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I don't understand the following step in my notes:

It is about filter roll-off.

$$H(\omega)={1\over 1+j(\omega/\omega_c)}$$

$$\Rightarrow|H(\omega)|= {1\over \sqrt{1+(\omega/\omega_c)^2}}$$

where $$H(\omega) = {v_o \over v_i}$$

$$\omega_c = \text{cut-off frequency} $$

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  • \$\begingroup\$ cross-posted: dsp.stackexchange.com/questions/4762/… \$\endgroup\$ – Ben Voigt Oct 23 '12 at 22:01
  • \$\begingroup\$ Filters become flat roll-off in the log-log scale and the magnitude of slope depends on the order of magnitude of the each power of ω. As this H(ω) has only one occurence of ω in the denominator, we say it has a power of -1 so the gain in the log display is 0dB at ω=0 {as log (1/1)=0} and has a slope of -1 order of magnitude where 20 log (H(ω) means it has a slope of -20dB per decade) which is a 1st order LPF. The intercept or break point is at ωc but we usually use fc instead to show Hz instead of Radians. (Sorry about brief explanation.) \$\endgroup\$ – Sunnyskyguy EE75 Oct 23 '12 at 23:30
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Roll-off is a realizable low-pass filter concept. The simplest low pass filter imaginable is a series resistor followed by a capacitor connected to ground.

              R
Vi -------+\/\/\/\+---------+--------- Vo
                            |
                           --- C
                           ---
                            |
                           GND

The current through the resistor, \$ \dfrac{V_i - V_o}{R}\$,

must be equal to the current through the capacitor, \$ \dfrac {V_o}{Z_C} \$.

ZC is the impedance of the capacitor, which is \$ \dfrac{1}{j \omega C} \$.

So:

\$ \dfrac{V_i - V_o}{R} = V_o j \omega C\$

... Solve for \$ \dfrac{V_o}{V_i} \$ (an exercise to the reader) and you get:

\$ \dfrac{V_o}{V_i} = \dfrac{1}{1 + j \omega R C} \$

Take the magnitude of the function in the standard ways for complex numbers, and you get the equation with the square root in the denominator.

You define the cutoff frequency as the point at which the magnitude of your response is attenuated by some amount. If you define that point such that

\$ \omega_c = \dfrac{1}{RC}\$, then

\$ \left\lvert\dfrac{V_o}{V_i}\right\rvert = \dfrac{1}{\sqrt{2}} ~= 0.707 \$, or about 70%.

I'm a little rusty here, but I think that if you take the second derivative of \$ \left\lvert\dfrac{V_o}{V_i}\right\rvert \$ you will also find that \$ \omega_c \$ is the point of maximum inflection.

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To calculate the magnitude, you multiply the expression by the complex conjugate (which you get by replacing j with -j) and take the square root

$$ \sqrt{\frac{1}{j + \frac{\omega}{\omega_{c}}}\: \: \:\cdot \frac{1}{-j + \frac{\omega}{\omega_{c}}}}\:\:\: = \sqrt{\frac{1}{-j^2 + \left(\frac{\omega}{\omega_{c}}\right )^2}} $$

and since j=sqrt(-1)

$$-j^2 = 1$$

This leaves you at the step I think you're having trouble with, and when reduced gives you the answer:

$$\Rightarrow|H(\omega)|= {1\over \sqrt{1+(\omega/\omega_c)^2}}$$

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