chengxu
小波变换进行变速箱故障诊断
首先将s1和s2信号时域谱（s1代表维修前的振动信号，s2代表维修后的振动信号），画出来进行对比，会发现s1信号的振动幅值会比较高；然后对s1和s2分别进行小波分解并画出各层的时域谱，只对比s1和s2分后的最高时域谱，s2更加的具有周期性，且幅值较小，s2周期性不够明显，且振动过于密集。分析完时域谱后开始分析频谱，首先对原信号s1和s2fft转换，得到频谱图，如图，然后再对小波分析后高频部分进行fft转换，做出频谱图。对比后我们发现，在f 2375HZ出（也就是齿轮的啮合频率），s1信号频谱幅值会有急剧的升高，然后，其谐波的幅值也会增大。所以可以推断出该维修前的增速箱的齿轮可能出现问题。(Gearbox fault diagnosis based on wavelet transform s1 and s2 first time domain signal spectrum (s1 behalf of the vibration signal before maintenance, vibration signal s2 representative of repaired), drawn comparison, you will find a signal s1 vibration amplitude will be higher then s1 and s2 respectively wavelet decomposition and draw the layers of time-domain spectrum to only a maximum time domain spectroscopy s1 and s2 points after, s2 is more cyclical and smaller magnitude, s2 cyclical enough obviously, and the vibration is too dense. End domain spectral analysis after the beginning of the spectrum analysis, the first of the original signal s1 and s2fft conversion to give the spectrum, as shown, and then after the high frequency part of wavelet analysis fft transformation to make spectrum. We found that after the comparison, at f 2375HZ out (i.e. the gear mesh frequency), there will be a sharp increase in the amplitude spectrum of the signal s1, and then, the amplitude of the ha)
- 2017-01-17 19:18:02下载
- 积分：1
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产生右图所示图像f1(m,n)，其中图像大小为256×256，中间亮条为128
×32，暗处=0，亮处=100。对其进行FFT：
① 同屏显示原图f1(m,n)和FFT(f1)的幅度谱图；
② 若令f2(m,n)=（-1）m+n f1(m,n)，重复以上过程，比较二者幅度
谱的异同，简述理由；
③ 若将f2(m,n)顺时针旋转90 度得到f3(m,n)，试显示FFT(f3)的幅
度谱，并与FFT(f2)的幅度谱进行比较；
④ 若将f1(m,n) 顺时针旋转90 度得到f4(m,n)，令f5(m,n)=f1(m,n)+f4(m,n)，试显
示FFT(f5)的幅度谱，并指出其与FFT(f1)和FFT(f4)的关系；
⑤ 若令f6(m,n)=f2(m,n)+f3(m,n)，试显示FFT(f6)的幅度谱，并指出其与FFT(f2)和
FFT(f3)的关系，比较FFT(f6)和FFT(f5)的幅度谱。(Generating an image f1 (m, n) shown in the figure, wherein the image size is 256 256, the intermediate light bar 128 32 0 = dark, bright Department 100. Its FFT: ① screen display picture f1 (m, n) and the FFT (f1) of the amplitude spectrum ② If so f2 (m, n) = (-1) m+n f1 (m, n), repeat The above process, comparing the amplitude spectrum of the similarities and differences between the two, brief reasons ③ If f2 (m, n) 90 degrees clockwise to get f3 (m, n), try to display FFT (f3) the amplitude spectrum and with the FFT (f2) comparing the amplitude spectrum ④ If f1 (m, n) obtained by 90 degrees clockwise f4 (m, n), so f5 (m, n) = f1 (m, n)+f4 (m, n ), try to display FFT (f5) amplitude spectrum, and pointed out its relationship with the FFT (f1) and FFT (f4) of ⑤ If so f6 (m, n) = f2 (m, n)+f3 (m, n) and try to display the FFT (f6) amplitude spectrum, and pointed out its relationship with the FFT (f2) and FFT (f3), comparing FFT (f6) and FFT (f5) amplitude spectrum.)
- 2013-11-26 16:24:18下载
- 积分：1
