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+++ "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex1/ex1.log"
@@ -0,0 +1,20 @@
+Single button test
+|length |fft-result |fft-time |grt-result |grt-time |time ratio |
+| ---- | ---- | ---- | ---- | ---- | ---- |
+|25695 | 0 |0.002852 | 0 |0.001258 | 2.267032 |
+|14988 | 1 |0.000599 | 1 |0.000753 | 0.796466 |
+|19985 | 2 |0.004619 | 2 |0.001288 | 3.586335 |
+|19271 | 3 |0.001731 | 3 |0.000950 | 1.822278 |
+|19985 | 4 |0.001775 | 4 |0.001007 | 1.762288 |
+|23018 | 5 |0.002294 | 5 |0.001170 | 1.959672 |
+|19271 | 6 |0.001924 | 6 |0.000983 | 1.956478 |
+|23018 | 7 |0.000978 | 7 |0.001117 | 0.875862 |
+|18736 | 8 |0.000715 | 8 |0.001087 | 0.657708 |
+|22483 | 9 |0.002365 | 9 |0.001126 | 2.100906 |
+|23019 | * |0.002585 | * |0.001172 | 2.205220 |
+|24625 | # |0.002101 | # |0.001710 | 1.228426 |
+telephone number
+FFT time_cost: 0.034963 s
+ans: ---8-22-1177-5-6699-0-99-99
+Goertzel time_cost: 0.021653 s
+ans: 1--8-22-1177-5-6699-0099-99
diff --git "a/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex1/main.m" "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex1/main.m"
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index 000000000..3de40e2ac
--- /dev/null
+++ "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex1/main.m"
@@ -0,0 +1,125 @@
+% 主函数
+function main
+ global keys; % 按键表
+ global freqs; % 频率表
+ keys = ['1', '2', '3', 'A';
+ '4', '5', '6', 'B';
+ '7', '8', '9', 'C';
+ '*', '0', '#', 'D';
+ '-', '-', '-', '-'];
+ freqs = [697, 770, 852, 941, 1209, 1336, 1477, 1633];
+
+ % 输出到文件 ex1.log
+ fout = fopen('./ex1.log', 'w');
+
+ % 单按键测试
+ fprintf(fout, "Single button test\n");
+ fprintf(fout, "|length |fft-result |fft-time |grt-result |grt-time |time ratio |\n");
+ fprintf(fout, "| ---- | ---- | ---- | ---- | ---- | ---- |\n");
+ for i = 0:11
+ [audio, fs] = audioread(['./data/',num2str(i),'.wav']);
+ audio = sum(audio, 2) / 2; % 合并左右声道
+ % audio = audio(1:4000);
+ fprintf(fout, "|%5d ", length(audio));
+ tic
+ ans = fft_test(audio, fs);
+ fft_time = toc;
+ fprintf(fout, "| %c |%f ", ans, fft_time);
+ tic
+ ans = goertzel(audio, fs);
+ grt_time = toc;
+ fprintf(fout, "| %c |%f ", ans, grt_time);
+ fprintf(fout, "| %f |", fft_time/grt_time);
+ fprintf(fout, "\n");
+ end
+
+ % 一组电话号码测试
+ [audio, fs] = audioread('./data/18217569099.wav');
+ patch_size = fs/3;
+ audio = sum(audio, 2) / 2;
+ total_len = size(audio, 1);
+
+ fprintf(fout, "telephone number\n");
+ fft_result = [];
+ tic
+ for i = 1:patch_size:total_len-patch_size
+ fft_result = [fft_result, fft_test(audio(i:i+patch_size), fs)];
+ end;
+ fprintf(fout, "FFT time_cost: %f s\n", toc);
+ fprintf(fout, "ans: %s\n", fft_result);
+
+ grt_result = [];
+ tic
+ for i = 1:patch_size:total_len-patch_size
+ grt_result = [grt_result, goertzel(audio(i:i+patch_size), fs)];
+ end;
+ fprintf(fout, "Goertzel time_cost: %f s\n", toc);
+ fprintf(fout, "ans: %s\n", grt_result);
+
+ fclose(fout);
+end
+
+% FFT Algorithm
+function ans = fft_test(audio, fs)
+ global keys;
+ global freqs;
+
+ y = abs(fft(audio));
+ len = length(y);
+ % 数组下标是数字频率，需要转换成模拟频率
+ l1 = round(650 / fs * len);
+ h1 = round(1000 / fs * len);
+ l2 = round(1150 / fs * len);
+ h2 = round(1700 / fs * len);
+ [px, col] = max(y(l2:h2));
+ [py, row] = max(y(l1:h1));
+ col = col + l2 - 1;
+ row = row + l1 - 1;
+ fx = col / len * fs;
+ fy = row / len * fs;
+
+ ans = '-';
+ if (px > 10 || py > 10) % 过滤噪声
+ [t, row] = min(abs(freqs(1:4) - fy));
+ [t, col] = min(abs(freqs(5:8) - fx));
+ ans = keys(row, col);
+ end;
+end
+
+% Goertzel Algorithm
+function ans = goertzel(audio, fs)
+ global keys;
+ global freqs;
+
+ t_freqs = freqs;
+ N = length(audio);
+ x = audio;
+ v = zeros(1,N+2);
+ y = zeros(1, 8);
+ t_freqs = round(t_freqs * N / fs);
+
+ % 差分方程
+ % v_k[n] = 2cos(2k\pi/N)v_k[n-1] - v_k[n-2] + x[n]
+ % y_k[n]^2 = v_k[n] - W_N^k*v_k[n-1]
+ for k = 1 : length(t_freqs)
+ factor = 2 * cos(2 * pi * t_freqs(k) / N);
+ for i = 1 : N
+ v(i+2) = factor*v(i+1) - v(i) + x(i); % v_k[n]
+ end
+ y(k) = v(N+2)^2 + v(N+1)^2 - factor*v(N+2)*v(N+1); % |y_k[N]|^2
+ end
+ y = sqrt(abs(y));
+
+ % 过滤噪声
+ [t, col] = max(y(5:8));
+ [t, row] = max(y(1:4));
+ px = y(col+4);
+ py = y(row);
+ trhld = max(px, py)*0.3;
+ if (sum((y-trhld) > 0) > 2)
+ row = 5;
+ end
+
+ ans = keys(row, col);
+end
+
diff --git "a/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex2/main.m" "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex2/main.m"
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+++ "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex2/main.m"
@@ -0,0 +1,98 @@
+function main
+ points = 400;
+ step = 100;
+ M = 1000; % length y
+ time_origin = zeros(1, points);
+ time_circle = zeros(1, points);
+ time_save = zeros(1, points);
+ time_add = zeros(1, points);
+
+ for i = 1 : points
+ N = M + (i-1) * step
+ fprintf("testing N = %d, M = %d\n", N, M);
+ x = rand([1, N]);
+ y = rand([1, M]);
+ tic;
+ ans1 = conv_origin(x, y);
+ time_origin(i) = toc;
+ tic;
+ ans2 = conv_circle(x, y);
+ time_circle(i) = toc;
+ tic;
+ ans3 = overlap_save(x, y);
+ time_save(i) = toc;
+ tic;
+ ans4 = overlap_add(x, y);
+ time_add(i) = toc;
+ end
+ x = M:step:M + step*(points-1);
+
+ plot(x, time_origin, '-', x, time_circle, '-', x, time_save, '-', x, time_add, '-');
+ legend({'origin conv', 'circle conv', 'overlap-save', 'overlap-add'}, 'Location', 'northwest');
+ xlabel('M');
+ ylabel('time_cost/s');
+end
+
+function ans = conv_origin(x, y)
+ N = length(x);
+ M = length(y);
+ L = M + N - 1;
+ ans = zeros(1, L);
+
+ for n = 1 : L
+ m0 = max(1, n + 1 - N);
+ m1 = min(M, n);
+ for m = m0 : m1
+ ans(n) = ans(n) + x(n - m + 1) * y(m);
+ end
+ end
+end
+
+function ans = conv_circle(x, y)
+ N = length(x);
+ M = length(y);
+ x = [x, zeros(1, M-1)]; % zero padding
+ y = [y, zeros(1, N-1)]; % zero padding
+ ans = ifft(fft(x) .* fft(y));
+end
+
+function y = overlap_add(x, h)
+ N = length(x);
+ M = length(h);
+ L = M + 1; % block size
+ len = L + M - 1;
+ fft_h = fft(h, len);
+ num = ceil(N / L); % split x into $num blocks
+ y = zeros(1, (num + 1) * L);
+ x = [x, zeros(1, (num + 1) * L - N)]; % zero padding to (num + 1) * L
+ ol = zeros(1, M - 1); % overlap-add
+ for i = 1:L:num * L + 1
+ y_k = ifft(fft(x(i:i + L - 1), len) .* fft_h); % calc circle conv
+ y_k(1:M - 1) = y_k(1:M - 1) + ol(1:M - 1); % overlap-add
+ ol(1:M - 1) = y_k(L + 1:len); % refresh
+ y(i:i + L - 1) = y_k(1:L); % save y
+ end
+ y = y(1:N + M - 1);
+end
+
+% conv by overlap-save
+function y = overlap_save(x, h)
+ % assume h is much shorter than x
+ N = length(x);
+ M = length(h);
+ L = M + 1; % block size
+ len = L + M - 1;
+ fft_h = fft(h, len);
+ num = ceil(N / L); % split x into $num blocks
+ y = zeros(1, (num + 1) * L);
+ x = [x, zeros(1, (num + 1) * L - N)]; % zero padding
+ ol = zeros(1, M - 1); % overlap-save
+ for i = 1:L:num * L + 1
+ subx = [ol, x(i:i + L - 1)];
+ ol = subx(L + 1 : L + M - 1); % refresh overlap-save
+ y_k = ifft(fft(subx, len) .* fft_h); % calc circle conv
+ y(i:i + L - 1) = y_k(M:M + L - 1); % save y
+ end
+ y = y(1:N + M - 1);
+end
+
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--- /dev/null
+++ "b/\345\244\247\344\270\211\344\270\212/\344\277\241\345\217\267\345\244\204\347\220\206\345\216\237\347\220\206/hw/matlab/2019/ex3/main.m"
@@ -0,0 +1,96 @@
+% 读取数据
+[audio1, fs] = audioread('./data/1.wav');
+[audio2, fs] = audioread('./data/2.wav');
+[audio3, fs] = audioread('./data/3.wav');
+fprintf("read freq: %dHz\n", fs);
+total_len = 150000; % 取前150000个点
+audio1 = audio1(1 : total_len);
+audio2 = audio2(1 : total_len);
+audio3 = audio3(1 : total_len);
+
+% 原始的时域信号
+t = 0 : total_len - 1;
+figure(1);
+subplot(311);
+stem(t, audio1, '.');
+subplot(312);
+stem(t, audio2, '.');
+subplot(313);
+stem(t, audio3, '.');
+
+% 原始的频域信号
+figure(2);
+subplot(311);
+stem(t, abs(fft(audio1)), '.');
+subplot(312);
+stem(t, abs(fft(audio2)), '.');
+subplot(313);
+stem(t, abs(fft(audio3)), '.');
+
+% 1/3抽取和三倍上采样的下标
+sample_idx = 1 : 3 : total_len - 1;
+
+% 在时域进行1/3抽取
+audio1 = audio1(sample_idx);
+audio2 = audio2(sample_idx);
+audio3 = audio3(sample_idx);
+
+% 时域上进行上采样
+a1 = zeros(1, total_len);
+a2 = zeros(1, total_len);
+a3 = zeros(1, total_len);
+a1(sample_idx) = audio1;
+a2(sample_idx) = audio2;
+a3(sample_idx) = audio3;
+
+% 变换到频域
+f1 = fft(a1);
+f2 = fft(a2);
+f3 = fft(a3);
+
+% f是频域编码信号
+len = total_len / 6;
+f = zeros(1, total_len);
+f(1:len*3) = [f1(1:len), f2(1:len), f3(1:len)];
+f(len*3+1:len*6) = [f3(len+1:len*2), f2(len+1:len*2), f1(len+1:len*2)];
+
+% y是编码完成后的时域信号
+y = ifft(f);
+t = 0:total_len-1;
+figure(3);
+subplot(211);
+stem(t, abs(f), '.');
+subplot(212);
+stem(t, real(y), '.');
+
+% 频域解码
+f = fft(y);
+f1 = [f(1:len), f(len*5+1:len*6)];
+f2 = [f(len+1:len*2), f(len*4+1:len*5)];
+f3 = [f(len*2+1:len*3), f(len*3+1:len*4)];
+t = 0:2*len-1;
+figure(4);
+subplot(311);
+stem(t, abs(f1), '.');
+subplot(312);
+stem(t, abs(f2), '.');
+subplot(313);
+stem(t, abs(f3), '.');
+
+% 恢复时域信号
+audio1 = real(ifft(f1));
+audio2 = real(ifft(f2));
+audio3 = real(ifft(f3));
+a1 = zeros(1, total_len);
+a2 = zeros(1, total_len);
+a3 = zeros(1, total_len);
+a1(sample_idx) = audio1;
+a2(sample_idx) = audio2;
+a3(sample_idx) = audio3;
+figure(5);
+subplot(311);
+stem(1:total_len, a1, '.');
+subplot(312);
+stem(1:total_len, a2, '.');
+subplot(313);
+stem(1:total_len, a3, '.');
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@@ -0,0 +1,135 @@
+## 信号处理原理实验报告
+
+> 陈海天 计73 2016010106
+
+### Ex1 双音频按键识别
+
+##### 1.1 算法与代码
+
+使用了两种不同的方法来识别按键音：
+
+1. 手动实现goertzel算法，在八个频率中找出最高的两个峰值作为预测结果（按一定条件过滤噪声）
+2. 使用matlab内置的fft函数，找出低频高频的两个峰值查表作为预测结果
+
+在main.m中写了两个函数`fft_test` `goertzel`分别实现上面两种方法，运行main.m即可得到结果。
+
+##### 1.2 实验结果
+
+1. 单按键测试
+
+用自己手机录了键盘上的12个按键音，放在data目录下（11--*，12--#），采样频率fs=48000，样本的长度在半秒左右。N取4000的结果如下：（其中time_ratio为fft_time/grt_time）
+
+|length |fft-result |fft-time |grt-result |grt-time |time ratio |
+| ---- | ---- | ---- | ---- | ---- | ---- |
+| 4000 | 0 |0.000207 | 0 |0.000319 | 0.649937 |
+| 4000 | 1 |0.000170 | 1 |0.000238 | 0.717172 |
+| 4000 | 2 |0.000146 | 2 |0.000237 | 0.617138 |
+| 4000 | 3 |0.000149 | 3 |0.000234 | 0.633532 |
+| 4000 | 4 |0.000151 | 4 |0.000237 | 0.640592 |
+| 4000 | 5 |0.000151 | 5 |0.000237 | 0.637785 |
+| 4000 | 6 |0.000146 | 6 |0.000293 | 0.498120 |
+| 4000 | 7 |0.000150 | 7 |0.000305 | 0.491148 |
+| 4000 | 8 |0.000148 | 8 |0.000236 | 0.628450 |
+| 4000 | 9 |0.000157 | 9 |0.000235 | 0.669365 |
+| 4000 | * |0.000147 | * |0.000236 | 0.620237 |
+| 4000 | # |0.000152 | # |0.000236 | 0.644161 |
+
+N取样本长度时的结果如下：
+|length |fft-result |fft-time |grt-result |grt-time |time ratio |
+| ---- | ---- | ---- | ---- | ---- | ---- |
+|25695 | 0 |0.002802 | 0 |0.001537 | 1.823246 |
+|14988 | 1 |0.000697 | 1 |0.000871 | 0.799656 |
+|19985 | 2 |0.001747 | 2 |0.001007 | 1.735525 |
+|19271 | 3 |0.001737 | 3 |0.001216 | 1.429006 |
+|19985 | 4 |0.001487 | 4 |0.000947 | 1.570976 |
+|23018 | 5 |0.001132 | 5 |0.001186 | 0.955048 |
+|19271 | 6 |0.001650 | 6 |0.000914 | 1.803937 |
+|23018 | 7 |0.001307 | 7 |0.001161 | 1.125560 |
+|18736 | 8 |0.000736 | 8 |0.000890 | 0.826947 |
+|22483 | 9 |0.002732 | 9 |0.001115 | 2.449646 |
+|23019 | * |0.002042 | * |0.001083 | 1.885400 |
+|24625 | # |0.001880 | # |0.001597 | 1.177458 |
+
+可见goertzel的准确性是没问题的。从性能上来说，当N大时，goertzel算法速度更快。这与理论结果相符，goertzel算法计算K个点的时间复杂度为O(KN)，而fft的时间复杂度为O(NlogN)，所以N越大，goertzel更有优势。
+
+2. 长音频测试
+
+录了一段连续的长音频（电话号码），N分别取fs=48000, fs/3=16000, fs/6=8000，可得到结果如下：
+
+```
+'-'表示噪声，噪声筛除方式：
+goertzel: 如果大于[0.3*最大能量]的点超过两个，则认为是噪声
+fft: 如果能量小于一定阈值，则认为是噪声
+
+N = 48000
+FFT time_cost: 0.044336 s
+ans: -82756#99
+Goertzel time_cost: 0.022292 s
+ans: 18--56-99
+
+N = 16000
+FFT time_cost: 0.034408 s
+ans: ---8-22-1177-5-6699-0-99-99
+Goertzel time_cost: 0.022509 s
+ans: 1--8-22-1177-5-6699-0099-99
+
+N = 8000
+FFT time_cost: 0.035877 s
+ans: ------88---22----11--777--55---66--999---0---99---999-
+Goertzel time_cost: 0.023650 s
+ans: 11----88---22----11--777--55---66#-999--000--99---999#
+```
+
+可见goertzel的准确性非常好，和fft不相上下，而且所用时间基本不变，因为它是线性关系。而fft的时间随batch_size的减小而减小。
+
+
+
+### Ex2 卷积计算方法的性能比较
+
+按照要求用以下四种方式计算卷积：`公式法` `圆卷积` `overlap-add` `overlap-save`
+
+所得的实验结果如下图所示（取了400个点，步长100，h的长度M=1000）：
+
+
+
+可见手搓的公式法求解速度最慢（其实内置函数的公式法是最快的，但是这应该是matlab在底层做了优化）；其次是圆卷积fft；两个overlap方法计算都很快，这是因为它们比较适合这种长短差异较大的序列的卷积求解。
+
+### Ex3 语音信号的频分复用
+
+本题有多种理解方式（感觉题目要求没太说清楚），我的理解方式是用同样长度的序列编码三条语音，并且恢复以后尽可能不失真。在这种理解方式下，有两种做法：
+
+1. 因为我们手机录的素材采样频率基本都在48k左右，而语音的带限是3.4k，所以可以直接在频域上截断三个信号然后拼起来。但是这种方式是有限制的，当采样频率小的时候就会严重失真。
+2. 第二种方式就是现在时域上进行1/3抽取，然后在按照ppt上的上采样--滤波--累加的方式。这样相当于在损失了一些音质，但是适用范围更广。
+
+两种方式都会损失信息，因为想要用一条语音的数据量存下三条语言，肯定是要损失一些东西的。我最后采用的方式是第二种（因为这样也可以按照ppt上介绍的流程来）。
+
+算法流程在代码注释中写的很详细了，大致为：
+
+1. 先将三条语音裁成同样长度，并进行1/3抽取
+2. 进行上采样，并fft得到上采样后的频域信号
+3. 在频域上编码
+4. 转换回时域
+
+解码过程就是上述的逆过程。
+
+#### 实验结果如下所示：
+
+
+
+上图为三条原始音频信号。
+
+
+
+上图为三条原始音频的频谱。
+
+
+
+上图为上采样之后三条音频的频谱，可见在一个fs内出现了三个周期。
+
+
+
+上图为编码之后的频谱与音频信号。可见频谱上的确是三条信号分开了。
+
+
+
+上图为解码之后恢复的音频信号。可见和原来的音频信号完全一致，试听效果也很好。
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