ANSWERS Sample decibel problems #1 for ASLP 334 Calculate the following dB SPL levels for the given measured levels:
1) 47 µP #dB SPL = 20 log (47/20) = 20 log 2.35 = 20 * .3711 = 7.42 dB SPL
2) 128 µP #dB SPL = 20 log (128/20) = 20 log 6.4 = 20 * .8062 = 16.12 dB SPL
3) 6,324,555.32 µP #dB SPL = 20 log (6324555.32/20) = 20 log 316227.766 = 20 * 5.5 = 110.0 dB SPL
4) 0.63246 µP #dB SPL = 20 log (0.63246/20) = 20 log .0316 = 20 * -1.5 = -29.9999 dB SPL or -30.00 dB
SPL Calculate the following dB IL levels for the given measured levels:
5) 10-5 watts/cm2 #dB IL = 10 log(10-5/10-16) = 10 log (10-5+16/10-16+16) = 10 log (1011/100) = 10 log (1011/1) = 10 log 1011 = 10 * 11 = 110.0 dB IL
6) 10-14 watts/cm2 #dB IL = 10 log (10-14/10-16) = 10 log (10-14+16/10-16+16) = 10 log 102/100 = 10 log = 10 log 102 = 10 * 2 = 20.0 dB IL
7) 16*(10-16) watts/cm2 Every time we double power we get a gain of +3 dB. If we start with 1*(10-16) watts/cm2 being 0 dB IL, then if we double 1 we get 2 (+ 3 dB), and if we double 2 we get 4 (+3 dB more), and if we double 4 we get 8 (+3 dB more), and finally if we double 8 we get 16 (+3 dB more) for a total increase of 12 dB. 0 dB IL + 12 dB = 12.0 dB IL
8) 10-18 watts/cm2 #dB IL = 10 log (10-18/10-16) = 10 log (10-18+16/10-16+16) = 10 log (10-2/100) = 10 log 10-2/1 = 10 log 10-2 = 10 * -2 = -20.0 dB IL
9) 10-1 watts/cm2 #dB IL = 10 log (10-1/10-16) = 10 log (10-1+16/10-16+16) = 10 log 1015/100 = 10 log 1015/1 = 10 log 1015 = 10 * 15 = 150.0 dB IL
10. A patient has a threshold of 45 dB HL at 4000 Hz. What is the sensation
level of a 4000 Hz tone presented to this patient at 82 dB SPL?
Answer: ANSI correction at 4000 Hz is 12.0 dB convert 82 dB SPL to HL:
82 dB SPL - 12.0 = 70.0 dB HL
70.0 dB HL - 45 dB HL = 25 dB sl. In other words, the 82 dB SPL tone is
equivalent to 70 dB HL which is 25 dB above the patient's threshold of 45 dB
HL.
11. If I am presenting a 8000 Hz tone to a patient at 53.5 dB SPL and this
tone is 22 dB sl, what is the patients threshold in dB SPL and in dB HL?
Answer: we are given the fact that the 53.5 dB SPL is 22 dB above threshold.
Therefore, we can find threshold by subtracting 22 dB from 53.5 dB SPL (53.5 dB
SPL - 22 dB = 31.5 dB SPL threshold) To find the patient's threshold in
dB HL we need to convert 31.5 dB SPL to dB HL by subtracting the ANSI
correction of 15.5 dB 31.5 dB SPL - 15.5 dB = 16 dB HL
12. A patient has a threshold at 125 Hz of 45 dB HL. What are the dB HL and
dB SPL levels of a 125 Hz tone presented at -10 dB sl?
Answer: 45 dB HL threshold is given. The tone is - 10 dB sl which is 10 dB
below threshold. Therefore, the dB HL level of the tone being presented to the
patient is 45 dB HL - 10 dB = 35 dB HL
To find this level in dB SPL we need to convert 35 dB HL into SPL by adding the
ANSI correction of 45.0 dB. 35 dB HL + 45.0 dB = 80.0 dB SPL
13. A patient is being presented with a 250 Hz tone at 65 dB HL. The level
of this tone is 25 dB sl. What is the patients 250 Hz threshold in dB SPL and
in dB HL?
Answer: Same method as used in #12. 65 dB HL - 25 dB = 40 dB HL
threshold.
Now we need to convert 40 dB HL to dB SPL by adding the ANSI correction value
of 27.0 dB 40 dB HL + 27.0 dB = 67.0 dB SPL
Goto additional decibel problems
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