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.



  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.
  • \$\begingroup\$ Could you take another look at your questions? There appears to be no difference between the two things...the original and the "become" version. \$\endgroup\$ – Elliot Alderson Dec 2 '18 at 16:42
  • \$\begingroup\$ @elliotAlderson I just edited it. \$\endgroup\$ – billyandriam Dec 2 '18 at 16:44
  • \$\begingroup\$ It looks all screwed up. \$\endgroup\$ – Andy aka Dec 2 '18 at 17:41
  • \$\begingroup\$ @Andyaka It really does. Here is a picture of the section for you to see: my.pcloud.com/publink/… \$\endgroup\$ – billyandriam Dec 2 '18 at 17:45
  • \$\begingroup\$ You should post that pic in your question to give it authenticity. \$\endgroup\$ – Andy aka Dec 2 '18 at 17:50

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?

  • \$\begingroup\$ Was just about to hit submit with a similar derivation. Basically I think the OP forgot the rules of exponential \$\endgroup\$ – JonRB Dec 2 '18 at 18:14
  • \$\begingroup\$ @jonk You completely understood the question. I also had the same line of logic as you. So I guess it's just a typo and I will email the editor about this. \$\endgroup\$ – billyandriam Dec 2 '18 at 18:15
  • \$\begingroup\$ So the extra j was wrong and the document linked was wrong. +1 for extended effort. \$\endgroup\$ – Andy aka Dec 2 '18 at 19:25
  • \$\begingroup\$ Hold on... are you saying the math looks kosher? I’m confused, you can’t agree with the error and imply the math is kosher at the same time hee hee \$\endgroup\$ – Andy aka Dec 2 '18 at 20:04
  • 1
    \$\begingroup\$ @Andyaka I'll remove that comment as confusing. Sorry about that. The derivation is sufficiently clear and there's no need for words that may muddle it. \$\endgroup\$ – jonk Dec 2 '18 at 20:12

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