- Thread starter
- #1

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?

- Thread starter dwsmith
- Start date

- Thread starter
- #1

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?

- Feb 13, 2012

- 1,704

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

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?

Kind regards

$\chi$ $\sigma$

- Thread starter
- #3

Then \((10^3, 80)\), and if I had a slope of -12 following, it would be \((10^4, 40)\), correct?A slope of - 6 dB/octave is equivalent to a slope of -20 dB/decade...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #5

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

$\chi$ $\sigma$