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Q. Discuss the output waveforms for a differentiating circuit.
The output waveform from a differentiating circuit depends upon the time constants and shape of the input wave.Three important cases will be considered. When input is a square wave.-When the input fed to a differentiating circuit is a square wave,output will consist of sharp pulses.During the OC part of the input wave, its amplitude changes abruptly and hence the differentiated wave will be sharp narrow pulse. However during the constant part CB of the input , the output will be zero because the derivative of a constant is zero.
Since time constant RC of the circuit is very small w.r.t time period of input wave and Xc >>R ,the capacitor will become fully half cycle, the output of the circuit will be zero because the capacitor voltage neutralizes the input voltage and there can be no current flow through R.Thus we get a sharp pulse at the output during the start of each half cycle of input wave while for the remainder part of the half cycle of input wave , the output will be zero.In this way , a symmetrical output wave with sharp positive and negative peaks is produced.Such pulses are used in many ways in electronic circuits e.g in television transmitters and receivers,in multivibrators to initiate action etc.
Consider the circuit of Figure (a) and obtain the complete solution for the voltage v C (t) across the 5-F capacitor and the voltage v x (t) across the 5- resistor.
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