Partial Fractions & Laplace

Cheatsheet Content

### Core Concept Partial Fraction Expansion (PFE) is a technique to decompose a complex rational function into simpler fractions. This simplifies integration, inverse Laplace/Z-transforms, and circuit analysis. - **Applicability:** Essential when performing inverse Laplace Transforms on functions $F(s) = \frac{N(s)}{D(s)}$ where $N(s)$ and $D(s)$ are polynomials in $s$. - **Key Condition:** PFE is directly applicable only if the degree of the numerator $N(s)$ is **less than** the degree of the denominator $D(s)$ (i.e., proper fraction). - **Improper Fractions:** If degree($N(s)$) $\ge$ degree($D(s)$), perform **polynomial long division** first to get: $$ \frac{N(s)}{D(s)} = Q(s) + \frac{R(s)}{D(s)} $$ where $Q(s)$ is the quotient and $\frac{R(s)}{D(s)}$ is a proper fraction for which PFE can be applied. ### PFE Cases & Formula Templates Let $F(s) = \frac{N(s)}{D(s)}$. The form of the partial fraction expansion depends on the factors of the denominator $D(s)$. #### Case 1: Distinct Linear Factors For each distinct linear factor $(s - a)$ in $D(s)$: $$ \frac{N(s)}{(s-a)(s-b)...} = \frac{A}{s-a} + \frac{B}{s-b} + ... $$ #### Case 2: Repeated Linear Factors For each repeated linear factor $(s - a)^k$ in $D(s)$: $$ \frac{N(s)}{(s-a)^k(...)} = \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + ... + \frac{A_k}{(s-a)^k} + ... $$ #### Case 3: Distinct Irreducible Quadratic Factors For each distinct irreducible quadratic factor $(s^2 + as + b)$ (where $a^2 - 4b ### Standard Expansion Forms | Denominator Factors | PFE Form | Example | |---------------------|----------|---------| | $(s-a)(s-b)$ | $\frac{A}{s-a} + \frac{B}{s-b}$ | $\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$ | | $(s-a)^2$ | $\frac{A}{s-a} + \frac{B}{(s-a)^2}$ | $\frac{s+3}{(s+1)^2} = \frac{A}{s+1} + \frac{B}{(s+1)^2}$ | | $(s-a)^3$ | $\frac{A}{s-a} + \frac{B}{(s-a)^2} + \frac{C}{(s-a)^3}$ | $\frac{1}{s(s+1)^3} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{(s+1)^3}$ | | $(s^2+as+b)$ | $\frac{As+B}{s^2+as+b}$ | $\frac{s}{(s^2+1)(s+2)} = \frac{As+B}{s^2+1} + \frac{C}{s+2}$ | | $(s^2+as+b)^2$ | $\frac{As+B}{s^2+as+b} + \frac{Cs+D}{(s^2+as+b)^2}$ | $\frac{1}{(s^2+1)^2} = \frac{As+B}{s^2+1} + \frac{Cs+D}{(s^2+1)^2}$ | ### Step-by-Step Procedure 1. **Check Degrees:** Ensure degree($N(s)$) ### Cover-Up Method (Heaviside's Method) - **When to Use:** Exclusively for **distinct linear factors** (simple poles) of the form $(s-a)$. - **Shortcut Formula:** To find the constant $A_k$ corresponding to a factor $(s-s_k)$, cover up that factor in the original function $F(s)$ and substitute $s=s_k$ into the remaining expression. $$ A_k = \left[ (s-s_k) \frac{N(s)}{D(s)} \right]_{s=s_k} $$ Or more generally, if $F(s) = \frac{N(s)}{(s-s_k)D'(s)}$, then $A_k = \left[ \frac{N(s)}{D'(s)} \right]_{s=s_k}$ #### Solved Examples: **Example 1:** $$ F(s) = \frac{s+3}{(s+1)(s-2)} = \frac{A}{s+1} + \frac{B}{s-2} $$ $$ A = \left[ \frac{s+3}{s-2} \right]_{s=-1} = \frac{-1+3}{-1-2} = \frac{2}{-3} = -\frac{2}{3} $$ $$ B = \left[ \frac{s+3}{s+1} \right]_{s=2} = \frac{2+3}{2+1} = \frac{5}{3} $$ $$ \implies F(s) = \frac{-2/3}{s+1} + \frac{5/3}{s-2} $$ **Example 2:** $$ F(s) = \frac{1}{s(s+2)(s-1)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s-1} $$ $$ A = \left[ \frac{1}{(s+2)(s-1)} \right]_{s=0} = \frac{1}{(0+2)(0-1)} = \frac{1}{-2} = -\frac{1}{2} $$ $$ B = \left[ \frac{1}{s(s-1)} \right]_{s=-2} = \frac{1}{(-2)(-2-1)} = \frac{1}{(-2)(-3)} = \frac{1}{6} $$ $$ C = \left[ \frac{1}{s(s+2)} \right]_{s=1} = \frac{1}{(1)(1+2)} = \frac{1}{3} $$ $$ \implies F(s) = -\frac{1/2}{s} + \frac{1/6}{s+2} + \frac{1/3}{s-1} $$ **Example 3:** (Mixed with a quadratic, cover-up for linear factor only) $$ F(s) = \frac{s+1}{s(s^2+4)} = \frac{A}{s} + \frac{Bs+C}{s^2+4} $$ $$ A = \left[ \frac{s+1}{s^2+4} \right]_{s=0} = \frac{0+1}{0^2+4} = \frac{1}{4} $$ (For $B$ and $C$, use equating coefficients or other methods) ### Repeated Poles (Linear Factors) When $D(s)$ has a factor $(s-a)^k$, the expansion includes terms $\frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + ... + \frac{A_k}{(s-a)^k}$. - **Finding $A_k$ (the highest power term):** Use the cover-up method for the highest power term. $$ A_k = \left[ (s-a)^k \frac{N(s)}{D(s)} \right]_{s=a} $$ - **Finding $A_j$ (lower power terms):** For $A_j$ where $j ### Quadratic Factors - **Why $(As+B)$?** When the denominator has an irreducible quadratic factor $(s^2+as+b)$, the numerator of its partial fraction term must be a general linear polynomial $(As+B)$ to account for all possible degrees in the numerator of the original proper fraction. If it were just $A$, it would imply that the original numerator's degree is at least 1 less than it truly is for that factor. - **Solving for Constants:** Typically involves equating coefficients after finding constants for any linear factors. Substituting complex roots (if applicable) can also work but is often more complex. #### Solved Examples: **Example 1:** $$ F(s) = \frac{s+1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1} $$ 1. **Find A (cover-up):** $$ A = \left[ \frac{s+1}{s^2+1} \right]_{s=0} = \frac{0+1}{0^2+1} = 1 $$ 2. **Find B, C (equating coefficients):** $$ \frac{s+1}{s(s^2+1)} = \frac{1}{s} + \frac{Bs+C}{s^2+1} $$ Multiply by $s(s^2+1)$: $$ s+1 = 1(s^2+1) + (Bs+C)s $$ $$ s+1 = s^2+1 + Bs^2 + Cs $$ $$ s+1 = (1+B)s^2 + Cs + 1 $$ Equating coefficients: - $s^2$: $0 = 1+B \implies B = -1$ - $s^1$: $1 = C \implies C = 1$ - $s^0$: $1 = 1$ (consistent) Thus, $F(s) = \frac{1}{s} + \frac{-s+1}{s^2+1} = \frac{1}{s} - \frac{s}{s^2+1} + \frac{1}{s^2+1}$. **Example 2:** $$ F(s) = \frac{2s^2+s+1}{(s+1)(s^2+2s+2)} = \frac{A}{s+1} + \frac{Bs+C}{s^2+2s+2} $$ 1. **Find A (cover-up):** $$ A = \left[ \frac{2s^2+s+1}{s^2+2s+2} \right]_{s=-1} = \frac{2(-1)^2+(-1)+1}{(-1)^2+2(-1)+2} = \frac{2-1+1}{1-2+2} = \frac{2}{1} = 2 $$ 2. **Find B, C (equating coefficients):** $$ \frac{2s^2+s+1}{(s+1)(s^2+2s+2)} = \frac{2}{s+1} + \frac{Bs+C}{s^2+2s+2} $$ Multiply by $(s+1)(s^2+2s+2)$: $$ 2s^2+s+1 = 2(s^2+2s+2) + (Bs+C)(s+1) $$ $$ 2s^2+s+1 = 2s^2+4s+4 + Bs^2+Bs+Cs+C $$ $$ 2s^2+s+1 = (2+B)s^2 + (4+B+C)s + (4+C) $$ Equating coefficients: - $s^2$: $2 = 2+B \implies B = 0$ - $s^0$: $1 = 4+C \implies C = -3$ - $s^1$: $1 = 4+B+C \implies 1 = 4+0-3 \implies 1 = 1$ (consistent) Thus, $F(s) = \frac{2}{s+1} + \frac{-3}{s^2+2s+2}$. Note: $s^2+2s+2 = (s+1)^2+1$, which is a standard form for inverse Laplace. ### Laplace Transform Connection - **Why PFE?** Inverse Laplace Transforms are typically found by matching $F(s)$ to standard transform pairs. Complicated $F(s)$ functions (rational polynomials) are not directly in these forms. PFE breaks $F(s)$ into simpler terms that *are* in the standard forms, allowing for easy inverse transformation term by term. - **Common Laplace Pairs for PFE:** | $F(s)$ | $f(t)$ | |---|---| | $\frac{1}{s}$ | $u(t)$ or $1$ | | $\frac{1}{s-a}$ | $e^{at}u(t)$ | | $\frac{1}{(s-a)^n}$ | $\frac{t^{n-1}}{(n-1)!}e^{at}u(t)$ | | $\frac{s}{s^2+\omega^2}$ | $\cos(\omega t)u(t)$ | | $\frac{\omega}{s^2+\omega^2}$ | $\sin(\omega t)u(t)$ | | $\frac{s-a}{(s-a)^2+\omega^2}$ | $e^{at}\cos(\omega t)u(t)$ | | $\frac{\omega}{(s-a)^2+\omega^2}$ | $e^{at}\sin(\omega t)u(t)$ | #### Example: Converting $F(s) \to f(t)$ Let's use the result from Example 1 in "Quadratic Factors": $$ F(s) = \frac{1}{s} - \frac{s}{s^2+1} + \frac{1}{s^2+1} $$ Using the table: - $\mathcal{L}^{-1}\left\{ \frac{1}{s} \right\} = 1$ - $\mathcal{L}^{-1}\left\{ \frac{s}{s^2+1} \right\} = \cos(t)$ - $\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1} \right\} = \sin(t)$ Therefore, $$ f(t) = 1 - \cos(t) + \sin(t) \quad \text{for } t \ge 0 $$ ### 10 Fully Worked Examples #### 1. Easy (Distinct Linear) $$ F(s) = \frac{5s-1}{s^2-s-2} = \frac{5s-1}{(s-2)(s+1)} = \frac{A}{s-2} + \frac{B}{s+1} $$ $A = \left[ \frac{5s-1}{s+1} \right]_{s=2} = \frac{5(2)-1}{2+1} = \frac{9}{3} = 3$ $B = \left[ \frac{5s-1}{s-2} \right]_{s=-1} = \frac{5(-1)-1}{-1-2} = \frac{-6}{-3} = 2$ $$ F(s) = \frac{3}{s-2} + \frac{2}{s+1} $$ #### 2. Moderate (Distinct Linear) $$ F(s) = \frac{s^2+2s+3}{s(s+1)(s+2)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2} $$ $A = \left[ \frac{s^2+2s+3}{(s+1)(s+2)} \right]_{s=0} = \frac{3}{(1)(2)} = \frac{3}{2}$ $B = \left[ \frac{s^2+2s+3}{s(s+2)} \right]_{s=-1} = \frac{(-1)^2+2(-1)+3}{(-1)(-1+2)} = \frac{1-2+3}{(-1)(1)} = \frac{2}{-1} = -2$ $C = \left[ \frac{s^2+2s+3}{s(s+1)} \right]_{s=-2} = \frac{(-2)^2+2(-2)+3}{(-2)(-2+1)} = \frac{4-4+3}{(-2)(-1)} = \frac{3}{2}$ $$ F(s) = \frac{3/2}{s} - \frac{2}{s+1} + \frac{3/2}{s+2} $$ #### 3. Repeated Poles $$ F(s) = \frac{s+3}{s(s+1)^2} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} $$ $A = \left[ \frac{s+3}{(s+1)^2} \right]_{s=0} = \frac{3}{(1)^2} = 3$ $C = \left[ \frac{s+3}{s} \right]_{s=-1} = \frac{-1+3}{-1} = \frac{2}{-1} = -2$ For $B$: Use $s=1$ (or any other non-pole value): $$ \frac{1+3}{1(1+1)^2} = \frac{A}{1} + \frac{B}{1+1} + \frac{C}{(1+1)^2} $$ $$ \frac{4}{4} = A + \frac{B}{2} + \frac{C}{4} \implies 1 = 3 + \frac{B}{2} - \frac{2}{4} = 3 + \frac{B}{2} - \frac{1}{2} $$ $$ 1 = \frac{5}{2} + \frac{B}{2} \implies 2 = 5+B \implies B = -3 $$ $$ F(s) = \frac{3}{s} - \frac{3}{s+1} - \frac{2}{(s+1)^2} $$ #### 4. Quadratic Terms $$ F(s) = \frac{s^2+1}{s(s^2+4s+5)} = \frac{A}{s} + \frac{Bs+C}{s^2+4s+5} $$ $s^2+4s+5 = (s+2)^2+1$, which is irreducible. $A = \left[ \frac{s^2+1}{s^2+4s+5} \right]_{s=0} = \frac{1}{5}$ Multiply by $s(s^2+4s+5)$: $$ s^2+1 = A(s^2+4s+5) + (Bs+C)s $$ $$ s^2+1 = \frac{1}{5}(s^2+4s+5) + Bs^2+Cs $$ $$ s^2+1 = (\frac{1}{5}+B)s^2 + (\frac{4}{5}+C)s + 1 $$ Equating coefficients: $s^2$: $1 = \frac{1}{5}+B \implies B = \frac{4}{5}$ $s^1$: $0 = \frac{4}{5}+C \implies C = -\frac{4}{5}$ $$ F(s) = \frac{1/5}{s} + \frac{(4/5)s - 4/5}{s^2+4s+5} = \frac{1}{5s} + \frac{4}{5} \frac{s-1}{s^2+4s+5} $$ #### 5. Mixed Factors (Linear, Repeated Linear, Quadratic) $$ F(s) = \frac{s^2+2s+1}{s(s+1)^2(s^2+1)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{Ds+E}{s^2+1} $$ Notice $s^2+2s+1 = (s+1)^2$. So, $F(s) = \frac{(s+1)^2}{s(s+1)^2(s^2+1)} = \frac{1}{s(s^2+1)}$ (for $s \ne -1$). This simplifies to Example 1 in "Quadratic Factors" section: $$ F(s) = \frac{1}{s} - \frac{s}{s^2+1} + \frac{1}{s^2+1} $$ *(Self-correction: Always simplify first! If the numerator has a common factor with the denominator, cancel it.)* Let's do a more complex mixed example that doesn't simplify: $$ F(s) = \frac{s+2}{s(s-1)(s^2+s+1)} = \frac{A}{s} + \frac{B}{s-1} + \frac{Cs+D}{s^2+s+1} $$ $A = \left[ \frac{s+2}{(s-1)(s^2+s+1)} \right]_{s=0} = \frac{2}{(-1)(1)} = -2$ $B = \left[ \frac{s+2}{s(s^2+s+1)} \right]_{s=1} = \frac{1+2}{1(1+1+1)} = \frac{3}{3} = 1$ Multiply by $s(s-1)(s^2+s+1)$: $$ s+2 = A(s-1)(s^2+s+1) + B s(s^2+s+1) + (Cs+D)s(s-1) $$ $$ s+2 = -2(s^3-1) + 1(s^3+s^2+s) + (Cs+D)(s^2-s) $$ $$ s+2 = -2s^3+2 + s^3+s^2+s + Cs^3-Cs^2+Ds^2-Ds $$ $$ s+2 = (-2+1+C)s^3 + (1-C+D)s^2 + (1-D)s + 2 $$ Equating coefficients: $s^3$: $0 = -2+1+C \implies C=1$ $s^2$: $0 = 1-C+D \implies 0 = 1-1+D \implies D=0$ $s^1$: $1 = 1-D \implies 1 = 1-0 \implies 1=1$ (consistent) $s^0$: $2 = 2$ (consistent) $$ F(s) = \frac{-2}{s} + \frac{1}{s-1} + \frac{s}{s^2+s+1} $$ #### 6. Improper Fraction (Long Division First) $$ F(s) = \frac{s^3+s^2+2s+1}{s^2+s} $$ Degree of numerator (3) $\ge$ degree of denominator (2). Perform long division: $$ (s^3+s^2+2s+1) / (s^2+s) = s + \frac{2s+1}{s^2+s} $$ So, $F(s) = s + \frac{2s+1}{s(s+1)}$. Now PFE for $\frac{2s+1}{s(s+1)}$: $$ \frac{2s+1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} $$ $A = \left[ \frac{2s+1}{s+1} \right]_{s=0} = \frac{1}{1} = 1$ $B = \left[ \frac{2s+1}{s} \right]_{s=-1} = \frac{2(-1)+1}{-1} = \frac{-1}{-1} = 1$ $$ F(s) = s + \frac{1}{s} + \frac{1}{s+1} $$ #### 7. Another Repeated Linear Factor Example $$ F(s) = \frac{1}{(s-1)^3} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{C}{(s-1)^3} $$ This is already in PFE form. $A=0, B=0, C=1$. Let's try a different one: $$ F(s) = \frac{s+1}{(s-1)^3} $$ $C = \left[ s+1 \right]_{s=1} = 2$ $B = \left[ \frac{d}{ds}(s+1) \right]_{s=1} = [1]_{s=1} = 1$ $A = \frac{1}{2!} \left[ \frac{d^2}{ds^2}(s+1) \right]_{s=1} = \frac{1}{2} [0]_{s=1} = 0$ $$ F(s) = \frac{1}{(s-1)^2} + \frac{2}{(s-1)^3} $$ #### 8. Laplace Inverse Problem (Distinct Linear) $$ F(s) = \frac{s+7}{s^2-3s-4} = \frac{s+7}{(s-4)(s+1)} = \frac{A}{s-4} + \frac{B}{s+1} $$ $A = \left[ \frac{s+7}{s+1} \right]_{s=4} = \frac{4+7}{4+1} = \frac{11}{5}$ $B = \left[ \frac{s+7}{s-4} \right]_{s=-1} = \frac{-1+7}{-1-4} = \frac{6}{-5} = -\frac{6}{5}$ $$ F(s) = \frac{11/5}{s-4} - \frac{6/5}{s+1} $$ $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{11}{5}e^{4t} - \frac{6}{5}e^{-t} $$ #### 9. Laplace Inverse Problem (Quadratic) $$ F(s) = \frac{2s+3}{s^2+2s+5} $$ Denominator is irreducible: $s^2+2s+5 = (s+1)^2+4$. $$ F(s) = \frac{2s+3}{(s+1)^2+4} = \frac{2(s+1)+1}{(s+1)^2+4} = 2 \frac{s+1}{(s+1)^2+2^2} + \frac{1}{2} \frac{2}{(s+1)^2+2^2} $$ $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = 2e^{-t}\cos(2t) + \frac{1}{2}e^{-t}\sin(2t) $$ *(This example shows direct manipulation without PFE, but PFE is used when there are other factors)*. Let's use a PFE example: $$ F(s) = \frac{s+1}{s(s^2+1)} = \frac{1}{s} - \frac{s}{s^2+1} + \frac{1}{s^2+1} $$ $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = 1 - \cos(t) + \sin(t) $$ #### 10. Laplace Inverse Problem (Repeated Linear) $$ F(s) = \frac{s+2}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} $$ $B = \left[ \frac{s+2}{s+1} \right]_{s=0} = \frac{2}{1} = 2$ $C = \left[ \frac{s+2}{s^2} \right]_{s=-1} = \frac{-1+2}{(-1)^2} = \frac{1}{1} = 1$ For A: $$ A = \left[ \frac{d}{ds} \left( \frac{s+2}{s+1} \right) \right]_{s=0} = \left[ \frac{(1)(s+1) - (s+2)(1)}{(s+1)^2} \right]_{s=0} $$ $$ A = \left[ \frac{s+1-s-2}{(s+1)^2} \right]_{s=0} = \left[ \frac{-1}{(s+1)^2} \right]_{s=0} = \frac{-1}{1^2} = -1 $$ $$ F(s) = -\frac{1}{s} + \frac{2}{s^2} + \frac{1}{s+1} $$ $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = -1 + 2t + e^{-t} $$ ### Common Mistakes in Exams - **Forgetting Long Division:** Not checking if degree($N(s)$) ### Exam Shortcuts Section - **Fast Solving Tips:** - **Cover-Up Method First:** Always use the cover-up method for all distinct linear factors. It's the fastest way to get some constants. - **Strategic Substitution:** After finding constants for distinct linear factors, substitute $s=0$ (if $s=0$ is not a pole) or $s=1$ (or any other easy non-pole value) into the full equation to quickly find remaining constants, especially for terms associated with repeated or quadratic factors. - **Equating Highest/Lowest Power Coefficients:** For quadratic or repeated factors, equate coefficients of the highest power of $s$ ($s^n$) and the constant term ($s^0$) first. These often give simpler equations. - **Pattern Recognition Tricks:** - If $N(s)$ is simple (e.g., constant or $s$), often many constants are zero or simple. - Recognize forms like $\frac{s}{(s^2+\omega^2)}$ or $\frac{\omega}{(s^2+\omega^2)}$ directly for inverse Laplace. - For terms like $\frac{s+k}{(s-a)^2+b^2}$, rewrite $s+k = (s-a) + (k+a)$ to match $\mathcal{L}^{-1}\{e^{at}\cos(\omega t)\}$ and $\mathcal{L}^{-1}\{e^{at}\sin(\omega t)\}$ forms. - **When to Avoid Equating Coefficients:** If you only have one or two constants left after using cover-up and strategic substitution, equating coefficients might be overkill. A single substitution of $s=0$ or $s=1$ is often faster. Only resort to full equating coefficients if you have a complex system or many unknown constants remaining. ### Final Summary Table: PFE Forms | Factor Type in $D(s)$ | Example Factor | PFE Term(s) | Notes | |-----------------------|----------------|-------------|-------| | Distinct Linear | $(s-a)$ | $\frac{A}{s-a}$ | Use Cover-Up Method | | Repeated Linear | $(s-a)^k$ | $\frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + ... + \frac{A_k}{(s-a)^k}$ | $A_k$ by Cover-Up. Others by derivative or substitution/equating. | | Distinct Irreducible Quadratic | $(s^2+bs+c)$ | $\frac{As+B}{s^2+bs+c}$ | $b^2-4c