Homework2
姓名:陳裕旻
學號:MA020129
1. Load the ex2_1.csv as attached
(1) Plot the original waveform(the base line adjust to 0.0)
Ans:如圖一是將時基線調整至 0.0 之原始訊號
原始訊號
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
4
4.5
5
圖一:原始訊號
(2) Add the noise ( -30 db) to the original waveform
Ans:
SNR=-30db
5
4
3
2
振幅
1
0
-1
-2
-3
-4
0
0.5
1
1.5
2
2.5
TIME(s)
3
3.5
圖二:加入-30DB 之雜訊
(3)Compute the heart rate( beat/min, mean +- standard deviation) of (2)
Ans:84 下/分鐘，平均值=0.66秒以及標準差=0.008
(4)Using the template technique to compute the heart rate of (2)
Ans:84 下/分鐘
(5)Compare the spectrum between (1) and (2)
Ans:
頻率在低頻時訊號比較好，在高頻時訊號較不好且會產生雜訊。
頻率響應圖
40
35
30
25
20
15
10
5
0
0
10
20
30
40
50
60
Frequency (Hz)
70
80
90
100
圖三:頻率響應圖
(6)Use the Welch method to estimate the power spectrum of (1) ( choice the
optimal window and overlap parameter)
Ans:我是使用 Kaiser Window 與 Harming 由圖三可知從 0HZ to 20HZ 的能量較
強，頻率越高時能量頻譜較低。
-3
2.5
Power Spectrum (Welch Method)
x 10
Power Spectrum
2
1.5
1
0.5
0
0
10
20
30
40
50
60
Frequency (Hz)
70
80
90
100
圖三:Kaiser 視窗
圖四:Harming 視窗
(7)Tell me your finding
Ans:原始訊號加入一雜訊後雜訊也跟著被放大且不易計算出心跳值。程式方面九
十六下是在MATLAB中加入以下這段%x=pl_Signal(o-70:o+71);
rxy=xcorr(pl_Signal,pl_Signal(614:755));
rxy=rxy(i1:(2*i1-1))/max(rxy);
L=[];j=0;
for i=1:141:length(rxy)-141
j=j+1;
rxy1=rxy(i:i+141);
L1=find(rxy1==max(rxy1));
L=[L,L1+i];
end
2. Use sig_noise to generate a 512 points array that contains two closely spaced
sinusoid at 150 and 180 Hz both with an SNR of -20 dB, assumes a sampling rate
of 1 kHz. Use the Welch method and find the spectrum of waveform for segment
length 512 (no overlap), and 128 point overlap.
Ans:
Power Spectrum (Welch Method)
0.01
0.009
0.008
Power Spectrum
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0
0
50
100
150
200
250
300
Frequency (Hz)
圖五
350
400
450
500
圖六