Logarithmic amplitude scaling of a signal

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Lorenzo Lellini
Lorenzo Lellini 2023 年 2 月 13 日
回答済み: Jeffrey Clark 2023 年 2 月 14 日
I need to logarithmically scale the amplitude a sinesweep signal from at the starting frequency to at the lastest frequency (it should be 12 dB per octave).
Is this code wrong? Why if I plot the exponential shape and the scaled signal the profiles does not overlap?
duration_sweep = 2; % Seconds
% Sine sweep frequency range
f_start = 100;
f_end = 10000;
%--------------------------------------------
% Settings sweeptone
duration_final_silence_sweep = 0.1; % Secondi
total_duration_sweep = duration_sweep + duration_final_silence_sweep ;
sweep = sweeptone(duration_sweep,duration_final_silence_sweep,Fs,"SweepFrequencyRange",[f_start f_end]);
% Time axis
t_input = (0:numel(sweep)-1);
% Scaling factor
amp = logspace(log10(0.1*10^-3), log10(10), length(sweep));
% amplitude scaling
out = sweep.*amp';
% Plot
figure(1)
plot (out, 'LineWidth', 1.5)
hold on
plot(t_input, amp)
grid minor
  3 件のコメント
Jeffrey Clark
Jeffrey Clark 2023 年 2 月 13 日
@Lorenzo Lellini, you are also using the length of sweep four your amp but sweep includes the silence duration as indicated in sweeptone documentation.
Lorenzo Lellini
Lorenzo Lellini 2023 年 2 月 13 日
ok, thank you! I forgot to specify Fs but I had already considered it...
How can I perform the procuct between amp and the sweeptone in a correct way since I must specify a silence duration in the sweep and vectors must have the same length?

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Jeffrey Clark
Jeffrey Clark 2023 年 2 月 14 日
@Lorenzo Lellini, note that your curve is assuming a full scale sweep of +/- 1 whereas sweeptone produces +/- 0.5 so you need to plot 0.5*amp. And as in my comments above, you need to exclude fix(duration_final_silence_sweep*Fs) bins from the end of sweep.

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