(assuming the feedback was moved exponentially down from 100% to 50% during the first run of the loop) |
(assuming the feedback was moved briskly down from 100% to 50% during the first run of the loop. Thanks to Andy Butler for this graphic and for making the point.) |
Parameter | Symbol | Unit | Range | Domain |
---|---|---|---|---|
feedback | F | % (percent) | 0-100 | integer |
loop duration | LD | s (seconds) | [0,... | real |
process duration | PD | s (seconds) | [0,... | real |
repetitions (number of) | R | [0,... | integer | |
volume | V | ? | ? | real |
initial volume | V0 | ? | ? | real |
Formula | Description | |
---|---|---|
(1) | V = V0 * (F/100)^{R} | Resulting volume V at the end of repetition R for a loop with a feedback setting of F |
(2) | R = PD / LD | Number of repetitions depending on process duration and loop duration |
What volume factor x will a 10-second-loop have decreased to after 60 seconds with a feedback setting of 80% ?
V0 * x = V0 * (F/100)^{(PD/LD)}; (using formula (1))
x = (F/100)^{(PD/LD)} = 80/100^{60/10} = 0.8^{6} = 0.26;
What feedback setting F will lead to a decrease in Volume by 90% within 30 seconds for a 10 second loop?
V = V0 * ( 100 - 90 ) / 100; (decrease by 90%)
V0 * 0.1 = V0 * (F/100)^{(PD/LD)}; (using formula (1))
F = 0.1^{(1/3)} * 100 = 46.
How many repetions R does it take until a loop's volume will have decreased to 10% with a feedback setting of 90% ?
V = V0 * 0.1;
V0 * 0.1 = V0 * (F/100)^{R}; (using formula (1))
0.1 = (9/10)^{R};
log (0.1) = R * log (0.9);
R = log(0.1) / log (0.9) = 22;
© Bernhard Wagner, 2006.