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Low-pass and high-pass filter circuits are used as special circuits in many applications. Low-pass filter (LPF) can work as an **Integrator**, whereas the high-pass filter (HPF) can work as a **Differentiator**. These two mathematical functions are possible only with these circuits which reduce the efforts of an electronics engineer in many applications.

At low frequencies, the capacitive reactance tends to become infinite and at high frequencies the reactance becomes zero. Hence at low frequencies, the LPF has finite output and at high frequencies the output is nil, which is same for an integrator circuit. Hence low pass filter can be said to be worked as an **integrator**.

For the LPF to behave as an integrator

$$\tau \gg T$$

Where $\tau = RC$ the time constant of the circuit

Then the voltage variation in C is very small.

$$V_{i}=iR+\frac{1}{C} \int i \:dt$$

$$V_{i}\cong iR$$

$$Since \:\: \frac{1}{C} \int i \:dt \ll iR$$

$$i=\frac{V_{i}}{R}$$

$$ Since \:\: V_{0}=\frac{1}{C}\int i dt =\frac{1}{RC}\int V_{i}dt=\frac{1}{\tau }\int V_{i} dt$$

$$Output \propto \int input$$

Hence a LPF with large time constant produces an output that is proportional to the integral of an input.

The Frequency response of a practical low pass filter, when it works as an Integrator is as shown below.

If the integrator circuit is given a sinewave input, the output will be a cosine wave. If the input is a square wave, the output wave form changes its shape and appears as in the figure below.

At low frequencies, the output of a differentiator is zero whereas at high frequencies, its output is of some finite value. This is same as for a differentiator. Hence the high pass filter is said to be behaved as a differentiator.

If time constant of the RC HPF is very much smaller than time period of the input signal, then circuit behaves as a differentiator. Then the voltage drop across R is very small when compared to the drop across C.

$$V_{i}=\frac{1}{C}\int i \:dt +iR$$

But $iR=V_{0}$ is small

$$since V_{i}=\frac{1}{C}\int i \:dt$$

$$i=\frac{V_{0}}{R}$$

$$Since \: V_{i} =\frac{1}{\tau }\int V_{0} \:dt$$

Where $\tau =RC$ the time constant of the circuit.

Differentiating on both sides,

$$\frac{dV_{i}}{dt}=\frac{V_0}{\tau }$$

$$V_{0}=\tau \frac{dV_{i}}{dt}$$

$$Since \:V_{0}\propto \frac{dV_{i}}{dt}$$

The output is proportional to the differential of the input signal.

The Frequency response of a practical high pass filter, when it works as a Differentiator is as shown below.

If the differentiator circuit is given a sinewave input, the output will be a cosine wave. If the input is a square wave, the output wave form changes its shape and appears as in the figure below.

These two circuits are mostly used in various electronic applications. A differentiator circuit produces a constant output voltage when the input applied tends to change steadily. An integrator circuit produces a steadily changing output voltage when the input voltage applied is constant.

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