I see a lot of examples saying that the falloff of a single pole RC circuit is -6dB/octave or -20/decade. I can see that thre frequency response of an RC lowpass filter shows this, but is there any mathematical prove that why it is -6dB/octave?
Maybe it is easy and I am just over complicating it !!
If Vout/Vin = R/[R^2 + (1/w^2*c^2)]^0.5, then is there some calculation that shows when w becomes (2w), then it Vout/Vin will be lower by about 6dB?
The output/input relationship for a low pass RC is: -
\$\dfrac{1}{1 + j\omega RC}\$
And once you get significantly past the 3dB threshold it tends to become this: -
\$\dfrac{1}{j\omega RC}\$
In other words a doubling of frequency (\$\omega\$) means a halving of output amplitude and of course doubling the frequency is an increase of 1 octave and a halving of amplitude is a 6dB attenuation therefore it's 6dB per octave.
The roll-off is the consequence of how a fucntion of the form \$\frac{1}{1 + s}\$ behaves as the frequency rises. \$s\$ is a complex variable, and the frequency domain is along the positive imaginary axis, in other words we substitute \$s = j\omega\$ to obtain \$\frac{1}{1 + i\omega}\$, and considering positive values of \$\omega\$.
The amplitude is the modulus: \$\left|\frac{1}{1 + j\omega}\right|\$, and the decibels are 20 times the base 10 log of that.
The modulus of a complex value is the square root of the product of it and its complex conjugate: \$|z| = \sqrt{z\bar z}\$. Thus:
$$\left|\frac{1}{1 + i\omega}\right| = \sqrt{\frac{1}{\left(1 + j\omega\right)\left(1 - j\omega\right)}} = \sqrt{\frac{1}{1 + w^2}}$$
For large values of \$\omega\$, that just reduces to approximately \$\displaystyle\frac{1}{w}\$, since the inside of the square root reduces to approximately \$\displaystyle\frac{1}{w^2}\$.
Of course, \$\displaystyle\frac{1}{w}\$ cuts in half with each doubling of frequency, which is approximately a -6db drop.
So it all boils down to the rate of shrinkage of an n-degree polynomial as the independent variable grows large. First order is \$1/x\$: amplitude cuts in half every octave, thus -6db. Second order is \$1/x^2\$: amplitude cuts four-fold every octave, thus -12dB.
"Order" is the order of the polynomial in the denominator. An n-order polynomial in the denominator gives rise to n poles, so that's where we get "one pole filter", "two pole", etc.
but is there any mathematical prove that why it is -6dB/octave?
For voltage ratios in dB, we have:
$$(\frac{V_o}{V_i})_{db} = 20\log\frac{V_o}{V_i}$$
Now, consider the function
$$f(\omega) = 20\log\frac{1}{\omega}$$
When \$\omega\$ is doubled, \$f(\omega)\$ decreases by
$$20\log\frac{1}{2} = -6dB$$
Thus, we conclude that if the frequency response of a filter is inversely proportional to frequency, the response decreases at the rate of \$-6dB\$ per octave.
Now, for a 1st order low-pass filter, we have
$$\frac{V_o}{V_i} = \frac{1}{1 + j\frac{\omega}{\omega_0}}$$
For \$\omega\$ much larger than \$\omega_0\$, we have
$$\frac{V_o}{V_i} \approx \frac{1}{j\frac{\omega}{\omega_0}} \propto \frac{1}{\omega}$$
Thus, well above the corner frequency, the frequency response is inversely proportional to frequency as required for a \$-6dB\$ roll-off.
