Physics, asked by Pjal2598, 11 months ago

If alternating triangular wave is applied to capacitor then output waveform is

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Answered by ssanjay15
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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.

Typical RC Waveform

Square Wave Signal

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 RCcircuit 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, Vcwould 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.

Frequency Response

The RC Integrator

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 RC Differentiator

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

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