I want to create a rocket shape by specifying a "cloud" of surface points and making MATLAB connect the points into rectangular patches (such that when there are many of these, it creates the illusion of a curved surface). How can I make MATLAB do this? I.e. in what order must I specify the points in order to make MATLAB patch them together correctly?
Here is what I have so far:
The above is using plot3; what I have is a long matrix of the following structure:
[x1 y1 z1
x2 y2 z2
.
.
.]
I would like to know how to make this into a surface. The full matrix is:
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-0.2470 0.0357 0.0130
-0.2470 0.0329 0.0190
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MATLAB plots the points in the same order as the vector that you supply. For example,
plot3([x1 x2 x3], [y1 y2 y3], [z1 z2 z3])
will connect [x1 y1 z1] to [x2 y2 z2], and connect [x2 y2 z2] to [x3 y3 z3]. If you want to create a rocket shape that is composed of several curves, then you will need to use one plot3 command for each curve.
Related
I want to carry out continuous wavelet transform of a signal. I tried to write a script but the script give output of very low resolution scalogram. So I need a high resolution output by changing some scales. The signal data and the script is attached below.
I tried the script
from scipy import signal
import numpy as np
import matplotlib.pyplot as plt
import obspy
w=np.loadtxt('signal')
t = np.arange(0,len(w))
fmin = 1 # Hz
fmax = 50 # Hz
df = 1./(t[-1]-t[0])
print(df)
fmin_samples = int(fmin/df)
fmax_samples = int(fmax/df)
extent = np.arange(1,10)
scalogram = signal.cwt(w, signal.morlet,extent)
fig, ax = plt.subplots(2, 1, sharex=True)
ax[0].plot(t, w)
ax[0].set(ylabel='amplitude')
ax[1].imshow(np.abs(scalogram), origin='lower')
ax[1].axis('tight')
ax[1].set(xlabel='time (s)', ylabel='frequency (Hz)')
plt.show()
signal data
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Using a library named PyWavelets (documentation here) you can do it in a straightforward way
import pywt
import pylab
[phi, psi, x] = pywt.Wavelet('db2').wavefun(level=4)
pylab.plot(x, psi)
pylab.show()
Here is my code:
sigma = 10;
sz = 20;
x = linspace(-sz / 2, sz / 2-1, sz);
gf = exp(-x .^ 2 / (2 * sigma ^ 2));
gf = gf / sum (gf); % normalize
f_filter = cconv(gf,f,length(f));
Basically I am Gaussian filtering original signal f. However, when I look at the filtered signal f_filter, there is a shift comparing the original signal f (See attached figure). I am not sure why this is happening. I would like to only smooth but not shift the orginal signal. Please help. Thanks.
my original signal f is here:
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0.1091
0.0423
0.0326
0.0343
0.0256
0.0231
0.0281
0.0298
0.0229
0.0283
0.0279
0.0270
0.0300
0.0245
0.0360
0.0280
0.0270
0.0232
0.0276
0.0270
0.0237
0.0197
0.0193
0.0172
0.0140
0.0093
0.0244
0.0226
0.0192
0.0145
0.0124
0.0167
0.0182
0.0111
0.0147
0.0081
0.0151
0.0130
0.0113
0.0131
0.0067
0.0028
0.0064
0.0069
0.0082
0.0075
0.0098
-0.0008
0.0037
0.0019
0.0060
0.0057
0.0033
0.0079
0.0122
0.0091
0.0067
-0.0038
0.0033
0.0013
0.0011
0.0034
0.0051
0.0009
-0.0001
-0.0005
0.0098
-0.0003
0.0067
0.0038
0.0106
0.0000
0.0126
0.0134
0.0090
0.0116
0.0083
0.0101
0.0152
0.0010
0.0068
0.0008
0.0053
0.0090
0.0087
0.0085
0.0054
0.0089
0.0077
0.0064
0.0046
0.0058
0.0025
0.0132
0.0088
0.0043
0.0052
0.0087
0.0122
0.0023
0.0066
0.0093
0.0042
0.0042
0.0138
0.0051
-0.0055
-0.0002
0.0048
0.0063
0.0076
0.0016
-0.0005
0.0086
0.0043
-0.0016
0.0100
0.0097
0.0042
0.0092
0.0051
0.0029
0.0044
0.0033
0.0073
0.0093
0.0077
0.0093
0.0021
0.0026
0.0093
0.0068
0.0039
0.0068
0.0041
0.0053
0.0037
0.0075
0.0016
0.0000
-0.0005
0.0073
0.0076
0.0049
0.0046
0.0087
0.0106
0.0072
0.0085
0.0036
0.0044
0.0043
0.0201
0.0076
0.0075
0.0134
0.0050
0.0071
0.0032
0.0055
0.0085
0.0046
0.0023
-0.0020
0.0027
0.0060
0.0066
0.0067
0.0014
0.0166
0.0067
0.0024
0.0072
0.0062
0.0081
0.0035
0.0077
0.0101
0.0045
0.0034
0.0144
0.0078
0.0065
0.0093
0.0181
0.0028
0.0050
0.0034
0.0063
0.0150
0.0035
0.0022
0.0079
0.0034
0.0110
0.0075
0.0058
0.0085
0.0152
0.0089
0.0060
0.0017
0.0041
0.0091
0.0072
-0.0109
0.0036
0.0063
0.0080
0.0037
0.0086
0.0097
0.0088
0.0016
0.0057
0.0059
0.0139
0.0061
0.0009
0.0059
0.0126
0.0117
0.0003
0.0060
0.0075
0.0073
0.0080
0.0154
0.0136
0.0121
0.0179
0.0150
0.0125
Instead of doing
f_filter = cconv(gf,f,length(f));
this does the trick:
f_filter = conv(gf,f);
f_filter = f_filter(sz/2+1:end-sz/2+1);
As suggested by #AnderBiguri you can use the option 'same' in your convolution fonction to preserve the original size of your array.
But if you apply a convolution with your normalized gaussian filter gf you will obtain a border effect.
To avoid the border effect you can apply the following tricks:
gf = exp(-x .^ 2 / (2 * sigma ^ 2)); %do not normalize gf now
f_filter = conv(f,gf,'same')./conv(ones(length(f),1),gf,'same') %normalization taking into account the lenght of the convolution
For example I've just transformed f into f = f+3
If we do not take into account the border effect we will obtain:
a sample probability matrix:
ans =
0.1444 0.0456 0.0138 0.0126 0.0116 0.0107 0.0052
0.1444 0.0456 0.0138 0.0126 0.0116 0.0107 0.0052
0.1222 0.0386 0.0116 0.0106 0.0098 0.0091 0.0044
0.1444 0.0456 0.0138 0.0126 0.0116 0.0107 0.0052
0.1222 0.0386 0.0116 0.0106 0.0098 0.0091 0.0044
0.1889 0.0596 0.0180 0.0164 0.0151 0.0140 0.0067
0.1333 0.0421 0.0127 0.0116 0.0107 0.0099 0.0048
I have used dataSample and randSample to sample 128 time from my data which has A=(7,7) size in matlab:
datasample(A,128)
ans =
0.1333 0.0421 0.0127 0.0116 0.0107 0.0099 0.0048
0.1222 0.0386 0.0116 0.0106 0.0098 0.0091 0.0044
0.1889 0.0596 0.0180 0.0164 0.0151 0.0140 0.0067
0.1889 0.0596 0.0180 0.0164 0.0151 0.0140 0.0067
0.1333 0.0421 0.0127 0.0116 0.0107 0.0099 0.0048
0.1444 0.0456 0.0138 0.0126 0.0116 0.0107 0.0052
0.1222 0.0386 0.0116 0.0106 0.0098 0.0091 0.0044
...
However, I am interested in having those 128 sample of 7 (128,7) in binary format with two discrete values of 0 and 1:
[1 1 1 0 1 0 1]
I can write a loop and round-down/up those values to 0 and 1 with certain thresholds (i.e. 0.5), but that for sure will be noisy. Is there a function that can output the sampling in binary (0,1) in Matlab ?
Is it possible to vectorize, and possibly run on a GPU, the following code
x = linspace(0,100,1000);
h = zeros(size(x));
for i = 1 : length(x)
exprho = expm(-x(i)*rho);
h(i) = trace(drho*exprho*drho*exprho);
end
out = 2 * trapz(x,h);
where rho and drho are two complex Hermitian square matrices of the same size. rho is in fact a quantum density matrix and drho is its derivative with respect to a parameter.
The size can range from 10 x 10 to 300 x 300 approximately but I would also like to reach bigger sizes.
Here are two sample matrices:
rho =
0.4046 0.3849 0.2589 0.1422 0.0676 0.0288 0.0112 0.0040 0.0014 0.0004 0.0001
0.3849 0.3661 0.2462 0.1352 0.0643 0.0274 0.0106 0.0038 0.0013 0.0004 0.0001
0.2589 0.2462 0.1656 0.0910 0.0433 0.0184 0.0071 0.0026 0.0009 0.0003 0.0001
0.1422 0.1352 0.0910 0.0500 0.0238 0.0101 0.0039 0.0014 0.0005 0.0002 0.0000
0.0676 0.0643 0.0433 0.0238 0.0113 0.0048 0.0019 0.0007 0.0002 0.0001 0.0000
0.0288 0.0274 0.0184 0.0101 0.0048 0.0020 0.0008 0.0003 0.0001 0.0000 0.0000
0.0112 0.0106 0.0071 0.0039 0.0019 0.0008 0.0003 0.0001 0.0000 0.0000 0.0000
0.0040 0.0038 0.0026 0.0014 0.0007 0.0003 0.0001 0.0000 0.0000 0.0000 0.0000
0.0014 0.0013 0.0009 0.0005 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000
0.0004 0.0004 0.0003 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
0.0001 0.0001 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
drho =
0.0366 0.0156 -0.0025 -0.0085 -0.0074 -0.0046 -0.0023 -0.0010 -0.0004 -0.0002 -0.0001
0.0156 -0.0035 -0.0147 -0.0148 -0.0103 -0.0057 -0.0028 -0.0012 -0.0005 -0.0002 -0.0001
-0.0025 -0.0147 -0.0181 -0.0145 -0.0091 -0.0048 -0.0022 -0.0009 -0.0004 -0.0001 -0.0000
-0.0085 -0.0148 -0.0145 -0.0105 -0.0062 -0.0031 -0.0014 -0.0006 -0.0002 -0.0001 -0.0000
-0.0074 -0.0103 -0.0091 -0.0062 -0.0035 -0.0017 -0.0008 -0.0003 -0.0001 -0.0000 -0.0000
-0.0046 -0.0057 -0.0048 -0.0031 -0.0017 -0.0008 -0.0004 -0.0001 -0.0001 -0.0000 -0.0000
-0.0023 -0.0028 -0.0022 -0.0014 -0.0008 -0.0004 -0.0002 -0.0001 -0.0000 -0.0000 -0.0000
-0.0010 -0.0012 -0.0009 -0.0006 -0.0003 -0.0001 -0.0001 -0.0000 -0.0000 -0.0000 -0.0000
-0.0004 -0.0005 -0.0004 -0.0002 -0.0001 -0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000
-0.0002 -0.0002 -0.0001 -0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000
-0.0001 -0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000
This question already has answers here:
Find local maximum value in the vector
(4 answers)
Closed 8 years ago.
I would like to find local minima and local maxima of a given vector. Let's assume that the given vector is as follow:
speed =
0.0002
0.0008
0.0014
0.0027
0.0037
0.0047
0.0054
0.0053
0.0053
0.0058
0.0060
0.0063
0.0062
0.0065
0.0062
0.0061
0.0060
0.0057
0.0062
0.0057
0.0053
0.0050
0.0047
0.0065
0.0049
0.0048
0.0033
0.0033
0.0041
0.0049
0.0063
0.0075
0.0085
0.0105
0.0108
0.0109
0.0105
0.0105
0.0099
0.0098
0.0099
0.0099
0.0105
0.0103
0.0112
0.0108
0.0088
0.0079
0.0066
0.0055
0.0058
0.0049
0.0049
0.0055
0.0060
0.0051
0.0055
0.0060
0.0053
0.0047
0.0058
0.0050
0.0044
0.0033
0.0022
0.0008
0.0015
0.0010
0.0011
0.0024
0.0028
0.0024
0.0016
0.0009
0.0009
0.0009
0.0015
0.0015
0.0025
0.0031
0.0030
0.0042
0.0051
0.0060
0.0065
0.0054
0.0012
0.0043
0.0059
0.0070
0.0078
0.0076
0.0082
0.0087
0.0088
0.0095
0.0101
0.0100
0.0110
0.0103
0.0111
0.0120
0.0118
0.0116
0.0115
0.0121
0.0120
0.0145
0.0107
0.0119
0.0110
0.0116
0.0102
0.0086
0.0076
0.0071
0.0055
0.0066
0.0063
0.0077
0.0052
0.0059
0.0061
0.0036
0.0047
0.0053
0.0027
0.0020
0.0011
0.0041
0.0034
0.0034
0.0019
0.0022
0.0008
0.0001
0.0007
0.0009
0.0010
0.0010
0.0001
0.0007
0.0014
0.0016
0.0016
0.0013
0.0008
0.0008
0.0005
0.0004
0.0002
0.0001
0.0004
0.0005
0.0006
0.0005
0.0006
0.0006
0.0004
0.0002
0.0000
0.0001
0.0001
0.0002
0.0003
0.0004
0.0004
0.0005
0.0007
0.0008
0.0007
0.0006
0.0005
0.0006
0.0006
0.0004
0.0002
0.0003
0.0006
0.0005
0.0005
0.0010
0.0012
0.0014
0.0020
0.0028
0.0039
0.0044
0.0061
0.0074
0.0082
0.0091
0.0102
0.0108
0.0110
0.0117
0.0128
0.0133
0.0148
0.0153
0.0155
0.0150
0.0146
0.0137
0.0130
0.0113
0.0110
0.0107
0.0112
0.0114
0.0113
0.0104
0.0101
0.0095
0.0088
0.0083
0.0076
0.0057
0.0047
0.0043
0.0046
0.0053
0.0063
0.0078
0.0070
0.0062
0.0053
0.0051
0.0055
0.0048
0.0053
0.0052
0.0055
0.0065
0.0075
0.0078
0.0081
0.0067
0.0044
0.0061
0.0047
0.0032
0.0033
0.0028
0.0019
0.0007
0.0017
0.0016
0.0025
0.0034
0.0037
0.0044
0.0039
0.0037
0.0029
0.0030
0.0025
0.0022
0.0025
0.0027
0.0028
0.0031
0.0029
0.0025
0.0025
0.0025
0.0024
0.0022
0.0021
0.0019
0.0020
0.0020
0.0016
0.0016
0.0015
0.0013
0.0011
0.0011
0.0010
0.0009
0.0008
0.0006
0.0005
0.0004
0.0002
0.0000
0.0002
0.0003
0.0004
0.0006
0.0005
0.0004
0.0003
0.0004
0.0003
0.0003
0.0004
0.0006
0.0004
0.0004
when I polt this vector in Matlab using command plot(speed) then I have the following figure:
How could I find the maxima and minima's of the given vector? For example, in this above picture my aim is to find the three minimums/maximums that are shown in the picture.
I have lots of such a vectors that I want to write a code for all to find local minimas and maximas as well.
First of all you need to define what you count as extremum (maximum or minimum), i.e. which scale is considered appropriate, as your curve in reality has much more local maxima and minima than 3 or 4. Therefore looking for zero-crossings of the first derivative with diff will give you a lots of spurious micro-peaks. One option is to smooth it before. However, it might be easier to resort to a standard tool.
Try findpeaks from Signal Processing Toolbox.
There you can specify the scale with various parameters, such as 'MinPeakDistance', 'MinPeakHeight', 'Threshold' etc.