3.4.1 Poles and Zeros of H

Chapter 3.4.1 Poles and Zeros of H

Physics Lecture Notes – Phys 395 Electronics Book
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Physics Lecture Notes – Phys 395 Electronics Book

  • CHAPTER 3. FILTER CIRCUITS50Our exponential function now becomesf (t)= Aest = Aeσtejωt(3.36)and we have a rich set of casesω =0,σ> 0,eσtexponential growth,ω =0,σ< 0,e−|σ|tdecay,ω =0,σ =0,ejωtoscillation,ω =0,σ> 0,eσ ejωtgrowing oscillationsandω =0,σ< 0,e−|σ|ejωt decaying oscillations.So we can not only describe oscillatory behavior but transient responses as well.3.4.1Poles and Zeros of HAs before, consider expanding the transfer function as the ratio of two polynomialsH(s)=P(s)Q(s).(3.37)If an are the roots of P(s) and bm are the roots of Q(s)wecan writeH(s)= A(s− a1)(s1− a2) ... (s− an)(s− b1)(s− b2) ... (s− bm),(3.38)where A is a real constant, an are zeros of H and bm are poles (infinities) of H. Knowledgeof an and bm determines H(s) everywhere.Lets now look at our two filter circuits. For a low-pass filterH(s)=11+ sRC=1/(RC)s +1/(RC)(3.39)and the filter has one pole at−1/(RC). For a high-pass filterH(s)=sRC1+ sRC=ss +1/(RC)(3.40)and it has one pole at−1/(RC) and one zero at 0. We refer to these two types of filters assingle-pole filters.There is a general rule that there must be at least as many reactive elements as poles.Based on the location of the poles we are able to deduce the general response properties ofthe filter. We will not do this here.Example: If a transfer function has poles at p1 =(−1, 2) and p2 =(−1, −2)and a zero at (0,0), as shown in figure 3.5,