Example of Inverse Laplace's Transform for repeated poles using residual method

Рет қаралды 10,932

RF Design Basics

Күн бұрын

Пікірлер: 13

@Kasun_Ish 2 ай бұрын

Well explained.

@romjanali5576 2 ай бұрын

Thank so much

@alivasss Жыл бұрын

Thank you Boss !!!!

@SaninSelimovic-zh8eq Жыл бұрын

can you solve this task?

@juanmaurin5253 7 ай бұрын

When u try to calculate C using residues, you cant just cancel the poles with the zeros and take that limit. you HAVE to derivate aftherwards

@SaninSelimovic-zh8eq Жыл бұрын

F(s)= s^2+2s+1 / s(s^2+2s+5)

@SaninSelimovic-zh8eq Жыл бұрын

can you solve this task?

@carultch 11 ай бұрын

@@SaninSelimovic-zh8eq Given: F(s) = (s^2+2*s+1)/(s*(s^2+2*s+5)) Complete the square on the denominator: F(s) = (s^2+2*s+1)/(s*((s+1)^2+4)) Set up partial fractions for a linear denominator term (A/s), and a quadratic denominator term (linear numerator), using the shifted value of s instead of just s. You'll see why this helps: F(s) = A/s + (B*(s + 1) + C)/((s+1)^2 + 4) Heaviside coverup finds A for us, at s=0: A = (0^2+2*0+1)/(covered*(0^2+2*0+5)) = 1/5 Reconstruct what remains: (s^2+2*s+1)/(s*((s+1)^2+4)) = 1/5/s + (B*(s + 1) + C)/((s+1)^2+4) Let s = -1, to solve for C. Notice that B cancels out: ((-1)^2+2*(-1)+1)/((-1)*((-1+1)^2+4)) = 1/5/(-1) + (B*((-1) + 1) + C)/((-1+1)^2+4) 0 = -1/5 + C/4 C = 4/5 Reconstruct what remains: (s^2+2*s+1)/(s*((s+1)^2+1)) = 1/5/s + (B*(s + 1) + 4/5)/((s+1)^2+4) Let s = -2, to solve for B. We choose -2, so all the (s+1)^2 terms become 1. ((-2)^2+2*(-2)+1)/(-2*((-2+1)^2+4)) = 1/5/(-2) + (B*(-2 + 1) + 4/5)/((-2+1)^2+4) -1/10 = -1/10 + (-B+ 4/5)/2 Solve for B: B = 4/5 Partial fraction result: F(s) = 1/5*[1/s + (4*(s + 1) + 4)/((s+1)^2+4)] Arrange to look like standard Laplace transforms: F(s) = 1/5*[1/s + 4*(s + 1)/((s+1)^2+4) + 2*2/((s+1)^2+4)] Inverse Laplace: f(t) = 1/5 + 1/5*e^(-t)*[4*cos(2*t) + 2*sin(2*t)]

@HusseinNaserddine Жыл бұрын

great thank you !!!!!

@alivasss Жыл бұрын

Okkkkkk

@godwincornelius201 Жыл бұрын

Doesn’t always work… talking about the partial fraction

@carultch 11 ай бұрын

It works as long as there aren't quadratic terms, or other irreducible polynomials beyond linear terms. Otherwise, you get interdependent unknown coefficients.