IIR FiltersThe General IIR Difference EquationBlock DiagramTime-Domain ResponseImpulse Response of a First-Order IIR SystemLinearity and Time Invariance of IIR FiltersStep Response of a First-Order Recursive SystemSystem Function of an IIR FilterThe General First-Order CaseSystem Functions and Block-Diagram StructuresThe Transposed StructuresRelation to the Impulse ResponsePoles and ZerosPoles or Zeros at the Origin or InfinityPole Locations and StabilityFrequency Response of an IIR Filter3D Surface Plot ofThe Inverse z-Transform and ApplicationsA General Procedure for Inverse z-TransformationSteady-State Response and StabilitySecond-Order FiltersPoles and ZerosImpulse ResponseFrequency ResponseExample of an IIR Lowpass FilterECE 2610 Signal and Systems 8–1IIR FiltersIn this chapter we finally study the general infiniteimpulse response (IIR) difference equation that was men-tioned back in Chapter 5. The filters will now include both feed-back and feedforward terms. The system function will be arational function where in general both the zeros and the polesare at nonzero locations in the z-plane.The General IIR Difference Equation• The general IIR difference equation described in Chapter 5was of the form(8.1)• In this chapter the text rearranges this equation so that ison the left and all of the other terms are on the right(8.2)• In so doing notice the sign change of the coefficients,and also we assume that • The total coefficient count is , meaning that thismany multiplies are needed to compute each new output fromthe difference equationalyn l–l 0=Nbkxn k–k 0=M=ynyn alyn l–l 1=Nbkxn k–k 0=M+=ala01=NM1++Chapter8Time-Domain ResponseECE 2610 Signals and Systems 8–2Block Diagram• As a special case consider • The above logically extends to any order• This structure is known as Direct-Form ITime-Domain Response• To get started with IIR time-domain analysis we will considera first-order filter ( ) with NM3==yn alyn l–l 1=3bkxn k–k 0=3+=ynxnz1–z1–z1–z1–z1–z1–yn 1–yn 2–yn 3–xn 1–xn 2–xn 3–b0b1b2b3a1a2a3Feed-forwardpartFeedbackpartDirect Form I StructureN 1=M 0=Time-Domain ResponseECE 2610 Signals and Systems 8–3(8.3)Impulse Response of a First-Order IIR System• The impulse response can be obtained by setting and insuring that the system is initially at rest• Definition: Initial rest conditions for an IIR filter means that:– (1) The input is zero prior to the start time , that is for – (2) The output is zero prior to the start time, that is for • We now proceed to find the impulse response of (8.3) viadirect recursion of the difference equation• In summary we have shown that the impulse response of a1st-order IIR filter is(8.4)where the unit step has been utilized to make it clearthat the output is zero for yn a1yn 1–b0xn+=xn n=n0xn 0=nn0yn 0=nn0y 0 a1y 1–b0 0+ b0==y 1 a1y 0 b0 1+ a1b0==y 2 a1y 1 b0 2+ a12b0==yn a1nb0n 0=000hn b0a1nun=unn 0Time-Domain ResponseECE 2610 Signals and Systems 8–4Example: First-Order IIR with • The impulse response isLinearity and Time Invariance of IIR Filters• Recall that in Chapter 5 the definitions of time invariance andlinearity were introduced and shown to hold for FIR filters• It can be shown that the general IIR difference equation alsoexhibits linearity and time invariance• Using linearity and time invariance we can find the output ofthe first-order system to a linear combination of time shiftedimpulses(8.5)• From the impulse response of (8.4)b01 a1 0.8==hn 0.8nun=5 10 15 20n0.20.40.60.81FirstOrder IIR a1 0.8, b0 1hnxn xknk–kN1=N2=Time-Domain ResponseECE 2610 Signals and Systems 8–5(8.6)Example: , and • Using the above result, it follows that• Plotting this function results in• Linearity and time-invariance can also be used to find theimpulse response of related IIR filters, e.g.,(8.7)• We can view this as the superposition of an undelayed anddelayed input to the filter(8.8)yn xkhn k–kN1=N2=xkb0a1nk–un k–kN1=N2=xn 2 n 2–n 4––=a10.5=b01=yn 2 0.5n 2–un 2–0.5n 4–un 4––=5 10 15 20n 0.50.511.52ynyn a1yn 1–b0xn b1xn 1–++=yn a1yn 1–xn+=Time-Domain ResponseECE 2610 Signals and Systems 8–6• Based on this observation, the impulse response is(8.9)Step Response of a First-Order Recursive System• The step response allows us to see how a filter (system)responds to an infinitely long input• We now consider the step response of • Via direct recursion of the difference equation• The summary form indicates a finite geometric series, whichhas solution(8.10)hn b0a1nun b1a1n 1–un 1–+=b0 n b0b1a11–+a1nun 1–+=yn a1yn 1–b0xn+=y 0 a1y 1–b0u 0+ b0==y 1 a1y 0 b0u 1+ a1b0b0+==y 2 a1y 1 b0u 2+ a1a1b0b0+b0+==yn b01 a1a1n+++b0a1kk 0=n==0rkk 0=L1 rL 1+–1 r–---------------------, r 1L 1,+ r = 1=Time-Domain ResponseECE 2610 Signals and Systems 8–7• Using (8.10) and assuming that , the step response ofthe first-order filter is(8.11)• Three conditions for exist1. When the term grows without bound as nbecomes large, resulting in an unstable condition2. When the term decays to zero as ,and we have a stable condition3. When we have the special case output of (8.10)where the output is of the form , which alsogrows without bound; with the output alternatessign, hence we have a marginally stable conditionExample: and • The step response of this filter isa11yn b01 a1n 1+–1 a1–---------------------un=a1a11a1n 1+a11a1n 1+n a11=b0n 1+a11–=a10.6 b0 1==xn un=yn1 0.6n 1+–1 0.6–-------------------------------un=Time-Domain ResponseECE 2610 Signals and Systems 8–8• The step response can also be obtained by direct evaluationof the convolution sum(8.4)• For the problem at hand(8.5)• To