Introduction to the Laplace Transform

The Laplace transform is an integral transformation of \(f(t)\) from the time domain into the complex frequency domain:

Several features of this definition deserve comment.

• The units of \(s\). The argument \(st\) of the exponential must be dimensionless, so \(s\) has units of inverse seconds — frequency. Hence the name complex frequency.

• The lower limit is \(0^{-}\), not 0. This includes the origin itself, so any discontinuity or impulse occurring at \(t = 0\) is captured. It also means the initial conditions used are those just before switching, which is exactly what the continuity conditions of Chapter 6 provide.

• The result is a definite integral, so \(t\) is integrated out and \(F(s)\) depends on \(s\) alone. Time has been eliminated, exactly as the phasor eliminated it.

• This is the one-sided or unilateral transform, which ignores \(f(t)\) for \(t < 0\) — equivalently, it transforms \(f(t)u(t)\). The two-sided transform runs from \(-\infty\), but for circuit analysis, where something is switched on at a definite instant, the one-sided version is what we need.

Four transforms, derived directly from the definition, generate almost everything else.

The unit step. With \(f(t) = u(t) = 1\) for \(t > 0\):

The exponential.

Note that the step is the special case \(a = 0\).

The impulse. Here the sifting property of Section 7.5 does all the work:

The impulse transforms to unity — the simplest possible \(F(s)\), and the reason the impulse response characterises a whole system. This is where the \(0^{-}\) lower limit earns its keep: with a lower limit of \(0^{+}\) the impulse would fall outside the range of integration and the transform would be zero.

The sinusoid. Writing the sine in exponential form via Euler's identity:

Each term is an exponential transform:

and the cosine follows the same way as \(s/\left(s^{2}+\omega^{2}\right)\).

The properties are what make the transform practical: they let almost any transform be built from the table without ever performing an integration.

The differentiation property is the one that matters most, and its general form for the \(n\)th derivative is

Notice what has happened. Differentiation becomes multiplication by \(s\), exactly as it became multiplication by \(j\omega\) for phasors. But now the initial conditions appear as extra terms in the same equation. They are not a separate problem to be solved afterwards; they are built in from the start. This is the single greatest advantage of the Laplace method over everything that preceded it.

Note also that \(f(0^{-})\) is used, matching the lower limit of the definition — and matching the pre-switching values that the continuity conditions of Chapter 6 supply.

The time-shift property deserves a warning. To use \(\mathcal{L}\left[f(t-a)u(t-a)\right] = e^{-as}F(s)\), the function must be expressed entirely in terms of \((t-a)\). A term such as \(t\,u(t-1)\) is not in that form and must first be rewritten as \((t-1)u(t-1) + u(t-1)\), as Example 15.3 demonstrates.

Two theorems extract the behaviour of \(f(t)\) at the extremes of time directly from \(F(s)\), without inverting it at all.

The correspondence is intuitive once noticed: large \(s\) corresponds to small \(t\) and vice versa, since \(s\) has units of inverse time.

These theorems are excellent checks. Having inverted an \(F(s)\) by partial fractions, evaluate the two limits and confirm they match the initial and final values of the answer. Any algebraic slip in the residues will usually show up immediately.

A caution about the final value theorem. It is valid only if \(f(\infty)\) actually exists — that is, if all poles of \(sF(s)\) lie in the left half-plane. Applying it to a function that oscillates forever or grows without bound gives a meaningless number. For \(F(s) = \omega/\left(s^{2}+\omega^{2}\right)\), for instance, the theorem returns 0, yet \(\sin\omega t\) has no final value at all. The initial value theorem carries no such restriction.

Inversion in practice means partial fraction expansion: break \(F(s)\) into a sum of terms each of which appears in the table, then invert term by term using linearity.

For distinct poles \(p_{i}\), and provided the degree of the numerator is less than that of the denominator,

The constants \(k_{i}\) are the residues, and each is found by the cover-up rule: multiply through by the corresponding factor and evaluate at that pole, which annihilates every other term.

Each term then inverts to an exponential:

Each pole contributes one exponential, whose decay rate is the pole's location. A pole far to the left decays fast; a pole near the origin decays slowly; a pole at the origin gives a constant. A pole in the right half-plane would give a growing exponential — which is why stability is a question about pole locations, as Chapter 16 will make explicit.

If the numerator degree is not less than the denominator's, divide out first: long division leaves a proper fraction plus a polynomial in \(s\), whose terms invert to \(\delta(t)\) and its derivatives.

A pole repeated \(n\) times requires \(n\) terms, one for each power up to \(n\):

where \(F_{1}(s)\) collects the remaining, non-repeated part. The highest-order residue comes from the cover-up rule as before, but the others require successive differentiation:

The differentiation is needed because simply covering up would leave the lower-order terms still present at \(s = -p\); differentiating first removes them.

The inversion uses the pair

so a repeated pole produces terms like \(te^{-at}\) and \(t^{2}e^{-at}\). These are exactly the forms met in the critically damped case of Chapter 8, where the repeated root of the characteristic equation forced an extra factor of \(t\). The Laplace method now explains why that factor appears: a repeated pole demands it.

A pair of complex conjugate poles is simple if not repeated, and could in principle be handled by the cover-up rule. In practice the complex algebra is cumbersome, and a better method is completing the square.

Write the quadratic factor and its numerator as

and complete the square in the denominator:

so that \(\alpha = a/2\) and \(\beta = \sqrt{b-\alpha^{2}}\). Then split the numerator to match the two standard forms:

Both terms are now in the table, giving

A complex pole pair produces a damped sinusoid, with \(\alpha\) — the real part — setting the decay and \(\beta\) — the imaginary part — setting the oscillation frequency. Comparing with Chapter 8, \(\alpha\) is the damping factor and \(\beta\) the damped natural frequency \(\omega_{d}\). The underdamped response is precisely what a complex pole pair looks like.

The cosine and sine terms can be merged into a single cosine with a phase angle, using the identity of Section 8.4:

If a system's response to an impulse is \(h(t)\), its response to an arbitrary input \(x(t)\) is given by the convolution integral. For a causal system driven by a signal starting at \(t=0\):

The limits deserve explanation. The lower limit is 0 because \(x(t) = 0\) for \(t < 0\). The upper limit is \(t\) because the system is causal: \(h(t) = 0\) for negative argument, so \(h(t-\lambda) = 0\) whenever \(\lambda > t\). A system cannot respond to an input it has not yet received, and that physical fact is what truncates the integral.

Properties. Convolution behaves much like multiplication:

That the impulse is the identity element explains why \(h(t)\) is called the impulse response: convolving any input with an impulse returns the input unchanged, so a system whose impulse response is an impulse passes everything through untouched.

The convolution of two signals is best understood graphically, and for piecewise-constant signals the graphical method is often the fastest route to an answer. There are four steps.

1. Folding. Take the mirror image of \(h(\lambda)\) about the vertical axis to obtain \(h(-\lambda)\).

2. Displacement. Shift \(h(-\lambda)\) to the right by \(t\) to obtain \(h(t-\lambda)\).

3. Multiplication. Form the product \(h(t-\lambda)\,x(\lambda)\).

4. Integration. For that value of \(t\), find the area under the product for \(0 < \lambda < t\). That area is \(y(t)\).

Repeating for every \(t\) traces out the whole response. The mnemonic is fold, shift, multiply, integrate.

In practice the work is organised by identifying the ranges of \(t\) over which the overlap pattern is qualitatively the same, then computing one integral per range. The number of ranges is determined by the breakpoints of the two signals.

The differentiation and integration properties together turn any linear integro-differential equation with constant coefficients into an algebraic equation. The procedure has three steps:

1. Transform the equation term by term, using \(\mathcal{L}\left[f'\right] = sF-f(0^{-})\) and \(\mathcal{L}\left[\int f\right] = F/s\). The initial conditions enter here automatically.

2. Solve algebraically for the transform of the unknown.

3. Invert by partial fractions.

Compare this with Chapters 7 and 8, where the natural response had to be found from the characteristic equation, the forced response from the steady state, the two added, and the arbitrary constants fitted to the initial conditions afterwards. The Laplace method does all of that in one pass, and it never requires the response to be classified as over-, critically or underdamped — the partial fraction expansion sorts that out by itself, according to whether the poles turn out to be real, repeated or complex.

• The Laplace transform maps \(f(t)\) into \(F(s)\) with \(s = \sigma+j\omega\), generalising the phasor's \(j\omega\) to include exponential growth and decay.

• The lower limit \(0^{-}\) captures impulses and discontinuities at the origin and matches the pre-switching initial conditions.

• The transform exists where the integral converges, in a right half-plane \(\mathrm{Re}(s) > \sigma_{c}\).

• Differentiation becomes multiplication by \(s\), with the initial conditions appearing automatically; integration becomes division by \(s\).

• A delay of \(a\) seconds contributes a factor \(e^{-as}\); a periodic function's transform is its first period divided by \(1-e^{-Ts}\).

• The initial and final value theorems give limiting behaviour without inversion, the latter valid only if the final value exists.

• Inversion is by partial fractions: simple poles give exponentials, repeated poles give \(t^{n}e^{-at}\) terms, and complex pairs give damped sinusoids.

• Complex poles are best handled by completing the square, with \(\alpha\) giving the decay and \(\beta\) the oscillation frequency.

• Convolution in time equals multiplication in \(s\); graphically it is fold, shift, multiply, integrate.

• Integro-differential equations become algebraic ones, solved in a single pass with no separate natural and forced parts.

• Applying the time-shift property to a function not expressed in \((t-a)\). \(t\,u(t-1)\) must first be rewritten as \((t-1)u(t-1)+u(t-1)\), as in Example 15.3.

• Omitting the initial-condition terms in the derivative property. \(\mathcal{L}\left[f'\right]\) is \(sF(s)-f(0^{-})\), not simply \(sF(s)\). Dropping them silently assumes zero initial conditions.

• Using the final value theorem when no final value exists. For an oscillating or growing function the limit returns a number that means nothing. Check the pole locations first.

• Forgetting the differentiation step for repeated poles. Only the highest-order residue comes from a plain cover-up; the rest need derivatives.

• Failing to make the sine numerator equal \(\beta\). The table entry is \(\beta/\left[\left(s+\alpha\right)^{2}+\beta^{2}\right]\), so a numerator of 2 with \(\beta = 3\) must be written as \(\tfrac{2}{3}\times3\).

• Applying partial fractions to an improper fraction. If the numerator degree is not less than the denominator's, divide out first.

• Using \(f(0^{+})\) where the property calls for \(f(0^{-})\). They differ whenever an impulse acts at the origin, which is exactly when it matters.

• Getting the convolution limits wrong. They are 0 to \(t\) for a causal system driven from rest — not \(-\infty\) to \(\infty\), and not 0 to \(\infty\).

• Forgetting to fold before shifting. The integrand is \(h(t-\lambda)\), which is \(h\) reversed and then displaced.

• Convolving when multiplying transforms would be easier. Unless the data are numerical or the product is intractable, transform first.

Work these before opening the answers. Where an inverse transform is found, check it with the initial and final value theorems.

1. P15.1 Find the Laplace transform of \(f(t) = 4\delta(t)-3u(t)+5e^{-4t}u(t)\).

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   \[F(s) = 4 - \frac{3}{s} + \frac{5}{s+4}\]

2. P15.2 Find the transform of a rectangular pulse of height 5 lasting from \(t = 1\) to \(t = 4\).

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   \[f(t) = 5\left[u(t-1)-u(t-4)\right] \;\Longrightarrow\; F(s) = \frac{5}{s}\left(e^{-s}-e^{-4s}\right)\]

3. P15.3 Find the initial and final values for \(F(s) = \dfrac{6\left(s+2\right)}{s\left(s+1\right)\left(s+4\right)}\).

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   \[f(0^{+}) = \lim_{s\to\infty}\frac{6s\left(s+2\right)}{s\left(s+1\right)\left(s+4\right)} = 0\]

   \[f(\infty) = \lim_{s\to0}\frac{6\left(s+2\right)}{\left(s+1\right)\left(s+4\right)} = \frac{12}{4} = 3\]

   Both poles of \(sF(s)\) are in the left half-plane, so the final value is legitimate.

4. P15.4 Invert \(F(s) = \dfrac{6\left(s+2\right)}{s\left(s+1\right)\left(s+4\right)}\) and verify against P15.3.

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   \[A = \frac{6\left(2\right)}{\left(1\right)\left(4\right)} = 3, \qquad B = \frac{6\left(1\right)}{\left(-1\right)\left(3\right)} = -2, \qquad C = \frac{6\left(-2\right)}{\left(-4\right)\left(-3\right)} = -1\]

   \[f(t) = \left(3 - 2e^{-t} - e^{-4t}\right)u(t)\]

   Check: \(f(0) = 3-2-1 = 0\) \(\checkmark\) and \(f(\infty) = 3\) \(\checkmark\)

5. P15.5 Invert \(F(s) = \dfrac{8}{s\left(s+2\right)^{2}}\).

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   \[F(s) = \frac{A}{s}+\frac{C}{\left(s+2\right)^{2}}+\frac{D}{s+2}\]

   \[A = \frac{8}{4} = 2, \qquad C = \left.\frac{8}{s}\right|_{s=-2} = -4, \qquad D = \left.\frac{d}{ds}\left(\frac{8}{s}\right)\right|_{s=-2} = \left.\frac{-8}{s^{2}}\right|_{s=-2} = -2\]

   \[f(t) = \left(2 - 4te^{-2t} - 2e^{-2t}\right)u(t)\]

   Check: \(f(0) = 2-0-2 = 0\) \(\checkmark\), \(f(\infty) = 2\) \(\checkmark\)

6. P15.6 Invert \(F(s) = \dfrac{10}{s^{2}+6s+25}\) by completing the square.

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   \[s^{2}+6s+25 = \left(s+3\right)^{2}+16 \;\Longrightarrow\; \alpha = 3, \; \beta = 4\]

   \[F(s) = \frac{10}{\left(s+3\right)^{2}+16} = \frac{10}{4}\cdot\frac{4}{\left(s+3\right)^{2}+16}\]

   \[f(t) = 2.5\,e^{-3t}\sin 4t\;u(t)\]

7. P15.7 Solve \(\dfrac{dx}{dt}+3x = 6u(t)\) with \(x(0) = 1\).

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   \[sX-1+3X = \frac{6}{s} \;\Longrightarrow\; X(s) = \frac{6}{s\left(s+3\right)}+\frac{1}{s+3} = \frac{s+6}{s\left(s+3\right)}\]

   \[A = \frac{6}{3} = 2, \qquad B = \frac{-3+6}{-3} = -1\]

   \[x(t) = \left(2 - e^{-3t}\right)u(t)\]

   Check: \(x(0) = 1\) \(\checkmark\) and \(x(\infty) = 2 = 6/3\) \(\checkmark\) — the universal first-order formula of Chapter 7 in disguise.

8. P15.8 Use the convolution theorem to find \(\mathcal{L}^{-1}\left[\dfrac{1}{s\left(s+1\right)}\right]\) two ways.

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   By convolution: the factors invert to \(u(t)\) and \(e^{-t}\), so

   \[f(t) = \int_{0}^{t}\left(1\right)e^{-(t-\lambda)}d\lambda = e^{-t}\left(e^{t}-1\right) = 1-e^{-t}\]

   By partial fractions: \(\dfrac{1}{s}-\dfrac{1}{s+1}\), giving the same result \(\checkmark\) The partial-fraction route is shorter, which is typical.

9. P15.9 Two rectangular pulses, each of unit height and width 2, are convolved. Sketch the result and state its peak value, duration and area.

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   Equal-width rectangles convolve to a triangle. Peak value 2 (the overlap length at full coincidence), duration \(2+2 = 4\), and area \(\left(1\times2\right)\times\left(1\times2\right) = 4\).
   Check: a triangle of base 4 and height 2 has area \(\tfrac{1}{2}\left(4\right)\left(2\right) = 4\) \(\checkmark\)

10. P15.10 A system has \(H(s) = \dfrac{2}{s+4}\). Find its impulse response and its step response.

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    Impulse response: \(h(t) = 2e^{-4t}u(t)\).
    Step response: multiply by \(1/s\) and invert:

    \[Y(s) = \frac{2}{s\left(s+4\right)} = \frac{0.5}{s}-\frac{0.5}{s+4} \;\Longrightarrow\; y(t) = 0.5\left(1-e^{-4t}\right)u(t)\]

    Note that the step response is the integral of the impulse response, since dividing by \(s\) is integration.