You honestly made this a whole lot easier, I hadn't done partial fractions since calc 2. Really made it easy to understand, thank you !

Your video series on the Laplace transform is just brilliant 👍🏼👍🏼

Ive watched dozens of videos on Laplace and inverse Laplace, this is the first video that really made me click and have that Aha! moment. Thank you. You demonstrated the material very well. Excited to watch your other videos.

What a grt lvl of confidence ,concept as well as good body language , comunication ....I hats off 🎓for you SIR with MASSIVE RESPECT❤!!!

Hi Trefor ! I watched your whole linear algebra series last year but when i saw that you only got like 30k sub i thinked that this channel was dead. But suprisingly, you constantly upload your video despite of little sub. Thanks for all your teaching. God bless you, Jesus protect from Virus ! Love you.

Thank you for showing ALL the steps

Thank you for this simple but masterly explanation!!!!

thank you so much these videos are so helpful , and the most special thing a bout your videos is that you explain How to think in every step and where did they come from, thanks.

Your way of teaching is awesome!!.. More power to you sir

Was actually able to hang on to this one a bit, nice

learned this in less than five minutes. you are awesome.

I started watching your videos Freshman year in Calc 1 at UC. (I had Dr. Knoll) So funny to stumble upon your video when I was trying to refresh on this topic! Awesome to see how your channel has grown! Great videos thanks!

@DrTrefor

3 жыл бұрын

Hey that’s cool! Hope you are all doing great at UC:)

even with number of poles > 3, this approach is much faster. Thank you so much.

Extraordinary!!!

Thank you sooo much Dr.

Thanks for the video! Highly appreciated :)

Damn u r better than my college professors... They didn't raise the bar too high tho 😂. I'll make sure to subscribe u are excellent at explaining things. Thanks

Thank you for the video

your the best ❤

Hi, do you think you can do a video on the sequence for laplace transform and also inverse transform? like should we scale first or translate first, how to decide the power etc. Thank you so much for the wonderful videos.

great work sir

Thank for your video. I need that shirt!

I was taught a cool trick to solve A at 4:23, I think its called the method of predetermined coefficients? I think it only works for the highest power coefficient. So in this case, you would have s^2 + 9s + 2 = A(s -1)(s+3) + B(s+3) + C(s-1)^2 So, A would have s^2 and C would have s^2 and the coefficient on the left hand side for s^2 is 1 Therefore, we would have S^2 = As^2 + Cs^2 Then, 1 = A + C Since we easily solved for C = -1 1 = A -1 Therefore, A = 2 This method comes in handy for more difficult equations and saves time when the rule applies.

@carultch

Жыл бұрын

Similar idea as the Heaviside cover-up method. Identify the roots of each linear term. Cover up each linear term as you enter in its corresponding roots to the numerator and remaining denominator terms. This then gives you the value of the coefficients on all terms that are viable for H's cover-up method to work. You can get the coefficients on all the distinct linear terms, and the highest power of the repeated linear terms, using Heaviside cover-up. Then you use trial values of the variable, to get the remaining term.

Beautiful

thank you very much

Thank u so much 🙂❤️

Hello sir I am from India. And I am preparing for GATE Exam. This video was really helpful. Thanks a lot sir

Say the denominator of our transfer function was s(s^2+1), would we have A/s + B/s^2+1 + C/s^2+1? Or would there only be one s^2+1 term?

For the sake of youtube algorithem I just did it in the first place

@prakruthiprajwal6439

11 ай бұрын

Really

Thanks sir

thanks sir

oh l like your shirt too !!! l'm kinda busy recently , l promise l'll watch all videos l missed in the past few months

for s = -3 , if you plug it in you should get -34 not -16 right ? because he says -3^2 = 9 but its -9 so -9+ -27 + 2 = -34 not 9 + -27 + 2 = -16

how does this process change, if we have double complex roots?

Great examples and pace!

thank you jacksfilms

i love ur t shirttttttt

7:52 This might be a late question, but I don't fully understand where this "t" comes from (after e^t)? Sure it's a "1 over something squared", but it's not clicking for me. What piece of knowledge am I missing?

Thanks, what happens if i got a value of a letter in terms of another letter lets say A=-B/2?

@adrianocamposalas7860

4 жыл бұрын

@@DrTrefor Thanks!

I don't understand how you got 2e^t. Thank you for your help.

Sir I am from India. I have a question... If F(x). =(X² +2 x +a)/(x² +4x+3a) Then what will be the range of (a ). If F(x) is surjective? Pls sir reply how to solve it

@lylechen8881

4 жыл бұрын

Let me guess the variables are all real numbers: 1) To make sure frac is reasonable, x² +4x+3a MUST NOT equal to zero (b² - 4ac = 4² - 12a ≠ =), so a ≠ 4/3; 2) Beg your pardon?

U(t) at the end??

just a small question, isn't it usually far easier to do convolution instead of partial fraction decomposition? great video as usual!

Duude you make things sooo much simpler. WHAAAT??

why s=0 how to find this

Can you explain laplase inverse of 1 please

@jeetmohanty

2 жыл бұрын

okay got it in the 'dirac delta' video

Can we take any value for s?

@carultch

Жыл бұрын

What you are ultimately doing, is trying to make the numerators on both sides equal. The numerator in the given function is a polynomial, and after you combine the partial fraction expansion in progress, the numerator is also a polynomial. You can plug in any value you prefer, as a data point to solve for the unknown coefficients in the partial fraction expansion. It usually helps you the most to plug in s=0 and s=1, if you haven't used them already, and a tertiary option is s=-1. Plugging in s=0, allows you to cancel terms like B*s from a numerator of (B*s + C), so that you can solve for C without other variables getting in the way.

@carultch

Жыл бұрын

If you've already used a value of s in the Heaviside cover-up method, you can't use it again to solve for the remaining coefficients. It will give you no new information. As an example: 10/((s - 1)*(s^2 + 2*s + 2) = A/(s - 1) + (B*s + C)/(s^2 + 2*s + 2) Plug in s=1 for the cover-up method, and get A=2. Now we cross-multiply to set up equations to solve for B&C: 2*(s^2 + 2*s + 2) + (B*s + C)*(s - 1) = 10 If you attempt to use s = 1, you will see that B*s + C will cancel out entirely, and you'll generate 2*5 = 10, which doesn't help you. So you need to use values of s that aren't already spoken-for. s=0: 2*(0 + 0 + 2) + (C)*(-1) = 10 C = -6 s = -1: 2*(1 - 2*1 + 2) + (B*(-1) + -6)*(-1 - 1) = 10 B = -2 Solution: 2/(s - 1) - (2*s + 6)/(s^2 + 2*s + 2)

sir ,if you don't mind , c(s-1)^3 will come ,or you right

@carultch

Жыл бұрын

The reason why C*(s - 1)^3 doesn't come up, is that you don't need a third copy of (s - 1) to make a common denominator. For the same reason that when adding 1/3 and 1/6, that you don't need to multiply 1/3 by 6/6 to get 6/18. You can multiply both terms by 6/6 if you want to, to get 6/18 + 3/18 = 9/18 = 1/2, but you can take credit for the fact that 3 and 6 already share a factor in common, and simplify your work in earlier stages. When trying to add up A/(s-1) + B/(s-1)^2 + C/(s+3), we only need to multiply each numerator by enough terms, to get them all to have (s+3)*(s-1)^2 as a common denominator. We only need 2 copies of (s-1) in each term's denominator, so at most, you would have 2 copies of (s - 1) in the numerator, as we do with the C-term.

You remind me of Dr kreiger from archer

like your shirt

eh chief it would be nice to post on ways for students to keep composure in these times since there's a lot of uncertainty all over the place

@mohamedridamahir1960

4 жыл бұрын

that would be mighty fine of you

How does he know to plug in s=1

@alessandras2963

3 жыл бұрын

It's one or technically two of the roots in the denominator... (s-1)=0 , s=1

@jebrentdamayon3279

2 жыл бұрын

you look in for the values that would cancel out the other two variables, so you only have to look for B. In his case, the variables A and C has (s-1); so if you plug in s=1 its gona result to 0 which will cancel them out after they are multiplied eg.(A*(1-1)(1+4)=A(0)(5) = 0 )

@carultch

Жыл бұрын

s=0 and s=1 are two of the easiest values to plug in, since s=0 cancels out all terms that depend on s, and using s=1 enables you to ignore all the exponents on s, since 1^anything = 1. So as long as s=0 and s=1 aren't already spoken-for, they are ideal choices to use for this method.

0:27 Wedding Ring ? 💔 Okay, you're just the Math Teacher I'll have forever a crush on and I'm a Math Teacher as well, damnit. 😅 Btw, jokes apart, I simply love your videos and thank you for the brilliant videos. Big cheers from India 🇮🇳 I'd only suggest you to write a little more clearly, and that's just all. Hope you don't mind this constructive criticism.

Fuck i love you man

.

hi😏

man sounds like Kermit the frog

booo, i have exam tomorrow why u dont make it easier?

@petersun3454

2 ай бұрын

nevermind probably the best video compared to the other ones...

thank you very much