Appendix-C

Complex Frequency and the Laplace Transform:

The transient analysis methods when applied to higher-order circuits cannot be solved without an alternate solution method based on the notions of complex frequency and of the Laplace transform. In this appendix we will demonstrate that the frequency response of linear circuits is but a special case of the general transient response of the circuit, when analyzed by means of Laplace methods. In addition, the use of the Laplace transform method allows the introduction of systems concepts, such as poles, zeros, and transfer functions that we frequently apply in designing FIR and IIR filters in digital signal processing.

Complex Frequency

A sinusoidal signal can be represented as

\[v(t) = A cos(\omega t + \phi)\space\space\space (C.1)\]

in the equivalent phasor form

\[V(j\omega) = Ae^{j\phi} = A\angle\phi\space\space\space (C.2)\] The above two expressions given above are related by

\[v(t) = Re(Ve^{j\omega t}) \space\space\space (C.3)\]

Phasor notation and analysis are very useful in solving AC steady-state circuits, in which the voltages and currents are steady-state sinusoids.

We will now consider a different class of waveforms, useful in the transient analysis of circuits, namely, damped sinusoids. The most general form of a damped sinusoid is

\[v(t) = Ae^{\sigma t} cos(\omega t + \phi)\space\space\space (C.4)\]

In eq. (C.4), we notice that, a damped sinusoid is a sinusoid multiplied by a real exponential \(e^{\sigma t}\). The constant \(\sigma\) is real and is usually zero or negative in most practical circuits. Figure C.1(a) and (b) depicts the case of a damped sinusoid with negative \(\sigma\) and with positive \(\sigma\), respectively. Note that the case of \(\sigma\) = 0 corresponds exactly to a sinusoidal waveform. The definition of phasor voltages and currents can easily be extended to account for the case of damped sinusoidal waveforms by defining a new variable s, called the complex frequency:

\[s = \sigma + j\omega \space\space\space(C.5)\] Note that the special case of = 0 corresponds to s = \(j\omega\), that is, the familiar steady-state sinusoidal (phasor) case. We shall now refer to the complex variable V(s) as the complex frequency domain representation of v(t). Laplace transform is analogous to phasor analysis in the circuit;by substituting the variable s wherever \(j\omega\) was used in a circuit.

Exercise C.1 Find the complex frequencies that are associated with

1. \(5e^{−4t}\) b. \(cos(2\omega t)\) c. \(sin(\omega t +2\theta)\) d. \(4e^{−2t}sin(3t −50^o)\) e. \(e^{−3t}(2+cos 4t)\)

Solutions: (a) Given \(5e^{−4t}\); s = -4 (Ans)

2. \(cos(2\omega t) = \frac{e^{2j\omega}+e^{-2j\omega}}{2}\); \(s =\pm 2j\omega\) (Ans)

3. \(sin(\omega t +2\theta)=\frac{{e^{j(\omega t+2\theta)}}+{e^{-j(\omega t+2\theta)}}}{2}\); \(s =\pm j\omega\) (Ans)

4. \(4e^{−2t}sin(3t−50^o)\); \(s = -2 \pm j3\)

5. \(e^{−3t}(2+cos 4t)\); \(s = -3, s = -3 \pm j4\)

Exercise C.2 Find s and V(s) if v(t) is given by

1. \(5e^{−2t}\) b. \(5e^{−2t} cos(4t + 10^o)\) c. \(4cos(2t− 20^o)\)

Solution:

Given (a) \(v(t) = 5e^{−2t}\) ; \(s=-2, V(s) = 5\angle 0^o\) (Ans)

2. \(5e^{−2t} cos(4t + 10^o)\) ; \(s = -2+4j, V(s) = 5\angle 10^o\)

3. \(4cos(2t− 20^o)\); \(s = 2j, V(s) = 4\angle(-20^o)\)

Exercise C.3 Find v(t) if

1. s = −2, \(V = 2\angle 0^o\) b. s = j2, \(V = 12\angle −30^o\) c. s = −4 + j3, \(V = 6\angle 10^o\)

Solution:

1. s = −2, \(V = 2\angle 0^o\) ; \(V(s)=2e^{-2t}\)

2. 2. s = j2, \(V = 12\angle −30^o\); \(V(s) = 12cos(2t-30^o)\)

3. s = −4 + j3, \(V = 6\angle 10^o\); \(V(s) = 6e^{-4t}cos(3t+10^o)\)

All the concepts and rules used in AC network analysis, such as impedance, admittance, KVL, KCL, and Thévenin’s and Norton’s theorems, carry over to the damped sinusoid case exactly. In the complex frequency domain, the current I(s) and voltage V(s) are related by the expression

\[V(s) = Z(s)I(s) \space\space\space (C.6)\]

where Z(s) is the familiar impedance, with s replacing \(j\omega\). We may obtain Z(s) from Z(jω) by simply replacing \(j\omega\) by s. For a resistance R, the impedance is

\[Z_R(s) = R \space\space\space (C.7)\]

For an inductance L, the impedance is

\[Z_L(s) = sL \space\space\space (C.8)\]

For a capacitance C, it is

ZC(s) = (C.9)

Impedances in series or parallel are combined in exactly the same way as in the AC steady-state case, since we only replace jω by s.

Exercise C.4_ Use complex impedance ideas to determine the response of a series RL circuit to a damped exponential voltage. The time-domain expression for the series current \(i_L(t)\). Given the \(v_s(t) = 10e^{−2t} cos(5t)\) V; R = 4 \(\Omega\); L = 2 H.

Solution:

The input voltage phasor can be represented by the expression

\(V(s) = 10\angle 0\) V

The impedance seen by the voltage source is

Z(s) = R + sL = 4 + 2s

Thus, the series current is

\[I(s) = \frac{V(s)}{Z(s)} = \frac{10}{4 + 2s} = \frac{10}{4 + 2(−2 + j5)} = \frac{10} {j10} = -j1 = 1∠\pi/2\] Finally, the time-domain expression for the current is \[i_L(t) = e^{−2t} cos(5t − \pi/2)\] A Similar to frequency response functions \(H(j\omega)\), it is possible to define a transfer function H(s). This can be a ratio of a voltage to a current, a ratio of a voltage to a voltage, a ratio of a current to a current, or a ratio of a current to a voltage. Using the transfer function and knowing the input (voltage or current) to a circuit, we can find an expression for the output either in the complex frequency domain or in the time domain.

As an example, suppose \(V_i(s)\) and \(V_o(s)\) are the input and output voltages to a circuit, respectively, in complex frequency notation. Then

\[H(s) = \frac{V_o(s)}{V_i(s)} \space\space\space (C.10)\]

from which we can obtain the output in the complex frequency domain by computing

\[V_o(s) = H(s)V_i(s) \space\space\space (C.11)\]

If \(V_i(s)\) is a known damped sinusoid, we can then proceed to determine \(v_o(t)\) by means of the method illustrated earlier in this section.

The Laplace Transform:

The Laplace transform, named after the French mathematician and astronomer Pierre Simon de Laplace, is defined by

\[\mathcal{L}[f(t)] = F(s) = \int_0^{\infty} f(t)e^{−st}dt \space\space\space (C.12)\] The function F(s) is the Laplace transform of f(t) and is a function of the complex frequency \(s = \sigma + j\omega\), considered earlier in this section. Note that the function f (t) is defined only for t ≥ 0. This definition of the Laplace transform applies to what is known as the one-sided or unilateral Laplace transform, since f (t) is evaluated only for positive t. To conveniently express arbitrary functions only for positive time, we introduce a special function called the unit step function u(t), defined by the expression

u(t) = 0 t < 0 1 t > 0 (C.13)

\[ u(t) = \begin{cases} 0\space\space t < 0 \\ 1 \space\space t > 0 \space\space\space \end{cases} \]

Exercise C.5_ Find the Laplace transform of \(f(t) = e^{−at}u(t)\).

Solution:

\[F(s)=\mathcal{L}[f(t)] =\int_0^{\infty}e^{-at}e^{-st}dt = \frac{1}{s+a}e^{-(s+a)t}\bigg|_0^{\infty}=\frac{1}{s+a}\] Exercise C.5_ Find the Laplace transform of \(f(t) = cos(\omega t)u(t)\).

Solution:

\[F(s)=\mathcal{L}[f(t)]=\mathcal{L}[cos(\omega t)u(t)]=\int_0^{\infty}\frac{1}{2}(e^{j\omega t}+e^{-j\omega t})e^{-st}dt\] \[F(s)=\frac{1}{2}\int_0^{\infty}e^{(j\omega -s)t}dt+\frac{1}{2}\int_0^{\infty}e^{(-j\omega -s)t}dt=\frac{1}{2}\frac{1}{(j\omega -s)}e^{(j\omega -s)t}\bigg|_0^{\infty}+\frac{1}{2}\frac{1}{(-j\omega -s)}e^{(j\omega -s)t}\bigg|_0^{\infty}\]

\[F(s)=\frac{1}{2}\frac{1}{(-j\omega +s)}+\frac{1}{2}\frac{1}{(j\omega +s)}=\frac{s}{s^2+\omega^2}\]

Table C.1 Some Common Laplace Transform Pairs.

f(t)	F(s)
\(\delta(t)\space (unit \space impulse)\)	1
\(u(t)\space (unit \space step)\)	\(\frac{1}{s}\)
\(e^{-at}u(t)\)	\(\frac{1}{s+\omega}\)
\(sin\omega tu(t)\)	\(\frac{\omega}{s^2+\omega^2}\)
\(cos\omega tu(t)\)	\(\frac{s}{s^2+\omega^2}\)
\(e^{-at}sin\omega tu(t)\)	\(\frac{\omega}{(s+a)^2+\omega^2}\)
\(e^{-at}cos\omega tu(t)\)	\(\frac{s+a}{(s+a)^2+\omega^2}\)
\(tu(t)\)	\(\frac{1}{s^2}\)

Exercise C.6 Find the inverse Laplace transform of

\[F(s) = \frac{2}{s+3}+\frac{4}{s^2+4}+\frac{4}{s}\] Solution:

Using Table C.1, we can find the inverse-transform each of the tersm of F(s) and then add together to get the result:

\[f(t) = \mathcal{L^{-1}}[F(s)]=\mathcal{L^{-1}}[\frac{2}{s+3}+\frac{4}{s^2+4}+\frac{4}{s}]=\mathcal{2L^{-1}}[\frac{1}{s+3}]+\mathcal{2L^{-1}}[\frac{2}{s^2+2^2}]+\mathcal{4L^{-1}}[\frac{1}{s}]\] \[f(t) =2e^{-3t}u(t)+2sin(2t)u(t)+4u(t)=[2e^{-3t}+2sin(2t)+4]u(t).\] Exercise C.7 Find the inverse Laplace transform of

\[F(s) = \frac{2s + 5}{s^2 + 5s + 6}\] Solution:

Table C.1 doesn’t have the direct results for this problem. We need to apply a partial fraction expansion of the function F(s) and then individually transform each term in the expansion. A partial fraction expansion is the inverse operation of obtaining a common denominator and is illustrated below.

\[F(s) = \frac{2s + 5}{s^2 + 5s + 6}=\frac{A}{s+2}+\frac{B}{s+3}\]

\[(s+2)F(s) =A+\frac{B(s+2)}{s+3}\] \[(s+3)F(s) =\frac{A(s+3)}{s+2}+B\] \[\frac{2s+5}{(s+2)(s+3)}=\frac{A}{s+2}+\frac{B}{s+3}\] \[\frac{2s+5}{(s+3)}\bigg|_{s=-2} =A+\frac{B(s+2)}{s+3}; A = 1\] \[\frac{2s+5}{(s+2)}\bigg|_{s=-3} =\frac{A(s+3)}{s+2}+B; B = 1\] \[F(s) = \frac{2s + 5}{s^2 + 5s + 6}=\frac{1}{s+2}+\frac{1}{s+3}\]

\[f(t) = \mathcal{L^{-1}}[F(s)]=\mathcal{L^{-1}}[\frac{1}{s+2}+\frac{1}{s+3}]=(e^{-2t}+e^{-3t})u(t)\space\space (Ans)\]