In the previous RC Charging and Discharging tutorials, we saw how a capacitor has the ability to both charge and discharges itself through a series connected resistor.

The time taken for this capacitor to either fully charge or fully discharge is equal to five RC time constants or 5T when a constant DC voltage is either applied or removed.

But what would happen if we changed this constant DC supply to a pulsed or square-wave waveform that constantly changes from a maximum value to a minimum value at a rate determined by its time period or frequency. How would this affect the output **RC waveform** for a given RC time constant value?

We saw previously that the capacitor charges up to 5T when a voltage is applied and discharges down to 5T when it is removed. In RC charging and discharging circuits this 5T time constant value always remains true as it is fixed by the resistor-capacitor (RC) combination. Then the actual time required to fully charge or discharge the capacitor can only be changed by changing the value of either the capacitor itself or the resistor in the circuit and this is shown below.

Useful wave shapes can be obtained by using RC circuits with the required time constant. If we apply a continuous *square wave* voltage waveform to the RC circuit whose pulse width matches that exactly of the 5RC time constant ( 5T ) of the circuit, then the voltage waveform across the capacitor would look something like this:

The voltage drop across the capacitor alternates between charging up to Vc and discharging down to zero according to the input voltage. Here in this example, the frequency (and therefore the resulting time period, ƒ = 1/T) of the input square wave voltage waveform exactly matches twice that of the 5RC time constant.

This (10RC) time constant allows the capacitor to fully charge during the “ON” period (0-to-5RC) of the input waveform and then fully discharge during the “OFF” period (5-to-10RC) resulting in a perfectly matched RC waveform.

If the time period of the input waveform is made longer (lower frequency, ƒ < 1/10RC) for example an “ON” half-period pulse width equivalent to say “8RC”, the capacitor would then stay fully charged longer and also stay fully discharged longer producing an RC waveform as shown.

If however we now reduced the total time period of the input waveform (higher frequency, ƒ > 1/10RC), to say “4RC”, the capacitor would not have sufficient time to either fully charge during the “ON” period or fully discharge during the “OFF” period. Therefore the resultant voltage drop across the capacitor, Vc would be less than its maximum input voltage producing an RC waveform as shown below.

Then by varying the RC time constant or the frequency of the input waveform, we can vary the voltage across the capacitor producing a relationship between Vc and time, t. This relationship can be used to change the shape of various waveforms so that the output waveform across the capacitor barely resembles that of the input.

The **Integrator** is a type of **Low Pass Filter** circuit that converts a square wave input signal into a triangular waveform output. As seen above, if the 5RC time constant is long compared to the time period of the input RC waveform the resultant output will be triangular in shape and the higher the input frequency the lower will be the output amplitude compared to that of the input.

From which we derive an ideal voltage output for the integrator as:

The **Differentiator** is a **High Pass Filter** type of circuit that can convert a square wave input signal into high frequency spikes at its output. If the 5RC time constant is short compared to the time period of the input waveform, then the capacitor will become fully charged more quickly before the next change in the input cycle.

When the capacitor is fully charged the output voltage across the resistor is zero. The arrival of the falling edge of the input waveform causes the capacitor to reverse charge giving a negative output spike, then as the square wave input changes during each cycle the output spike changes from a positive value to a negative value.

from which we have an ideal voltage output for the Differentiator as:

If we now change the input RC waveform of these RC circuits to that of a sinusoidal **Sine Wave** voltage signal the resultant output RC waveform will remain unchanged and only its amplitude will be affected. By changing the positions of the Resistor, R or the Capacitor, C a simple first order Low Pass or a High Pass filters can be made with the frequency response of these two circuits dependant upon the input frequency value.

Low-frequency signals are passed from the input to the output with little or no attenuation, while high-frequency signals are attenuated significantly to almost zero. The opposite is also true for a High Pass filter circuit. Normally, the point at which the response has fallen 3dB (cut-off frequency, ƒc) is used to define the filters bandwidth and a loss of 3dB corresponds to a reduction in output voltage to 70.7 percent of the original value.

where RC is the time constant of the circuit previously defined and can be replaced by tau, T. This is another example of how the *Time Domain* and the *Frequency Domain* concepts are related.

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What happens if time constant is greater than input square wave frequency in first order low pass filter

In digital communication, in receiver block diagram, multiplier output is given to integrator. I want to know how integrator converts sinusoidal signal into DC signal.

thank you!ðŸ˜Š

What will be the waveform across capacitor in differentiator ckt?

well explained

That’s excellent and marvelous. You explained the things in much simple easy and I found out in a much needed time. Thank you so much….

Thanks alot

Thanks alot for the knowledge you gave

This website is very helpful to me. I have suggested my friends to have a look at this website since all concepts are explained in simple language.

Thank you very much!!!!!!!!!

Hello.

In regards to the third example labeled “A Shorter 4RC Input Waveform” … the diagram seems to represent a “A Shorter 2RC Input Waveform”, not a 4RC time interval like the title suggests. (Or am I missing something?)

Hello John, The example shows that if the frequency is increased so that the time constant of the input waveform is equivalent to 4RC, then the half-period of the input signal will be much less than the required 5RC half-period required changing the shape of input waveform.