Lecture 12: Filter Networks (continued)Vijay Singh∗February 26, 2003AbstractDiscussion of Variable Frequency Response Analysis. Introduction to the con-cept of transfer functions (also called network functions), poles and zeros, and pre-lude to Bode plots.1 Overview1.1 What we have seen so far1. Though our lectures have been titled “Filter Networks”, they fall under the broadercategory of “Variable Frequency Response Analysis”. That is, we have so far stud-ied circuits whose response to external stimuli depend on the frequencies of thosestimuli.2. We have seen the concepts of gain, magnitude of the gain , phase of the gain,etc of a frequency dependent network.3. We have learnt some ways to identify whether a given electric circuit is low pass/bandpass/etc.4. We have also done some exercises in plotting/sketching the magnitude of the gainand the phase of the gain versus frequency. Such plotting/sketching has involvedpains-taking calculation of several points along the curves.1.2 What we will see in the next few lectures1. We will use a tool called Bode plot to get a quick and fairly accurate estimate ofhow the frequency response of a network (or electric circuit) looks like.2. We will look at the concept of transfer functions.3. We will do a mid-term project that involves transfer functions, Bode plots, andfilters.By the way, Bode plot can be used to quickly identify if a certain transfer functionthat is given to us is a low pass filter, high pass filter, band pass filter or a band rejectfilter.∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky,Lexington, KY, USA. E-mail: [email protected] Lecture to be presented on February 26, 2003 byRamprasad Potluri (substituting for Prof. Vijay Singh). Typeset in LATEX12 Frequency response of some simple circuits2.1 Example 1ZR = R<0RZL = jwL = wL <90oZR LZL ZC = 1/(jwC) = 1/(wC) <-90oCZC Rf00Mag(ZR)f00Phase(ZR)2pi f L00Mag(ZL)f0Phase(ZL)f090o00Mag(ZC)f0Phase(ZC)f-90o1/(2pi f C)Figure 1:22.2 Example 2RLCZeq Figure 2:Zeq= R+jωL+1jωC=(jω)2LC + jωRC + 1jωC(1)Let,s = jω (2)Zeq=s2LC + sRC + 1sC(3)|Zeq| =p(1 − ω2LC)2+ ω2R2C2)ωCMag(Zeq )w1/sqrt(LC)Phase(Zeq )w90o-90o00Figure 3:From Equation (1):ω −→ 0 Zeq=1jωC6(Zeq) = −90◦ω −→ ∞ Zeq−→ jωL6(Zeq) = +90◦32.3 Example 3 R15 ohm0.1 H2.53 mFLCVO+-+-10<0Figure 4:VO=ÃRR + jωL +1jωC!VS(1)VO=µjωCR(jω)2LC + jωRC + 1¶VS(2)VO=jω(37.95 × 10−3)106(0)(jω)2(2.53 × 10−4) + jω(37.95 × 10−3) + 1(3)Mag(VO)00250fFigure 5:Phase(VO)00f90o-90oFigure 6:Mag(VO)1log10f210 100100046Figure 7:Phase(VO)log10f-90o90oFigure 8:We see that the imp edances, voltage ratios, current ratios or some other ratios thatwe have come across are, in general, the ratio of two polynomials in jω. This ratio hasthe general form:H(jω) =N(jω)D(jω)=am(jω)m+ am−1(jω)m−1+ . . . + a1(jω) + a0bn(jω)n+ bn−1(jω)n−1+ . . . + b1(jω) + b0(4)Polynomials of order m and n.43 Transfer FunctionsSubstituting jω = s in Equation (4), we get the following form:H(s) =N(s)D(s)=amsm+ am−1sm−1+ . . . + a1s + a0bnsn+ bn−1sn−1+ . . . + b1s + b0(5)Polynomials of order m and n.Actually, s has a special significance which will be seen when we get to Laplace Trans-forms.• Transfer functions are also called network functions.• Network functions define the ratio of the response of a network to an input stimulus.• The term “transfer” in the name because the response can be at one point in acircuit while the stimulus (or input) can be at another point.• The output to input ratios can be1.Vout(s)Vin(s)called voltage gain2.Vout(s)Iin(s)called transimpedance3.Iout(s)Iin(s)called current gain4.Iout(s)Vin(s)called transadmittance• When the response and the stimulus are at the same point, then they are calleddriving point functions. E.g.: input impedance, output impedance.3.1 Important assumptions when working with transfer func-tions1. ai’s and bi’s in Equation (5) are real.This assumption is satisfied by default in electrical circuits because R, C, L, andcontrolled sources are always real.2. The circuits are linear time-invariant.This assumption may not always be satisfied by electrical circuits. In such a case,either the circuit may not have a transfer function, or we may have to linearizethe circuit description about some operating point and then develop the transferfunction of this circuit for that operating point.3.2 Poles and ZerosAssumption 1 of Subsection 3.1 allows us to write Equation (5) as follows:H(s) =K0(s − z1)(s − z2) · · · (s − zm)(s − p1)(s − p2) · · · (s − pn)where, K0is constant, zi’s are the roots of N(s) and pj’s are the roots of D(s).zi’s and pj’s may be complex as well as real. If they are complex, then they shouldoccur in conjugate pairs since the ai’s and bj’s are real.5Question 1 zi’s are called the zeros of the transfer function and pj’s are called the polesof the transfer function. Why?4 Next lectureThe next lecture will be on Bode plots.5 If time permitsWe will solve an example on deriving the transfer function of a filter built around an
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