DSP Notes 2007-08 MRV p43-52

IIR Filter Design using Analogue PrototypesIn many applications it is desirable to design a digital filter which has a response identical(or similar) to a known analogue filter. Hence a filter may be specified and designed in the s-domain (i.e. an ‘analogue’ filter), and an appropriate method applied to find a corresponding z-domain (i.e. a ‘digital’) filter. The analogue filter, when used in this way, is referred to as an analogue prototype. Analogue filters commonly used include Butterworth, Chebyshev and Bessel filters - for the purpose of these notes we will concentrate on only one example: Butterworth filters. A low-pass prototype filter may be used to design high-pass, band-pass or band-stop filters using appropriate substitutions, and we will therefore initially investigate the transfer function of a Butterworth low-pass analogue prototype.The magnitude squared transfer function of a normalised (i.e. c = 1) low-passButterworth filter is given by:N sG where n = order of filter complexity.If s is replaced by j / c (where c is the required cut-off frequency of the low-pass filter in rad/s), then the transfer function corresponds to a denormalised low-pass Butterworth filter in the frequency domain:G It is clear that the attenuation of this filter at a frequency may be found using10G log 10(dB)n Attenuatio At this stage it is helpful to consider normalised frequencies, i.e. to express any frequencyof interest as a value relative to c . This may be achieved using /c = N . This leads toAttenuation (dB)A specified required attenuation at a given frequency may be used to determine theminimum order of the filter which would fulfil this specification. Hence a Butterworth transfer function may be derived and the poles of the normalised filter may be located in the s-plane. Note that the poles of the squared transfer function s G 2N are situated at the roots of the equation1102nn swhich are all located on a unit circle in the s-plane at an angle k , wherekn oddn evenThe poles of the transfer function s G N may be found by discarding the poles of s G 2N which lie in the right-hand half of the s-plane. ExampleA Butterworth low-pass filter is required for which f c = 2 kHz. The order of the filter should be chosen such that the filter gives at least 25 dB attenuation at 5 kHz. Determine the order of the filter to be used and find its s-plane pole locations. Hence derive the transfer function of the filter. SolutionDerivation of a z-domain transfer functionHaving derived an analogue (i.e. s-domain) transfer function for a filter, it is then necessary to derive a corresponding z-domain transfer function to realise a digital filter. Various techniques may be used: here we will briefly investigate impulse invariance and then look in more detail at the bilinear z-transform. Impulse InvarianceOne procedure for deriving a digital transfer function from an analogue transfer function is to determine the coefficients in the z-domain such that the impulse response of the digital filter is a sampled version of the impulse response of the analogue prototype. If the analogue filter has a transfer function H a (s) (for example as derived above), then the continuous-time impulse response is denoted by h a (t). In order for the digital filter to have an identical impulse response, it should be designed such that its impulse response, h(n), is given bynTh nh a where T is the sampling period for the digital filter.A general continuous-time transfer function may be expressed, using partial fraction expansion, asH sA s s a k kk N1and its impulse response is given byh tA e u ta k s t k Nk 1where u(t) is a continuous-time step function introduced so as to describe a causal system.If this impulse response is then sampled, the discrete-time impulse response values aregiven by h(n), whereN 1k n T s k N1k nTs k a nu e A nu e A nTh nh k kwhich leads to the corresponding transfer function H(z) given byN1k zH Hence a pole at s k in the s-plane transforms to a pole at e s k T in the z-plane. However, thezeros of H(z) determined by this method are a function of the poles and the A k coefficients, and will not be mapped directly from the s-plane to the z-plane.Since any practical analogue prototype will not have a bandwidth limited to f s /2 in the continuous frequency domain, this technique will generally lead to aliasing errors, and hence its frequency response will not generally correspond to that required.Bilinear z-transformA more widely used technique for deriving a digital IIR filter from an analogue prototype is to use the bilinear transform. As an example, consider the simple s-domain transfer function defined bysH a which can be seen to correspond to the first order differential equationty B t y tx A a a (1)MRV0708We may writett aa 0a 0a a a 0dy t y t y t y t y t y Putting t = nT and t 0 = (n-1)T, we see thatnTT1n aa a dy T 1n y nT y The integral in the above equation may be approximated using the trapezoidal rule, givingT1n y nT y T 1n y nT y a a 2Ta a (2)and from equation (1) we see thatty B tx A ty a a (3)Substituting (3) into (2) givesnTy B nTx A T1n y B T1n x T1n y nT y a a a a Here we may take z-transforms and replace y a (nT) by Y(z), y a ((n-1)T) by Y(z)z -1, etc. leading to111z 1z X 2ATz 12BTz1z YMRV0708which may be rearranged to givezHComparing this to the original analogue transfer function it can be seen that H(z) may be obtained by substitutingsThis may be shown to be true in general. Hence a digital filter may be derived from an analogue (s-domain) transfer function using this substitution. ExampleDerive a digital filter transferfunction from the analogue prototype defined by the equationH sa using the bilinear transformation. SolutionDiscussion of Bilinear z-TransformIt was seen above that the bilinear z-transform is implemented by substitutingsinto the expression for an s-domain transfer function.What are the implications of this with regard to the frequency axis in the continuous s-domain (i.e. analogue frequency a ) and the frequency axis in the discrete z-domain (digital frequency d )?We may investigate this by replacing s by j a , and z by exp(j d T), givingajwhich leads totan T 2aThus it can be seen that the relationship between a and d is non-linear due to the tan function.Since we know from our filter specification the required values (for cut-off frequencies etc.) of the digital filter, it is necessary tocalculate the corresponding analogue frequencies using the above equation. This is known as pre-warping the frequency axis. The relationship between analogue and digital frequency axes for the bilinear transformation is shown in the sketch below.An alternative view of the bilinear transform is discussed below.Consider an s-domain transfer function H(s) which is represented in the s-plane wheres = + jThis may be warped into a different space, which still uses a continuous-time and continuous-frequency representation, i.e. we may warp H(s) into an ‘alternative’ s-plane. We will call this ‘alternative’ s-plane the R-plane. R is complex and may be writtenR = u + jvThe transformation which we will use to achieve this warping is given bytanhT 2s where T is the sampling period to be used in the digital filter implementation. Consider whenRthen it can be seen thattanhT 2Now consider whenskj Rwhere k is an integer. Here it can be seen that the corresponding value of s iss =Hence we see that the entire frequency range from zero to in the s-plane transforms tothe range 0 to (/T) in the R-plane. We also see that non-overlapping complementary frequency bands exist in the R-plane.We may now use the standard z-transform to take us from an R-plane representation to a corresponding z-plane representation - in this case we useRTe zIf we are considering the frequency axis, then we may use R = jv, leading tovTsin j vTcos e zjvT Considering the point v = 0 in the R-plane (corresponding to a = 0), then it can be seen that this transforms to the point (1 + j0) in the z-plane.It can also be seen that the point in the R-plane corresponding to a = , i.e.vtransforms to the point (-1 + j0) in the z-plane. Thus we see that the entire s-plane frequency range from zero to is mapped to the semi-circumference of the unit circle in the z-plane.We are usingtanhT 2s andRTe zand it can be shown (how?) that this leads to the substitution used earlier for the bilinear z-transform, i.e.sMRV0708These concepts may be visualised as shown below.s-plane to R-plane to z-plane transformationR-plane z-planes-planeFrequency Range Mappings-plane z-plane。