Summary: AE_Lecture4_PartD4 describes the time domain response of high pass filter and gives the method of determining lower -3dB frequency from 10% sag frequency.
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Figure 14. The circuit diagram of C-R high Pass Filter and square wave response.
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Figure 15. The Square Wave Response or Time Domain Response of HPF.
In Figure 14, the square wave response as well as the step response of C-R High Pass Filter is shown.
The slope of the tangent drawn to the exponentially decaying curve at t=o is:
d(vR)/dt|t=0 = (1V/τ sec) ;
Two triangles shown are similar hence:
∆V/(T/2) = (1V/τ sec) ;
Or ∆V/1V = fractional sag = T/(2τ) ;
We have to adjust the frequency of the periodic square wave to achieve 10% sag;
Say the frequency of 10% sag is f*= 1/T where T is the period of repetition.
Therefore T=(2τ) = 0.1= (2πfL)/(2f*); since 1/τ = ωL = 2πfL.
Therefore fL = (0.1f*)/π
In figure 15, the square wave response is shown.
By inspection we find that
when prf >> fL then we have faithful reproduction of the square wave except that the dc component is blocked by the series capacitance;
when prf ~ fL then we have sag effect at the top and bottom of the square wave;
when prf << fL then we get the differentiation of the leading and lagging edge. Since the leading edge and lagging edge are positive step and negative step, we get positive impulse and negative impulse at the instants where step voltages had occurred.
Hence at frequencies much lower than the cutoff frequency of HPF, the filter behaves like a differentiator.
5.1 Frequency Response of High Pass Filter.
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Equation (2) has zero at zero frequency and a pole at
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Figure 16. Magnitude of Transfer Function vs Frequency response of HPF.