Music & Noise
Practice
practice problem 1
Identify this interval on the equal tempered scale.Especially harsh is the dissonance between notes whose frequencies are incommensurable; such a case occurs when one has two strings in unison and sounds one of them open, together with a part of the other which bears the same ratio as the side of a square bears to the diagonal….
solution
The ratio of the diagonal of a square to a side is √2:1. (Galileo stated the order of the ratio the other way around, but that's a minor detail.) Each half step (a semitone) up the equal tempered scale multiplies the previous note by the twelfth root of two, two half steps (a whole tone) multiplies the note by the twelfth root of two squared, three half steps by the twelfth root of two cubed, and so on…
interval name  interval size  

1 semitone  ^{12}√2 =  ^{12}√2  
2 semitones  = whole tone  ^{12}√2^{12}√2 =  ^{6}√2 
3 semitones  ^{12}√2^{12}√2^{12}√2 =  ^{4}√2  
4 semitones  = ditone  ^{12}√2^{12}√2^{12}√2^{12}√2 =  ^{3}√2 
5 semitones  ^{12}√2^{12}√2^{12}√2^{12}√2^{12}√2 =  ^{2.4}√2  
6 semitones  = tritone  ^{12}√2^{12}√2^{12}√2^{12}√2^{12}√2^{12}√2 =  ^{2}√2 
Six semitones is equal to the twelfth root of two to the sixth power, which is equal to the square root of two. This interval is called a tritone, an augmented fourth, or a diminshed fifth — C and F♯, for example, or F and B. Had I given you a more complete quote from Galileo you would have already known this.
Especially harsh is the dissonance between notes whose frequencies are incommensurable; such a case occurs when one has two strings in unison and sounds one of them open, together with a part of the other which bears the same ratio as the side of a square bears to the diagonal. This yields a dissonance similar to the augmented fourth or diminished fifth.
practice problem 2
solution
On an equal tempered scale, C is 9 semitones below A and G is 2 semitones below A.
ƒ_{C} =  ƒ_{A}/2^{9/12} = 261.6256 Hz 
ƒ_{G} =  ƒ_{A}/2^{2/12} = 391.9954 Hz 
A perfect fifth is the ratio 3:2. Therefore three times the tonic should equal twice the fifth, but on an equal tempered scale they don't. The difference in these multiples results in a beat.
ƒ_{beat} =  3ƒ_{C} − 2ƒ_{G} 
ƒ_{beat} =  3ƒ_{A}/2^{9/12} − 2ƒ_{A}/2^{2/12} 
ƒ_{beat} =  3(440 Hz)/2^{9/12} − 2(440 Hz)/2^{2/12} 
ƒ_{beat} =  784.877 Hz − 783.991 Hz 
ƒ_{beat} =  0.886 Hz 
On a properly tuned piano, when C and A were played, one would expect a beat every…
T_{beat} = ƒ_{beat}^{−1} = 1.1 s
practice problem 3
 A sawtooth wave "contains odd and even harmonics that fall off at −6 dB/octave."




 A square wave "contains odd harmonics that fall off at −6 dB/octave."




 A triangle wave "contains odd harmonics that fall off at −12 dB/octave."




solution
The Fourier series equations given above use y and x as variables. When sound waves are received by an ear or a microphone, they're detecting fluctuations in pressure (P) over time (t). The claim we're testing using decibel. This means we should use the pressure level equation in decibels.
L_{P} = 20 log  ⎛ ⎝  ∆P  ⎞ ⎠ 
∆P_{0} 
 The second term in the sawtooth wave series has twice the frequency of the first. Plug the ratio of the two amplitudes into the equation derived above.
 No term in the square wave series has a frequency that is a whole number of octaves above another. Pick the amplitudes on the terms that are slightly larger than an octave apart. For example 3 is slightly greater than 2 × 1 = 2, 7 is slightly greater than 2 × 3 = 6, etc. Check the limit of the level difference for these values.
 The situation for the triangle wave is the same as the square wave. We only have odd frequencies, but this time the amplitudes are squared.
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{2}  ⎞ ⎠ 
= −6.0206 dB 
^{1}/_{1} 
We could repeat this for other harmonics, but we'd always wind up with the same answer: −6 dB per octave. We're done with this part.
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{3}  ⎞ ⎠ 
= −9.54 dB 
^{1}/_{1}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{7}  ⎞ ⎠ 
= −7.34 dB 
^{1}/_{3}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{11}  ⎞ ⎠ 
= −6.85 dB 
^{1}/_{5}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{15}  ⎞ ⎠ 
= −6.62 dB 
^{1}/_{7}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{19}  ⎞ ⎠ 
= −6.49 dB 
^{1}/_{9}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{23}  ⎞ ⎠ 
= −6.41 dB 
^{1}/_{11} 
This sequence is converging too slowly. Let's try some bigger numbers.
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{223}  ⎞ ⎠ 
= −6.0596 dB 
^{1}/_{111}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{2,223}  ⎞ ⎠ 
= −6.0245 dB 
^{1}/_{1,111}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{22,223}  ⎞ ⎠ 
= −6.0210 dB 
^{1}/_{11111}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{222,223}  ⎞ ⎠ 
= −6.0206 dB 
^{1}/_{111,111} 
I see our limit approaching. It's the same as the result we got with the sawtooth wave. The harmonics fall off by −6 dB per octave.
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{9}  ⎞ ⎠ 
= −19.08 dB 
^{1}/_{1}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{49}  ⎞ ⎠ 
= −14.72 dB 
^{1}/_{9}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{121}  ⎞ ⎠ 
= −13.70 dB 
^{1}/_{25}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{225}  ⎞ ⎠ 
= −13.24 dB 
^{1}/_{49}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{361}  ⎞ ⎠ 
= −12.98 dB 
^{1}/_{81}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{529}  ⎞ ⎠ 
= −12.81 dB 
^{1}/_{121} 
This is also converging too slowly for me. Use larger values — like the square of those we used in part b.
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{49,729}  ⎞ ⎠ 
= −12.12 dB 
^{1}/_{12,321}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{4,941,729}  ⎞ ⎠ 
= −12.05 dB 
^{1}/_{1,234,321}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{493,861,729}  ⎞ ⎠ 
= −12.04 dB 
^{1}/_{123,454,321}  
L_{P} = 20 log  ⎛ ⎝ 
^{1}/_{49,383,061,729}  ⎞ ⎠ 
= −12.04 dB 
^{1}/_{12,345,654,321} 
This is twice the power level drop of the sawtooth wave, but of course you already knew this. Squaring a quantity in a logarithm is the same as doubling the logarithm of the unsquared quantity. All of the calculations in this last part are interesting, but unnecessary. I guess I just didn't feel like being clever today.
practice problem 4
solution
Answer it.