# How can I compute envelope for three phase current?

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Rajeev Kumar il 24 Set 2022
Modificato: Rajeev Kumar il 26 Set 2022
I stuck in a very small step i.e. the envelope identification of three phase current signal. In matlab when I used the predefined envelope function, it computed the three individual envelope plots instead of one plot for three phase current signal.
Please guide me, how can I plot the single envelope for all the three phase wave?.
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### Risposta accettata

Star Strider il 24 Set 2022
I am not certain what you want.
Perhaps this —
Fs = 0.0001;
t = linspace(0, 1, 1/Fs)/Fs;
s = sin(2*pi*t*60 + deg2rad([0; 120; 240]))
s = 3×10000
0 0.0377 0.0753 0.1129 0.1502 0.1874 0.2243 0.2609 0.2971 0.3329 0.3682 0.4029 0.4372 0.4707 0.5037 0.5359 0.5673 0.5980 0.6277 0.6566 0.6846 0.7116 0.7376 0.7625 0.7863 0.8091 0.8307 0.8510 0.8702 0.8882 0.8660 0.8466 0.8259 0.8041 0.7811 0.7570 0.7318 0.7056 0.6784 0.6502 0.6211 0.5911 0.5603 0.5287 0.4963 0.4632 0.4295 0.3952 0.3603 0.3248 0.2890 0.2527 0.2160 0.1791 0.1418 0.1044 0.0669 0.0292 -0.0085 -0.0462 -0.8660 -0.8843 -0.9012 -0.9169 -0.9313 -0.9444 -0.9561 -0.9665 -0.9755 -0.9831 -0.9893 -0.9941 -0.9975 -0.9994 -1.0000 -0.9991 -0.9968 -0.9931 -0.9880 -0.9815 -0.9736 -0.9643 -0.9536 -0.9416 -0.9282 -0.9135 -0.8975 -0.8803 -0.8618 -0.8420
[seu,sel] = envelope(s, 1, 'peak')
seu = 3×10000
0 0.0377 0.0753 0.1129 0.1502 0.1874 0.2243 0.2609 0.2971 0.3329 0.3682 0.4029 0.4372 0.4707 0.5037 0.5359 0.5673 0.5980 0.6277 0.6566 0.6846 0.7116 0.7376 0.7625 0.7863 0.8091 0.8307 0.8510 0.8702 0.8882 0.8660 0.8466 0.8259 0.8041 0.7811 0.7570 0.7318 0.7056 0.6784 0.6502 0.6211 0.5911 0.5603 0.5287 -0.2482 -0.2316 -0.2148 -0.1976 -0.1801 -0.1624 -0.1445 -0.1263 -0.1080 -0.0895 -0.0709 -0.0522 -0.0334 -0.0146 0.0042 0.0231 -0.8660 -0.8843 -0.9012 -0.9169 -0.9313 -0.9444 -0.9561 -0.9665 -0.9755 -0.9831 -0.9893 -0.9941 -0.9975 -0.9994 -1.0000 -0.9991 -0.9968 -0.9931 -0.9880 -0.9815 -0.9736 -0.9643 -0.9536 -0.9416 -0.9282 -0.9135 -0.8975 -0.8803 -0.8618 -0.8420
sel = 3×10000
0 0.0377 0.0753 0.1129 0.1502 0.1874 0.2243 0.2609 0.2971 0.3329 0.3682 0.4029 0.4372 0.4707 0.5037 0.5359 0.5673 0.5980 0.6277 0.6566 0.6846 0.7116 0.7376 0.7625 0.7863 0.8091 0.8307 0.8510 0.8702 0.8882 -0.4330 -0.4233 -0.4129 -0.4020 -0.3905 -0.3785 -0.3659 -0.3528 -0.3392 -0.3251 -0.3106 -0.2956 -0.2802 -0.2643 -0.2482 -0.2316 -0.2148 -0.1976 -0.1801 -0.1624 -0.1445 -0.1263 -0.1080 -0.0895 -0.0709 -0.0522 -0.0334 -0.0146 0.0042 0.0231 -0.8660 -0.8843 -0.9012 -0.9169 -0.9313 -0.9444 -0.9561 -0.9665 -0.9755 -0.9831 -0.9893 -0.9941 -0.9975 -0.9994 -1.0000 -0.9991 -0.9968 -0.9931 -0.9880 -0.9815 -0.9736 -0.9643 -0.9536 -0.9416 -0.9282 -0.9135 -0.8975 -0.8803 -0.8618 -0.8420
figure
plot(t, s)
hold on
plot(t, max(seu), '-r', 'LineWidth',2)
plot(t, min(sel), '-g', 'LineWidth',2)
hold off
xlim([0 250])
figure
plot(t, s)
hold on
plot(t, max(s), '-r', 'LineWidth',2)
plot(t, min(s), '-g', 'LineWidth',2)
hold off
xlim([0 250])
.
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Star Strider il 24 Set 2022
The type of envelope result you get depends on the arguments to the envelope function. I chose the 'peak' option with the number of points equalling 1, although larger numbers (for example, 10) also worked. Experiment with the function to get the result you want, since I am not certain what that is.
Fs = 0.0001;
t = linspace(0, 1, 1/Fs)/Fs;
s = sin(2*pi*t*60 + deg2rad([0; 120; 240]))
s = 3×10000
0 0.0377 0.0753 0.1129 0.1502 0.1874 0.2243 0.2609 0.2971 0.3329 0.3682 0.4029 0.4372 0.4707 0.5037 0.5359 0.5673 0.5980 0.6277 0.6566 0.6846 0.7116 0.7376 0.7625 0.7863 0.8091 0.8307 0.8510 0.8702 0.8882 0.8660 0.8466 0.8259 0.8041 0.7811 0.7570 0.7318 0.7056 0.6784 0.6502 0.6211 0.5911 0.5603 0.5287 0.4963 0.4632 0.4295 0.3952 0.3603 0.3248 0.2890 0.2527 0.2160 0.1791 0.1418 0.1044 0.0669 0.0292 -0.0085 -0.0462 -0.8660 -0.8843 -0.9012 -0.9169 -0.9313 -0.9444 -0.9561 -0.9665 -0.9755 -0.9831 -0.9893 -0.9941 -0.9975 -0.9994 -1.0000 -0.9991 -0.9968 -0.9931 -0.9880 -0.9815 -0.9736 -0.9643 -0.9536 -0.9416 -0.9282 -0.9135 -0.8975 -0.8803 -0.8618 -0.8420
[seu,sel] = envelope(s, 150, 'analytic')
seu = 3×10000
0.8216 0.8186 0.8166 0.8158 0.8161 0.8175 0.8200 0.8237 0.8284 0.8341 0.8407 0.8482 0.8566 0.8658 0.8756 0.8860 0.8970 0.9084 0.9202 0.9322 0.9445 0.9569 0.9694 0.9818 0.9942 1.0064 1.0184 1.0302 1.0416 1.0526 1.1025 1.1101 1.1171 1.1234 1.1290 1.1338 1.1379 1.1412 1.1438 1.1455 1.1464 1.1465 1.1458 1.1444 1.1421 1.1390 1.1352 1.1306 1.1252 1.1191 1.1123 1.1049 1.0968 1.0880 1.0787 1.0689 1.0585 1.0477 1.0365 1.0249 0.8216 0.8290 0.8352 0.8404 0.8445 0.8474 0.8492 0.8499 0.8495 0.8479 0.8452 0.8414 0.8365 0.8305 0.8234 0.8152 0.8059 0.7956 0.7843 0.7719 0.7586 0.7443 0.7290 0.7129 0.6958 0.6780 0.6593 0.6398 0.6197 0.5989
sel = 3×10000
-0.8216 -0.8186 -0.8166 -0.8158 -0.8161 -0.8175 -0.8200 -0.8237 -0.8284 -0.8341 -0.8407 -0.8482 -0.8566 -0.8658 -0.8756 -0.8860 -0.8970 -0.9084 -0.9202 -0.9322 -0.9445 -0.9569 -0.9694 -0.9818 -0.9942 -1.0064 -1.0184 -1.0302 -1.0416 -1.0526 -1.1025 -1.1101 -1.1171 -1.1234 -1.1290 -1.1338 -1.1379 -1.1412 -1.1438 -1.1455 -1.1464 -1.1465 -1.1458 -1.1444 -1.1421 -1.1390 -1.1352 -1.1306 -1.1252 -1.1191 -1.1123 -1.1049 -1.0968 -1.0880 -1.0787 -1.0689 -1.0585 -1.0477 -1.0365 -1.0249 -0.8216 -0.8290 -0.8352 -0.8404 -0.8445 -0.8474 -0.8492 -0.8499 -0.8495 -0.8479 -0.8452 -0.8414 -0.8365 -0.8305 -0.8234 -0.8152 -0.8059 -0.7956 -0.7843 -0.7719 -0.7586 -0.7443 -0.7290 -0.7129 -0.6958 -0.6780 -0.6593 -0.6398 -0.6197 -0.5989
figure
plot(t, s)
hold on
plot(t, max(seu), '-r', 'LineWidth',2)
plot(t, min(sel), '-g', 'LineWidth',2)
hold off
xlim([0 500])
If you want to duplicate the result in the posted image exactly, you will need to experiment. I am using calculated representations of the three-phase signal, while the image appears to be a sampled signal with an uneven baseline and slightly varying amplitude. I am not certain how that was calculated, however using polyfit (and polyval) on the peak amplitudes (using the max or abs functions to determine the peak amplitudes) might be an approach that would duplicate it, since it does not appear to be an envelope function result.
.
Rajeev Kumar il 26 Set 2022
Modificato: Rajeev Kumar il 26 Set 2022
I have a two array
A = [1,2,3,4,0,6,7,0,9,10,11,0,13,14,15];
t=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
Now skip the zeros form the array A and also skip the crossponding index form the time (t) array. and make a plot in between them?

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### Più risposte (1)

Sam Chak il 24 Set 2022
You can try if the envelope function works for you.
https://www.mathworks.com/help/signal/ref/envelope.html
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