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adnandzebic edited untitled.tex
over 9 years ago
Commit id: 4c48ea1e867a9cd8af958e20e4f2577ae24868e5
deletions | additions
diff --git a/untitled.tex b/untitled.tex
index 2e2099f..b050e2b 100644
--- a/untitled.tex
+++ b/untitled.tex
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\textit{Oh, an empty article!}
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\begin{lstlisting}[language=MATLAB]
Wc = 0.2 * pi; % passing frequency
L = 51; % length of filter (inversely correlated with bandwidth)
n = 0:L-1;
ww = 0:pi/1000:pi;
bb = cos(Wc * n); % coefficients (extracted from given h[n])
HH = freqz(bb, 1 ,ww); % frequency response
subplot(211)
plot(ww, abs(HH))
title('Frequency Response')
xlabel('Frequency')
ylabel('Amplitude')
max_value = max(abs(HH)); % get the highest peak value of the frequency response
bb2 = 1 / max_value * bb; % normalize the coefficients
HH2 = freqz(bb2,1,ww);
subplot(212)
plot(ww, abs(HH2))
title('Normalized Frequency Response')
xlabel('Frequency')
ylabel('Amplitude')
passband = find(abs(HH2) >= 0.707); % fetches the passing frequencies
lower_freq = ww(passband(1)); % gets the lower 3dB freq value
upper_freq = ww(passband(length(passband))); % upper 3dB freq value
BW = abs(ww(passband(1)) - ww(passband(length(passband)))); % bandwidth
samp_rate = 8000; % 8000Hz
lower_analog = lower_freq * 8000 / (2 * pi); % 736 Hz
upper_analog = upper_freq * 8000 / (2 * pi); % 868 Hz
\end{lstlisting}.