你给的代码运行错误,提示矩阵维度必须一致。 出错 Untitled13 (line 23) echo = x2_watermarked - x1;
时间: 2024-02-26 10:58:54 浏览: 96
非常抱歉,我的回答有误。由于矩阵维度不一致,导致代码出现了错误。以下是一个修正过的代码,可以通过将较短的音频信号插值为与较长的音频信号维度一致,使得两个音频信号的维度相同,从而进行减法运算。
```matlab
% 导入回声音频和原始音频
[x1, Fs1] = audioread('D:/school/毕业设计/音频回声3.wav');
[x2, Fs2] = audioread('D:/school/毕业设计/1/fastICA--2.wav');
% 比较两个音频信号的长度,将长度较短的信号插值为与长度较长的信号维度一致
if length(x1) > length(x2)
x2 = resample(x2, length(x1), length(x2));
elseif length(x1) < length(x2)
x1 = resample(x1, length(x2), length(x1));
end
% 将音频进行FFT变换
X1 = fft(x1);
X2 = fft(x2);
% 计算音频中嵌入的水印信息
watermark = 'HELLO WORLD';
watermark_bits = reshape(dec2bin(watermark, 8).' - '0', 1, []);
N = length(watermark_bits);
% 将水印信息嵌入到回声音频中
x2_watermarked = x2;
x2_watermarked(watermark_bits == 1) = x2_watermarked(watermark_bits == 1) + 0.1;
x2_watermarked(watermark_bits == 0) = x2_watermarked(watermark_bits == 0) - 0.1;
% 将原始音频和带有水印的回声音频进行相减,得到回声信号
echo = x2_watermarked - x1;
% 对回声信号进行时域和频域分析,分别得到回声信号的时延和衰减系数
[~, I] = max(abs(echo));
tau = I - 1;
H = X2 ./ X1;
h = ifft(H);
% 利用水印信息的受损抹零,从而达到消除回声的目的
echo_cleaned = echo;
echo_cleaned(watermark_bits == 1) = echo_cleaned(watermark_bits == 1) - 0.1;
echo_cleaned(watermark_bits == 0) = echo_cleaned(watermark_bits == 0) + 0.1;
echo_cleaned(tau + (1:N)) = echo_cleaned(tau + (1:N)) + 0.1 * (watermark_bits == 0) - 0.1 * (watermark_bits == 1);
% 重构带有水印的回声音频
x2_watermarked_cleaned = x1 + echo_cleaned;
% 输出重构的音频
audiowrite('echo_removed_audio.wav', x2_watermarked_cleaned, Fs2);
```
在修正后的代码中,我们比较两个音频信号的长度,将长度较短的信号插值为与长度较长的信号维度一致。这样,两个音频信号的维度就一致了,可以进行减法运算。
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