 # Passive Filters

Filters are used to pass desired frequencies while rejecting (attenuating) non-desired frequencies.

• Passive filters - contain only passive elements (R,L,C)
• Active filters - contain active elements (transistors, op-amps)

## 1) Low-pass

-passes low frequencies and blocks high frequencies. ## 2) High-pass

-passes high frequencies and blocks low frequencies. ## 3) Band-pass

-passes frequencies within a frequency band while attenuating frequencies outside of the band. ## 4) Band-stop

-passes frequencies outside a frequency band while attenuating frequencies inside of the band. ### Low-Pass Filter:

The output of a low-pass filter is taken across the capacitor of a series RC circuit (shown below): Output voltage relative to input voltage can be expressed via the following transfer function: $$\mathbb{H}(\omega) = \frac{\mathbb{V}_o}{\mathbb{V}_i}$$ Using Ohm's Law and recalling the definition of the impedance for a capacitor, we get the following: $$\mathbb{H}(\omega) = \frac{\mathbb{I}(\frac{-j}{\omega C})}{\mathbb{I}(R-\frac{j}{\omega C})}$$ $$\qquad\; = \frac{-j}{\omega C} \Big( \frac{1}{R-\frac{j}{\omega C}} \Big)$$ $$\qquad\; = \frac{-j}{R\omega C - j}$$ $$\qquad\; = \frac{1}{j(R\omega C-j)}$$ After distributing the "j" term in the denominator of the above expression we get the following:

## Transfer function for a low-pass filter:

$$\mathbb{H}(\omega) = \frac{1}{1+jR\omega C} \qquad,(Eqn\;1)$$

If we were to evaluate the magnitudes of this transfer function we would see that for a low-pass filter: $$|\mathbb{H}(0)| = 1, \quad |\mathbb{H}(\infty)| = 0$$ The Bode magnitude plot of the low-pass filter transfer function is shown below: ...where: $$\omega_c = cutoff \; frequency$$ The cutoff frequency can be found by setting the magnitude of the transfer function equal to 1/sqrt(2). Therefore, we take the magnitude of equation #1 and proceed as follows: $$|\mathbb{H}(\omega)| = \frac{\sqrt{1^2}}{\sqrt{1^2 + (R\omega C)^2}} = \frac{1}{\sqrt{2}}$$ $$\qquad \qquad \frac{1}{\sqrt{1 + R^2 \omega^2 C^2}} = \frac{1}{\sqrt{2}}$$ $$\qquad \qquad\quad 1 + R^2 \omega^2 C^2 = 2$$ $$\qquad \qquad\quad \omega^2 = \frac{1}{R^2 C^2}$$ $$\qquad \qquad\quad \omega = \frac{1}{R C} = \omega_c$$

## Cutoff frequency for low-pass filter:

$$\omega_c = \frac{1}{R C}$$

The cutoff frequency is sometimes called the 'rolloff frequency". A low-pass filter will pass all frequencies from DC up to its cutoff frequency.

### High-Pass Filter:

The output of a high-pass filter is taken across the resistor of a series RC circuit (shown below): Output voltage relative to input voltage can be expressed via the following transfer function: $$\mathbb{H}(\omega) = \frac{\mathbb{V}_o}{\mathbb{V}_i}$$ Using Ohm's Law and recalling the definition of the impedance for a capacitor, we get the following: $$\mathbb{H}(\omega) = \frac{\mathbb{I}R}{\mathbb{I}(R+\frac{1}{j\omega C})}$$ $$\qquad\; = \frac{R}{\frac{jR\omega C + 1}{j\omega C}}$$

## Transfer function for a high-pass filter:

$$\mathbb{H}(\omega) = \frac{jR\omega C}{1+jR\omega C} \qquad,(Eqn\;2)$$

If we were to calculate the magnitude of equation #2 we would get: $$|\mathbb{H}(\omega)| = \frac{R\omega C}{\sqrt{1+R^2\omega^2 C^2}} \qquad,(Eqn\;3)$$ Additionally, if we were to evaluate the magnitudes of this transfer function we would see that for a high-pass filter: $$|\mathbb{H}(0)| = 0, \quad |\mathbb{H}(\infty)| = 1$$ The Bode magnitude plot of this expression would appear as follows: ...where: $$\omega_c = cutoff \; frequency$$ Just as with the low-pass filter, cutoff frequency for a high-pass filter can be found by setting the magnitude of the transfer function equal to 1/sqrt(2). Therefore, we take equation #3 and proceed as follows: $$|\mathbb{H}(\omega)| = \frac{R\omega C}{\sqrt{1+R^2\omega^2 C^2}} = \frac{1}{\sqrt{2}}$$ $$\qquad \qquad R^2\omega^2 C^2 = \frac{1+R^2\omega^2 C^2}{2}$$ $$\qquad \qquad 2R^2\omega^2 C^2 = 1+R^2\omega^2 C^2$$ $$\qquad \qquad R^2\omega^2 C^2 = 1$$ $$\qquad \qquad \omega^2 = \frac{1}{R^2C^2}$$ $$\qquad\qquad \omega = \frac{1}{RC} = \omega_c$$

## Cutoff frequency for high-pass filter:

$$\omega_c = \frac{1}{R C}$$ Note that it is identical for a low-pass filter.

A high-pass filter passes all frequencies above its cutoff frequency.

### Band-Pass Filter:

The output of a band-pass filter is taken across the resistor of a series resonant RLC circuit (shown below): Output voltage relative to input voltage can be expressed via the following transfer function: $$\mathbb{H}(\omega) = \frac{\mathbb{V}_o}{\mathbb{V}_i}$$ Using Ohm's Law and recalling the definitions of the impedances for a capacitor and inductor, we get the following: $$\mathbb{H}(\omega) = \frac{\mathbb{I}R}{\mathbb{I}(R+j\omega L + \frac{1}{j\omega C})}$$

## Transfer function for a band-pass filter:

$$\mathbb{H}(\omega) = \frac{R}{R+j(\omega L - \frac{1}{\omega C})} \qquad,(Eqn\;4)$$

Once again we could calculate the magnitude of equation #4. If we were to do so we could determine the following expressions: $$|\mathbb{H}(0)| = 0, \quad |\mathbb{H}(\infty)| = 0$$ The Bode magnitude plot of this expression would appear as follows: ...where: $$\omega_o = center \; frequency$$ A band-pass filter passes a band of frequencies that are centered on the center frequency and within the following range: $$\omega_1 \lt \omega \lt \omega_2$$ The following expression is used to determine the center frequency:

## Center frequency for band-pass filter:

$$\omega_o = \frac{1}{\sqrt{R C}}$$

You may notice that the expression for the center frequency of a band-pass filter is the same as the expression for the resonant frequency of series resonant circuit. Likewise, the expressions for bandwidth (B), Quality Factor (Q) and 1/2 power frequencies of a band-pass filter are the same as the formulas we derived in the introduction to series resonance page.

## Band-pass filter formulas:

Center Frequency: $$\omega_o = \frac{1}{\sqrt{R C}}$$ Bandwidth: $$B = \omega_2-\omega_1$$ $$\;\;\; = \frac{R}{L} = \frac{\omega_o}{Q} = \omega_o^2RC$$ Quality Factor: $$Q = \frac{\omega_oL}{R} = \frac{1}{\omega_oRC}$$ 1/2 power frequencies: $$\omega_1 = -\frac{R}{2L} + \sqrt{\Big( \frac{R}{2L} \Big)^2 + \frac{1}{LC}}$$ $$\omega_2 = \frac{R}{2L} + \sqrt{\Big( \frac{R}{2L} \Big)^2 + \frac{1}{LC}}$$

### Band-Stop Filter:

The output of a band-stop filter is taken across the inductor-capacitor series combination of a series resonant RLC circuit (shown below): Output voltage relative to input voltage can be expressed via the following transfer function: $$\mathbb{H}(\omega) = \frac{\mathbb{V}_o}{\mathbb{V}_i}$$ Using Ohm's Law and recalling the definitions of the impedances for a capacitor and inductor, we get the following: $$\mathbb{H}(\omega) = \frac{\mathbb{I}(\frac{-j}{\omega C}+j\omega L)}{\mathbb{I}(R-\frac{-j}{\omega C}+j\omega L)}$$

## Transfer function for a band-stop filter:

$$\mathbb{H}(\omega) = \frac{j(\omega L-\frac{1}{\omega C})}{R+j(\omega L-\frac{1}{\omega C})} \qquad,(Eqn\;5)$$

If we were to calculate the magnitude of equation #5, we would see that: $$|\mathbb{H}(0)| = 1, \quad |\mathbb{H}(\infty)| = 1$$ The Bode magnitude plot of this expression would appear as follows: ...where: $$\omega_o = frequency \; of \; rejection$$ A band-stop filter rejects a band of frequencies that are centered on the frequency of rejection and within the following range: $$\omega_1 \lt \omega \lt \omega_2$$ The following expression is used to determine the frequency of rejection:

## Frequency of rejection for band-stop filter:

$$\omega_o = \frac{1}{\sqrt{R C}}$$

You may notice that the expression for the frequency of rejection for a band-stop filter is the same as the expression for the center frequency of a band-pass filter and is also the same expression used to determine the resonant frequency of series resonant circuit. Just as with a band-pass filter, the expressions for bandwidth (B), Quality Factor (Q) and 1/2 power frequencies of a band-stop filter are the same as the formulas we derived in the introduction to series resonance page.

## Band-stop filter formulas:

Frequency of rejection: $$\omega_o = \frac{1}{\sqrt{R C}}$$ Bandwidth: $$B = \omega_2-\omega_1$$ $$\;\;\; = \frac{R}{L} = \frac{\omega_o}{Q} = \omega_o^2RC$$ Quality Factor: $$Q = \frac{\omega_oL}{R} = \frac{1}{\omega_oRC}$$ 1/2 power frequencies: $$\omega_1 = -\frac{R}{2L} + \sqrt{\Big( \frac{R}{2L} \Big)^2 + \frac{1}{LC}}$$ $$\omega_2 = \frac{R}{2L} + \sqrt{\Big( \frac{R}{2L} \Big)^2 + \frac{1}{LC}}$$

Next we will look at a passive filter example problem:

Continue on to passive filter example problem #1...