frequency Responses For Low, High, Band Pass Active Filters
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frequency Responses For Low, High, Band Pass Active Filters

Active Low Pass Filter 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 R1 = R2 = 100 kΩ and C = 0.1 μF, determine the attenuation in decibels at ω = 1,0000 rad/s.

The strings: S7P5A51 (change - physical).

The math:
Pj Problem of Interest is of type change (physical - change).

Active Low Pass Filter

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 ZF = RF || 1/jωCF
So, ZF = (RF(1/jωCF))/(RF + 1/jωC)
So, ZF = RF/(1 + jωCFRF)

Input impedance Zs = Rs

So, closed-loop frequency response (closed-loop gain):
A(jω) = -ZF/Zs = -(RF/Rs)/(1 + jωCFRF)---------(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 RF/Rs, and the passive filtering component 1/(1 + jωCRF).

Active High Pass Filter

(b) Consider figure 16.2:
Feedback Impedance ZF = RF

Input Impedance Zs = Rs + 1/jωCs = (jωCsRs + 1)/jωCs

So, closed-loop frequency response (closed-loop gain):
A(jω) = -ZF/Zs = -RF/(1 + jωCsRs)/jωCs
So, A(jω) = -(jωCsRF)/(1 + jωCsRs) -------(2)
A(jω) -> 0 as ω -> 0; A(jω) -> -RF/Rs as ω -> ∞

Active Band Pass Filter

(c) Consider figure 16.3:
Feedback impedance ZF = RF || 1/jωCF = RF/(1 + jωCFRF)

Input impedance Zs = Rs + 1/jωCs = (jωCsRs + 1)/jωCs

So, closed loop frequency response (closed loop gain):
A(jω) = -ZF/Zs = - jωCsRF/[(1 + jωCsRs)(1 + jωCFRF)]-------(3)
Equation(3) is the frequency response of an active band pass filter.

Active Low Pass Filter Problem

(d) Consider figure 16.4:
Feedback impedance ZF = R2 + 1/jωC = (1 + jωCR2)/jωC

Input impedance Zs = Rs

So, frequency response of circuit:
A(jω) = -ZF/Zs = - (1 + jωCR2)/jωCRs.
Circuit is an active low pass filter.

(ii) Attenuation in decibels = 20log10|A(jω)|
So, |A(jω)| = |- (1 + jωCR2)/jωCRs|
So, |A(jω)| = (1 +j(103)(0.1)(10-6)(105))/j(103)(0.1)(10-6)(105) = (1.01)1/2
So, Attenuation in decibels = 20log10(1.004881) = 20(0.00215) = 0.043dB.

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