If you would like to use your own algorithm, that's great. Otherwise, you might try coding an algorithm that works something like this:
- For this discussion, "power of two" means both +2^k and –2^k where k is some integer.
- Assuming the number is not a power of two, start with the powers of two just smaller and just larger than the input number. For example, if the input is 56, consider 32 and 64.
- Choose whichever number is closer. In this example, choose 64.
- Subtract that power of two and repeat the procedure with the new number until the remainder is reduced to zero. For example, choose 56 – 64 = –8 so now use –8.
a) [15 pts] Write the described function in matlab.
b) [5 pts] Assuming your function is called "numppterms", run the following bit of matlab code (a few points need fixing), and report the Total Sum for all numbers 0.5 – 100.00 .
StepSize = 0.5;
NumTermsArrayPos = zeros(1, 100/StepSize); % small speedup if init first
for k = StepSize : StepSize : 100,
NumTermsArrayPos(k/StepSize) = numppterms(k);
end
fprintf('Total sum for +0.5 - +100.00 = %i\n', sum(NumTermsArrayPos));
figure(1); clf;
plot(StepSize:StepSize:100, NumTermsArrayPos, 'x');
axis([0 101 0 1.1* max(NumTermsArrayPos)]);
xlabel('Input number');
ylabel('Number of partial product terms');
coeff = [13 41 85 298 85 41 13];Assume the coefficients cannot be scaled larger, but they can be scaled smaller (up to 50% smaller) and the gain change can be compensated elsewhere. This implementation works with integers only, so round() scaled coefficients.
a) [5 pts] How many partial products are necessary to implement the FIR filter with the given coefficients?
b) [5 pts] See if a scaling for the coefficients exists such that the filter can be built with fewer partial products. Find the scaling that yields the minimum number of partial products.
c) [5 pts] Turn in a plot of the number of required partial products vs. the scaling factor. The plot should look something like the bogus results this matlab code generates.
figure(1); clf;
plot(0.5:0.001:1.0, round(5* rand(1,501)+1), 'x');
axis([0.45 1.05 0 6.5]);
xlabel('Scaling factor');
ylabel('Number of partial product terms');
title('EEC 281, Hwk/proj 2, Problem 3, Plot of bogus results');
d) [15 pts] Draw dot diagrams showing how the partial products would be
added (include sign extension) for the optimized coefficients you
found in (b), using the FIR architecture shown below and an 8-bit
2's complement input word. Use 4:2, 3:2, and half adders as necessary
and no need to design the final stage carry-propagate adder.
- Passband below 5.0 MHz: no more than 4dB ripple from minimum to maximum levels
- Passband below 7.0 MHz: no more than 4dB attenuation below gain level at DC
- Stopband above 10.0 MHz: at least 24dB attenuation below gain level at DC
- Stopband above 15.0 MHz: at least 28dB attenuation below gain level at DC
- a) [10 pts]
Write a matlab function lpfirstats(H) that takes a
frequency response vector H (or a vector of filter coefficients)
as an input and returns the four critical values listed above
(passband ripple, etc.).
- b) [10 pts] Write matlab code which repeatedly calls lpfirstats(H) and finds the smallest area filter. There is no need to write a sophisticated optimization algorithm, just something reasonable that does more than simple coefficient scaling. For example, making small perturbations to the frequency and amplitude values that remez() uses such as using 0.01 and other small values instead of 0.00 in the stopband.
It may be helpful to use the following matlab code. Remember that matlab vectors start at index=1 so H(1) is the magnitude at frequency=0.coeffs1 = remez(numtaps-1, freqs, amps); coeffs2 = coeffs1*scale; coeffs = round(coeffs2); [H,W] = freqz(coeffs); H_norm = abs(H) ./ abs(H(1)); [ripple, minpass, maxstoplo, maxstophi] = lpfirstats(H_norm);-
Assume area is:
Total_num_partial_products + 2*Num_filter_taps- c) [20 pts] i) List your smallest-area coefficients in your paper submission and in a matlab-readable vector in your electronic submission.
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ii) List the number of taps,
number of required partial products,
area estimate, and the
attained values for the four filter criteria.
Results from a recent year for a different filter show there is lots of room for optimization: 109 area, 31 taps, 47 PPs 109 area, 31 taps, 47 PPs 113 area, 33 taps, 47 PPs 114 area, 33 taps, 48 PPs 123 area, 33 taps, 57 PPs 128 area, 34 taps, 60 PPs 182 area, 55 taps, 72 PPs 221 area, 59 taps, 103 PPs 221 area, 59 taps, 103 PPs 250 area, 61 taps, 128 PPs- d) [10 pts] i) Include a paper copy of a plot by: plot_one_lpfir.m that shows the correct filter specifications.
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ii) Show a stem() plot of the filter's impulse response.
- b) [10 pts] Write matlab code which repeatedly calls lpfirstats(H) and finds the smallest area filter. There is no need to write a sophisticated optimization algorithm, just something reasonable that does more than simple coefficient scaling. For example, making small perturbations to the frequency and amplitude values that remez() uses such as using 0.01 and other small values instead of 0.00 in the stopband.