DIGITAL FIR FILTERS-III DESIGN OF LOW-PASS FILTERS WITHOUT USING WINDOW FUNCTIONS

EDITOR: B. SOMANATHAN NAIR

Example 1:   Design a low-pass FIR filter for the following specifications:

·         Cut-off frequency                                  :           500 Hz

·         Sampling frequency                              :           2000 Hz

·         Order of the filter N                               :           10

·         Filter length required L = N + 1             :           11                                       

Solution:

STEP 1: NORMALIZATION OF CUT-OFF FREQUENCY

As in the case of IIR filters, here also we normalize the cut-off frequency as

                                                                  

                                                ω_c = 2π(f_c/f_s) = 2π(500/2000) = π/2   (1)

STEP 2: FIXING THE TRANSFER FUNCTION TO BE USED

As stated previously, we fix the transfer function as

                                               

H(ω) = 1,  - π/2  £  ω  £  π/2

                                                          =  0,   elsewhere                  (2)                                        

In (2), we have neglected the phase part of the transfer function. If the phase factor is also to be taken into account, we then use the expression

where q_o = w_oT  = w_o,  (assuming T = 1)  is the desired phase angle. We begin our design with the first set of given specifications.

STEP 3: DETERMINING THE IMPULSE RESPONSE OF THE FILTER

Since the transfer function H(w) is specified as the discrete-time Fourier transformation (DTFT) of the impulse response h(n) of the filter, h(n) can be obtained by taking the inverse discrete-time Fourier transformation (IDFT) of H(w). Thus, we find

Substituting the given value of H(w) from (2) into (4), we get

Equation (5) may be simplified to

                                    h(n) = (1/nπ)(e^jnπ/2‒ e^‒jnπ/2)/2j = (1/nπ) sin(nπ/2)

                                          

       = 0.5 sin(nπ/2)/(nπ/2)   (6)                                      

STEP 4: DETERMINING THE COEFFICIENTS OF THE IMPULSE-RESPONSE

In (6), we substitute various values of n and determine the corresponding values of h(n). Thus, for n = 0, we have

                         h(0) =  0.5 sin(0)/(0) = 0.5  (7)                                            

where we have used the L’Hospital’s (to be pronounced as lopithal) rule. Now, when n = 1, we get                      

                  h(1) = 0.5 sin(π/2)/(π/2) = 1/ π = 0.3183  (8)             

Similarly, for n = 2,

                        h(2) = 0.5 sin(π)/(π) = 0   (9)                                                  

and for n = 3,

                                                h(3) = 0.5 sin(1.5π)/(1.5π) = ‒0.1061   (10)                     

                                           

and for n = 4,

                                                h(4) = 0.5 sin(2π)/(2π) = 0   (11)

                    

                                                   

Finally, for n = 5,

                                           h(5) = 0.5 sin(2.5π)/(2.5π) = 0.0637   (12)

         

We stop our computation at this point, since the required length of the filter L = N + 1 is only 11, and we can achieve this length by truncating the number of samples at n = 5.  It may be noted that, since sine functions are odd symmetric, we have

                                     h(n) = h(-n)  (13)                                                             

This means that h(-1) = h(1), h(-2) = h(2) = 0, and so on. Thus we get the impulse-response

   h(n) = (0.0637, 0, ‒0.1061, 0, 0.3183, 0.5, 0.3183, 0, ‒0.1061, 0 , 0.0637) (14)                     

The coefficients for negative values of n [i.e., h(-1), h(-2), etc.] appear in the value of h(n) because we are determining the Fourier series expansion of H(w), and in this, we have to determine the values of the coefficients from -¥ to +¥.

STEP 5: DETERMINING THE TRANSFER FUNCTION FROM IR

 We can now obtain the transfer function back from the impulse-response sequence by attaching appropriate value of z. Thus we find

H(z) = 0.064z⁵, ‒0.106 z³, 0.318 z¹, 0.5, 0.318 z^‒1, ‒0.106 z^‒3, 0.064 z^‒5  (15)

                                                                                                         

There exists one problem with this computation. The negative values of frequencies (i.e., terms containing positive powers of z, such as z¹, z², etc.) indicate that we are going to realize a noncausal filter. This means that we can not physically construct the filter that we have just now designed. To make the filter causal and physically realizable, we multiply all the coefficients with an appropriate power of z^-n (in this case, with z ^-5). This results in the equation

  H(z) = 0.064, ‒0.106 z^‒2, 0.318z^‒4, 0.5z^‒5, 0.318z^‒6, ‒0.106z^‒8, 0.064z^‒10 (16)

                                                                                                                                    

Equation (16) shows the transfer function of a physically realizable low-pass FIR filter. Figure 1 shows the circuit implementation of this filter.