# TF EIM Chapt3

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# Differentiator circuit

Consider the effect of a high-pass filter on a rectangular shaped input pulse with a width and period .

The first thing to consider is what happens to the different parts of the square pulse as it travels to the circuit.

You know that the circuit, as a high pass filter, will tend to attenuate low frequency (slow changing) voltages and not high frequency (quick changing) voltages.

This means no DC voltage values pass beyond the capacitor.

This means that points and in the pulse shown below should pass through the circuit.

The voltage changes in between point and DC are passed according to what you select for the break frequency ()

In the above pulse picture it was assumed that or in other words the RC network time constant is a lot smaller than the pulse width.

If the RC time constant is larger than the pulse width then you will see the voltage stay high. The rectangular pulse will be output as a square like pulse which is "sagging".

## Loop Theorem

How does compare to  ?

The maximum value of is determined by the pulse width while is the period that corresponds the lowest frequency (other than DC).

When

may be ignored.

taking the derivative with respect to time yields

or

The Ohm's Law for  :

The output of the above circuit when will be proportional to the derivative of the input AC source with respect to time.

# Integrator

The above circuit is a low pass filter. When a square pulse hits it all of the low frequency components will pass through and the high frequency components will be attenuated. This essentially makes the pulse smooth as shown below.

The above is referred to as an integration circuit.

## Loop theorem

This differential equation may be written as

If

then

The circuit appears to be integrating the input voltage.

If then very little integration is done because the pulse is changing at a low frequency compared to the RC constant.

# Pulse Sharpening

The circuit below is a way to decrease the rise time of an input pulse at the expense of attenuating the pulse. The output pulse is a "sharpened" version of the input pulse and attenuated.

## Junction Rule

If we apply the junction rule for the currents in the above circuit

Then

Collecting terms

= constant =

assuming

If

Then

The above is a first order non-homogeneous differential equation

To solve you first find a solution to the homogeneous equation ( and then form a particular solution based on the homogeneous solution

### Apply Boundary Conditions

where

Note
is always less than

or in other words the rise time of the output pulse is short than the rise time of the input pulse

or in other words the output pulse is sharper (rising faster) than the input pulse

So the output pulse is faster by a factor of and attenuated by the same factor!