Impulse Invariance & Bilinear Transform © Ammar Abu-Hudrouss -Islamic University Gaza 1 Impulse invariance By impulse invariance method, the analogue impulse response h(t) is sampled to get the discreet sample response h(n). If the analogue filter has a system function Ha(s) then the system function of the digital filter can be achieved from the sampling theorem as H ( f ) Fs H a (( f kFs )) k Or equivalently H () Fs H a (( 2kFs )) k The period of the sampling T should be small enough to avoid or minimize the effect of the aliasing. 2 Digital Signal Processing Slide 2 1 Impulse invariance From the equation H () Fs H a (( 2kFs )) k We can generalise H (z) F H ((s j2kF )) zesT s a s k Then the mapping is characterized by z e sT We substitute by s j z eT e jT 3 Digital Signal Processing Slide 3 Impulse invariance z eT e jT It is clear that If = 0. The imaginary axis (s =j ) in the s plane map into the unity circle (z = ej T in the z-plane. If < 0, the left hand plane in the s plane map into the inside of the unity circle in the z-plane. 4 Digital Signal Processing Slide 4 2 Impulse invariance On the assumption that the poles of the analogue filter are distinct, we can write N ck H a (s) k 1 s pk pk are the poles of the analogue filter and ck’s are the coefficients of the partial expansion. Consequently N pk t ha (t) ck e k 1 If we sample the analogue impulse response, we get N pk nT h(n) ha (nT) ck e k 1 5 Digital Signal Processing Slide 5 Impulse invariance So the system function of the IIR filter n H a (z) h(n)z n0 N pkT 1 n ck (e z ) k 1 n0 For pk < 0 N c H(z) k pkT 1 k 1 1 e z We observe that the digital filter has poles at pkT zk e k 1,2,....., N 6 Digital Signal Processing Slide 6 3 Impulse invariance Example: Convert the analogue filter with the system function s 0.1 H (s) a (s 0.1) 2 9 Solution 1/ 2 1/ 2 H (s) a s 0.1 j3 s 0.1 j3 Then the transfer function of the digital filter 1/ 2 1/ 2 H (z) 1 e0.1T e j3T z1 1 e0.1T e j3T z 1 1 (e 0.1T cos 3T )z 1 H (z) 1 (2e0.1T cos 3T)z 1 e 0.2T z 2 7 Digital Signal Processing Slide 7 Bilinear Transformation Let us consider an analogue filter with the system function b H (s) s a The system can be also described by the differential equation dy(t) ay(t) bx(t) dt Starting from the integral t y(t) y( )d y(t0 ) t 0 By the trapezoidal formula at t = nT and t0 = nT-T y(nT) (T / 2)[y(nT) y(nT T )] y(nT T ) 8 Digital Signal Processing Slide 8 4 Bilinear Transformation But as y(nT) ay(nT ) bx(nT ) The integral become aT aT bT (1 )y(n) (1 )y(n 1) [x(n) x(n 1)] 2 2 2 Convert the equation into z domain aT aT bT (1 )Y(z) (1 )z 1Y(z) [X (z) z 1 X (z)] 2 2 2 Then the transfer function b H (z) 2 1 z 1 a 1 T 1 z 9 Digital Signal Processing Slide 9 Bilinear Transformation Clearly the mapping from s-plane to z-plane is 2 1 z 1 s 1 T 1 z To investigate the characteristics of the bilinear transformation, let z re jw s j Then the mapping formula can be expressed as 2 z 1 s T z 1 2 re j 1 T re j 1 10 Digital Signal Processing Slide 10 5 Bilinear Transformation or 2 r 2 1 2r sin s j 2 2 T 1 r 2r cos 1 r 2r cos j if r < 1 leads to < 0 or the left hand plane in the s plane maps into the inside of the unity circle in the z-plane. When r = 1, =0 and 2 sin T 1 cos 2 tan T 2 T 2 tan 1 2 11 Digital Signal Processing Slide 11 Bilinear Transformation Example: Convert the analogue filter with the system function s 0.1 H (s) (s 0.1) 2 16 Into a IIR digital filter by means of bilinear transformation. The digital filter is to have a resonant frequency r = /2 The analogue filter has a resonance frequency of =4. this is to be mapped into r = /2 by using 2 tan T 2 This leads that T = 0.5 12 Digital Signal Processing Slide 12 6 Bilinear Transformation Thus the desired mapping is 1 z 1 s 4 1 1 z And the resulting digital filter has the system function 0.128 0.006z 1 0.122z 2 H(z) 1 0.0006z 1 0.975z 2 0.128 0.006z 1 0.122z 2 H(z) 1 0.975z 2 j / 2 With poles at p1,2 0.987e With zero at z1,2 10.95 13 Digital Signal Processing Slide 13 Bilinear Transformation Example: Design a single-pole lowpass filter with 3-dB bandwidth of 0.2 using the bilinear transformation to analogue filter H (s) c s c The digital filter is specified to have -3dB gain at c = 0.2. In the frequency domain of the analogue filter 2 tan 0.65 /T c T 2 Thus the analogue filter has the system function 0.65/T H (s) s 0.65/T 14 Digital Signal Processing Slide 14 7 Bilinear Transformation Now, we apply the bilinear transformation to get 0.245(1 z 1 ) H (z) 1 0.509z 1 The frequency response of the digital filter 0.245(1 e j ) H () 1 0.509e j Which applies H (0) 1 and H(0.2 ) 0.707 15 Digital Signal Processing Slide 15 8.