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Approximation of Derivatives

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Approximation of Derivatives Approximation of Derivatives Spring 2012 © Ammar Abu-Hudrouss -Islamic ١ University Gaza Introduction IIR filter design techniques depend mainly on converting analogue filter to a digital one. Analogue filter can be described by three methods 1) Its system function. M k sk k 0 H a (s) N k sk k 0 2) By its impulse response which is related to the system function by st H a (s) ha (t)e dt ٢ Digital Signal Processing ٢ Slide ١ Introduction 3) By linear constant-coefficient differential equation d k y(t) M d k x(t) k k k k dt k 0 dt Each of these three characterizations leads to alternative method for converting the analog filter to its digital equivalents ٣ Digital Signal Processing ٣ Slide Introduction The system function Ha(s) is stable if all its poles lie in the left half of the s-plane. Hence, the conversion is effective if 1. The j axis in the s-plane map into the unit circle in the z-plane. 2. The left-half plane LHP of the s-plane map into the inside of a unit circle in the z-plane (i.e stable analogue filter will be converted to stable digital filter. ٤ Digital Signal Processing ٤ Slide ٢ Approximation of derivatives The simplest methods to convert an analogue filter to a digital filter is to convert the differential equation into equivalent difference equation. For the first derivative dy(t) y(nT) y(nT T) y(n) y(n 1) dt tnT T T y(t) H (s) s dy(t) dt 1 z 1 y(n) H (z) y(n) y(n 1) T T ٥ Digital Signal Processing ٥ Slide Approximation of derivatives For the analog derivative, the system function is given by H (s) s Whereas, the digital system function is H (z) (1 z 1 ) /T Which leads to 1 z 1 s T ٦ Digital Signal Processing ٦ Slide ٣ Approximation of derivatives The 2nd derivative is replaced by the second difference d 2 y(t) d dy(t) [y(nT ) y(nt T)]/ T [y(nT T ) y(nT 2T)]/T 2 dt dt dt t nT T y(n) 2y(n 1) y(n 2) T 2 In frequency domain this is equivalent to 2 1 2z 1 z 2 1 z 1 s 2 2 T T Then the kth derivative is equivalent frequency-domain relationship 1 k k 1 z s T ٧ Digital Signal Processing ٧ Slide Approximation of derivatives The system function for IIR filter obtained as a result of approximation of the derivatives is H (z) H (s) a s(1z 1 ) /T Let us investigate the implication of the mapping from the s-plane to the z-plane 1 z 1 sT Then the j axis is mapped into 1 z 1 jT 1 T j 1 2T 2 1 2T 2 ٨ Digital Signal Processing ٨ Slide ٤ Approximation of derivatives The y -axis in the s -plane is mapped into a circle of radius ½ and with centre z = ½. Any point in the left hand plane will be mapped to points inside that circle. So the resultant filter is stable ٩ Digital Signal Processing ٩ Slide Approximation of derivatives The possible locations of poles of the digital filter are confined to small frequencies. As consequence, the mapping is restricted to low-pass and band-pass filters with relatively small resonant frequencies. It is not possible to transform a high-pass analog filter into a corresponding high-pass digital filter. In attempt to overcome this limitations, more complex substitutions for the derivatives have been proposed such as dy(t) 1 L y(nT kT) y(nT kT) tnT k dt T k 1 T ١٠ Digital Signal Processing ١٠ Slide ٥ Approximation of derivatives The resulting mapping between the s-plane and the z-plane is now L 1 k k s k z z T k0 When z = ej j2 L s k sink j T k1 Which is pure imaginary, which means by carefully selecting ks, The j axis can be transformed into the unity circle. This can be done by using optimization ١١ Digital Signal Processing ١١ Slide Approximation of derivatives Example: Convert the analogue bandpass filter with system function 1 H s a (s 0.1)2 9 Into a digital IIR filter by substituting for the derivatives method 1 H (z) ((1 z 1 ) / T 0.1) 2 9 T 2 /(1 0.2T 9.01T 2 ) H (z) 2(1 0.1T ) 1 1 z 1 z 2 1 0.2T 9.01T 2 1 0.2T 9.01T 2 T should be less than or equal 0.1 for poles near the unity circle ١٢ Digital Signal Processing ١٢ Slide ٦ Approximation of derivatives Example: Convert the analogue bandpass filter in the previous example by use of the mapping 1 s z z 1 T Solution: by substituting for s in H(s), we obtain 1 H (z) ((z z 1 ) /T 0.1) 2 9 z 2T 2 H (z) z 4 0.2Tz 3 (2 9.01T 2 )z 2 0.2Tz 1 H (z) has four poles while H (s) has two poles which means that the conversion has led to more complex system ١٣ Digital Signal Processing ١٣ Slide ٧.
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