Easy Analog Circuit: Integrator & Differentiator-B Tech university Question & answer

Q1. Explain the working of RC low pass filter, how it can be act as an integrator.

Answer key: circuit(2), working+fre response(4), act as integrator(5), output(1) (12 marks)

To know, how to prepare answer key, visit link.
Answer:
Consider a first order RC circuit. It can be act as a low pass filter and allows to pass only the low frequency signals.

Circuit diagram:

fig 1

where,
AC – AC source that provides the input(Vin)
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.

Working of a low pass filter:

The reactance of capacitor is Xc = $\frac{1}{{2\pi fc}}$

where, f – frequency of input AC c - capacitance of the capacitor C

Equation reveals that Xc is inversely proportional to frequency f. So,

At low frequency input signal, Xc become very large. So Xc >> R. Therefore voltage drop across the capacitor- which is the output, is very large. In a similar way, at high frequency, Xc << R. So output is low.

Frequency response of first order Low pass filter

Summarising :
For low frequency AC / for DC -- Vc >> VR -- larger output voltage

For high frequency AC -- Vc << VR -- smaller output voltage

This circuit produces a high output voltage for low frequency inputs and it attenuate the output voltage for high frequency input signals. That why it has the name ‘low pass filter’.

Integrator

An integrator produces an output which is proportional to integral of its input.

Low pass RC circuit itself is acting as an integrator provided it obey one condition. Lets check that condition.
Equation for output voltage in fig1 is    $Vout = Vin\frac{{1/CS}}{{R + 1/CS}}$
                                       1/CS –reactance offered by C in terms of Laplace transform

Therefore,              $\frac{{Vout}}{{Vin}} = \frac{{1/CS}}{{R + 1/CS}} = \frac{1}{{1 + RCS}}$

Laplace transform s=jw and w=2πf (w-omega)

                   $\frac{{Vout}}{{Vin}} = \frac{1}{{1 + RCjw}}$

Low pass circuit act as Integrator:

In the above equation,if we are adopting a condition as wRC >> 1, then , in denominator 1+jwRC is nearly equal to jwRC. Then equation become

                  $\frac{{Vout}}{{Vin}} = \frac{1}{{jwRC}} = \frac{1}{{RCS}}$

Re arranging,

                  $Vout = \frac{1}{{RC}}\frac{1}{S}Vin$ Here 1/RC is taken as a constant

Taking inverse Laplace transform,

                 $Vout = \frac{1}{{RC}}\int {Vindt}$

Above equation shows that Vout is proportional to integral of input voltage. So if we have a low pass RC circuit which obeys the condition wRC >> 1, it will act as integrator.
In order to achieve a good integration, the following conditions must be satisfied:

The time constant RC of the circuit should be very large as compared to the time period of the input signal.
The value of R should be 10 or more times larger than Xc.

Output of integrator for square wave input is a triangular wave.

Q2. Explain the working of RC high pass filter, how it can be act as an differentiator.

Answer key: circuit(2), working+fre response(4), act as integrator(5), output(1) (12 marks)

To know, how to prepare answer key, visit link.
Answer:
Consider a first order RC circuit. It can be act as a high pass filter and allows to pass only the high frequency signals.

Circuit diagram:

Input is an AC source, Output is taken across resistor. Contains one pair of R & C.

where,
AC – AC source that provides the input(Vin)
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.

Working of high pass filter:

Voltage across R is taken as output voltage. Let’s see what happened to this voltage at low and high frequencies.

For high frequency input:

Xc is very low(Xc inversely proportional to frequency). So Vc is also low. So Vc << V_R,where V_R is the output voltage. In conclusion, for high frequency input, output is higher.

For low frequency input:
Xc is very high.Vc is also high. Therefore for lower frequencies V_R<< Vc. So output voltage V_R is much less for low frequency input.

Frequency response of High pass filter

Summarising :
for high frequency input signal ----- output voltage is larger
for low frequency input signal ----- output voltage is lesser
Then we can call it as a “high pass”.

High-pass RC circuit as Differentiator:

In differentiator output voltage, Vout is proportional to the rate of change of input.

$Vout\infty \frac{d}{{dt}}Vin$

When the high pass circuit obey some condition, it act as a differentiator.

Lets write the equation for output voltage.

$Vout = Vin\frac{R}{{R + 1/CS}} = \frac{{RCS}}{{1 + RCS}}$

Laplace transform s=jw (where w=2πf) and RC is the time constant.Substitute for 's'.

$\frac{{Vout}}{{Vin}} = \frac{{RCjw}}{{1 + RCjw}}$ Rearranging,we get

$\frac{{Vout}}{{Vin}} = \frac{1}{{1 + \frac{1}{{RCjw}}}}$

Take the condition that RCw << 1, then

$\frac{{Vout}}{{Vin}} = jwRC$

We know jw = S, Laplace transform. So $\frac{{Vout}}{{Vin}} = RCs$ therefore, Vout = RC S Vin .

The above equation is a Laplace equation and we take inverse Laplace, then equation become,

$Vout = RC\frac{d}{{dt}}Vin$

Equation shows that output voltage is proportional to derivative of input voltage. So the circuit represents a differentiator circuit. Output of the differentiator circuit for different inputs is given below.

Easy Analog Circuit

Friday 11 September 2015

Integrator & Differentiator-B Tech university Question & answer

Q1. Explain the working of RC low pass filter, how it can be act as an integrator.

Circuit diagram:

Working of a low pass filter:

Integrator

Low pass circuit act as Integrator:

Q2. Explain the working of RC high pass filter, how it can be act as an differentiator.

Circuit diagram:

Working of high pass filter:

High-pass RC circuit as Differentiator:

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