Music & Noise

Practice

practice problem 1

According to Galileo …

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 ….

Identify this interval on the equal tempered scale.

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 this is 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 …

1 semitone		minor second	¹²√2	¹²√2
2 semitones	= 1 tone (whole tone)	major second	¹²√2 ¹²√2	⁶√2
3 semitones		minor third	¹²√2 ¹²√2 ¹²√2	⁴√2
4 semitones	= 2 tones (ditone)	major third	¹²√2 ¹²√2 ¹²√2 ¹²√2	³√2
5 semitones		perfect fourth	¹²√2 ¹²√2 ¹²√2 ¹²√2 ¹²√2	^2.4√2
6 semitones	= 3 tones (tritone)	augmented fourth	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; for example, C and F♯ (G♭) 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

Determine the beat frequency between C₄ and G₄ (a perfect fifth) when played on an equal tempered scale where A₄ = 440 Hz.

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

Verify the following statements about these Fourier series made on Wikipedia.

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


y = ∑	(− 1)^n +1	sin nx	= sin x −	1	sin 2x +	1	sin 3x −	1	sin 4x +	1	sin 5x − ⋯
	n			2		3		4		5

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


y = ∑	1	sin(2n − 1)x	= sin x +	1	sin 3x +	1	sin 5x +	1	sin 7x +	1	sin 9x + ⋯
	2n − 1			3		5		7		9

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


y = ∑	1	cos(2n − 1)x	= cos x +	1	cos 3x +	1	cos 5x +	1	cos 7x +	1	cos 9x + ⋯
	(2n − 1)²			9		25		49		81

solution

Intensity is proportional to the square of amplitude.

I = 2π²ρf²vΔx²_max

Power level is the logarithm of the ratio of intensity to a reference intensity.

β[dB] = 10 log	⎛ ⎝	I	⎞ ⎠
		I₀

Take the ratio of the square of the amplitudes of a term and another term corresponding to half the frequency. Or instead of squaring, multiply it by 20 instead of 10.

β[dB] = 10 log	⎛ ⎝	x²	⎞ ⎠	= 20 log	⎛ ⎝	x	⎞ ⎠
		x²₀				x₀

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.


β[dB] = 20 log	⎛ ⎝	1/2	⎞ ⎠	= −6.0206 dB
		1

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 these values.


β[dB] = 20 log	⎛ ⎝	1/3	⎞ ⎠	= −9.54 dB
		1
β[dB] = 20 log	⎛ ⎝	1/7	⎞ ⎠	= −7.34 dB
		1/3
β[dB] = 20 log	⎛ ⎝	1/11	⎞ ⎠	= −6.85 dB
		1/5
β[dB] = 20 log	⎛ ⎝	1/15	⎞ ⎠	= −6.62 dB
		1/7
β[dB] = 20 log	⎛ ⎝	1/19	⎞ ⎠	= −6.49 dB
		1/9
β[dB] = 20 log	⎛ ⎝	1/23	⎞ ⎠	= −6.41 dB
		1/11


β[dB] = 20 log	⎛ ⎝	1/223	⎞ ⎠	= −6.0596 dB
		1/111
β[dB] = 20 log	⎛ ⎝	1/2223	⎞ ⎠	= −6.0245 dB
		1/1111
β[dB] = 20 log	⎛ ⎝	1/22223	⎞ ⎠	= −6.0210 dB
		1/11111
β[dB] = 20 log	⎛ ⎝	1/222223	⎞ ⎠	= −6.0206 dB
		1/111111

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.


β[dB] = 20 log	⎛ ⎝	1/9	⎞ ⎠	= −19.08 dB
		1
β[dB] = 20 log	⎛ ⎝	1/49	⎞ ⎠	= −14.72 dB
		1/9
β[dB] = 20 log	⎛ ⎝	1/121	⎞ ⎠	= −13.70 dB
		1/25
β[dB] = 20 log	⎛ ⎝	1/225	⎞ ⎠	= −13.24 dB
		1/49
β[dB] = 20 log	⎛ ⎝	1/361	⎞ ⎠	= −12.98 dB
		1/81
β[dB] = 20 log	⎛ ⎝	1/529	⎞ ⎠	= −12.81 dB
		1/121


β[dB] = 20 log	⎛ ⎝	1/49729	⎞ ⎠	= −12.1193 dB
		1/12321
β[dB] = 20 log	⎛ ⎝	1/4941729	⎞ ⎠	= −12.0490 dB
		1/1234321
β[dB] = 20 log	⎛ ⎝	1/493861729	⎞ ⎠	= −12.0420 dB
		1/123454321
β[dB] = 20 log	⎛ ⎝	1/49383061729	⎞ ⎠	= −12.0413 dB
		1/12345654321

practice problem 4

Write something completely different.

solution

Answer it.

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