Chapter 9 · Section 9.4

9.4 Discontinuous Forcing: Step Functions and the Delta Function

Real inputs are often switched on and off. A motor receives voltage for a few seconds; a valve opens at one moment and closes at another; a hammer strikes a beam for a tiny fraction of a second. With the methods of Chapter 4, such forcing is awkward: you must solve the equation separately on every interval and glue the pieces together so that $y$ and $y^{'}$ are continuous. The Laplace transform does the gluing for you.

Two new tools make this possible. The unit step function lets us write any piecewise-defined forcing as a single formula, and the second shifting theorem tells us how a time delay looks in the $s$ -domain. At the end of the section we model very short, very strong inputs with the Dirac delta function, and we meet the impulse response, the single function that describes a linear system completely.

The unit step function

Definition 9.4.1 (Unit step function)

For a constant $a \geq 0$ , the unit step function (or Heaviside function) switched on at $t = a$ is

u (t - a) = {0, 1, 0 \leq t < a, t \geq a .

With $a = 0$ we write simply $u (t)$ ; on $t \geq 0$ it is identically $1$ .

Multiplying a function by $u (t - a)$ switches it on at $t = a$ : before $a$ the product is $0$ , after $a$ it equals the function. Multiplying by $1 - u (t - a)$ switches it off at $t = a$ . The value exactly at the jump does not matter for transforms (Theorem 9.2.1).

The transform comes straight from the definition: the integrand is zero up to $t = a$ , so for $s > 0$

L {u (t - a)} = \int_{a}^{\infty} e^{- s t} d t = [- \frac{e ^{- s t}}{s}]_{a}^{\infty} = \frac{e ^{- a s}}{s} .

Compare with $L {1} = 1/ s$ : the delay by $a$ seconds produced the factor $e^{- a s}$ . This is a first glimpse of the second shifting theorem.

Pulses and piecewise functions

A rectangular pulse that is on for $a \leq t < b$ is the difference of two steps (Figure 9.4.1):

u (t - a) - u (t - b) = ⎩ ⎨ ⎧ 0, 1, 0, 0 \leq t < a, a \leq t < b, t \geq b .

Figure 9.4.1. A rectangular pulse is a step up followed by a step down:

u (t - 1) - u (t - 4)

equals

1

1 \leq t < 4

and

0

elsewhere.

More generally, any piecewise function can be written with steps. Suppose

f (t) = ⎩ ⎨ ⎧ f_{1} (t), f_{2} (t), f_{3} (t), 0 \leq t < a, a \leq t < b, t \geq b .

Start with $f_{1}$ . At $t = a$ we must replace $f_{1}$ by $f_{2}$ , so we add the change $f_{2} - f_{1}$ , switched on at $a$ . At $t = b$ we add the change $f_{3} - f_{2}$ , switched on at $b$ :

f (t) = f_{1} (t) + [f_{2} (t) - f_{1} (t)] u (t - a) + [f_{3} (t) - f_{2} (t)] u (t - b) .

Check: for $t < a$ both steps are $0$ , leaving $f_{1}$ ; for $a \leq t < b$ the first step is on, giving $f_{1} + f_{2} - f_{1} = f_{2}$ ; for $t \geq b$ both are on, giving $f_{3}$ . The rule "add the jump in the formula, switched on where it happens" extends to any number of pieces.

Example 9.4.1 (Writing piecewise functions with steps)

Write each function with unit steps and find its transform.
(a) $f (t) = 2$ for $0 \leq t < 1$ , $f (t) = - 1$ for $1 \leq t < 4$ , $f (t) = 0$ for $t \geq 4$ .
(b) The saturating ramp $g (t) = 2 t$ for $0 \leq t < 2$ , $g (t) = 4$ for $t \geq 2$ .

Solution

(a) Changes. At $t = 1$ the value changes by $- 1 - 2 = - 3$ ; at $t = 4$ by $0 - (- 1) = + 1$ . So $f (t) = 2 - 3 u (t - 1) + u (t - 4) .$
(a) Transform. By linearity and $L {u (t - a)} = e^{- a s} / s$ , $F (s) = \frac{2}{s} - \frac{3 e ^{- s}}{s} + \frac{e ^{- 4 s}}{s} = \frac{2 - 3 e ^{- s} + e ^{- 4 s}}{s} .$
(b) Changes. At $t = 2$ the formula changes from $2 t$ to $4$ , a change of $4 - 2 t = - 2 (t - 2)$ . So $g (t) = 2 t - 2 (t - 2) u (t - 2) .$ Notice that the change came out as a function of $t - 2$ . That is no accident: the ramp bends at $t = 2$ , and the correction is a ramp that starts there.
(b) Transform. $L {2 t} = 2/ s^{2}$ . For the second term we need the transform of a ramp that starts at $t = 2$ ; the second shifting theorem below gives $L {(t - 2) u (t - 2)} = e^{- 2 s} / s^{2}$ . Hence $G (s) = \frac{2}{s ^{2}} - \frac{2 e ^{- 2 s}}{s ^{2}} = \frac{2 ( 1 - e ^{- 2 s} )}{s ^{2}} .$

Translated functions and the second shifting theorem

Let $f (t)$ be defined for $t \geq 0$ . The function

u (t - a) f (t - a) = {0, f (t - a), 0 \leq t < a, t \geq a,

is the graph of $f$ moved $a$ units to the right, with zeros filling the gap on $[0, a)$ . It describes the same signal, delayed by $a$ seconds. Do not confuse it with $u (t - a) f (t)$ , which is the original $f$ with its first part cut away (Figure 9.4.2).

Figure 9.4.2. Left: the unit step

u (t - 2)

. Right:

f (t) = t^{2}

(dashed), the delayed copy

u (t - 2) f (t - 2) = u (t - 2) (t - 2)^{2}

, and the cut-off function

u (t - 2) f (t) = u (t - 2) t^{2}

, which jumps from

0

4

t = 2

Theorem 9.4.1 (Second shifting theorem: translation in

t

)

If $L {f (t)} = F (s)$ for $s > α$ and $a \geq 0$ , then

L {u (t - a) f (t - a)} = e^{- a s} F (s), s > α .

Equivalently, in the inverse direction,

L^{- 1} {e^{- a s} F (s)} = u (t - a) f (t - a) .

Why this works

The integrand vanishes for $t < a$ . On $t \geq a$ substitute $τ = t - a$ :

\int_{a}^{\infty} e^{- s t} f (t - a) d t = \int_{0}^{\infty} e^{- s (τ + a)} f (τ) d τ = e^{- a s} \int_{0}^{\infty} e^{- s τ} f (τ) d τ = e^{- a s} F (s) .

Compare the two shifting theorems; they mirror each other:

First: multiplying by $e^{a t}$ in $t$ shifts $s$ : $L {e^{a t} f (t)} = F (s - a)$ .
Second: shifting $t$ (a delay) multiplies by $e^{- a s}$ in $s$ : $L {u (t - a) f (t - a)} = e^{- a s} F (s)$ .

Often the function multiplying the step is not written in terms of $t - a$ , as in $u (t - 2) t$ . There are two ways to proceed. Either rewrite the function in powers of $t - a$ , or use the following equivalent form of the theorem (apply Theorem 9.4.1 to $f (t) = g (t + a)$ ):

L {u (t - a) g (t)} = e^{- a s} L {g (t + a)} .

Example 9.4.2 (Transforms of translated and switched functions)

Find (a) $L {(t - 2)^{2} u (t - 2)}$ , (b) $L {t u (t - 2)}$ , (c) $L {e^{t} u (t - 3)}$ , (d) the transform of the "sine burst" $h (t) = sin 2 t$ for $0 \leq t < π$ , $h (t) = 0$ for $t \geq π$ .

Solution

(a) This is already in translated form with $f (t) = t^{2}$ , $F (s) = 2/ s^{3}$ , $a = 2$ . So $L {(t - 2)^{2} u (t - 2)} = \frac{2 e ^{- 2 s}}{s ^{3}}$ .
(b) By rewriting. $t = (t - 2) + 2$ , so $t u (t - 2) = (t - 2) u (t - 2) + 2 u (t - 2)$ , and $L {t u (t - 2)} = \frac{e ^{- 2 s}}{s ^{2}} + \frac{2 e ^{- 2 s}}{s} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) .$ By the alternative form. $g (t) = t$ , $g (t + 2) = t + 2$ , and $L {t + 2} = 1/ s^{2} + 2/ s$ . Same result.
(c) $g (t) = e^{t}$ , so $g (t + 3) = e^{t + 3} = e^{3} e^{t}$ with transform $\frac{e ^{3}}{s - 1}$ . Hence $L {e^{t} u (t - 3)} = e^{- 3 s} \cdot \frac{e ^{3}}{s - 1} = \frac{e ^{- 3 (s - 1)}}{s - 1}, s > 1.$
(d) Steps. $h (t) = sin 2 t - u (t - π) sin 2 t$ . Since $sin 2 t$ has period $π$ , $sin 2 t = sin (2 (t - π))$ , so the second term is already a translated function: $h (t) = sin 2 t - u (t - π) sin (2 (t - π)) .$
(d) Transform. $H (s) = \frac{2}{s ^{2} + 4} - e^{- π s} \frac{2}{s ^{2} + 4} = \frac{2 ( 1 - e ^{- π s} )}{s ^{2} + 4} .$

Common mistake

$L {u (t - a) g (t)}$ is not $e^{- a s} G (s)$ . The factor $e^{- a s}$ multiplies the transform of the shifted function $g (t + a)$ . For example, $L {t u (t - 2)} \neq = e^{- 2 s} / s^{2}$ ; the correct answer, from Example 9.4.2(b), contains the extra term $2 e^{- 2 s} / s$ .

Inverse transforms containing $e^{- a s}$

When a transform contains a factor $e^{- a s}$ , set it aside, invert what remains to get $f (t)$ , and then replace $t$ by $t - a$ and multiply by $u (t - a)$ .

Example 9.4.3 (Inverting delayed transforms)

Find (a) $L^{- 1} {\frac{2 e ^{- s}}{s ( s + 2 )}}$ and (b) $L^{- 1} {\frac{s e ^{- 2 s}}{s ^{2} + 4}}$ .

Solution

(a) Without the exponential. $\frac{2}{s ( s + 2 )} = \frac{1}{s} - \frac{1}{s + 2}$ (cover-up: $2/2 = 1$ at $s = 0$ , $2/ (- 2) = - 1$ at $s = - 2$ ). Its inverse is $f (t) = 1 - e^{- 2 t}$ .
(a) Delay by $1$ . $L^{- 1} {\frac{2 e ^{- s}}{s ( s + 2 )}} = u (t - 1) (1 - e^{- 2 (t - 1)}) .$ Be careful to replace every $t$ by $t - 1$ , including the one in the exponent.
(b) $L^{- 1} {s / (s^{2} + 4)} = cos 2 t$ , delayed by $2$ : $L^{- 1} {\frac{s e ^{- 2 s}}{s ^{2} + 4}} = u (t - 2) cos (2 (t - 2)) = {0, cos (2 t - 4), t < 2, t \geq 2.$

Initial-value problems with discontinuous forcing

Now we combine everything. The method is the one of Section 9.3; the only new step is that the forcing is written with unit steps, so its transform contains exponentials $e^{- a s}$ . These exponentials simply ride along: we decompose the rational part, invert, and delay.

Method in a nutshell

Write the forcing $f (t)$ with unit steps, each change in the form $u (t - a) g (t - a)$ .
Transform the equation and solve for $Y (s)$ . It has the form $Y = R_{0} (s) + e^{- a s} R_{1} (s) + e^{- b s} R_{2} (s) + \dots$ with rational $R_{k}$ .
Invert each rational $R_{k}$ separately (partial fractions), giving $r_{k} (t)$ .
Apply the second shifting theorem: $y (t) = r_{0} (t) + u (t - a) r_{1} (t - a) + u (t - b) r_{2} (t - b) + \dots$
If you want, write $y$ interval by interval. For forcing with finite jumps, $y, y^{'}, \dots, y^{(n - 1)}$ are all continuous; only $y^{(n)}$ jumps where $f$ jumps. Checking this at each switching time is a good test of your answer.

Example 9.4.4 (An undamped oscillator pushed for half a period)

A mass on a spring, at rest, is pushed by a constant force that is removed at $t = π /3$ : $y^{''} + 9 y = 9 [1 - u (t - \frac{π}{3})], y (0) = 0, y^{'} (0) = 0.$ Find $y (t)$ .

Solution

Transform. With zero initial values, $(s^{2} + 9) Y = \frac{9}{s} - \frac{9 e ^{- π s /3}}{s}$ , so $Y = \frac{9}{s ( s ^{2} + 9 )} (1 - e^{- π s /3}) .$
Invert the rational part. $\frac{9}{s ( s ^{2} + 9 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 9}$ . Cover-up at $s = 0$ : $A = 1$ . Then $9 = (s^{2} + 9) + (B s + C) s$ , so $B = - 1$ , $C = 0$ : $\frac{9}{s ( s ^{2} + 9 )} = \frac{1}{s} - \frac{s}{s ^{2} + 9}, g (t) = L^{- 1} {\frac{9}{s ( s ^{2} + 9 )}} = 1 - cos 3 t .$
Second shifting theorem. $y (t) = g (t) - u (t - \frac{π}{3}) g (t - \frac{π}{3}) = 1 - cos 3 t - u (t - \frac{π}{3}) [1 - cos (3 t - π)] .$
Interval by interval. For $t \geq π /3$ use $cos (3 t - π) = - cos 3 t$ : $y (t) = {1 - cos 3 t, (1 - cos 3 t) - (1 + cos 3 t) = - 2 cos 3 t, 0 \leq t < π /3, t \geq π /3.$
Check continuity at $t = π /3$ . Left: $1 - cos π = 2$ ; right: $- 2 cos π = 2$ . Derivatives: left $3 sin π = 0$ , right $6 sin π = 0$ . Both match.
Interpretation. While the force acts, the mass oscillates between $0$ and $2$ around the new equilibrium $y = 1$ . The force is removed exactly when the mass is at its farthest point and momentarily at rest. From then on it oscillates around $y = 0$ with amplitude $2$ , twice the amplitude it had while being pushed. Timing matters!

Example 9.4.5 (A damped system and a rectangular pulse)

Solve $y^{''} + 2 y^{'} + 2 y = 2 [u (t - 1) - u (t - 4)]$ , $y (0) = 0$ , $y^{'} (0) = 0$ .

Solution

Transform. $(s^{2} + 2 s + 2) Y = \frac{2 e ^{- s}}{s} - \frac{2 e ^{- 4 s}}{s}$ , so $Y = (e^{- s} - e^{- 4 s}) \frac{2}{s ( s ^{2} + 2 s + 2 )} .$
Partial fractions. The quadratic has discriminant $4 - 8 < 0$ . Write $\frac{2}{s ( s ^{2} + 2 s + 2 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 2 s + 2}$ . Cover-up at $s = 0$ : $A = 2/2 = 1$ . From $2 = (s^{2} + 2 s + 2) + (B s + C) s$ : the $s^{2}$ terms give $B = - 1$ , the $s$ terms give $2 + C = 0$ , $C = - 2$ : $\frac{2}{s ( s ^{2} + 2 s + 2 )} = \frac{1}{s} - \frac{s + 2}{s ^{2} + 2 s + 2} .$
Complete the square. $s^{2} + 2 s + 2 = (s + 1)^{2} + 1$ and $s + 2 = (s + 1) + 1$ : $g (t) = L^{- 1} {\frac{2}{s ( s ^{2} + 2 s + 2 )}} = 1 - e^{- t} (cos t + sin t) .$
Second shifting theorem. $y (t) = u (t - 1) g (t - 1) - u (t - 4) g (t - 4) .$
Interval by interval. $y (t) = ⎩ ⎨ ⎧ 0, 1 - e^{- (t - 1)} (cos (t - 1) + sin (t - 1)), e^{- (t - 4)} (cos (t - 4) + sin (t - 4)) - e^{- (t - 1)} (cos (t - 1) + sin (t - 1)), 0 \leq t < 1, 1 \leq t < 4, t \geq 4.$
Interpretation (Figure 9.4.3). Nothing happens until the pulse starts. During the pulse the system rises toward the level $1$ and slightly overshoots it: at the end of the pulse $y (4) = g (3) \approx 1.042$ . After the pulse it decays back to $0$ , again with a small undershoot. The graph is smooth: $y$ and $y^{'}$ are continuous at $t = 1$ and $t = 4$ ; only $y^{''}$ jumps there.

Figure 9.4.3. Response of

y^{''} + 2 y^{'} + 2 y

to the pulse

2 [u (t - 1) - u (t - 4)]

(Example 9.4.5). The forcing jumps, but the response is smooth: integration by the system irons out the corners.

Use the solver to explore pulse responses. Enter the coefficients, the height $A$ of the pulse, and the times $a$ and $b$ when it is switched on and off (leave $b$ empty for a step that stays on). The preset is Example 9.4.5. Try making the pulse shorter and higher with the same area $A (b - a)$ , for instance $A = 6$ , $a = 1$ , $b = 2$ , then $A = 60$ , $a = 1$ , $b = 1.1$ ; the responses approach a limit. That limit is the subject of the next part.

The Dirac delta function

A hammer blow, a lightning strike, a sudden electrical surge: these inputs are very large but last a very short time. What matters is not their exact shape but their total effect, the integral of the force over time, called the impulse. To model them, consider a rectangular pulse of width $ϵ$ and height $1/ ϵ$ , starting at $t = a$ :

δ_{ϵ} (t - a) = \frac{1}{ϵ} [u (t - a) - u (t - a - ϵ)] .

Whatever $ϵ > 0$ is, its area (impulse) is $ϵ \cdot \frac{1}{ϵ} = 1$ . As $ϵ$ shrinks, the pulse becomes taller and narrower (Figure 9.4.4).

Figure 9.4.4. Pulses

δ_{ϵ} (t - 1)

of width

ϵ

and height

1/ ϵ

, all of area

1

. The delta function

δ (t - 1)

is their limit as

ϵ \to 0

, drawn as an arrow.

Let us transform $δ_{ϵ}$ and then let $ϵ \to 0$ :

L {δ_{ϵ} (t - a)} = \frac{1}{ϵ} (\frac{e ^{- a s}}{s} - \frac{e ^{- (a + ϵ) s}}{s}) = e^{- a s} \frac{1 - e ^{- ϵs}}{ϵs} .

As $ϵ \to 0$ , the fraction $\frac{1 - e ^{- ϵs}}{ϵs}$ tends to $1$ (by L'Hôpital's rule in $ϵ$ , or from $e^{- x} \approx 1 - x$ for small $x$ ). The transforms therefore converge to $e^{- a s}$ .

Definition 9.4.2 (Dirac delta function)

The unit impulse or Dirac delta function $δ (t - a)$ , $a \geq 0$ , is the limit of the pulses $δ_{ϵ} (t - a)$ as $ϵ \to 0$ . It is characterised by the properties

δ (t - a) = 0 for t \neq = a, \int_{0}^{\infty} g (t) δ (t - a) d t = g (a)

for every function $g$ continuous at $t = a$ (for $a > 0$ ). Its Laplace transform is

L {δ (t - a)} = e^{- a s}, in particular L {δ (t)} = 1.

The second property, the sifting property, says that integrating against $δ (t - a)$ "picks out" the value of $g$ at $a$ . For example $\int_{0}^{\infty} cos t δ (t - π) d t = cos π = - 1$ . The transform is the special case $g (t) = e^{- s t}$ .

Is it really a function?

No ordinary function can be zero everywhere except at one point and still have integral $1$ . The delta is a generalised function (a distribution): it is defined by what it does inside integrals. That is also why its transform $e^{- a s}$ does not tend to $0$ as $s \to \infty$ , which Theorem 9.1.1 would require of an ordinary function. For the purposes of this course, the definition above and the rule $L {δ (t - a)} = e^{- a s}$ are all we need.

What does an impulse do to a system? Take $a_{0} y^{''} + a_{1} y^{'} + a_{2} y = A δ (t - t_{0})$ with $t_{0} > 0$ . Integrate both sides over a tiny interval around $t_{0}$ . The $y$ and $y^{'}$ terms contribute almost nothing (they stay bounded), while the right side contributes $A$ . So $a_{0} y^{'}$ must jump by $A$ : the velocity changes instantaneously by $A / a_{0}$ , while the position $y$ stays continuous. For a mass $m$ this is just Newton's law in impulse form: an impulse $A$ changes the momentum $m y^{'}$ by $A$ .

Example 9.4.6 (A hammer blow to a vibrating system)

A damped oscillator, released from $y (0) = 1$ with $y^{'} (0) = - 2$ , receives an impulse of strength $6$ at $t = 1$ : $y^{''} + 4 y^{'} + 13 y = 6 δ (t - 1), y (0) = 1, y^{'} (0) = - 2.$ Find $y (t)$ .

Solution

Transform. $(s^{2} Y - s + 2) + 4 (s Y - 1) + 13 Y = 6 e^{- s} .$
Solve for $Y$ . The constants on the left are $- s + 2 - 4 = - (s + 2)$ . Moving them to the right, $(s^{2} + 4 s + 13) Y = s + 2 + 6 e^{- s}, Y = \frac{s + 2}{s ^{2} + 4 s + 13} + e^{- s} \frac{6}{s ^{2} + 4 s + 13} .$
Complete the square. $s^{2} + 4 s + 13 = (s + 2)^{2} + 3^{2}$ . Conveniently, the first numerator is already $s + 2$ , and $6 = 2 \cdot 3$ : $Y = \frac{s + 2}{( s + 2 ) ^{2} + 3 ^{2}} + 2 e^{- s} \frac{3}{( s + 2 ) ^{2} + 3 ^{2}} .$
Invert. The first term gives $e^{- 2 t} cos 3 t$ . Without the exponential, the second gives $2 e^{- 2 t} sin 3 t$ ; the factor $e^{- s}$ delays it by $1$ : $y (t) = e^{- 2 t} cos 3 t + 2 u (t - 1) e^{- 2 (t - 1)} sin (3 (t - 1)) .$
Check the jump. At $t = 1$ the added term is $0$ , so $y$ is continuous. Its derivative at $t = 1^{+}$ is $2 \cdot 3 cos 0 = 6$ : the velocity jumps by exactly $6 = A / a_{0}$ , as predicted.

Figure 9.4.5. Example 9.4.6. The impulse at

t = 1

leaves the position continuous but makes the velocity jump by

6

, producing a corner in the graph. The dashed curve is the motion without the blow.

The delta preset of the solver loads Example 9.4.6. Change the strength $A$ and the time $a$ of the impulse. Can you find a time and a strength that stop the motion completely? (Exercise 11 does this for an undamped spring.)

Impulse response and the convolution formula

Consider $a_{0} y^{''} + a_{1} y^{'} + a_{2} y = f (t)$ with zero initial conditions (a system "at rest"). Transforming gives $p (s) Y (s) = F (s)$ , with the characteristic polynomial $p (s) = a_{0} s^{2} + a_{1} s + a_{2}$ . Hence

Y (s) = \frac{1}{p ( s )} F (s) = H (s) F (s), H (s) = \frac{1}{p ( s )} .

The function $H (s)$ is called the transfer function of the system: output transform $=$ transfer function $\times$ input transform. Its inverse $h (t) = L^{- 1} {H (s)}$ has a clear meaning. If the input is $f (t) = δ (t)$ , then $F (s) = 1$ and $Y = H$ : so $h (t)$ is the response of the system at rest to a unit impulse at $t = 0$ . It is called the impulse response. (Equivalently, $h$ solves the homogeneous equation with $h (0) = 0$ , $h^{'} (0) = 1/ a_{0}$ : the impulse at time $0$ produces exactly this initial velocity.)

By the convolution theorem, $Y = H F$ means

Theorem 9.4.2 (Response of a system at rest)

Let $a_{0}, a_{1}, a_{2}$ be constants, $a_{0} \neq = 0$ , and let $f$ be piecewise continuous and of exponential order. The solution of

a_{0} y^{''} + a_{1} y^{'} + a_{2} y = f (t), y (0) = 0, y^{'} (0) = 0,

y (t) = (h * f) (t) = \int_{0}^{t} h (t - τ) f (τ) d τ, h (t) = L^{- 1} {\frac{1}{a _{0} s ^{2} + a _{1} s + a _{2}}} .

The same statement holds for equations of any order $n$ with zero initial values.

Read the formula physically: think of the input as a long chain of tiny impulses, of size $f (τ) d τ$ at each time $τ$ . Each one produces the response $h (t - τ) f (τ) d τ$ at the later time $t$ , and by linearity the total response is the sum (integral) of all of them. Once you know $h$ , you know how the system responds to every input. Engineers measure $h$ (or $H$ ) experimentally for exactly this reason.

Example 9.4.7 (Impulse response and step response)

For the system $y^{''} + 3 y^{'} + 2 y = f (t)$ at rest, (a) find the impulse response $h (t)$ , (b) write the solution for an arbitrary input as a convolution, and (c) evaluate it for the unit step input $f (t) = 1$ .

Solution

(a) $H (s) = \frac{1}{s ^{2} + 3 s + 2} = \frac{1}{( s + 1 ) ( s + 2 )} = \frac{1}{s + 1} - \frac{1}{s + 2}$ (cover-up: $1$ at $s = - 1$ , $- 1$ at $s = - 2$ ). So $h (t) = e^{- t} - e^{- 2 t} .$ Check: $h (0) = 0$ and $h^{'} (0) = - 1 + 2 = 1 = 1/ a_{0}$ .
(b) For any input $f$ , $y (t) = \int_{0}^{t} (e^{- (t - τ)} - e^{- 2 (t - τ)}) f (τ) d τ .$
(c) With $f = 1$ , integrate each exponential in $τ$ : $\int_{0}^{t} e^{- (t - τ)} d τ \int_{0}^{t} e^{- 2 (t - τ)} d τ = [e^{- (t - τ)}]_{τ = 0}^{τ = t} = 1 - e^{- t}, = [\frac{1}{2} e^{- 2 (t - τ)}]_{τ = 0}^{τ = t} = \frac{1}{2} (1 - e^{- 2 t}) .$ So $y (t) = 1 - e^{- t} - \frac{1}{2} + \frac{1}{2} e^{- 2 t} = \frac{1}{2} - e^{- t} + \frac{1}{2} e^{- 2 t}$ .
Check. $y (0) = \frac{1}{2} - 1 + \frac{1}{2} = 0$ , $y^{'} (0) = 1 - 1 = 0$ , and $y^{''} + 3 y^{'} + 2 y = (- e^{- t} + 2 e^{- 2 t}) + 3 (e^{- t} - e^{- 2 t}) + (1 - 2 e^{- t} + e^{- 2 t}) = 1$ . Correct.

Exercises

In Exercises 1–3 write the function with unit steps and find its Laplace transform.

$f (t) = 3$ for $0 \leq t < 2$ , $f (t) = 5$ for $t \geq 2$ .
Answer
$f = 3 + 2 u (t - 2)$ , $F (s) = \frac{3}{s} + \frac{2 e ^{- 2 s}}{s}$
$f (t) = t^{2}$ for $0 \leq t < 2$ , $f (t) = 4$ for $t \geq 2$ .
Answer
$f = t^{2} - (t^{2} - 4) u (t - 2)$ and $t^{2} - 4 = (t - 2)^{2} + 4 (t - 2)$ ; $F (s) = \frac{2}{s ^{3}} - e^{- 2 s} (\frac{2}{s ^{3}} + \frac{4}{s ^{2}})$
$f (t) = cos t$ for $0 \leq t < π$ , $f (t) = 0$ for $t \geq π$ .
Answer
$f = cos t + u (t - π) cos (t - π)$ ; $F (s) = \frac{s ( 1 + e ^{- π s} )}{s ^{2} + 1}$
Find $L^{- 1} {\frac{e ^{- 4 s}}{s ^{3}}}$ .
Answer
$\frac{1}{2} (t - 4)^{2} u (t - 4)$
Find $L^{- 1} {\frac{1 - e ^{- 2 s}}{s ( s + 1 )}}$ .
Answer
$(1 - e^{- t}) - u (t - 2) (1 - e^{- (t - 2)})$
Find $L^{- 1} {\frac{3 e ^{- π s /2}}{s ^{2} + 9}}$ and simplify.
Answer
$u (t - \frac{π}{2}) sin (3 (t - \frac{π}{2})) = u (t - \frac{π}{2}) cos 3 t$
Find $L {δ (t - 3) + 2 δ (t - 1)}$ and evaluate $\int_{0}^{\infty} e^{- t} cos t δ (t - π) d t$ .
Answer
$e^{- 3 s} + 2 e^{- s}$ ; $e^{- π} cos π = - e^{- π}$
Solve $y^{'} + 3 y = 6 u (t - 2)$ , $y (0) = 1$ .
Answer
$y = e^{- 3 t} + 2 u (t - 2) (1 - e^{- 3 (t - 2)})$
Solve $y^{''} + 9 y = 18 [u (t - 1) - u (t - 2)]$ , $y (0) = y^{'} (0) = 0$ .
Answer
$y = 2 u (t - 1) (1 - cos 3 (t - 1)) - 2 u (t - 2) (1 - cos 3 (t - 2))$
Solve $y^{''} + 2 y^{'} + y = 2 δ (t - 1)$ , $y (0) = 1$ , $y^{'} (0) = - 1$ .
Answer
$y = e^{- t} + 2 u (t - 1) (t - 1) e^{- (t - 1)}$
Solve $y^{''} + 16 y = 2 δ (t - π /4)$ , $y (0) = 0$ , $y^{'} (0) = 2$ , and describe the motion.
Answer
$y = \frac{1}{2} sin 4 t - \frac{1}{2} u (t - \frac{π}{4}) sin 4 t$ : the spring oscillates as $\frac{1}{2} sin 4 t$ until $t = π /4$ , when it passes $y = 0$ with velocity $- 2$ ; the impulse $+ 2$ cancels this velocity and the mass stays at rest, $y = 0$ , for $t \geq π /4$ .
Find the impulse response of $y^{''} + 5 y^{'} + 6 y = f (t)$ and write the solution with $y (0) = y^{'} (0) = 0$ as a convolution integral.
Answer
$h (t) = e^{- 2 t} - e^{- 3 t}$ ; $y (t) = \int_{0}^{t} (e^{- 2 (t - τ)} - e^{- 3 (t - τ)}) f (τ) d τ$

The unit step function

Pulses and piecewise functions

Translated functions and the second shifting theorem

Inverse transforms containing e−as

Initial-value problems with discontinuous forcing

The Dirac delta function

Impulse response and the convolution formula

Exercises

Inverse transforms containing $e^{- a s}$