Normalized frequency π radsample magnitude db single

Normalized Frequency ( ×π rad/sample)
Magnitude (dB)
Single-Stage Decimator Filter Response
Figure 26: Single-Stage Decimator Filter Response. 4-(a)
(b) Use the method discussed in class to implement a 2-stage decimator. Set M 1 = 10
and M 2 = 10 for the corresponding stages. Calculate the number of multiplica-
tions per second needed to implement this filter. Comment on the efficiency of
this method versus the single stage implementation.

ECE 513 - Homework 9 Solution 24
[N_H1,Fo1,Ao1,w1] = firpmord([Fp_1 Fstop_1], [1 0], [delta_p1 delta_s1], F0);
h1 = firpm(N_H1,Fo1,Ao1,w1);
% design stage 2 filter
[N_H2,Fo2,Ao2,w2] = firpmord([Fp_2 Fstop_2], [1 0], [delta_p2 delta_s2], F1);
h2 = firpm(N_H2,Fo2,Ao2,w2);
numSamp1 = F1; % number of samples processed at first stage per second
numSamp2 = F1/M2; % number of samples processed at second stage per second
numMult1 = (N_H1+1)*numSamp1; % number of mult per second at first stage
numMult2 = (N_H2+1)*numSamp2; % number of mult per second at second stage
fprintf(1, ’\n---------------------2 Stage Implementation---------------\n’)
fprintf(1, ’Number of filter coefficients for first stage: %d\n’, N_H1);
fprintf(1, ’Number of filter coefficients for second stage: %d\n’, N_H2);
fprintf(1, ’Number of samples per second processed at output ...
of first stage decimation of %d: %d\n’, M1, numSamp1);
fprintf(1, ’Number of multiplications per second at first stage: %d\n’, numMult1);
fprintf(1, ’Number of samples per second processed at output ...
of second stage decimation of %d: %d\n’, M2, numSamp2);
fprintf(1, ’Number of multiplications per second at second stage: %d\n’, numMult2);
fprintf(1, ’Total number of multiplications per second: %d\n’, numMult1+numMult2);
fprintf(1, ’---------------------------------------------------------------\n’)
% check the overall response of the filters
figure(2)
h2_up = upsample(h2, M1);
h_casc = conv(h1, h2_up);
freqz(h_casc, 1)
title(’Two-Stage Decimator Filter Response’)
---------------------2 Stage Implementation---------------
Number of filter coefficients for first stage: 22
Number of filter coefficients for second stage: 404
Number of samples per second processed at output of first stage decimation of 10: 1000
Number of multiplications per second at first stage: 23000
Number of samples per second processed at output of second stage decimation of 10: 100
Number of multiplications per second at second stage: 40500
Total number of multiplications per second: 63500
---------------------------------------------------------------
Figure 27 show Two-Stage Decimator Filter Response.

ECE 513 - Homework 9 Solution 25
0 0.2 0.4 0.6 0.8 1
-4
-3
-2
-1
0
x 10
4
Normalized Frequency ( ×π rad/sample)
Phase (degrees)
0 0.2 0.4 0.6 0.8 1
-200
-150
-100
-50
0
Normalized Frequency ( ×π rad/sample)
Magnitude (dB)
Two-Stage Decimator Filter Response
Figure 27: Two-Stage Decimator Filter Response. 4-(b)