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Lecture 6: Analogue filters Analogue Filters To design a recursive filter, we begin with a good analogue filter and approximate it with a digital filter. It is the best way at hand. This requires us to design an analogue filter first. These filters are described by a constant coefficient differential equation – output is the weighted sum of the derivatives of the input and output signals, defined as N d k x (t ) M d k y (t ) y t ak bk (1) k 0 dt k k 1 dt k e.g. y t x (t ) with a0=1, ak=0 for k=1, ...,N, and bk=0 for k=1, ...,M. dx (t ) y t with a0=0, a1=1, ak=0 for k=2, ...,N, and bk=0 for k=1, ...,M. dt Let e st is an eigenfunction for the analogue filter likewise e jk is an eigenfunction st of a digital filter. e is an eigenfunction that can be shown as follows d st dt e s e st d 2 st dt 2 e s 2 e st (2) d k st dt k e s k e st Eigen Eigen function value Analogue H ( s )e st e st filter Figure 1: Eigen function and eigen value Eigenfunction and eigenvalue play an important role. Eigen value H(s) is called the transfer function of the analogue filter http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters y(t ) H ( s)est where x(t ) e st (3) Replacing y(t) in the general equation (9), we get N d k e st M d k H ( s )e st H ( s )e ak st bk k 0 dt k k 1 dt k N M (4) H ( s )e ak s e H ( s ) bk s e st k st k st k 0 k 1 Solving equation (12) for H(s) yields N a s k e st k 0 k H ( s) (5) M 1 bk s k e st k 1 The transfer function of the recursive analogue filter is as follows N ak s k H ( s) k 0 M (6) 1 bk s k k 1 Since H(s) is a ratio of polynomials, it can be represented by the coefficients of the polynomials ak and bk. a 0 a1 s a 2 s 2 a k s k H ( s) 1 b1 s b2 s 2 bk s k (7) H(s) can also be represented by poles, zeros, and gains. H ( s) G s z1 s z2 s z N s p1 s p2 s pM (8) Where G is the gain, N is the number of zeros and M is the number of poles of the filter. The poles and zeros can be plotted in complex s-palne shown in Figure 5. http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters Im{s} zero pole Re{s} Figure 2: Pole-zero diagram Example 1: An analogue filter is described by the following differential equation dx (t ) dy (t ) y (t ) x (t ) 2 dt dt Find the transfer function of the analogue filter. Example 2: An analogue filter is described by the transfer function 1 s H ( s) 1 2s Find the pole-zero-gain form of the of the transfer function of the filter and plot the pole-zero diagram. http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters Stability of analogue filter An analogue filter is stable if and only if its impulse response goes to zero, i.e. Lt h(t ) 0 (9) t If the impulse response increases with time, the filter is unstable. In general, the computation of impulse response is very tedious. It requires the solution of the filter’s differential equation or the evaluation of the inverse Laplace transform of the transfer function. Fortunately, to test the stability, we don’t need to compute filter’s impulse response. It is possible to determine the stability by the position of the poles of the filter’s transfer function in the s-plane. The filter is stable if the poles lie in the left half of the s-plane. If the poles lie in the right half of the s-plane, filters are unstable. Im{s} s-plane unstable region stable region Re{s} Figure 3: Stability from Pole-zero diagram Example 3: An analogue filter is described by the differential equation y (t ) a0 x (t ) b1 y (t ) Show the stability of the filter using pole-zero plot. http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters We know the stability for a single pole filter and this idea can be extended to realisation of more complicated filters by expressing an arbitrary filter as a collection of single-pole filters. Lets consider an analogue filter with M poles. The transfer function can be expressed in terms of poles, zeros and gain H ( s) G s z1 s z2 s z N (10) s p1 s p2 s pM Where N is the number of zeros and M is the number of poles. These single-pole filters can be realised in two ways: Cascade combination Parallel combination By cascade combination it is difficult to visualise the relationship between individual impulse response and response of the original filter. For this reason, we turn to parallel combination of single-pole filters. Pole-zero gain representation of the transfer function can be expressed as the parallel combination of M-pole filters by performing a partial fraction expansion g1 g2 gM H ( s) (11) s p1 s p2 s pM g1 s p1 g2 x(t) y(t) s p2 gM s pM Figure 4: Parallel combination of single-poles filters. In this representation, the impulse response of the original filter is the sum of the impulse responses of single-pole filters. If any of these filter is unstable, the original filter will also be unstable. http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters In other words, M-pole filter is stable, only if each of its single-pole filter is stable. Therefore, the m-pole filter is stable iff (if and only if) all of its poles lie in the left half of s-plane. Example 4: A low-pass analogue filter has three poles p1 = -5 p2 = -10 p3 = -5 Calculate the gains of the single-pole filters using partial fraction expansion. Give a parallel realisation of the filter. http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters Butterworth filter design The general form of the analogue filter is d k x (t ) M d k y (t ) N y t ak bk (12) k 0 dt k k 1 dt k st Where the complex exponential e is applied to an analogue filter. The output st becomes H ( s)e where H (s ) is the eigenvalue or the transfer function. The transfer function of the general form becomes N ak s k H ( s) k 0 M (13) 1 bk s k k 1 The frequency response is the filter’s transfer function at s j . Therefore, the frequency response of the filter can be rewritten as N d k jt M dk H ( )e jt a k k e bk k H ( )e jt k 0 dt k 1 dt N M H ( )e jt e jt a k ( j ) k e jt H ( ) bk ( j ) k (14) k 0 k 1 N a k ( j ) k H ( ) k 0 M 1 bk ( j ) k k 1 Butterworth filter is very popular because of its passband and stopband are without ripples. But it has the widest transition band and cutoff frequency. The Butterworth filter has two parameters Order of the filter order of the denominator of H (s ) that corresponds to ˆ the number of poles of the filter’s transfer function Cutoff frequency denoted as c http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters The Butterworth filter is defined, in general, by the squared transfer function H B ( s) H B ( s) H B ( s) 2 * (15) If the order of the filter is m , then the transfer function square becomes 1 H B ( s) 2 2m (16) s j 1 c We know that the frequency response is the filter’s transfer function at s j 1 H B ( ) 2 2m j 1 j c 1 H B ( ) 2 2m (17) 1 c Poles of the squared transfer function are values of s for which 2m s j 1 0 c 2m s j 1 c s 2 m ( 1) j c 1 s ( 1) 2m j c (18) Now, (1) and j can be expressed in exponential form as http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters 1 e j ( 2 k 1) (19) j je 2 NOTE: cos ,3,5,... 1 , sin ,3,5,... 0 sin 1 1 1, and cos 0 2 2 Therefore, the poles become 1 s ( 1) 2m j c s e 1 j j ( 2 k 1) 2m e 2 c ( 2 k 1) j se j 2m e 2 c ( 2 k 1) se j 2m 2 c ( 2 k m 1) se j 2m c (20) The Butterworth filter’s complex pole s has the magnitude and angle as | s | c 2k m 1 (21) s 2m They are distributed around a circle of radius c centred at origin at a distance of 360 from each other. 2m http://www.infm.ulst.ac.uk/~siddique Lecture 6: Analogue filters Im{s} 360/2m | Re{s} |c| Figure 5: Butterworth filter’s poles Butterworth filter design can be summarised in a three-step design procedure Step 1: Select the order m and cutoff frequency c of Butterworth filter Step 2: Plug the values into the equation ( 2 k m 1) se j 2m c 360 and compute the 2m poles. The poles are equally spaced at around the circle 2m with radius of c . Step 3: Select m stable poles (those are in the left half of s-plane) Example 5: Find the poles of a third-order Butterworth low-pass filter with a cutoff frequency 10 rad. http://www.infm.ulst.ac.uk/~siddique