1 What is a rational function?
1.1 Front
What is a rational function?
1.2 Back
Is a function that is the ratio of two polynomials
\({\displaystyle \frac{s + }{s^2 + 7s + 9}}\)
2 When can we say that a rational function is proper?
2.1 Front
When can we say that a rational function is proper?
2.2 Back
A rational function is called proper if the degree of the numerator is strictly smaller than the degree of the denominator
For example: \({\displaystyle \frac{s + 1}{s^2 + 7s + 9}}\)
3 What is the objective of long-division of 2 polynomials?
3.1 Front
What is the objective of long-division of 2 polynomials?
3.2 Back
Using long-division we can always write an improper rational function as a polynomial plus a proper rational function.
The partial fraction decomposition only applies to proper functions.
4 When can we do the partial fraction decomposition?
4.1 Front
When can we do the partial fraction decomposition?
4.2 Back
When you have a proper rational function
5 Use long-division to rewrite this rational function?
5.1 Front
Use long-division to rewrite this rational function?
\({\displaystyle \frac{s^3 + 2s + 1}{s^2 + s - 2}}\)
5.2 Back
\({\displaystyle \frac{s^3 + 2s + 1}{s^2 + s - 2} = s - 1 + \frac{5s - 1}{s^2 + s - 2}}\)
6 Compute the partial fraction for this rational function
6.1 Front
Compute the partial fraction for this rational function
\({\displaystyle \frac{3}{(s - 1)(s + 1)(s - 3)}}\)
6.2 Back
\({\displaystyle \frac{3}{(s - 1)(s + 1)(s - 3)} = \frac{A}{s - 1} + \frac{B}{s + 1} + \frac{C}{s - 3}}\)
\({\displaystyle 3 = A(s + 1)(s - 3) + B(s - 1)(s - 3) + C (s - 1)(s + 1)}\)
- For \(s = 1\), gives \({\displaystyle A = \frac{-3}{4}}\)
- For \(s = -1\), gives \({\displaystyle B = \frac{3}{8}}\)
- For \(s = 3\), gives \({\displaystyle C = \frac{3}{8}}\)
\({\displaystyle \frac{3}{(s - 1)(s + 1)(s - 3)} = - \frac{3}{4(s-1)} + \frac{3}{8(s + 1)} + \frac{3}{8 (s - 3)}}\)
7 Compute the partial fraction for this rational function
7.1 Front
Compute the partial fraction for this rational function
\({\displaystyle \frac{s - 1}{(s + 1)(s^2 + 4)}}\)
7.2 Back
\({\displaystyle \frac{s - 1}{(s + 1)(s^2 + 4)} = \frac{A}{s + 1} + \frac{Bs + C}{s^2 + 4}}\)
\({\displaystyle s - 1 = A(s^2 + 4) + (Bs + C)(s + 1) = (A + B)s^2 + (B + C)s + (4A + C)}\)
- \(s^2\): \(0 = A +B\)
- \(s\): \(1 = B + C\)
- \(1\): \(-1 = 4A + C\)
Solving, we get \(A=-2/5\), \(B = 2/5\) and \(C = 3/5\)
8 Compute the partial fraction for this rational function
8.1 Front
Compute the partial fraction for this rational function
\({\displaystyle \frac{1}{(s + 1)(s^2 - 4)}}\)
Only set the partial fractions
8.2 Back
\({\displaystyle \frac{1}{(s + 1)(s^2 - 4)} = \frac{1}{(s + 1)(s + 2)(s - 2)} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{s - 2}}\)
9 Compute the partial function for this rational function
9.1 Front
Compute the partial function for this rational function
\({\displaystyle \frac{s^3 + 2s + 1}{s^2 + s - 2}}\)
Only set the partial fractions
9.2 Back
\({\displaystyle \frac{s^3 + 2s + 1}{s^2 + s - 2} = s - 1 + \frac{5s - 1}{s^2 + s - 2} = s - 1 + \frac{A}{s + 2} + \frac{B}{s - 1}}\)
10 How is notated the inverse Laplace transform?
10.1 Front
How is notated the inverse Laplace transform?
10.2 Back
\(\mathcal{L}^{-1}\)
11 What is the inverse Laplace transform of this expression?
11.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{1}{s - 2} \biggr)}\)
11.2 Back
\({\displaystyle e^{2t}}\)
12 What is the inverse Laplace transform of this expression?
12.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{1}{s^2 + 9} \biggr)}\)
12.2 Back
Use the table entry \({\displaystyle \mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}}\) and linearity
\({\displaystyle \frac{1}{3} \sin(3t)}\)
13 What is the inverse Laplace transform of this expression?
13.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{4}{(s - 2)^2} \biggr)}\)
13.2 Back
\({\displaystyle e^{2t} 4t}\)
14 What is the inverse Laplace transform of this expression?
14.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{1}{s^2 + 4s + 13} \biggr)}\)
14.2 Back
Complete the square: \({\displaystyle s^2 + 4s + 13 = s^2 + 4s + 4 + 9 = (s + 2)^2 + 9}\)
\(F(s + 2)\) where \(F(s) = \frac{1}{s^2 + 9}\), where \({\displaystyle f(t) = \frac{\sin(3t)}{3}}\), so using the s-shift rule
\({\displaystyle \mathcal{L}^{-1} (F(s-2)) = e^{-2t} \frac{\sin(3t)}{3}}\)
15 What is the inverse Laplace transform of this expression?
15.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{s-1}{(s+1)(s^2 + 4)} \biggr)}\)
Don’t compute the coefficients
15.2 Back
\({\displaystyle \frac{s-1}{(s + 1)(s^2 + 4)} = \frac{A}{s+1} + \frac{Bs + C}{s^2 + 4}}\)
\({\displaystyle \mathcal{L}^{-1} \biggl( \frac{A}{s+1} + \frac{Bs + C}{s^2 + 4} \biggr) = A e^{-t} + B \cos(2t) + \frac{C}{2} \sin(2t)}\)
16 Which is the partial fraction of this rational function?
16.1 Front
Which is the partial fraction of this rational function?
\({\displaystyle \frac{2s}{s^3(s + 1)^2 (s + 2)}}\)
Don’t compute the coefficients
16.2 Back
\({\displaystyle \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s + 1} + \frac{E}{(s + 1)^2} + \frac{F}{s + 2}}\)
17 Which is the partial fraction of this rational function?
17.1 Front
Which is the partial fraction of this rational function?
\({\displaystyle \frac{2s}{s(s^2 + 1)^2(s^2 + 4s + 2)}}\)
Don’t compute the coefficients
17.2 Back
\({\displaystyle \frac{A}{s} + \frac{Bs + C}{s^2 + 1} + \frac{Ds +E}{(s^2 + 1)^2} + \frac{Fs + G}{s^2 + 4s + 6}}\)
18 What is the inverse Laplace transform of this expression?
18.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl( \frac{Ds}{(s^2 + 1)^2} \biggr)}\)
18.2 Back
\({\displaystyle D \frac{t}{2} \sin(t)}\)
19 What is the inverse Laplace transform of this expression?
19.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl( \frac{E}{(s^2 + 1)^2} \biggr)}\)
19.2 Back
\({\displaystyle E \frac{1}{2} (\sin(t) - t \cos(t))}\)
20 What is the inverse Laplace transform of this expression?
20.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{Fs + G}{s^2 + 4s + 6} \biggr)}\)
20.2 Back
Complete the square: \(s^2 + 4s + 6 = (s + 2)^2 + 2\)
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{F(s + 2) + G - 2F}{(s + 2)^2 + 2} \biggr) = Fe^{-2t} \cos(\sqrt{2}t) + \frac{1}{\sqrt{2}} (G - 2F)e^{-2t} \sin(\sqrt{2}t)}\)
21 Describe the process of cover-up to decompose this partial fraction
21.1 Front
Describe the process of cover-up to decompose this partial fraction
\({\displaystyle \frac{s-7}{(s-1)(s+2)}}\)
Why does this method work?
21.2 Back
\({\displaystyle \frac{s-7}{(s-1)(s+2)} = \frac{A}{s-1} + \frac{B}{s+2}}\)
To determine \(A\) by the cover-up method, on the left-hand side we mentally remove (or cover up with a finger) the factor \(s - 1\) associated with \(A\), and substitute \(s = 1\) into what’s left; this gives \(A\)
\({\displaystyle \frac{s-7}{(s + 2)} \bigg|_{s=1} = \frac{1 - 7}{1 + 2} = -2 = A}\)
Similarly, \(B\) is found by covering up the factor \(s + 2\) on the left, and substituting \(s = -2\) into what’s left
\({\displaystyle \frac{s - 7}{(s - 1)} \bigg|_{s = -2} = \frac{-2 - 7}{-2 -1} = 3 = B}\)
\({\displaystyle \frac{s-7}{(s-1)(s+2)} = \frac{-2}{s-1} + \frac{3}{s+2}}\)
The reason is simple. The “right” way to determine \(A\) from the equation would be to multiply both sides by \((s - 1)\)
\({\displaystyle \frac{s-7}{(s + 2)} = A + \frac{B}{s+2}(s - 1)}\)
Now if we substitute \(s = 1\), what we get is exactly \({\displaystyle A = \frac{1 - 7}{1 + 2}}\), since the term on the right disappears
22 What is the inverse Laplace transform of this expression?
22.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{e^{-5s}}{s^2 - 4} \biggr)}\)
22.2 Back
\({\displaystyle u(t - 5) \biggl(\frac{1}{4} e^{2(t-5)} - \frac{1}{4} e^{-2(t - 5)} \biggr)}\)
23 What is the inverse Laplace transform of this expression?
23.1 Front
What is the inverse Laplace transform of this expression?
\({\displaystyle \mathcal{L}^{-1} \biggl( \frac{5 - 2s}{s^2 + 7s + 10} \biggr)}\)
23.2 Back
\({\displaystyle \frac{5 - 2s}{(s + 2)(s + 5)} = \frac{3}{s+2} - \frac{5}{s + 5}}\)
\({\displaystyle \mathcal{L}^{-1} \biggl(\frac{3}{s+2} - \frac{5}{s + 5} \biggr) = 3 e^{-2t} - 5e^{-5t}}\)