Applications of the Laplace Transform

Each element's time-domain relation is transformed using the differentiation property of Section 15.4.

Resistor. Ohm's law contains no calculus, so it transforms unchanged:

Inductor. Applying \(\mathcal{L}\left[di/dt\right] = sI(s)-i(0^{-})\):

or, solving for the current instead,

Capacitor. Similarly, from \(i = C\,dv/dt\):

Read those equations as circuits. The inductor's voltage equation says: an impedance \(sL\) in series with a voltage source of value \(Li(0^{-})\). The current equation says the same thing in Norton form: an admittance \(1/sL\) in parallel with a current source \(i(0^{-})/s\). Either model may be used, and the choice is made to suit the analysis — series models for mesh analysis, parallel models for nodal.

Every problem in this chapter follows three steps.

1. Transform the circuit from the time domain to the \(s\)-domain. Replace each source by its transform, each element by its impedance, and each initial condition by the appropriate source.

2. Solve the circuit by any technique — nodal, mesh, superposition, source transformation, Thévenin, Norton. The algebra is in \(s\); the methods are those of Chapters 3 and 4.

3. Invert the solution by partial fractions to return to the time domain.

The procedure is deliberately identical in shape to the phasor procedure of Section 10.2. What has changed is its reach: a step, a pulse, an exponential or a sinusoid can all be handled, and a circuit with stored energy needs no special treatment.

One point of technique. Because a transformed source such as \(1/s\) is itself a function of \(s\), the algebra produces fractions within fractions. Multiplying numerator and denominator by \(s\) at the earliest opportunity keeps the expressions manageable, and it is worth doing before rather than after collecting terms.

Kirchhoff's laws applied to a circuit containing both an inductor and a capacitor produce an equation with both a derivative and an integral. The transform handles both at once, using \(\mathcal{L}\left[f'\right] = sF - f(0^{-})\) and \(\mathcal{L}\left[\int f\right] = F/s\).

With impedances defined and Kirchhoff's laws intact, mesh and nodal analysis proceed exactly as in Chapters 3 and 10. The equations look the same; only the entries are functions of \(s\).

Here the method shows its real advantage. A circuit with stored energy needs no separate transient analysis: the initial conditions simply appear as extra sources and the algebra proceeds unchanged.

Thévenin's and Norton's theorems, superposition and source transformation all apply in the \(s\)-domain, for the same reason they applied in the phasor domain: the circuit is linear and Kirchhoff's laws hold.

Note that \(Z_{Th}\) is now a function of \(s\) rather than a number — it may contain terms in \(s\) and \(1/s\), representing inductance and capacitance in the equivalent. A Thévenin equivalent in the \(s\)-domain is valid at all frequencies and for all time, unlike the phasor equivalent of Section 10.10 which held at one frequency only.

As in Chapter 4, when dependent sources are present the independent sources are deactivated but the dependent ones are kept, and \(Z_{Th}\) is found from the ratio above or by a test source.

The transfer function of Chapter 14 returns, now as a function of \(s\) rather than \(\omega\):

defined for zero initial conditions, since it is meant to describe the network rather than its state. As before there are four kinds — voltage gain, current gain, transfer impedance and transfer admittance — depending on whether input and output are voltages or currents.

The impulse response. Consider the special case \(x(t) = \delta(t)\). Then \(X(s) = 1\) by Section 15.3, so

The impulse response is the inverse transform of the transfer function. They are the same object viewed in two domains, and this is why \(h(t)\) characterises the entire network: once it is known, the response to any input follows by convolution, or equivalently by multiplying transforms.

The relation \(Y(s) = H(s)X(s)\) is the convolution theorem of Section 15.9 read in the other direction. Convolution in time, multiplication in \(s\) — and the transfer function is what does the multiplying.

Any network function can be written as a ratio of polynomials, and then factored:

where \(K = a_{m}/b_{n}\) is the scale factor.

• The roots of \(\mathbf{N}(s) = 0\) are the zeros, where the function vanishes. They are marked \(\bigcirc\) on a diagram.

• The roots of \(\mathbf{D}(s) = 0\) are the poles, where the function becomes infinite. They are marked \(\times\).

Properties of the poles and zeros of a realisable network function:

• If not repeated they are called simple; otherwise multiple.

• Complex poles and zeros must occur in conjugate pairs, because the polynomial coefficients are real.

• The real parts of all poles must be negative, and any pole on the \(j\omega\) axis must be simple.

The last property is the stability condition of the next section, stated in advance.

Why the diagram matters. Plotting the poles and zeros in the complex \(s\)-plane gives an immediate picture of the network's behaviour. A pole's real part sets an exponential decay rate; its imaginary part sets an oscillation frequency; a pole close to the \(j\omega\) axis decays slowly and dominates the response. The pole–zero diagram is used both in analysis and in synthesis, where a network is designed by first choosing where its poles should be.

A question that could not even be asked with phasors can now be answered precisely.

A network is stable if its response to a bounded input remains bounded. Since each pole \(p\) contributes a term \(e^{pt}\) to the impulse response, the condition is immediate:

A pole in the left half-plane gives a decaying exponential; a pole in the right half-plane gives one that grows without bound. The dividing line is the \(j\omega\) axis itself.

Note why a repeated pole on the axis is unstable while a simple one is not: the repetition introduces a factor of \(t\), as Section 15.7 showed, and \(t\sin\omega t\) grows without limit.

A passive RLC network is always stable, because its resistance always dissipates energy and pushes the poles into the left half-plane. A purely LC network is marginally stable — the case of Example 16.8. Instability requires an active element: an amplifier, a dependent source, or feedback. This is why oscillators are built with amplifiers, and why amplifier design is largely the art of keeping poles out of the right half-plane.

All the methods so far produce a single high-order equation in one unknown. The state variable approach instead produces a set of first-order equations in several unknowns — a form far better suited to computation, and the standard language of modern control theory.

The state variables of a circuit are its capacitor voltages and inductor currents — precisely the continuous quantities identified in Chapter 6. They are chosen because they represent the circuit's stored energy, and hence its complete internal condition at any instant. Knowing them at \(t_{0}\), together with the inputs thereafter, determines the entire future.

The equations are written in matrix form:

where \(\mathbf{x}\) is the state vector, \(\mathbf{z}\) the input vector and \(\mathbf{y}\) the output. The first is the state equation, the second the output equation.

The procedure. Apply KCL at every node containing a capacitor to obtain \(dv/dt\), and KVL around every loop containing an inductor to obtain \(di/dt\). Then rearrange so that each derivative appears alone on the left.

• In the \(s\)-domain, \(Z_{R} = R\), \(Z_{L} = sL\) and \(Z_{C} = 1/sC\) — the phasor impedances with \(j\omega \to s\).

• Initial conditions appear as independent sources: \(Li(0^{-})\) or \(i(0^{-})/s\) for an inductor, \(v(0^{-})/s\) or \(Cv(0^{-})\) for a capacitor.

• The procedure is transform, solve, invert — with every technique from Chapters 3, 4 and 10 available unchanged.

• Integro-differential equations become algebraic, with initial conditions built in.

• The transfer function \(H(s) = Y(s)/X(s)\) assumes zero initial conditions, and \(h(t) = \mathcal{L}^{-1}\left[H(s)\right]\) is the impulse response.

• Poles make a function infinite, zeros make it vanish; complex ones occur in conjugate pairs.

• A network is stable if all poles lie in the left half-plane, marginally stable with simple poles on the \(j\omega\) axis, and unstable otherwise.

• Passive RLC networks are always stable; instability requires an active element.

• The Routh criterion tests stability from the coefficients, without factoring.

• State variables are the capacitor voltages and inductor currents; the eigenvalues of \(\mathbf{A}\) are the network's poles.

• Omitting the initial-condition sources. A charged capacitor is not just \(1/sC\) — it is \(1/sC\) plus a source. Forgetting the source silently assumes zero stored energy.

• Mixing the series and parallel models within one element. Use \(Li(0^{-})\) or \(i(0^{-})/s\), never both.

• Getting the polarity of the initial-condition source wrong. Derive it from the transformed equation rather than guessing; the sign follows from \(V(s) = sLI(s)-Li(0^{-})\).

• Defining a transfer function for a circuit with stored energy. \(H(s)\) requires zero initial conditions; otherwise the ratio depends on the state as well as the network.

• Inverting an improper fraction without dividing first. If the numerator degree is not less than the denominator's, long division is required — and the constant term inverts to an impulse, as in Example 16.6.

• Expecting every node voltage to be continuous. Only capacitor voltages and inductor currents are. Example 16.3's node voltage jumps from 0 to 25 V.

• Confusing poles with zeros on a diagram. Poles are crosses and make the function infinite; zeros are circles and make it vanish.

• Judging stability from the numerator. Only the poles matter; zeros affect the shape of the response but never its stability.

• Treating a simple \(j\omega\)-axis pole as unstable. It gives a sustained oscillation, which is marginally stable. A repeated axis pole is genuinely unstable.

• Choosing the wrong state variables. They must be capacitor voltages and inductor currents — the continuous quantities — not resistor voltages or arbitrary node voltages.

Work these before opening the answers. State initial conditions explicitly wherever they appear.

1. P16.1 Write the \(s\)-domain model of a 2 H inductor carrying an initial current of 3 A, in both series and parallel forms.

   Show answer

   Series: impedance \(2s\) with a voltage source \(Li(0^{-}) = 6~\mathrm{V}\).
   Parallel: impedance \(2s\) with a current source \(i(0^{-})/s = 3/s\).

2. P16.2 Solve \(\dfrac{dy}{dt}+4y+3\displaystyle\int_{0}^{t}y\,d\tau = 0\) with \(y(0) = 3\).

   Show answer

   \[sY-3+4Y+\frac{3}{s}Y = 0 \;\Longrightarrow\; Y\left(s^{2}+4s+3\right) = 3s\]

   \[Y = \frac{3s}{\left(s+1\right)\left(s+3\right)}, \qquad A = \frac{-3}{2} = -1.5, \qquad B = \frac{-9}{-2} = 4.5\]

   \[y(t) = \left(-1.5e^{-t}+4.5e^{-3t}\right)u(t)\]

   Check: \(y(0) = -1.5+4.5 = 3\) \(\checkmark\)

3. P16.3 A series circuit of 4 \(\Omega\) and 0.5 H is driven by \(12u(t)~\mathrm{V}\) with \(i(0) = 0\). Find \(i(t)\) in the \(s\)-domain.

   Show answer

   \[I(s) = \frac{12/s}{4+0.5s} = \frac{24}{s\left(s+8\right)} = \frac{3}{s}-\frac{3}{s+8}\]

   \[i(t) = 3\left(1-e^{-8t}\right)u(t)~\mathrm{A}\]

   Final value 3 A \(= 12/4\) \(\checkmark\), time constant \(L/R = 0.125~\mathrm{s}\) \(\checkmark\)

4. P16.4 Repeat P16.3 with \(i(0) = 2~\mathrm{A}\).

   Show answer

   The inductor now contributes a source \(Li(0) = 1~\mathrm{V}\):

   \[I(s) = \frac{12/s+1}{4+0.5s} = \frac{2\left(12+s\right)}{s\left(s+8\right)} = \frac{3}{s}-\frac{1}{s+8}\]

   \[i(t) = \left(3-e^{-8t}\right)u(t)~\mathrm{A}\]

   Check: \(i(0) = 3-1 = 2\) \(\checkmark\) — the initial condition emerged from the algebra.

5. P16.5 Find the impulse response of a system with \(H(s) = \dfrac{5s}{s^{2}+6s+8}\).

   Show answer

   \[H(s) = \frac{5s}{\left(s+2\right)\left(s+4\right)}, \qquad A = \frac{-10}{2} = -5, \qquad B = \frac{-20}{-2} = 10\]

   \[h(t) = \left(-5e^{-2t}+10e^{-4t}\right)u(t)\]

6. P16.6 Locate the poles and zeros of \(H(s) = \dfrac{4\left(s+3\right)}{s\left(s^{2}+2s+5\right)}\) and classify the network.

   Show answer

   Zero: \(s = -3\).
   Poles: \(s = 0\) and, from \(s = \dfrac{-2\pm\sqrt{4-20}}{2}\), the pair \(s = -1 \pm j2\).
   The complex pair is in the left half-plane, but the pole at the origin lies on the \(j\omega\) axis and is simple, so the network is marginally stable.

7. P16.7 For what range of \(k\) is \(H(s) = \dfrac{1}{s^{3}+2s^{2}+ks+6}\) stable?

   Show answer

   Routh with \(a_{2} = 2\), \(a_{1} = k\), \(a_{0} = 6\):

   \[\frac{a_{2}a_{1}-a_{0}}{a_{2}} = \frac{2k-6}{2} > 0 \;\Longrightarrow\; k > 3\]

   At \(k = 3\) the network is marginally stable, oscillating at \(\omega = \sqrt{3} = 1.732~\mathrm{rad/s}\).

8. P16.8 Write the state equations of a parallel RLC circuit with \(R = 1~\Omega\), \(L = 1~\mathrm{H}\), \(C = 1~\mathrm{F}\) driven by a current source \(i_{s}\).

   Show answer

   KCL at the node and the inductor relation give

   \[C\frac{dv}{dt} = i_{s}-\frac{v}{R}-i, \qquad L\frac{di}{dt} = v\]

   \[\begin{bmatrix}\dot{v}\\\dot{i}\end{bmatrix} = \begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix}\begin{bmatrix}v\\i\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix}i_{s}\]

   Eigenvalues: \(\lambda^{2}+\lambda+1 = 0\), giving \(\lambda = -0.5 \pm j0.866\) — complex, so the circuit is underdamped, and both have negative real part, so it is stable.

9. P16.9 A system's step response is \(y(t) = \left(2-2e^{-5t}\right)u(t)\). Find \(H(s)\) and the impulse response.

   Show answer

   \[Y(s) = \frac{2}{s}-\frac{2}{s+5} = \frac{10}{s\left(s+5\right)}, \qquad X(s) = \frac{1}{s}\]

   \[H(s) = \frac{Y}{X} = \frac{10}{s+5} \;\Longrightarrow\; h(t) = 10e^{-5t}u(t)\]

   Note that the impulse response is the derivative of the step response, since multiplying by \(s\) is differentiation.

10. P16.10 Explain why the transfer function of any passive RLC network can never have a pole in the right half-plane.

    Show answer

    A right half-plane pole gives a term \(e^{pt}\) with \(\mathrm{Re}(p) > 0\), so the response would grow without bound and the energy delivered to the network would increase indefinitely. A passive network contains no energy source, and its resistance can only dissipate — so unbounded growth is impossible. Instability therefore requires an active element capable of supplying energy.