Operations on intervals: Difference between revisions
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m →Stacking and unstacking: Put log space operations on the left (more perceptually infinitive) |
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Stacking two intervals feels perceptually like we are ''adding'' two distances, but it corresponds to ''multiplying'' two frequency ratios. The logarithm function is the bridge between frequency space and pitch space: | Stacking two intervals feels perceptually like we are ''adding'' two distances, but it corresponds to ''multiplying'' two frequency ratios. The logarithm function is the bridge between frequency space and pitch space: | ||
<math> | <math>\log_2 a + \log_2 b = \log_2 (ab)</math> | ||
for any two frequency ratios ''a'' and ''b''. The above equation tells us that the sum of the size (in octaves) of ''a'' and the size (in octaves) of ''b'' is equal to the size in octaves of the product of two frequency ratios ''ab''. Hence, ''stacking corresponds to multiplying frequency ratios in linear (frequency) space, and adding cent values in logarithmic (pitch) space.'' | for any two frequency ratios ''a'' and ''b''. The above equation tells us that the sum of the size (in octaves) of ''a'' and the size (in octaves) of ''b'' is equal to the size in octaves of the product of two frequency ratios ''ab''. Hence, ''stacking corresponds to multiplying frequency ratios in linear (frequency) space, and adding cent values in logarithmic (pitch) space.'' | ||
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To convert octaves to cents, we simply multiply both sides by 1200: | To convert octaves to cents, we simply multiply both sides by 1200: | ||
<math>1200 \log_2 | <math>1200 \log_2 a + 1200 \log_2 b = 1200 \log_2 (ab).</math> | ||
As a consequence, since the inverse operation of multiplication is division, dividing frequency ratios corresponds to subtracting their perceptual sizes: | As a consequence, since the inverse operation of multiplication is division, dividing frequency ratios corresponds to subtracting their perceptual sizes: | ||
<math>1200 \log_2 | <math>1200 \log_2 a - 1200 \log_2 b = 1200 \log_2 (a/b).</math> | ||
=== Examples === | === Examples === | ||
# <math>1200 \log_2 ( | # <math>1200 \log_2 (3/2) - 1200 \log_2 (5/4) = 1200 \log_2 (6/5)</math> | ||
# <math>1200 \log_2 ( | # <math>1200 \log_2 (3/2) + 1200 \log_2 (3/2) = 1200 \log_2 (9/4)</math> | ||
# <math>1200 \log_2 ( | # <math>1200 \log_2 (16/9) - 1200 \log_2 (3/2) = 1200 \log_2 (32/27)</math> | ||
== Logarithmic multiplication and division == | == Logarithmic multiplication and division == | ||
Revision as of 22:15, 18 December 2025
The following are common arithmetic operations on musical intervals. If you're unfamiliar with the operations, you're encouraged to verify the examples for interval operation rules with a calculator.
Stacking and unstacking
Stacking two intervals feels perceptually like we are adding two distances, but it corresponds to multiplying two frequency ratios. The logarithm function is the bridge between frequency space and pitch space:
for any two frequency ratios a and b. The above equation tells us that the sum of the size (in octaves) of a and the size (in octaves) of b is equal to the size in octaves of the product of two frequency ratios ab. Hence, stacking corresponds to multiplying frequency ratios in linear (frequency) space, and adding cent values in logarithmic (pitch) space.
To convert octaves to cents, we simply multiply both sides by 1200:
As a consequence, since the inverse operation of multiplication is division, dividing frequency ratios corresponds to subtracting their perceptual sizes:
