我想知道下面发布的 spectrogram 是否是给定非平稳信号的真实表示 .
如果它是真实的表示，我对图中的特定功能有很多疑问......
对于水平轴上的0 - > .25，为什么它会显示最高频率的信号分量？我假设，鉴于第一次持续时间 t1 ，我应该只看到信号 x1 的频率 . 此外，给定第二个持续时间 t2 ，我应该只看到信号 x2 的频率，依此类推 . 但是，这不是我在下面发布的 spectrogram .
你能解释为什么我们在频谱图中看到这些功能吗？
Spectrogram With equations
Code ：
% Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 1; % seconds
t = (0:dt:StopTime-dt); % seconds
t1 = (0:dt:.25);
t2 = (.25:dt:.50);
t3 = (.5:dt:.75);
t4 = (.75:dt:1);
x1 = (10)*sin(2*pi*10*t1);
x2 = (10)*sin(2*pi*20*t2) + x1;
x3 = (10)*sin(2*pi*30*t3) + x2;
x4 = (10)*sin(2*pi*40*t4) + x3;
NFFT = 2 ^ nextpow2(length(t)); % Next power of 2 from length of y
Y = fft(x4, NFFT);
f = Fs / 2 * linspace(0, 1, NFFT/2 + 1);
%{
figure;
plot(f(1:200), 2 * abs( Y( 1:200) ) );
%}
T = 0:.01:1;
spectrogram(x4,10,9,NFFT);
ylabel('Frequency');
axis(get(gcf,'children'), [0, 1, 1, 50]);
Update_1 ：当我尝试建议的答案时，我收到了以下内容 .
??? Out of memory. Type HELP MEMORY for your options.
Error in ==> spectrogram at 168
y = y(1:length(f),:);
Error in ==> stft_1 at 36
spectrogram(x,10,9,NFFT);
使用的代码：
% Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 1; % seconds
t = (0:dt:StopTime-dt); % seconds
%get a full-length example of each signal component
x1 = (10)*sin(2*pi*10*t);
x2 = (10)*sin(2*pi*20*t);
x3 = (10)*sin(2*pi*30*t);
x4 = (10)*sin(2*pi*40*t);
%construct a composite signal
x = zeros(size(t));
I = find((t >= t1(1)) & (t <= t1(end)));
x(I) = x1(I);
I = find((t >= t2(1)) & (t <= t2(end)));
x(I) = x2(I);
I = find((t >= t3(1)) & (t <= t3(end)));
x(I) = x3(I);
I = find((t >= t4(1)) & (t <= t4(end)));
x(I) = x4(I);
NFFT = 2 ^ nextpow2(length(t)); % Next power of 2 from length of y
Y = fft(x, NFFT);
f = Fs / 2 * linspace(0, 1, NFFT/2 + 1);
%{
figure;
plot(f(1:200), 2 * abs( Y( 1:200) ) );
%}
T = 0:.01:1;
spectrogram(x,10,9,NFFT);
ylabel('Frequency');
axis(get(gcf,'children'), [0, 1, 1, 50]);
Update_2
% Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 1; % seconds
t = (0:dt:StopTime-dt); % seconds
t1 = ( 0:dt:.25);
t2 = (.25:dt:.50);
t3 = (.5:dt:.75);
t4 = (.75:dt:1);
%get a full-length example of each signal component
x1 = (10)*sin(2*pi*100*t);
x2 = (10)*sin(2*pi*200*t);
x3 = (10)*sin(2*pi*300*t);
x4 = (10)*sin(2*pi*400*t);
%construct a composite signal
x = zeros(size(t));
I = find((t >= t1(1)) & (t <= t1(end)));
x(I) = x1(I);
I = find((t >= t2(1)) & (t <= t2(end)));
x(I) = x2(I);
I = find((t >= t3(1)) & (t <= t3(end)));
x(I) = x3(I);
I = find((t >= t4(1)) & (t <= t4(end)));
x(I) = x4(I);
NFFT = 2 ^ nextpow2(length(t)); % Next power of 2 from length of y
Y = fft(x, NFFT);
f = Fs / 2 * linspace(0, 1, NFFT/2 + 1);
%{
figure;
plot(f(1:200), 2 * abs( Y( 1:200) ) );
%}
T = 0:.001:1;
spectrogram(x,10,9);
ylabel('Frequency');
axis(get(gcf,'children'), [0, 1, 1, 100]);
Dpectrogram_2 ：
1 回答
我不认为你正在策划你认为你正在策划的事情 . 您应该在时域中绘制信号，以确保它看起来像您期望的那样......
plot(x4). 我认为你认为你的信号是x1后跟x2后跟x3后跟x4 . 但是，根据您显示的Matlab代码，情况并非如此 .相反，你的信号是x1 x2 x3 x4的总和 . 因此，由于信号的瞬时启动，您应该看到10,20,30,40 Hz的信号分量以及其他分量 .
要获得您想要的信号，您应该执行以下操作：
然后，您可以在时域（
plot(x)）中绘制新信号x，以确保它是您想要的 . 最后，您可以执行spectrogram.请注意，您将在每个时间段（t1和t2之间，t2和t3之间，然后t3和t4之间）的过渡处看到频谱图中的伪像 . 信号伪像将反映在不同信号频率之间的不连续转换期间信号复杂 .