You probably dont give a shit but does someone know a method to log back into an Instagram account?? I was stupid forgot the account password. I love any tips you can offer me
Yeah, I did it in a completely different way! I completed the square at the bottom: s^2 + 2s - 3 = (s + 1)^2 - 2^2. Then I split the inverse laplace transform into: L^-1{(s + 1)/((s+1)^2 - 2^2)} + L^-1{3/((s+1)^2 - 2^2)}. The s+1 gives us a common factor of e^-t, so it becomes e^-5 * (L^-1{s/(s^2 - 2^2)} + 3/2 L^-1{2/(s^2 - 2^2}) = e^-t * (cosh(2t) + 3/2 sinh(2t)). If you substitute the cosh and sinh for their e-power forms, it results in the same answer, obviously.
where were these videos 5 years ago when i was in school. studying for my FE and these are a Lifesaver (make sure you take your FE before you graduate kids)
Yes, you will arrive at the same answer but it'll be a long and tedious solution since you'll be using trigonometric identities a lot until you will end up at e.
Hi, perfect. Is laplace inverse of (2/p^2-2p)= 2e^2t. or we apply Convolution theorem of inverse laplace. or 2 L'(1/(p-1)^2-1)=2/1 (Sinht). sir pls reply.
An irreducible quadratic denominator with a constant numerator, is the Laplace transform of an exponential multiplied by a sine wave. From the standard table of Laplace transforms: £{e^(a*t) * sin(b*t)} = b/((s - a)^2 + b^2) To get your expression to look like this, we identify that a=2 and b=5. We then need to multiply by 1 in a fancy way (i.e. 5/5), to make your numerator also equal 5. 3/((s-2)^2 +25) * 5/5 = 3/5 * 5/(s-2)^2 +25) Pull the 3/5 out in front as a constant, and evaluate, and we have our solution: 3/5 * £^(-1) {5/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t) £^(-1) {3/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t)
I had never seen this method for solving partial fractions in the 4 years I have spent in college and it's a life saver, thank you!
You probably dont give a shit but does someone know a method to log back into an Instagram account??
I was stupid forgot the account password. I love any tips you can offer me
BEST explanation I have seen so far. Thank you very much.
Thank you I couldn't find any comprehensive explanation of the inverse laplace apart from these videos
this is going to save my butt on my signals and systems test tommorow. thank you
Yeah, I did it in a completely different way! I completed the square at the bottom: s^2 + 2s - 3 = (s + 1)^2 - 2^2. Then I split the inverse laplace transform into: L^-1{(s + 1)/((s+1)^2 - 2^2)} + L^-1{3/((s+1)^2 - 2^2)}. The s+1 gives us a common factor of e^-t, so it becomes e^-5 * (L^-1{s/(s^2 - 2^2)} + 3/2 L^-1{2/(s^2 - 2^2}) = e^-t * (cosh(2t) + 3/2 sinh(2t)).
If you substitute the cosh and sinh for their e-power forms, it results in the same answer, obviously.
I did it in the same way as you did. Thanks for the cross check
Thanks dude Youre really good at explaining in a good pace Really helped me comprehend
otheamazing my pleasure!
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Oh my gosh this is why you are my favorite. Took something scary and made digestible.
This was more than helpful. Thanks a ton!
where were these videos 5 years ago when i was in school. studying for my FE and these are a Lifesaver (make sure you take your FE before you graduate kids)
What FE ?
@@lawtraf8008 electrical
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Is it wrong to use completing the square with this problem? Do you have to use partial fraction?
Hi
thx saved me an hour to study this topic
I used completing the square on the denominator and got a different answer: e^(-t)*[cos(2t)+3sin(2t)] - did I do something wrong?
Yes, you will arrive at the same answer but it'll be a long and tedious solution since you'll be using trigonometric identities a lot until you will end up at e.
What concept did you apply on getting the numerator of the right side?
amazing
What if you have 3 terms as the denominator? Do you do faction over fraction over fraction?
rather a "masochistic "way to find couple of constants. try partial fraction, less painful.
what if the denominator for both is equal?
Thanks for helping, and deliver the issue of the issue if it executed F (s) = s-1 / s3 + 2s2 + s
thank you for this! based from inverse laplace table
Can you do this with completing the square on the bottom and using the translation theorem please
Never mind that'd be with complex numbers
why with complex numbers only?
You don't need complex numbers, you need sinh and cosh.
Hi,
perfect.
Is laplace inverse of (2/p^2-2p)= 2e^2t.
or we apply Convolution theorem of inverse laplace.
or
2 L'(1/(p-1)^2-1)=2/1 (Sinht).
sir pls reply.
I did it with another concept and my answer comes epow(-t)[cos2t+1.5sin2t] Is it right?
You’re perfect! Thanks :)
Thank you mate. Now I can make my plates. 😂❤
Excellent job
Wait...how
If we substitute s = 1 to get B as 0 to grt A value...then we get A = 1
How is it 5/4 what
This guy just covered a partial fraction that will take 4lines with zero lines
that shit blew my mind no cap
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Good shortly and clear
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I got an equivalent answer:
(e^-t)(cosh(2t)+(3/2)sinh(2t))
I calculatored it with : (e^-t)(cosh(2t)+(3/2)sinh(2t)) = (5/4)e^t-(1/4)e^(-3t)
Yeah bro got the same answer
Thank you soooo much
i didn't understood that how s square plus 2 equals to s-1?
Saurabh Jha he's using a trick
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Woww soo fast, thanks
plz ,i need inverse Laplace for (3/(s-2)^2 +25)
An irreducible quadratic denominator with a constant numerator, is the Laplace transform of an exponential multiplied by a sine wave. From the standard table of Laplace transforms:
£{e^(a*t) * sin(b*t)} = b/((s - a)^2 + b^2)
To get your expression to look like this, we identify that a=2 and b=5. We then need to multiply by 1 in a fancy way (i.e. 5/5), to make your numerator also equal 5.
3/((s-2)^2 +25) * 5/5 = 3/5 * 5/(s-2)^2 +25)
Pull the 3/5 out in front as a constant, and evaluate, and we have our solution:
3/5 * £^(-1) {5/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t)
£^(-1) {3/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t)
translite indonesia ??
thank you bro
thank you so much
thanks man!
What a shortttttt cuttttt
24000 GBP uni tuition and failed after attending 2 hours class.
3 min youtube video and i have completely understood the process.
same here
Thank you
Im ready for my Final....i hope
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Tq bro
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blue pen?!?!?!
Thanks
My pleasure!
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frac{2}{3}
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tq
what's tq?
+blackpenredpen First time I've seen it but I read it like
ty = thank you
tq = than q (say it fast)
thank you T_T
Thanks