# The Laplace Transform   Whatsapp  Our frequency-domain analysis has been limited to circuits with sinusoidal inputs. In other words, we have assumed sinusoidal time-varying excitations in all our non-dc circuits. This chapter introduces the Laplace transform, a very powerful tool for analyzing circuits with sinusoidal or non-sinusoidal inputs.
The idea of transformation should be familiar by now. When using phasors for the analysis of circuits, we transform the circuit from the time domain to the frequency or phasor domain. Once we obtain the phasor result, we transform it back to the time domain. The Laplace transform method follows the same process: we use the Laplace transformation to transform the circuit from the time domain to the frequency domain, obtain the solution, and apply the inverse Laplace transform to the result to transform it back to the time domain.
The Laplace transform is significant for a number of reasons. First, it can be applied to a wider variety of inputs than phasor analysis. Second, it provides an easy way to solve circuit problems involving initial conditions, because it allows us to work with algebraic equations instead of differential equations. Third, the Laplace transform is capable of providing us, in one single operation, the total response of the circuit comprising both the natural and forced responses.
We begin with the definition of the Laplace transform and use it to derive the transforms of some basic, important functions. We consider some properties of the Laplace transform that are very helpful in circuit analysis. We then consider the inverse Laplace transform, transfer functions, and convolution. Finally, we examine how the Laplace transform is applied in circuit analysis, network stability, and network synthesis.|

#### DEFINITION OF THE LAPLACE TRANSFORM

Given a function $f(t)$, its Laplace transform, denoted by $F(s)$ or $\mathcal{L}[f(t)]$, is given by
$$\mathcal{L}[f(t)]=F(s)=\int_{0-}^{\infty} f(t) e^{-s t} d t \tag{1}$$
where $s$ is a complex variable given by $$s=\sigma+j \omega$$ Since the argument $s t$ of the exponent $e$ in Eq. (1) must be dimensionless, it follows that $s$ has the dimensions of frequency and units of inverse seconds $\left(\mathrm{s}^{-1}\right)$. In Eq. (1), the lower limit is specified as $0^{-}$to indicate a time just before $t=0$. We use $0^{-}$as the lower limit to include the origin and capture any discontinuity of $f(t)$ at $t=0$; this will accommodate functions - such as singularity functions - that may be discontinuous at $t=0$.
The Laplace transform is an integral transformation of a function f (t) from the time domain into the complex frequency domain, giving F(s).
We assume in Eq. (1) that $f(t)$ is ignored for $t<0$. To ensure that this is the case, a function is often multiplied by the unit step. Thus, $f(t)$ is written as $f(t) u(t)$ or $f(t), t \geq 0$.
The Laplace transform in Eq. (1) is known as the one-sided (or unilateral) Laplace transform. The two-sided (or bilateral) Laplace transform is given by $$F(s)=\int_{-\infty}^{\infty} f(t) e^{-s t} d t$$ The one-sided Laplace transform in Eq. (1), being adequate for our purposes, is the only type of Laplace transform that we will treat in this book.
A function $f(t)$ may not have a Laplace transform. In order for $f(t)$ to have a Laplace transform, the integral in Eq. (1) must converge to a finite value. Since $\left|e^{j \omega t}\right|=1$ for any value of $t$, the integral converges when $$\int_{0^{-}}^{\infty} e^{-\sigma t}|f(t)| d t<\infty \tag{2}$$ for some real value $\sigma=\sigma_{c}$. Thus, the region of convergence for the Laplace transform is $\operatorname{Re}(s)=\sigma>\sigma_{c}$, as shown in Fig. 1. In this region, $|F(s)|<\infty$ and $F(s)$ exists. $F(s)$ is undefined outside the region of convergence. Fortunately, all functions of interest in circuit analysis satisfy the convergence criterion in Eq. (2) and have Laplace transforms. Therefore, it is not necessary to specify $\sigma_{c}$ in what follows. Fig. 1: Region of convergence for the Laplace transform.
A companion to the direct Laplace transform in Eq. (1) is the inverse Laplace transform given by
$$\mathcal{L}^{-1}[F(s)]=f(t)=\frac{1}{2 \pi j} \int_{\sigma_{1}-j \infty}^{\sigma_{1}+j \infty} F(s) e^{s t} d s \tag{3}$$
where the integration is performed along a straight line $\left(\sigma_{1}+j \omega,-\infty<\right.$ $\omega<\infty)$ in the region of convergence, $\sigma_{1}>\sigma_{c}$. See Fig. 1. The direct application of Eq. (3) involves some knowledge about complex analysis beyond the scope of this book. For this reason, we will not use Eq. (3) to find the inverse Laplace transform. The functions $f(t)$ and $F(s)$ are regarded as a Laplace transform pair where $$f(t) \quad \Longleftrightarrow \quad F(s)$$ meaning that there is one-to-one correspondence between $f(t)$ and $F(s)$. The following examples derive the Laplace transforms of some important functions.
Example 1: Determine the Laplace transform of each of the following functions:(a) $u(t)$, (b) $e^{-a t} u(t), a \geq 0$, and (c) $\delta(t)$.
Solution:
(a) For the unit step function $u(t)$, shown in Fig. 2(a), the Laplace transform is
\begin{aligned}\mathcal{L}[u(t)] &=\int_{0^{-}}^{\infty} 1 e^{-s t} d t=-\left.\frac{1}{s} e^{-s t}\right|_{0} ^{\infty} \\&=-\frac{1}{s}(0)+\frac{1}{s}(1)=\frac{1}{s}\end{aligned}
(b) For the exponential function, shown in Fig. 2(b), the Laplace transform is
\begin{aligned}\mathcal{L}\left[e^{-a t} u(t)\right] &=\int_{0^{-}}^{\infty} e^{-a t} e^{-s t} d t \\&=-\left.\frac{1}{s+a} e^{-(s+a) t}\right|_{0} ^{\infty}=\frac{1}{s+a}\end{aligned}
(c) For the unit impulse function, shown in Fig. 2(c),
$$\mathcal{L}[\delta(t)]=\int_{0^{-}}^{\infty} \delta(t) e^{-s t} d t=e^{-0}=1$$
since the impulse function $\delta(t)$ is zero everywhere except at $t=0$.   Fig. 2: For Example 1: (a) unit step function, (b) exponential function, (c) unit impulse function.
Example 2: Determine the Laplace transform of $f(t)=\sin \omega t u(t)$.
Solution: Using Eq. (1), we obtain the Laplace transform of the sine function as
\begin{aligned}F(s)=\mathcal{L}[\sin \omega t] &=\int_{0}^{\infty}(\sin \omega t) e^{-s t} d t=\int_{0}^{\infty}\left(\frac{e^{j \omega t}-e^{-j \omega t}}{2 j}\right) e^{-s t} d t \\&=\frac{1}{2 j} \int_{0}^{\infty}\left(e^{-(s-j \omega) t}-e^{-(s+j \omega) t}\right) d t \\&=\frac{1}{2 j}\left(\frac{1}{s-j \omega}-\frac{1}{s+j \omega}\right)=\frac{\omega}{s^{2}+\omega^{2}}\end{aligned}

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