Some simple digital filters

1. One zero at the origin and one positive real pole makes a low pass filter.
   

   

   The steady state frequency response is
   as shown in the diagram for the case p=0.9.
   The digital frequency is the angle ( ). As the frequency increases decreases - hence it is a low pass filter.

   The unit sample response of this filter is :

   

   The time domain DE that gives this response can be deduced as follows :

   

   

   Taking the inverse z-transform and using the shifting theorem :

   

   Note : The magnitude response doesn't look much like that of the ideal low pass filter (M not constant over any finite range).
   

2. One zero at the origin and one negative real pole makes a high pass filter.
   

   

   

   Similarly for the case p=-0.9 we see from the diagram that a high pass filter results.
   

   The unit impulse response is
   

   The DE giving the same filtering effect as H(z) is :

   

3. Apair of zeros at the origin and a complex conjugate pair of poles makes a band pass filter.
   

   Let the poles be at e.g. and . ( ).

   

   (The coefficient of z in the denominator is :

   

   From the diagram : as goes from :

   

   and is a minimumn at .
   

   The unit step response h(n) is found by inverting the z-transform.

   

   num=[1 0];
   den=[1 -0.9 0.81];
   [r,p,k]=residue(num,den);

   

   

   

   

   

   

   Invert the z-transform :

   

   

   

   

   The DE which would give the same band pass filter response is :

   

   

   Take the inverse z-transform, applying the shifting theorem :

   

4. A band rejection filter can be made using :
   
   • A pair of complex conjugate zeros on the unit circle
   • Together with a pair of complex conjugate poles of magnitude < 1 at the same phase i.e. with .

   e.g let and

   

   

   We see from the diagram that and goes to zero at . By having the poles close to the zeros the ratios and don't vary much with frequency except near .
   

   The unit impulse response :

   

   num=[1 0 1];
   den=[1 0 0.81 0];
   [r,p,k]=residue(num,den);

   r=-0.1173,-0.1173,1.2346 p=0+0.9i,0-0.9i,0=
   

   

   

   Invert the z-transform :

   

   

   The DE to apply to get the same band rejection filter :

   

   

   Invert the z-transform :