Impulse Invariance & Bilinear Transform

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 ((  2kFs )) k 

The period of the sampling T should be small enough to avoid or minimize the effect of the aliasing.

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1 Impulse invariance

From the equation  H ()  Fs  H a ((  2kFs )) k  We can generalise

 H (z)  F H ((s  j2kF )) zesT s  a s k 

Then the mapping is characterized by z  e sT

We substitute by s    j z  eT e jT

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Impulse invariance

z  eT e jT

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.

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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

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Impulse invariance

So the system function of the IIR filter n H a (z)   h(n)z n0 N  pkT 1 n  ck (e z ) k 1 n0

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

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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 e0.1T e j3T z1 1 e0.1T e j3T z 1 1  (e 0.1T cos 3T )z 1 H (z)  1  (2e0.1T 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 )

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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

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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

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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

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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

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