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.
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:
Rfig 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 =
where, f – frequency of input AC c - capacitance of the capacitor C
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 |
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
1/CS –reactance offered by C in terms of Laplace transform
Therefore,
Laplace transform s=jw and w=2πf (w-omega)
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 becomeRe arranging,
Here 1/RC is taken as a constant
Taking inverse Laplace transform,
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.
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.
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 << VR,
where VR 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 VR<< Vc. So output voltage VR is much less for low frequency input.
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”.
Xc is very high.Vc is also high. Therefore for lower frequencies VR<< Vc. So output voltage VR is much less for low frequency input.
Frequency response of High pass filter |
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.
When the high pass circuit obey some condition, it act as a differentiator.
Lets write the equation for output voltage.
Laplace transform s=jw (where w=2πf) and RC is the time constant.Substitute for 's'.
Rearranging,we get
Take the condition that RCw << 1, then
We know jw = S, Laplace transform. So therefore, Vout = RC S Vin .
The above equation is a Laplace equation and we take inverse Laplace, then equation become,
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.