TF EIM Chapt3
Consider the effect of a high-pass filter on a rectangular shaped input pulse with a widthand 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 pointsand in the pulse shown below should pass through the circuit.
The voltage changes in between pointand DC are passed according to what you select for the break frequency ( )
In the above pulse picture it was assumed thator 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".
- How does compare to ?
The maximum value ofis determined by the pulse width while is the period that corresponds the lowest frequency (other than DC).
- may be ignored.
taking the derivative with respect to time yields
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.
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.
This differential equation may be written as
- The circuit appears to be integrating the input voltage.
Ifthen very little integration is done because the pulse is changing at a low frequency compared to the RC constant.
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.
If we apply the junction rule for the currents in the above circuit
- = constant =
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
- 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!