Determine Y of New Coordinate with -6 db/octave Slope on Log Plot

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Dustinsfl
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On a log plot in the x axis, if I have a slope of -6 db/octave from 100db, what would be the location of the new coordinate?

So the y-axis is db and the x-axis is in log.

First coordinate is \((10^2, 100)\) then a slope of -6. The second coordinate is \((10^3, ?)\)?

How does one determine the y location?
 
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dwsmith said:
On a log plot in the x axis, if I have a slope of -6 db/octave from 100db, what would be the location of the new coordinate?

So the y-axis is db and the x-axis is in log.

First coordinate is \((10^2, 100)\) then a slope of -6. The second coordinate is \((10^3, ?)\)?

How does one determine the y location?

A slope of - 6 dB/octave is equivalent to a slope of -20 dB/decade...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
A slope of - 6 dB/octave is equivalent to a slope of -20 dB/decade...

Kind regards

$\chi$ $\sigma$

Then \((10^3, 80)\), and if I had a slope of -12 following, it would be \((10^4, 40)\), correct?
 
dwsmith said:
Then \((10^3, 80)\), and if I had a slope of -12 following, it would be \((10^4, 40)\), correct?...

Not exactly... $\displaystyle (10^{3}, 80\ \text{dB})$ is correct and -20 dB\decade means $\displaystyle (10^{4}, 60\ \text{dB})$, $\displaystyle (10^{5}, 40\ \text{dB})$, etc...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Not exactly... $\displaystyle (10^{3}, 80\ \text{dB})$ is correct and -20 dB\decade means $\displaystyle (10^{4}, 60\ \text{dB})$, $\displaystyle (10^{5}, 40\ \text{dB})$, etc...

Kind regards

$\chi$ $\sigma$

You said -6 is -20 so wouldn't -12 be -40?