Step1:

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$\"\"$

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The amplitude of the triangular pulse is 5 V.

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The average value of pulse is $\"\"$ .

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The freequency is $\"\"$ .

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The time period of the pulse is $\"\"$ .

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Substitute $\"\"$ .

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$\"\"$ .

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The time period of the triangular pulse is 0.001 sec.

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An op-amp differentiator with triangular waveform as input, the output waveform is a rectangular waveform.

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Hence, for an triangular pulse as input for one cycle , we have two cycles of rectangular waveform.

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Draw a triangular pulse with time period of 0.0005 sec:

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$\"\"$ \ \

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Find the voltage $\"\"$ using waveform given.

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Observe the graph:

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Two points in the interval $\"\"$ are $\"\"$ and $\"\"$ . \ \

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Find the voltage $\"\"$ in $\"\"$ .

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Slope = $\"\"$ .

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Slope point form of the equation is $\"\"$ .

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Slope = 20000 and the point is $\"\"$ .

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$\"\"$

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Voltage in $\"\"$ is $\"\"$ V. \ \

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The output voltage of the differentiator is $\"\"$ .

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$\"\"$

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$\"\"$

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Step 2:

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Similarly for $\"\"$ .

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Two points in the interval $\"\"$ are $\"\"$ and $\"\"$ . \ \

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Slope = $\"\"$ .

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Slope = $\"\"$ and the point is $\"\"$ .

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$\"\"$

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Voltage in $\"\"$ is $\"\"$ .

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The output voltage of the differentiator is $\"\"$ .

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$\"\"$

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$\"\"$ \ \ Peak to peak voltage $\"\"$

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Peak to peak voltage is 8V. \ \

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Solution:

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Peak to peak voltage is 8V.

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