Inverse Laplace Transform & Partial Fractions

DaysHoursMinSec

This offer has expired!

SubscribePREMIUM PROMO: Remove Ads!! Get Freebies!!! and Get EXTRA MONTHS before Time EXPIRED!

Going from $f(t)$ to $F(s)$ is a lookup. Going back from $F(s)$ to $f(t)$ is not — because most expressions you encounter on the boards do not match a basic pair directly. You have to break them apart first using partial fraction decomposition, and that decomposition step is where the most points are lost in the entire Laplace Transform topic. This post covers all three decomposition cases — distinct linear factors, repeated linear factors, and irreducible quadratic factors — with 10 fully worked problems at exam-level difficulty. This is Part 2 of the Laplace Transform Series.

📋 BOARD EXAM RELEVANCE

EE (Electrical Engineer) — Very high frequency. Partial fractions appear every time a circuit transient problem needs to be inverted back to the time domain, which is every circuit problem in this series.
ECE (Electronics Engineer) — Very high frequency. Signals problems often hand you $F(s)$ directly and ask for $f(t)$ — partial fractions is the bridge between them in almost every case.
ME (Mechanical Engineer) — Moderate frequency. Vibration and control system IVPs solved by Laplace all require inversion, which comes back here.
CE, ChE, GeE, MetE, MinE, Naval Architect and Marine Engineer — Low to rare. Inversion is tested only when a broader Laplace Transform problem requires it, not as a standalone exam topic for these boards.

Bottom line: If you only have time to master two things in this series, make it the basic pairs from Part 1 and the three decomposition cases from this post. These two skills together cover the majority of Laplace Transform points on the EE and ECE boards.

The Inverse Laplace Transform

The inverse Laplace Transform recovers the original time-domain function from its $s$ -domain transform. Written formally:

THE INVERSE LAPLACE TRANSFORM

$\mathcal{L}^{-1}\{F(s)\} = f(t)$

In practice, you rarely compute this using the formal Bromwich integral. Instead, you rewrite $F(s)$ as a sum of terms that each match one of the basic pairs from Part 1, then invert term by term using linearity. Partial fraction decomposition is the tool that makes that rewriting possible.

The Three Partial Fraction Cases

Every denominator you will encounter on the boards falls into one of three structural types. Recognizing which case applies before you start is the single most important habit to build for this topic.

CASE 1 — DISTINCT LINEAR FACTORS

$\dfrac{N(s)}{(s-a)(s-b)} = \dfrac{A}{s-a} + \dfrac{B}{s-b}$

One constant per distinct linear factor. Solve by substituting the roots $s = a$ and $s = b$ to isolate each constant individually.

CASE 2 — REPEATED LINEAR FACTORS

$\dfrac{N(s)}{(s-a)^2} = \dfrac{A}{s-a} + \dfrac{B}{(s-a)^2}$

One term for every power of the repeated factor from 1 up to $n$ . A single constant for the whole repeated group is the most common structural error on this case.

CASE 3 — IRREDUCIBLE QUADRATIC FACTORS

$\dfrac{N(s)}{s^2+k^2} \rightarrow \dfrac{As+B}{s^2+k^2}$

A quadratic factor that will not factor over the reals gets a linear numerator $As+B$ , not a constant. This case always produces sine and cosine terms in the final time-domain answer.

One rule before you start decomposing: Check whether the degree of $N(s)$ is less than the degree of the denominator. If it is equal or higher, do polynomial long division first — partial fractions only work on proper rational functions.

10 Worked Board Exam Problems

Problem 1. Invert a Transform With Two Distinct Linear Factors

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{7s-1}{(s+1)(s-2)}\right\}$ .

Given: $F(s) = \dfrac{7s-1}{(s+1)(s-2)}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Set up Case 1 decomposition, one constant per distinct linear factor.

$\dfrac{7s-1}{(s+1)(s-2)} = \dfrac{A}{s+1} + \dfrac{B}{s-2}$

Step 2: Clear denominators, then substitute $s = -1$ to isolate $A$ and $s = 2$ to isolate $B$ .

$7s-1 = A(s-2) + B(s+1)$

$s=-1: \quad -8 = A(-3) \implies A = \dfrac{8}{3}$

$s=2: \quad 13 = B(3) \implies B = \dfrac{13}{3}$

Step 3: Rewrite $F(s)$ with the found constants.

$F(s) = \dfrac{8/3}{s+1} + \dfrac{13/3}{s-2}$

Step 4: Invert each term using $\mathcal{L}^{-1}\left\{\dfrac{1}{s-a}\right\} = e^{at}$ .

✓ ANSWER: $f(t) = \dfrac{8}{3}e^{-t} + \dfrac{13}{3}e^{2t}$

Examiner note: Substitution at the roots is always faster than expanding and matching coefficients for distinct linear factors. Save the coefficient-matching method for cases where substitution does not isolate a single constant cleanly.

Problem 2. Invert a Transform With Three Distinct Linear Factors

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{10}{s(s+1)(s+2)}\right\}$ .

Given: $F(s) = \dfrac{10}{s(s+1)(s+2)}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Set up Case 1 with three constants, one per factor.

$\dfrac{10}{s(s+1)(s+2)} = \dfrac{A}{s} + \dfrac{B}{s+1} + \dfrac{C}{s+2}$

Step 2: Clear denominators, then substitute roots $s = 0$ , $s = -1$ , and $s = -2$ .

$10 = A(s+1)(s+2) + Bs(s+2) + Cs(s+1)$

$s=0: \quad 10 = A(1)(2) \implies A = 5$

$s=-1: \quad 10 = B(-1)(1) \implies B = -10$

$s=-2: \quad 10 = C(-2)(-1) \implies C = 5$

Step 3: Rewrite and invert term by term.

$F(s) = \dfrac{5}{s} - \dfrac{10}{s+1} + \dfrac{5}{s+2}$

✓ ANSWER: $f(t) = 5 - 10e^{-t} + 5e^{-2t}$

Examiner note: $\mathcal{L}^{-1}\{1/s\} = 1$ , not $t$ . A bare $1/s$ term inverts to a constant, not a ramp. This is the most common inversion slip when the denominator contains a plain $s$ factor.

Problem 3. Invert a Transform With a Repeated Linear Factor

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{s+3}{(s-1)^2}\right\}$ .

Given: $F(s) = \dfrac{s+3}{(s-1)^2}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Set up Case 2 decomposition — two terms for the squared factor.

$\dfrac{s+3}{(s-1)^2} = \dfrac{A}{s-1} + \dfrac{B}{(s-1)^2}$

Step 2: Clear denominators.

$s+3 = A(s-1) + B$

Step 3: Substitute $s = 1$ to isolate $B$ , then match coefficients of $s$ to find $A$ .

$s=1: \quad 4 = B \implies B = 4$

$\text{Coefficient of } s: \quad 1 = A \implies A = 1$

Step 4: Invert each term. Recall $\mathcal{L}^{-1}\left\{\dfrac{1}{(s-a)^2}\right\} = te^{at}$ .

$F(s) = \dfrac{1}{s-1} + \dfrac{4}{(s-1)^2}$

✓ ANSWER: $f(t) = e^{t} + 4te^{t}$

Examiner note: $\mathcal{L}^{-1}\left\{1/(s-a)^2\right\} = te^{at}$ , not just $e^{at}$ . The squared denominator introduces a factor of $t$ in the time domain, which comes from the $n!/s^{n+1}$ pair shifted by $a$ .

Problem 4. Invert a Transform With a Repeated Factor and a Simple Factor

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{4s}{(s+1)^2(s-2)}\right\}$ .

Given: $F(s) = \dfrac{4s}{(s+1)^2(s-2)}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Mixed case — repeated factor $(s+1)^2$ needs two terms, and the simple factor $(s-2)$ needs one.

$\dfrac{4s}{(s+1)^2(s-2)} = \dfrac{A}{s+1} + \dfrac{B}{(s+1)^2} + \dfrac{C}{s-2}$

Step 2: Clear denominators.

$4s = A(s+1)(s-2) + B(s-2) + C(s+1)^2$

Step 3: Substitute roots to isolate $B$ and $C$ .

$s=-1: \quad -4 = B(-3) \implies B = \dfrac{4}{3}$

$s=2: \quad 8 = C(9) \implies C = \dfrac{8}{9}$

Step 4: Expand and match the coefficient of $s^2$ to find $A$ .

$s^2 \text{ coefficient}: \quad 0 = A + C \implies A = -\dfrac{8}{9}$

Step 5: Invert all three terms.

✓ ANSWER: $f(t) = -\dfrac{8}{9}e^{-t} + \dfrac{4}{3}te^{-t} + \dfrac{8}{9}e^{2t}$

Examiner note: When substitution cannot isolate $A$ directly, always fall back to expanding and matching coefficients. Never guess — one wrong constant contaminates the entire final answer.

Problem 5. Invert a Transform With an Irreducible Quadratic Factor

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{3s+5}{s^2+4}\right\}$ .

Given: $F(s) = \dfrac{3s+5}{s^2+4}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: The denominator $s^2+4$ does not factor over the reals — this is Case 3. Split the numerator to match the cosine and sine pair shapes.

$\dfrac{3s+5}{s^2+4} = 3\left(\dfrac{s}{s^2+4}\right) + 5\left(\dfrac{1}{s^2+4}\right)$

Step 2: The cosine pair is ready. For the sine pair, force the $k = 2$ factor into the numerator by multiplying and dividing.

$\dfrac{1}{s^2+4} = \dfrac{1}{2}\left(\dfrac{2}{s^2+4}\right)$

Step 3: Invert each piece separately.

$\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+4}\right\} = \cos(2t), \quad \mathcal{L}^{-1}\left\{\dfrac{2}{s^2+4}\right\} = \sin(2t)$

✓ ANSWER: $f(t) = 3\cos(2t) + \dfrac{5}{2}\sin(2t)$

Examiner note: The sine pair requires $k$ in the numerator — $k/(s^2+k^2)$ . If your numerator is just 1 or some other constant, multiply and divide by $k$ to force the correct form before inverting.

Problem 6. Invert a Transform With Mixed Linear and Quadratic Factors

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{2s^2+5s+3}{(s+1)(s^2+4)}\right\}$ .

Given: $F(s) = \dfrac{2s^2+5s+3}{(s+1)(s^2+4)}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Mixed Cases 1 and 3. The linear factor gets a constant, the quadratic gets a linear numerator.

$\dfrac{2s^2+5s+3}{(s+1)(s^2+4)} = \dfrac{A}{s+1} + \dfrac{Bs+C}{s^2+4}$

Step 2: Clear denominators.

$2s^2+5s+3 = A(s^2+4) + (Bs+C)(s+1)$

Step 3: Substitute $s = -1$ to isolate $A$ .

$s=-1: \quad 2-5+3 = A(5) \implies A = 0$

Step 4: With $A = 0$ , expand the right side and match coefficients.

$2s^2+5s+3 = Bs^2 + (B+C)s + C$

$s^2: B = 2 \quad s^1: B+C = 5 \implies C = 3$

Step 5: Invert each term. Note $A = 0$ so the exponential term drops out entirely.

$F(s) = \dfrac{2s}{s^2+4} + \dfrac{3}{s^2+4} = 2\cdot\dfrac{s}{s^2+4} + \dfrac{3}{2}\cdot\dfrac{2}{s^2+4}$

✓ ANSWER: $f(t) = 2\cos(2t) + \dfrac{3}{2}\sin(2t)$

Examiner note: $A = 0$ is a valid answer — do not second-guess it. When a constant comes out as zero it means that particular factor has no contribution to the final answer. Rechecking the arithmetic on that zero is good practice, but forcing a non-zero value instead is not.

Problem 7. Handle an Improper Rational Function Before Decomposing

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{s^2+2s+3}{s^2+2s}\right\}$ .

Given: $F(s) = \dfrac{s^2+2s+3}{s^2+2s}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: The numerator and denominator have equal degree — this is improper. Do polynomial long division first.

$\dfrac{s^2+2s+3}{s^2+2s} = 1 + \dfrac{3}{s^2+2s}$

Step 2: Factor the remainder’s denominator and decompose.

$\dfrac{3}{s^2+2s} = \dfrac{3}{s(s+2)} = \dfrac{A}{s} + \dfrac{B}{s+2}$

$s=0: A = \dfrac{3}{2}, \quad s=-2: B = -\dfrac{3}{2}$

Step 3: The full expression is now in invertible form.

$F(s) = 1 + \dfrac{3/2}{s} - \dfrac{3/2}{s+2}$

Step 4: Invert each term. The leading $1$ is a constant and inverts to $\delta(t)$ (the Dirac delta, from Part 4), but for most board exam contexts this constant is just noted as a coefficient.

✓ ANSWER: $f(t) = \delta(t) + \dfrac{3}{2} - \dfrac{3}{2}e^{-2t}$

Examiner note: Partial fractions only works when the degree of the numerator is strictly less than the degree of the denominator. If not, long division first — skipping this step and trying to decompose directly will give wrong constants that will not check out when recombined.

Problem 8. Invert a Transform With a Repeated Quadratic Factor

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{1}{(s^2+1)^2}\right\}$ .

Given: $F(s) = \dfrac{1}{(s^2+1)^2}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: This does not decompose by simple substitution. Use the known transform pair for this specific form.

$\mathcal{L}\left\{\dfrac{1}{2}(\sin t - t\cos t)\right\} = \dfrac{1}{(s^2+1)^2}$

Step 2: This is a standard board exam pair that should be recognized directly.

✓ ANSWER: $f(t) = \dfrac{1}{2}(\sin t - t\cos t)$

Examiner note: $(s^2+k^2)^2$ in the denominator is a repeated quadratic factor that appears on the EE and ECE boards specifically in RLC resonance problems. If it appears, check whether it matches the standard pair for $(\sin kt - kt\cos kt)/(2k^3)$ directly before attempting long decomposition.

Problem 9. Identify the Value of a Constant From a Partial Fraction Identity

Problem: Given that $\dfrac{5s-2}{(s-1)(s+2)} = \dfrac{A}{s-1} + \dfrac{B}{s+2}$ , find the value of $A$ .

Given: Partial fraction identity as stated

Find: $A$

Solution:

Step 1: Clear denominators on both sides.

$5s-2 = A(s+2) + B(s-1)$

Step 2: Substitute $s = 1$ to zero out the $B$ term.

$5(1)-2 = A(1+2) + B(0)$

$3 = 3A$

✓ ANSWER: $A = 1$

Examiner note: This “find the constant” format is the most direct version of the partial fraction skill — the board gives you the decomposition structure already and asks for a single constant. Substitution at the root that eliminates every other term is the fastest path to the answer.

Problem 10. Full Inversion Problem in an EE Circuit Context

Problem: The impedance-transformed current in an RLC circuit is given by $I(s) = \dfrac{5s}{(s^2+4)(s+3)}$ . Find $i(t)$ .

Given: $I(s) = \dfrac{5s}{(s^2+4)(s+3)}$

Find: $i(t) = \mathcal{L}^{-1}\{I(s)\}$

Solution:

Step 1: Mixed Cases 1 and 3. Quadratic factor gets $As+B$ , linear factor gets $C$ .

$\dfrac{5s}{(s^2+4)(s+3)} = \dfrac{As+B}{s^2+4} + \dfrac{C}{s+3}$

Step 2: Clear denominators.

$5s = (As+B)(s+3) + C(s^2+4)$

Step 3: Substitute $s = -3$ to find $C$ .

$-15 = C(9+4) = 13C \implies C = -\dfrac{15}{13}$

Step 4: Expand the right side and match coefficients to find $A$ and $B$ .

$5s = As^2 + (3A+B)s + (3B + 4C) + Cs^2$

$s^2: \quad 0 = A + C \implies A = \dfrac{15}{13}$

$s^0: \quad 0 = 3B + 4(-15/13) \implies B = \dfrac{20}{13}$

Step 5: Rewrite with all constants, then invert term by term.

$I(s) = \dfrac{15}{13}\cdot\dfrac{s}{s^2+4} + \dfrac{20}{13}\cdot\dfrac{1}{s^2+4} - \dfrac{15}{13}\cdot\dfrac{1}{s+3}$

$= \dfrac{15}{13}\cdot\dfrac{s}{s^2+4} + \dfrac{10}{13}\cdot\dfrac{2}{s^2+4} - \dfrac{15}{13}\cdot\dfrac{1}{s+3}$

✓ ANSWER: $i(t) = \dfrac{15}{13}\cos(2t) + \dfrac{10}{13}\sin(2t) - \dfrac{15}{13}e^{-3t}$

Examiner note: This problem type — $F(s)$ from a physical system, three-term mixed decomposition, invert to get the time-domain response — is exactly what appears in EE and ECE circuit transient problems. The math is identical whether the problem says “current in an RLC circuit” or just hands you the fraction directly.

Common Mistakes and Examiner Traps

❌ Common Mistake	✅ Correct Approach
Trying to decompose an improper fraction directly without long division first	Always check degree of numerator vs denominator first. If numerator degree $\geq$ denominator degree, do long division before setting up partial fractions.
Writing only one term for a repeated factor $(s-a)^n$ instead of $n$ terms	A factor $(s-a)^2$ needs both $A/(s-a)$ and $B/(s-a)^2$ . Every power from 1 to $n$ gets its own term.
Using a constant numerator for a quadratic factor instead of a linear one	An irreducible quadratic factor always gets $As+B$ as its numerator, not just $A$ . A constant alone cannot reproduce the correct partial fraction identity.
Inverting $1/(s-a)^2$ as $e^{at}$ instead of $te^{at}$	$\mathcal{L}^{-1}\{1/(s-a)\} = e^{at}$ , but $\mathcal{L}^{-1}\{1/(s-a)^2\} = te^{at}$ . The squared denominator always introduces a factor of $t$ .
Inverting $1/s$ as $t$ instead of $1$	$\mathcal{L}^{-1}\{1/s\} = 1$ (a constant). $\mathcal{L}^{-1}\{1/s^2\} = t$ (a ramp). The power of $s$ in the denominator matters.
Forgetting to force $k$ into the numerator before inverting the sine pair	The sine pair is $k/(s^2+k^2)$ . If you have $1/(s^2+k^2)$ , multiply and divide by $k$ to produce $\frac{1}{k} \cdot \frac{k}{s^2+k^2}$ before inverting.

Board Exam Quick Tips

Classify the denominator before writing anything. Is it distinct linear, repeated, or quadratic? The case determines the structure of the decomposition, and writing the wrong structure means the constants you find will not check out.
Substitution at the roots is almost always the fastest method for Case 1. For Cases 2 and 3, combine substitution with coefficient matching for the constants that substitution cannot reach.
Always verify your constants by recombining. Mentally add the partial fractions back together and check that you get the original numerator — one wrong sign here fails the entire problem silently.
An irreducible quadratic in the denominator always means sine and cosine in the final answer. If you get only exponentials from a problem whose denominator has a quadratic factor, something went wrong.
If a constant comes out as zero, accept it. A zero constant is a valid answer and means that particular factor contributes nothing to the final $f(t)$ . Do not try to force a non-zero value.

Frequently Asked Questions

Q1. How do I know which decomposition case to use?

Look at the denominator alone, before the numerator. If all factors are linear and distinct, use Case 1. If any linear factor is repeated (raised to a power), use Case 2 for that factor. If the denominator contains a quadratic that has no real roots (negative discriminant), use Case 3 for that factor. A denominator can mix cases — a repeated linear factor and a quadratic factor in the same denominator requires both Case 2 and Case 3 structure simultaneously.

Q2. What is the discriminant test for an irreducible quadratic?

For $as^2 + bs + c$ , compute $b^2 - 4ac$ . If this is negative, the quadratic has no real roots and is irreducible — use Case 3 with a linear numerator $As+B$ . If the discriminant is zero or positive, the quadratic factors into real linear terms and you should factor it first, then use Cases 1 or 2.

Q3. Can I always find all the constants by substituting roots?

For Case 1 (all distinct linear factors), yes — you get exactly one useful substitution per constant. For Case 2 (repeated factors) and Case 3 (quadratic factors), substitution can isolate some constants but not all. You will need to expand the cleared-denominator equation and match coefficients of powers of $s$ to find the remaining ones.

Q4. What happens after I complete the partial fraction decomposition?

Each term in the decomposition matches one of the basic inverse pairs from Part 1. Apply linearity to invert term by term. Every Case 1 term becomes an exponential, every Case 2 term involving $(s-a)^n$ involves $t^{n-1}e^{at}$ , and every Case 3 term involving $s^2+k^2$ becomes sine or cosine.

Q5. What is the difference between $\mathcal{L}^{-1}\{1/s\}$ and $\mathcal{L}^{-1}\{1/s^2\}$ ?

$\mathcal{L}^{-1}\{1/s\} = 1$ (a constant equal to 1). $\mathcal{L}^{-1}\{1/s^2\} = t$ (a ramp). In general, $\mathcal{L}^{-1}\{1/s^{n+1}\} = t^n/n!$ . The power of $s$ in the denominator directly controls the power of $t$ in the time-domain result.

What is Next

You can now move from $F(s)$ back to $f(t)$ for any rational function the boards are likely to put in front of you. In Part 3, the direction of the problem shifts again — instead of inverting a given $F(s)$ , you will use the Laplace Transform to solve initial value problems directly. The derivative transform formulas turn the ODE into an algebraic equation for $Y(s)$ , you solve that algebra, and then you apply exactly what you learned in this post to invert $Y(s)$ back to $y(t)$ . Parts 1 and 2 together are the complete toolkit for Part 3.

→ Continue to Part 3 — Laplace Transform of Derivatives, Integrals, and Solving ODEs

→ Back to the Laplace Transform Series Index

Published by PinoyBIX.org — Engineering Education for Every Filipino Student. Electronics · Mathematics · Board Exam Review · Free for Everyone.

Please do Subscribe on YouTube!

P inoyBIX educates thousands of reviewers and students a day in preparation for their board examinations. Also provides professionals with materials for their lectures and practice exams. Help me go forward with the same spirit.

“Will you subscribe today via YOUTUBE?”

What You Also Get: FREE ACCESS & DOWNLOAD via GDRIVE

TIRED OF ADS?

Become Premium Member and experienced complete ads-free content browsing.
Full Content Access to Premium Solutions Exclusive for Premium members
Access to PINOYBIX FREEBIES folder
Download Reviewers and Learning Materials Free
Download Content: You can see download/print button at the bottom of each post.

PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

CIVIL ENGINEERING REVIEWER
CIVIL SERVICE EXAM REVIEWER
CRIMINOLOGY REVIEWER
ELECTRONICS ENGINEERING REVIEWER (ECE/ECT)
ELECTRICAL ENGINEERING & RME REVIEWER
FIRE OFFICER EXAMINATION REVIEWER
LET REVIEWER
MASTER PLUMBER REVIEWER
MECHANICAL ENGINEERING REVIEWER
NAPOLCOM REVIEWER
Additional upload reviewers and learning materials are also FREE

FOR A LIMITED TIME

If you subscribe for PREMIUM today!

You will receive an additional 1 month of Premium Membership FREE.

For Bronze Membership an additional 2 months of Premium Membership FREE.

For Silver Membership an additional 3 months of Premium Membership FREE.

For Gold Membership an additional 5 months of Premium Membership FREE.

Join the PinoyBIX community.

DaysHoursMinSec

This offer has expired!

SubscribeMembership Plan

Prev Article Next Article

Inverse Laplace Transform & Partial Fractions | Board Exam Reviewer

The Inverse Laplace Transform

The Three Partial Fraction Cases

10 Worked Board Exam Problems

Problem 1. Invert a Transform With Two Distinct Linear Factors

Problem 2. Invert a Transform With Three Distinct Linear Factors

Problem 3. Invert a Transform With a Repeated Linear Factor

Problem 4. Invert a Transform With a Repeated Factor and a Simple Factor

Problem 5. Invert a Transform With an Irreducible Quadratic Factor

Problem 6. Invert a Transform With Mixed Linear and Quadratic Factors

Problem 7. Handle an Improper Rational Function Before Decomposing

Problem 8. Invert a Transform With a Repeated Quadratic Factor

Problem 9. Identify the Value of a Constant From a Partial Fraction Identity

Problem 10. Full Inversion Problem in an EE Circuit Context

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

TIRED OF ADS?

PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

FOR A LIMITED TIME

Add Comment Cancel Reply

FOR A LIMITED TIME

The Inverse Laplace Transform

The Three Partial Fraction Cases

10 Worked Board Exam Problems

Problem 1. Invert a Transform With Two Distinct Linear Factors

Problem 2. Invert a Transform With Three Distinct Linear Factors

Problem 3. Invert a Transform With a Repeated Linear Factor

Problem 4. Invert a Transform With a Repeated Factor and a Simple Factor

Problem 5. Invert a Transform With an Irreducible Quadratic Factor

Problem 6. Invert a Transform With Mixed Linear and Quadratic Factors

Problem 7. Handle an Improper Rational Function Before Decomposing

Problem 8. Invert a Transform With a Repeated Quadratic Factor

Problem 9. Identify the Value of a Constant From a Partial Fraction Identity

Problem 10. Full Inversion Problem in an EE Circuit Context

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

Related Content

201 Terms and Definitions in Algebra for the Mathematics Engineering Board Exam

201 Plane Trigonometry Terms and Definitions | Mathematics Board Exam Review

301 Advanced Engineering Mathematics Terms and Definitions | Mathematics Board Exam Review

Electronics Engineering Mastery Test 2: ECE Pre-Board Practice Exam

TIRED OF ADS?

PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

FOR A LIMITED TIME

Add Comment Cancel Reply