Medium W to Medium P to Medium G.

Medium W    ¦    Medium P    ¦    Medium G


Now W denotes water, P denotes some polymer and G denotes a type of glass. I would think that the transmission from water to glass would be low because of the high acoustic impedance mismatch. I would also think that the transmission from water to the polymer would be high, because of the low acoustic impedance mismatch.

However, using

$$T = \frac{2Z_1}{Z_1 + Z_0}$$

where Z1 is the medium into which the field is propagating, I get that the transmission in scenario 1 is 88.56% and in scenario 2 it is 42.35%.

I using an acoustic impedance (units of Pa.s.m-1) of:

water 1,480,000
polymer 2,758,460
glass 23,676,565


Would anyone care to hazard a guess as to why the transmission in scenario 2 is so much lower?

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First, I believe you have a mistake in your equations. The transmission should be $$T = \frac{2Z_0}{Z_1 + Z_0}.$$

The transmission in Scenario 1 is the transmission through the one boundary:

$$T_1 = \frac{2Z_W}{Z_G + Z_W},$$

while the total transmission in Scenario 2 is the product of two boundaries:

$$T_2 = \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P}.$$

Using the corrected equations, the transmissions are 11.8% in Scenario 1 and 14.6% in Scenario 2.

I was curious if Scenario 2 could ever give a lower transmission, so I worked it out. If we assume $T_1$ is bigger, we can figure out under what circumstances that is satisfied as follows:

$$T_1 \stackrel{?}{>} T_2,$$

$$\frac{2Z_W}{Z_G + Z_W} \stackrel{?}{>} \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P},$$

$$Z_W(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_W Z_P(Z_G+Z_W),$$

$$(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_P (Z_G+Z_W),$$

$$Z_P Z_G + Z_W Z_G + Z_W Z_P + Z_P^2 \stackrel{?}{>} 2 Z_P Z_G + 2 Z_P Z_W,$$

$$Z_W Z_G + Z_P^2 \stackrel{?}{>} Z_P Z_G + Z_PZ_W,$$

$$Z_P^2 - Z_P (Z_G + Z_W) + Z_W Z_G \stackrel{?}{>} 0,$$

$$(Z_P - Z_G)(Z_P - Z_W) \stackrel{?}{>} 0.$$

By inspection we see that $Z_P = Z_W$ and $Z_P = Z_G$ are roots. This parabola is upward facing (positive coefficient of $Z_P^2$), so at $Z_P = \{Z_W,Z_P\}$ it will cross zero and go positive outside of there and be negative between those two points.

Thus the transmission of Scenario 2 will be lower only if $Z_P < Z_W$ or $Z_P > Z_G$. In other words, the transmission would be worse if you put the glass between the water and polymer or sandwich water between polymer and glass.

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If anyone is looking, equation 9.23 in the book available at goo.gl/FSu2j confirms this correction. Thank you david.smith :) –  user714852 Dec 21 '12 at 17:35