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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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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. 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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