How do we get the complex frequency of an exponentially damped sinusoidal function?

Question

In my textbook, Engineering Circuit Analysis 8.Ed by William H. Hayt (Chapter 14, Section 14.1 page 536) I stumbled across a section explaining the complex frequency. The goal is to cover the concept of complex frequency when it comes to exponentially damped sinusoidal functions.

HOW DID THIS HAPPEN?

1. How did \$e^{\sigma t}\$ become \$e^{j\sigma t}\$ out of the blue?
2. And why did \$e^{j(\omega t)}\$ become \$e^{j(j\omega t)}\$ ? Any help would be much appreciated.

Answer 1

Let's first start by proving the first replacement. Do take note that \$\operatorname{cos}\left(-x\right)=\operatorname{cos}\left(x\right)\$ and that \$\operatorname{sin}\left(-x\right)=-\operatorname{sin}\left(x\right)\$:

$$\begin{align*} e^{j\left(\omega\,t+\theta\right)}&+e^{-j\left(\omega\,t+\theta\right)}\\ e^{j\left(\omega\,t+\theta\right)}&+e^{j\left(-\omega\,t-\theta\right)}\\ \left[\operatorname{cos}\left(\omega\, t+\theta\right)+j\cdot\operatorname{sin}\left(\omega\, t+\theta\right)\right]&+\left[\operatorname{cos}\left(-\omega\, t-\theta\right)+j\cdot\operatorname{sin}\left(-\omega\, t-\theta\right)\right]\\ \left[\operatorname{cos}\left(\omega\, t+\theta\right)+j\cdot\operatorname{sin}\left(\omega\, t+\theta\right)\right]&+\left[\operatorname{cos}\left(\omega\, t+\theta\right)-j\cdot\operatorname{sin}\left(\omega\, t+\theta\right)\right]\\ \operatorname{cos}\left(\omega\, t+\theta\right)&+\operatorname{cos}\left(\omega\, t+\theta\right)\\\\&= 2\operatorname{cos}\left(\omega\, t+\theta\right) \end{align*}$$

Obviously, this equivalent is true:

$$\operatorname{cos}\left(\omega\, t+\theta\right)=\frac12\left[ e^{j\left(\omega\,t+\theta\right)}+e^{-j\left(\omega\,t+\theta\right)}\right]$$

So, this means the first two lines of the quoted text appears correct:

$$\begin{align*} v\left(t\right)&=V_m\:e^{\sigma\,t}\operatorname{cos}\left(\omega\, t+\theta\right)\\\\ &=\frac12 V_m\:e^{\sigma\,t}\left[e^{j\left(\omega\,t+\theta\right)}+e^{-j\left(\omega\,t+\theta\right)}\right] \end{align*}$$

Now I will go in tiny steps below, so there is no mistaking the simple algebra involved:

$$\begin{align*} v\left(t\right)&=\frac12 V_m\:e^{\sigma\,t}\left[e^{j\left(\omega\,t+\theta\right)}+e^{-j\left(\omega\,t+\theta\right)}\right]\\\\ &=\frac12 V_m\:e^{\sigma\,t}\:e^{j\left(\omega\,t+\theta\right)}+\frac12 V_m\:e^{\sigma\,t}\:e^{-j\left(\omega\,t+\theta\right)}\\\\ &=\frac12 V_m\:e^{\sigma\,t+j\left(\omega\,t+\theta\right)}+\frac12 V_m\:e^{\sigma\,t-j\left(\omega\,t+\theta\right)}\\\\ &=\frac12 V_m\:e^{\sigma\,t+j\,\omega\,t+j\,\theta}+\frac12 V_m\:e^{\sigma\,t-j\,\omega\,t-j\,\theta}\\\\ &=\frac12 V_m\:e^{j\,\theta}\:e^{\sigma\,t+j\,\omega\,t}+\frac12 V_m\:e^{-j\,\theta}\:e^{\sigma\,t-j\,\omega\,t}\\\\ &=\frac12 V_m\:e^{j\,\theta}\:e^{\left(\sigma+j\,\omega\right)t}+\frac12 V_m\:e^{-j\,\theta}\:e^{\left(\sigma-j\,\omega\right)t} \end{align*}$$

This seems too easy. So did I completely misunderstand your question?