Inverse Laplace transform (first shifting theorem)
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This is my textbook from what I am studying and the âÂÂgreenâ highlighted part is where I am questioning myself.
LetâÂÂs pull out the expression-
$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$
From here, I distribute it to become
$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $
Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?
calculus laplace-transform
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up vote
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favorite
This is my textbook from what I am studying and the âÂÂgreenâ highlighted part is where I am questioning myself.
LetâÂÂs pull out the expression-
$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$
From here, I distribute it to become
$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $
Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?
calculus laplace-transform
the green part is just added to cancel the added part in orange part (that+2/denom
). If you add green and orange parts, you gets/denom
.
â Alla Tarighati
Aug 17 at 7:08
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is my textbook from what I am studying and the âÂÂgreenâ highlighted part is where I am questioning myself.
LetâÂÂs pull out the expression-
$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$
From here, I distribute it to become
$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $
Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?
calculus laplace-transform
This is my textbook from what I am studying and the âÂÂgreenâ highlighted part is where I am questioning myself.
LetâÂÂs pull out the expression-
$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$
From here, I distribute it to become
$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $
Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?
calculus laplace-transform
edited Aug 17 at 7:04
asked Aug 17 at 5:49
user185692
1466
1466
the green part is just added to cancel the added part in orange part (that+2/denom
). If you add green and orange parts, you gets/denom
.
â Alla Tarighati
Aug 17 at 7:08
add a comment |Â
the green part is just added to cancel the added part in orange part (that+2/denom
). If you add green and orange parts, you gets/denom
.
â Alla Tarighati
Aug 17 at 7:08
the green part is just added to cancel the added part in orange part (that
+2/denom
). If you add green and orange parts, you get s/denom
.â Alla Tarighati
Aug 17 at 7:08
the green part is just added to cancel the added part in orange part (that
+2/denom
). If you add green and orange parts, you get s/denom
.â Alla Tarighati
Aug 17 at 7:08
add a comment |Â
1 Answer
1
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up vote
1
down vote
accepted
Let us go back one line, then we have three parts:
$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$
Now let us focus on the second part (containing orange and green parts):
$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$
And adding a zero to the nomerator will not change anything:
$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$
Or:
$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let us go back one line, then we have three parts:
$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$
Now let us focus on the second part (containing orange and green parts):
$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$
And adding a zero to the nomerator will not change anything:
$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$
Or:
$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$
add a comment |Â
up vote
1
down vote
accepted
Let us go back one line, then we have three parts:
$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$
Now let us focus on the second part (containing orange and green parts):
$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$
And adding a zero to the nomerator will not change anything:
$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$
Or:
$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let us go back one line, then we have three parts:
$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$
Now let us focus on the second part (containing orange and green parts):
$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$
And adding a zero to the nomerator will not change anything:
$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$
Or:
$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$
Let us go back one line, then we have three parts:
$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$
Now let us focus on the second part (containing orange and green parts):
$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$
And adding a zero to the nomerator will not change anything:
$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$
Or:
$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$
edited Aug 20 at 6:18
answered Aug 17 at 7:18
Alla Tarighati
2353
2353
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add a comment |Â
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the green part is just added to cancel the added part in orange part (that
+2/denom
). If you add green and orange parts, you gets/denom
.â Alla Tarighati
Aug 17 at 7:08