[英]What is the interpretation of this STFT diagram?
This is a STFT diagram, I have got from my signal.这是一个 STFT 图,我从我的信号中得到的。 I have used these command below:
我在下面使用了这些命令:
figure(1)
stft(signals,500000,'Window',kaiser(256,5),'OverlapLength',220,'FFTLength',512);
I have used matlab 2019b.我用过matlab 2019b。
My confusion arises because on the STFT diagram, I am seeing negetive magnitude -20DB.我之所以感到困惑,是因为在 STFT 图上,我看到负震级 -20DB。 Usually on the examples of matlab central I have found all with positive magnitude and made senses to me.
通常在 matlab central 的示例中,我发现所有的都具有正幅度并且对我有意义。 However, with my sample signal, I have confused.
但是,对于我的示例信号,我感到困惑。 the sampling frequency is 500000.
采样频率为 500000。
I have added the我已经添加了
file for usage.文件供使用。
It will be really helpful if some one can guide me to understand what is going on over here.如果有人可以指导我了解这里发生的事情,那将非常有帮助。
In addition I have added the raw time domain signal.此外,我还添加了原始时域信号。
The magnitude is what you sometimes call amplitude in standard FFTs -- but usually given without scaling it logarithmically.幅度就是您有时在标准 FFT 中所说的幅度——但通常给出时没有按对数比例缩放。 In you case it is displayed in dB, ie 10*lg(A).
在您的情况下,它以 dB 显示,即 10*lg(A)。 So, a negative dB-value indicates a very low amplitude (-70dB = 10^(-70/10) = 0,0000001 and -20dB =0,01).
因此,负 dB 值表示非常低的振幅(-70dB = 10^(-70/10) = 0,0000001 和 -20dB =0,01)。 So it is just a matter of scaling.
所以这只是缩放的问题。 How does your original signal look like?
你的原始信号是什么样子的? I guess that it is also very low in amplitude…
我猜它的振幅也很低……
Now to your graph: It seems that you a time-invariant signal and thus, that there is no need for performing an STFT.现在看你的图表:你似乎是一个时不变信号,因此不需要执行 STFT。 Do a FFT and have a look at the amplitudes.
进行 FFT 并查看振幅。 They should be in the same – very small – range.
它们应该在相同的——非常小的——范围内。
EDITED Add to updated post:编辑添加到更新的帖子:
Alright.好的。 You see that your signal is
你看你的信号是
However, the FFT of the whole signals suggests amplitudes up to 16dB.然而,整个信号的 FFT 表明幅度高达 16dB。 This only at the first glance a contradiction.
乍一看,这是矛盾的。 If you sum up the amplidues of the STFT, you will have the same amplitudes... almost because you have a worse frequency resolution because your signal-length is smaller (frequency resolution = 1/T_signal)
如果你总结 STFT 的振幅,你将有相同的振幅......几乎是因为你的频率分辨率更差,因为你的信号长度更小(频率分辨率 = 1/T_signal)
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