frequency Responses For Low, High, Band Pass Active Filters

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frequency Responses For Low, High, Band Pass Active Filters

Figure 16.1 illustrates an active low pass filter. Assume each op-amp is an ideal op-amp.
**(a)** Determine its frequency response
**(b)** Illustrate an active high pass filter and determine its frequency response
**(c)** Illustrate an active band pass filter and determine its frequency response
**(di)** Figure 16.4 is an active filter. Determine its frequency response.

(ii) If R_{1} = R_{2} = 100 kΩ and C = 0.1 μF, determine the attenuation in decibels at ω = 1,0000 rad/s.

**The strings**:
S_{7}P_{5}A_{51} (change - physical).
**The math**:

Pj Problem of Interest is of type *change* (physical - change).

*Active filters*, are the group of filters whose filtering functions are derived from operational amplifier designs.
They amplify their input signals in addition to filtering the frequencies of their input signals. *Active filters* are different from *passive filters* in which there are no operational amplifiers and therefore no amplification of input signals.
**(a)** Consider figure 16.1. There are two impedances: feedback impedance and input impedance.

Feedback impedance Z_{F} = R_{F} || 1/jωC_{F}

So, Z_{F} = (R_{F}(1/jωC_{F}))/(R_{F} + 1/jωC)

So, Z_{F} = R_{F}/(1 + jωC_{F}R_{F})

Input impedance Z_{s} = R_{s}

So, closed-loop frequency response (closed-loop gain):

A(jω) = -Z_{F}/Z_{s} = -(R_{F}/R_{s})/(1 + jωC_{F}R_{F})---------(1)

It is evident that the active filter is a low pass filter because in eq(1) A(jω) -> 0 with increasing frequency ω.

Equation (1) also indicates the amplification component R_{F}/R_{s}, and the passive filtering component 1/(1 + jωCR_{F}).

**(b)** Consider figure 16.2:

Feedback Impedance Z_{F} = R_{F}

Input Impedance Z_{s} = R_{s} + 1/jωC_{s} = (jωC_{s}R_{s} + 1)/jωC_{s}

So, closed-loop frequency response (closed-loop gain):

A(jω) = -Z_{F}/Z_{s} = -R_{F}/(1 + jωC_{s}R_{s})/jωC_{s}

So, A(jω) = -(jωC_{s}R_{F})/(1 + jωC_{s}R_{s}) -------(2)

A(jω) -> 0 as ω -> 0; A(jω) -> -R_{F}/R_{s} as ω -> ∞

**(c)** Consider figure 16.3:

Feedback impedance Z_{F} = R_{F} || 1/jωC_{F} = R_{F}/(1 + jωC_{F}R_{F})

Input impedance Z_{s} = R_{s} + 1/jωC_{s} = (jωC_{s}R_{s} + 1)/jωC_{s}

So, closed loop frequency response (closed loop gain):

A(jω) = -Z_{F}/Z_{s} = - jωC_{s}R_{F}/[(1 + jωC_{s}R_{s})(1 + jωC_{F}R_{F})]-------(3)

Equation(3) is the frequency response of an active band pass filter.

**(d)** Consider figure 16.4:

Feedback impedance Z_{F} = R_{2} + 1/jωC = (1 + jωCR_{2})/jωC

Input impedance Z_{s} = R_{s}

So, frequency response of circuit:

A(jω) = -Z_{F}/Z_{s} = - (1 + jωCR_{2})/jωCR_{s}.

Circuit is an active low pass filter.

(ii) Attenuation in decibels = 20log_{10}|A(jω)|

So, |A(jω)| = |- (1 + jωCR_{2})/jωCR_{s}|

So, |A(jω)| = (1 +j(10^{3})(0.1)(10^{-6})(10^{5}))/j(10^{3})(0.1)(10^{-6})(10^{5}) = (1.01)^{1/2}

So, Attenuation in decibels = 20log_{10}(1.004881) = 20(0.00215) = 0.043dB.

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