Laplace transforms: practice problems with step-by-step solutions

1What the Laplace transform does

Circuits, motors and tanks are described by differential equations: the output depends on its own rate of change. Solving these directly is slow.

The Laplace transform turns such a problem into algebra. You swap every function of time f(t) for a function F(s) of a new variable s.

You solve the problem with ordinary algebra in s, and then translate the result back to time. That last step is the inverse transform.

You rarely use this integral yourself. In practice you look up transform pairs in a table, like words in a dictionary between two languages.

The figure shows three common signals: a constant 1 (a step), a decaying exponential and a sine. Below each one is its transform.

2The table of transform pairs

Learn these five pairs. Together with two rules they cover almost every exam problem.

• Linearity: constants stay in front, and a sum is transformed term by term. So 3/(s + 2) ↔ 3·e^−2t.
• Shift: replacing s by s + a multiplies the time function by e^−at. So (s + 2)/((s + 2)² + 9) ↔ e^−2t·cos 3t.

Watch the sine pair: the numerator must be ω itself. For 12/(s² + 9) you have ω = 3, so write 12 = 4·3. The result is 4·sin 3t.

In sin ωt and cos ωt the angle ωt is in radians. Set your calculator to radians before you evaluate.

3Start and end values without inverting

Sometimes you only need where f(t) starts or where it ends. Two theorems give these straight from F(s).

• Initial value (s → ∞): for very large s only the highest power of s in the numerator and in the denominator counts.
• Final value (s → 0): cancel the factor s, then set s = 0 in every remaining factor, zeros included.
• The final value theorem only works if f(t) really settles: all poles of s·F(s) must have a negative real part.

Final value: for F(s) = 12/(s(s + 2)(s + 3)) you get s·F(s) = 12/((s + 2)(s + 3)). At s = 0 this is 12/6 = 2, so f(∞) = 2.

Initial value: for F(s) = (3s + 10)/(s² + 5s + 6) you get s·F(s) = (3s² + 10s)/(s² + 5s + 6). For large s this behaves like 3s²/s², so f(0⁺) = 3.

For the initial slope f′(0⁺), use the transform of the derivative, ℒ{f′} = s·F(s) − f(0⁺). Then apply the initial value theorem to it:

For the same F(s): s·F(s) − 3 = (−5s − 18)/(s² + 5s + 6). Multiply by s and let s → ∞. The highest powers give −5s²/s², so f′(0⁺) = −5.

4Partial fractions, step by step

Most F(s) are not in the table as they stand. You split them into simple fractions that are. This is called partial fractions.

Each root of the denominator is a pole. A simple pole at s = −a gives one term A/(s + a), which becomes A·e^−at in time.

The cover-up rule finds each coefficient. To find A, cover the factor (s + a) in F(s) and put s = −a in everything that remains.

Example: F(s) = (s + 7)/((s + 1)(s + 4)). Then A = (−1 + 7)/(−1 + 4) = 6/3 = 2 and B = (−4 + 7)/(−4 + 1) = −1. So f(t) = 2e^−t − e^−4t.

The position of a pole tells you the shape. A real pole at −3 gives a pure decay e^−3t. A complex pair −1 ± j2 gives a decaying oscillation, e^−t times sin 2t or cos 2t.

• Repeated pole (s + a)²: write A/(s + a) + B/(s + a)². The second term gives B·t·e^−at. So (s + 4)/(s + 1)² = 1/(s + 1) + 3/(s + 1)² ↔ e^−t + 3t·e^−t.
• Complex poles: complete the square, s² + 4s + 8 = (s + 2)² + 2². Then 10/((s + 2)² + 2²) = 5·2/((s + 2)² + 2²) ↔ 5·e^−2t·sin 2t.

1. Factor the denominator and list the poles. For a quadratic, find its roots or complete the square.
2. Write one fraction per pole: A/s, B/(s + a) and so on. A repeated pole also gets a squared term.
3. Find each coefficient with the cover-up rule. Keep the signs of the remaining factors.
4. Translate every term back to time with the table.
5. Fill in the time t (radians for sine and cosine). Check f(0⁺) and f(∞) with the value theorems.

5A worked example

This is a typical medium question. Try each step yourself before you open it.

6Common mistakes

• Mixing up the theorems: s → ∞ gives the start value f(0⁺), and s → 0 gives the end value f(∞).
• Forgetting the factor s in s·F(s), or dropping a zero factor such as (s + 2) when you set s = 0.
• Sign errors in the cover-up rule: at s = −3 the factor (s + 2) equals −1, not 5.
• Using the sine pair without dividing by ω: ω/(s² + ω²) needs ω in the numerator.
• Evaluating sin ωt or cos ωt with the calculator in degrees. The angle ωt is in radians.
• Using the final value theorem when f(t) never settles, for example F(s) = 1/(s² + 4), which is a pure sine.

7How exam questions look

• Easy: table pairs and value theorems. You invert a constant, an exponential, t·e^−at, a sine or a cosine and evaluate it, or you find f(0⁺), f(∞) or the initial slope f′(0⁺).
• Medium: inverse transforms by partial fractions. You find one coefficient A, invert three simple poles s(s + a)(s + b), a repeated pole, or complex poles by completing the square.
• Hard: differential equations and transfer functions, for example y″ + 7y′ + 6y = 15·u(t) with given starting values. You find y at a given time.

For the hard level, transform each derivative together with its initial conditions:

Then solve the algebraic equation for Y(s), and invert it with partial fractions as in the worked example.