RESISTANCES AND CAPACITANCES 



From A, draw a dotted straight line to the left and downwards, of slope 



6* dB per octave, that is, 6 dB's for a twofold change in frequency. 

 From A, mark the point B, 3 dB's below A. 

 Draw a smooth curve, through B, asymptotic to the dotted hnes. 

 This is the required characteristic curve {Figure 3.25). The transmission curve 



A 



Transmission 

 factor 

 1 Slope : 



f 6 B's /octave,^ 



dB's down 



Frequency — — »- 



Figure 3.25 



for the low-pass case is, of course, the mirror image of this and the technique 

 for sketching it will be obvious. 



Finding the phase shift when using j operator — To plot the modulus of the 

 transmission factor for a filter is not to tell the whole story, for we are saying 

 nothing about the relative phase angle between input and output. To find 

 the 'phase shift' when using j operator we employ a process called 'rationaliza- 

 tion'. In general the final expression for Fout/^in has the form {A +jB)l 

 (C+jD) 



To rationalize this, we multiply numerator and denominator by C —jD, 

 giving 



Fout _ (A -f JB)iC - jD) _ {AC -V BC) -^ j{BC - AD) 

 Fin ~{C+jD){C-jD) C^-^r D^ 



The tangeni of the phase shift is now given by they part of the numerator 

 divided by the real part, i.e. 



, BC-AD 

 ^ - *"^ AC-VBC 



If the phase of the output leads on that of the input, the phase angle will 

 come out positive ; if it lags, negative. 

 Thus, for the high-pass filter, we had 



Fout 1 



rationalizing, 



Fout 



Fm 



Fin 



1 + 



1 



J 



coCR 



J 



coCR 



1 + 



coCR 



1 - 



J 



mCR, 



1 + 



^r 1 + (— V 



ayCR] ^ [coCRj 



♦ 6 dB's loss is almost exactly a transmission factor of ^ (Graph 6). 



36 



