FOUNDATION · CHAPTER 2 OF 8

The Laplace transform as a working tool

This chapter shows how the Laplace transform turns a differential equation into algebra, why the transfer function is the transform of the impulse response, and how partial fractions and the two limit theorems let you read a response. You need it to move between s and t with confidence.

Updated 13 Oct 2026

What you will be able to do

You will turn a linear differential equation with zero initial conditions into a transfer function $G (s)$ , and say why $G (s)$ is the transform of the impulse response. You will expand a response into partial fractions, including repeated poles and complex pairs, and read the behaviour each pole sets. You will derive the initial and final value theorems and say when each one fails.

Before you start

Integration by parts; the integral of an exponential
Complex numbers, the real part $Re (z)$ , conjugate pairs and Euler's formula $e^{j θ} = cos θ + j sin θ$
Polynomials: roots, long division, completing the square, comparing coefficients
Poles and zeros: the roots of a ratio's denominator and of its numerator
Linear constant-coefficient differential equations; the unit step and the unit impulse $δ (t)$
The Laplace transform and its derivative rule (recapped below)

From the equation to G(s)

The transform trades calculus for algebra. The signals here switch on at $t = 0$ , so we integrate from $0^{-}$ , just before any impulse at the origin. For a signal that is zero before $t = 0$ , the bilateral transform, which integrates from $- \infty$ , gives the same result. The block below recalls the definition and the one rule we need.

TRANSFORM AND DERIVATIVE RULE

F (s) L {f^{'}} L {f^{''}} = \int_{0^{-}}^{\infty} f (t) e^{- s t} d t = s F (s) - f (0^{-}) = s^{2} F (s) - s f (0^{-}) - f^{'} (0^{-})

Valid where the integral converges, that is for $Re (s)$ large enough. $f (t)$ is the signal with $t$ in seconds, $F (s)$ its transform, and $s = σ + j ω$ in s $^{- 1}$ , where $j = - 1$ .

Now take a plant with input $u$ and output $y$ , with transforms $U (s)$ and $Y (s)$ . Transform each term of its equation. Every derivative leaves behind a term in $y (0^{-})$ , $y^{'} (0^{-})$ and so on. If all initial conditions are zero, those terms vanish and each derivative is just a factor of $s$ . The equation becomes a product, and the ratio of output to input is a function of $s$ alone.

TRANSFER FUNCTION FROM THE EQUATION

a_{2} y^{''} + a_{1} y^{'} + a_{0} y = b_{0} u \Rightarrow (a_{2} s^{2} + a_{1} s + a_{0}) Y (s) = b_{0} U (s) \Rightarrow G (s) = \frac{Y ( s )}{U ( s )} = \frac{b _{0}}{a _{2} s ^{2} + a _{1} s + a _{0}}

Conditions: the plant is linear with constant coefficients, $y (0^{-}) = y^{'} (0^{-}) = 0$ , and $u$ is applied at $t = 0$ . $a_{i}$ and $b_{0}$ are constants. Non-zero initial conditions add terms with no $U (s)$ in them: the response with no input, which adds to $G (s) U (s)$ because the equation is linear.

Why is $G (s)$ also the transform of the impulse response? Feed in $u = δ (t)$ . Its transform is $\int_{0^{-}}^{\infty} δ (t) e^{- s t} d t = 1$ , because the impulse sits inside an interval that starts at $0^{-}$ . That is the reason for the lower limit. Then $Y (s) = G (s) \cdot 1$ .

TRANSFER FUNCTION IS THE TRANSFORM OF THE IMPULSE RESPONSE

U (s) = 1 \Rightarrow Y (s) = G (s) \Rightarrow g (t) = L^{- 1} {G (s)}

Conditions: linear time-invariant plant, zero initial conditions. $g (t)$ is the output for $u = δ (t)$ . Any other input gives $Y (s) = G (s) U (s)$ .

WATCH OUT FOR

Using $G (s)$ when initial conditions are not zero. Test: after transforming, a term with neither $Y (s)$ nor $U (s)$ in it is an initial condition, and then $Y \neq = G U$ .

WORKED EXAMPLE

A plant obeys $y^{'} + 4 y = 8 u$ , with time in seconds. It is at rest, $y (0^{-}) = 0$ , and $u$ is a unit step applied at $t = 0$ . Find $y (t)$ .

Solution

Transform both sides. Rest means $y (0^{-}) = 0$ , so $y^{'} \to s Y (s)$ and we get $s Y + 4 Y = 8 U$ .
The unit step has $U (s) = 1/ s$ . So $G (s) = \frac{8}{s + 4}$ and $Y (s) = \frac{8}{s ( s + 4 )}$ .
The integral of $e^{pt} e^{- s t}$ from 0 is $1/ (s - p)$ , so $1/ s$ ( $p = 0$ ) and $1/ (s + 4)$ ( $p = - 4$ ) invert to $1$ and $e^{- 4 t}$ . Split into $\frac{A}{s} + \frac{B}{s + 4}$ .
Multiply both sides by $s$ and set $s = 0$ : the $B$ term keeps a factor $s$ and vanishes, so $A = \frac{8}{0 + 4} = 2$ . Multiply by $s + 4$ and set $s = - 4$ : $B = \frac{8}{- 4} = - 2$ .
Invert term by term: $2/ s \to 2$ and $- 2/ (s + 4) \to - 2 e^{- 4 t}$ .

Answer: $y (t) = 2 (1 - e^{- 4 t})$ for $t \geq 0$ . It starts at 0 and settles at 2.

Partial fractions: each pole is a mode

The responses here are ratios of polynomials in $s$ . We cannot invert a ratio directly, but we can invert $1/ (s - p)$ , because $\int_{0}^{\infty} e^{pt} e^{- s t} d t = \frac{1}{s - p}$ for $Re (s) > Re (p)$ , with $p$ real or complex. So we rewrite the ratio as a sum of such terms. Each pole $p$ then gives one mode: a term in time such as $e^{pt}$ , whose growth or decay that pole alone fixes.

DISTINCT POLES

F (s) = \frac{N ( s )}{( s - p _{1} ) \dots ( s - p _{n} )} = k = 1 \sum n \frac{r _{k}}{s - p _{k}}, r_{k} = s \to p_{k} lim (s - p_{k}) F (s), f (t) = k = 1 \sum n r_{k} e^{p_{k} t}

Conditions: the $p_{k}$ (in s $^{- 1}$ ) are all different, the degree of $N$ is below $n$ (otherwise divide first; a polynomial part gives impulses and their derivatives), and $t \geq 0$ . Multiply both sides by $(s - p_{k})$ : every other term keeps a factor $(s - p_{k})$ and vanishes at $s = p_{k}$ , leaving $r_{k}$ , the residue at $p_{k}$ .

REPEATED POLE

F (s) = \frac{N ( s )}{( s - p ) ^{2} ( s - q )} = \frac{A}{( s - p ) ^{2}} + \frac{B}{s - p} + \frac{D}{s - q}, A = [(s - p)^{2} F]_{s = p}, B = [\frac{d}{d s} (s - p)^{2} F]_{s = p}, f (t) = (A t + B) e^{pt} + D e^{q t}

Conditions: $p \neq = q$ , degree of $N$ below 3, $t \geq 0$ . $(s - p)^{2} F = A + B (s - p) + D (s - p)^{2} / (s - q)$ : its value at $s = p$ is $A$ and its slope there is $B$ , as the $D$ term keeps a factor $(s - p)^{2}$ . $D = [(s - q) F]_{s = q}$ . By parts, $\int_{0}^{\infty} t e^{- (s - p) t} d t = \frac{1}{s - p} \int_{0}^{\infty} e^{- (s - p) t} d t = \frac{1}{( s - p ) ^{2}}$ for $Re (s) > Re (p)$ , so $L {t e^{pt}} = \frac{1}{( s - p ) ^{2}}$ .

COMPLEX PAIR

L^{- 1} {\frac{b ( s + α ) + d}{( s + α ) ^{2} + ω _{d}^{2}}} = e^{- α t} (b cos ω_{d} t + \frac{d}{ω _{d}} sin ω_{d} t)

Conditions: $b$ , $d$ real, $α$ real in s $^{- 1}$ , $ω_{d} > 0$ in rad/s, $t \geq 0$ . The poles are $- α \pm j ω_{d}$ : $α$ sets the decay, $ω_{d}$ the oscillation. By Euler's formula, half the sum of $e^{(- α \pm j ω_{d}) t}$ is $e^{- α t} cos ω_{d} t$ and their difference over $2 j$ is $e^{- α t} sin ω_{d} t$ ; the same sums of $1/ (s - p)$ give $\frac{s + α}{( s + α ) ^{2} + ω _{d}^{2}}$ and $\frac{ω _{d}}{( s + α ) ^{2} + ω _{d}^{2}}$ . Complete the square to reach this form.

The pole sets how a mode behaves; its coefficient only says how much of the mode is present. The poles of $G (s)$ give the plant's own modes, and the poles of $U (s)$ add forced modes that follow the input. A pole drops out only if a zero cancels it. In the table, $ϕ$ is a phase angle fixed by the coefficients.

Pole of the response	Mode in time	Behaviour as $t$ grows
Real, $p = - a$ , $a > 0$	$e^{- a t}$	Decays
Repeated real, $p = - a$ , $a > 0$	$t e^{- a t}$ , with $e^{- a t}$	Decays; alone, $t e^{- a t}$ first rises to a peak at $t = 1/ a$
Simple, at the origin	$1$	Stays constant
Real, $p = a > 0$	$e^{a t}$	Grows without bound
Pair $- α \pm j ω_{d}$ , $α > 0$	$e^{- α t} cos (ω_{d} t + ϕ)$	Damped sinusoid, envelope $e^{- α t}$
Simple pair $\pm j ω_{d}$	$cos (ω_{d} t + ϕ)$	Sustained oscillation

WATCH OUT FOR

Expecting only the plant's poles in the response. Test: write $Y = G U$ and list every pole of the product, including those of $U (s)$ .

WORKED EXAMPLE

A plant has $G (s) = \frac{10}{s ^{2} + 2 s + 10}$ and is at rest, $y (0^{-}) = y^{'} (0^{-}) = 0$ . The input $u$ is a unit step at $t = 0$ , with time in seconds. Find $y (t)$ .

Solution

Form $Y (s) = G (s) / s = \frac{10}{s ( s ^{2} + 2 s + 10 )}$ . The quadratic has roots $- 1 \pm j 3$ , so use one real term and one complex-pair term: $\frac{A}{s} + \frac{B s + D}{s ^{2} + 2 s + 10}$ .
Multiply by $s$ and set $s = 0$ : $A = \frac{10}{10} = 1$ .
With $A = 1$ , clear the denominators and compare coefficients: $10 = (s^{2} + 2 s + 10) + (B s + D) s$ . The $s^{2}$ terms give $0 = 1 + B$ , so $B = - 1$ . The $s$ terms give $0 = 2 + D$ , so $D = - 2$ .
Complete the square: $s^{2} + 2 s + 10 = (s + 1)^{2} + 3^{2}$ , so $α = 1 s^{- 1}$ and $ω_{d} = 3$ rad/s. Rewrite the numerator $- s - 2 = - (s + 1) - 1$ , so $b = - 1$ and $d = - 1$ .
Apply the complex-pair result: that term inverts to $e^{- t} (- cos 3 t - \frac{1}{3} sin 3 t)$ , and $A / s$ inverts to 1.

Answer: $y (t) = 1 - e^{- t} (cos 3 t + \frac{1}{3} sin 3 t)$ for $t \geq 0$ . Check: $y (0) = 1 - 1 = 0$ , as a plant at rest requires.

Initial and final value theorems

Sometimes you need only where a response starts or ends, not the whole curve. Both numbers come from one identity: transform the derivative from $0^{+}$ , then let $s$ go to its limits. We take $f$ with no impulse at the origin, so $F (s)$ is the same whether the integral starts at $0^{-}$ or $0^{+}$ .

INITIAL VALUE THEOREM

\int_{0^{+}}^{\infty} f^{'} (t) e^{- s t} d t = s F (s) - f (0^{+}) \Rightarrow f (0^{+}) = s \to \infty lim s F (s)

Conditions: $F$ is a ratio of polynomials, numerator degree below denominator degree, so $f$ has no impulse at $t = 0$ . The identity is integration by parts; the boundary term $f (t) e^{- s t}$ vanishes as $t \to \infty$ for $Re (s)$ large enough. For $t > 0$ , $f^{'}$ is a sum of terms $t^{k} e^{pt}$ , with transforms $k! / (s - p)^{k + 1}$ (repeat the by-parts step above), which go to 0 as $s \to \infty$ ; so does the integral. $f (0^{+})$ is the value just after switch-on.

FINAL VALUE THEOREM

s \to 0 lim [s F (s) - f (0^{+})] = \int_{0^{+}}^{\infty} f^{'} (t) d t = f (\infty) - f (0^{+}) \Rightarrow f (\infty) = s \to 0 lim s F (s)

Conditions: $F$ as above, and every pole of $s F (s)$ lies in the LHP, the open left half-plane $Re (s) < 0$ (the RHP is $Re (s) > 0$ ; the $j ω$ axis is in neither). The factor $s$ cancels a simple pole of $F$ at the origin, not a double one. Then $f^{'}$ decays exponentially; for real $s > 0$ , $∣ f^{'} (t) e^{- s t} ∣ \leq ∣ f^{'} (t) ∣$ , whose integral is finite, so the limit can be taken inside the integral. $f (\infty)$ means $lim_{t \to \infty} f (t)$ .

If a pole of $s F (s)$ lies in the RHP, $f$ grows. If a pair lies on the $j ω$ axis, $f$ keeps oscillating, and a pole at the origin makes $f$ grow without bound. Either way $f$ has no limit. Row 2 below is a unit step into the unstable plant $G = 1/ (s - 2)$ . Row 3 is a unit sine input, $sin t$ , into the stable plant $G = 1/ (s + 1)$ .

$F (s)$	$lim_{s \to 0} s F (s)$	What $f (t)$ does	Theorem applies?
$\frac{8}{s ( s + 4 )}$	$2$	Settles at 2	Yes
$\frac{1}{s ( s - 2 )}$	$- \frac{1}{2}$	$f = \frac{1}{2} (e^{2 t} - 1)$ grows without bound	No, pole at $s = 2$
$\frac{1}{( s + 1 ) ( s ^{2} + 1 )}$	$0$	$f = \frac{1}{2} e^{- t} + \frac{1}{2} (sin t - cos t)$ oscillates for ever	No, poles at $\pm j$
$\frac{1}{s ^{2}}$	does not exist	$f = t$ grows without bound	No, $s F = 1/ s$ has a pole at 0

WATCH OUT FOR

Trusting the formula because it returns a number. Test: cancel common factors, then list the poles of $s F (s)$ . One on the $j ω$ axis or in the RHP means there is no final value.

Check yourself

CHECK YOURSELF 1

A plant at rest has transfer function $G (s) = \frac{4}{s + 2}$ . What is its impulse response for $t \geq 0$ ?

(A)
$4 e^{- 2 t}$
(B)
$4 e^{2 t}$
(C)
$2 e^{- 4 t}$
(D)
$2 (1 - e^{- 2 t})$

Answer and solution

Answer: (A)

The impulse response is $g (t) = L^{- 1} {G (s)}$ , since an impulse has transform 1.
$L {e^{- 2 t}} = 1/ (s + 2)$ , so $g (t) = 4 e^{- 2 t}$ .
Option D is the step response: $\frac{4}{s ( s + 2 )} = \frac{2}{s} - \frac{2}{s + 2}$ .

CHECK YOURSELF 2

A response has transform $Y (s) = \frac{5 s + 8}{s ( s + 2 )}$ . Find $y (0^{+})$ .

Answer and solution

Answer: $5$ (accepted range $\pm 0.01$ )

The numerator degree is below the denominator degree, so $y$ has no impulse at $t = 0$ and $y (0^{+}) = lim_{s \to \infty} s Y (s)$ .
$s Y (s) = \frac{5 s + 8}{s + 2}$ .
Divide top and bottom by $s$ : $\frac{5 + 8/ s}{1 + 2/ s} \to 5$ .

CHECK YOURSELF 3One or more options are correct

For which of these response transforms $Y (s)$ may the final value theorem be used? Select all that apply.

(A)
$\frac{3}{s ( s + 1 )}$
(B)
$\frac{1}{s ( s - 1 )}$
(C)
$\frac{2}{s ^{2} + 4}$
(D)
$\frac{s + 2}{s ( s ^{2} + 3 s + 2 )}$

Answer and solution

Answer: (A), (D)

A: $s Y = 3/ (s + 1)$ , pole at $- 1$ in the LHP. Allowed.
B: $s Y = 1/ (s - 1)$ , pole at $+ 1$ in the RHP. Not allowed.
C: $s Y = 2 s / (s^{2} + 4)$ , poles at $\pm j 2$ on the $j ω$ axis. Not allowed.
D: $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ , so $s Y = (s + 2) / ((s + 1) (s + 2)) = 1/ (s + 1)$ after cancelling. Pole at $- 1$ . Allowed.

CHECK YOURSELF 4

A response has transform $Y (s) = \frac{s + 5}{( s + 1 ) ( s + 3 )}$ . What is $y (t)$ for $t \geq 0$ ?

(A)
$2 e^{- t} - e^{- 3 t}$
(B)
$2 e^{- t} + e^{- 3 t}$
(C)
$e^{- t} - 2 e^{- 3 t}$
(D)
$4 e^{- t} - 2 e^{- 3 t}$

Answer and solution

Answer: (A)

The poles $- 1$ and $- 3$ are distinct, so $r_{k} = lim_{s \to p_{k}} (s - p_{k}) Y (s)$ .
At $s = - 1$ : $r_{1} = \frac{- 1 + 5}{- 1 + 3} = \frac{4}{2} = 2$ .
At $s = - 3$ : $r_{2} = \frac{- 3 + 5}{- 3 + 1} = \frac{2}{- 2} = - 1$ .
So $y = 2 e^{- t} - e^{- 3 t}$ .

The Laplace transform as a working tool

What you will be able to do

Before you start

From the equation to G(s)

Partial fractions: each pole is a mode

Initial and final value theorems

Check yourself

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