数字图像处理作业(2)
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(a) Load the signal by the command “load TestSig”. You will find two variables t and y. t is the time label with the unit “s”, and y is the corresponding amplitude of the signal. Use plot() command to plot the signal, and show the x (time) and y (amplitude) labels. Tell what the sampling frequency is. Answer:
The code of the program is: >> load TestSig; >> plot(t,y); >> xlabel('time'); >> ylabel('amplitude');
The sampling frequency is 1/0.005s=200Hz
0.5
1
1.52
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-1-0.8-0.6-0.4-0.200.20.40.60.8
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(b) Use the time-domain method to analyze how many frequency components contained in the signal, and what these frequencies are. Answer:
由图时域图可知，三秒内分别是三个不同频率的信号：第一秒内信号经历5个周期，所以频率为5Hz;第二秒内信号经历10个周期，所以频率为10；第三秒内，信号经历15个周期，所以频率为15。综上可知：信号包含3个频率分量，分别是5Hz，10Hz，15Hz。
(c) Use the frequency-domain method to analyze how many frequency components contained in the signal, and what these frequencies are. Please use fft() command to calculate the spectrum. Use plot() command to plot the spectrum, and show the x (frequency) and y (amplitude) labels. (Review the role of sampling frequency)
Answer:
The code of the program is:
>> load TestSig;
>>Fs=200;
>>L=length(y);
>>NFFT = 2^nextpow2(L);
>>Y = fft(y,NFFT)/L;
>>f = Fs/2*linspace(0,1,NFFT/2+1);
>>plot(f,2*abs(Y(1:NFFT/2+1)));
>>title('Single-Sided Amplitude Spectrum of y(t)');
>>xlabel('Frequency (Hz)'); >>ylabel('|Y(f)|');
由图可知，信号包含3个频率分量，频率分别为：5Hz ，10Hz ，15Hz 。 (d) Use the wavelet method to analyze the signal. Figure out how many frequency components contained in the signal, and what these frequencies are. Please use wavedec() command to perform the digital wavelet transform. Use plot() command to plot the wavelet coefficients (detailed and approximate). Answer ：
The code of program is as follows: >> load TestSig
>> [coefs,sgram] = cwt(y,[1:32],'haar','scal')；
0102030
405060708090100
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0.1
0.15
0.20.25
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0.35
Single-Sided Amplitude Spectrum of y(t)
Frequency (Hz)
|Y (f )|
Plot_3D=cwt(y,[1:32],'haar','3Dplot');
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1Analyzed Signal
Scalogram
Percentage of energy for each wavelet coefficient
Time (or Space) b
S c a l e s a
100200
300400
500600
1 3 5
7 911131517192123252729310.005
0.010.0150.020.0250.030.0350.04
可以看出有三个频率分量。 [C,L] = wavedec(y,3,'haar'); A=appcoef(C,L,’haar’); time_a3=[0:0.4:3]; plot(time_a3,A);
傅里叶分析： Fs=25; LG=76;
NFFT = 2^nextpow2(LG); Y = fft(A,NFFT)/LG;
f = Fs/2*linspace(0,1,NFFT/2+1); plot(f,2*abs(Y(1:NF ·FT/2+1)));
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1.52
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