I was learning to work with RLC circuits with sinusoidal excitations and the book I was dealing with showed me that the responses of these passive elements to sinusoidal excitation were themselves purely sinusoidal and hence phasor diagrams were the best way to deal with these elements.

When I tried solving a simple series RLC circuit using Laplace/complex frequency analysis, I landed myself a sine term, a cosine term, and a real exponential term.

I don't know why I am getting that real exponential term. I tried to do it over and over and I got the same results. I am pretty sure my calculations are correct but the book mentions nothing about the exponential term. Kindly pitch in...

This is the circuit I was talking about:


simulate this circuit – Schematic created using CircuitLab

Here is an image for a KVL and Laplace transformation:

KVL and Laplace

Here is an image for the inverse after separation. I don't think there is a mistake.

variable separation and inverse

  • 1
    \$\begingroup\$ Why don't you show us the circuit you are trying to analyze and the work that you have done? My internet connection is too slow to read your mind. \$\endgroup\$
    – Joe Hass
    Mar 22, 2014 at 13:04
  • \$\begingroup\$ I uploaded the circuit, will upload my equations in a minute... \$\endgroup\$
    – Ghosal_C
    Mar 22, 2014 at 13:23
  • \$\begingroup\$ Where is the "L" in your circuit? The circuit you provided doesn't have an inductor, but you said it was a series RLC circuit. \$\endgroup\$
    – Joe Hass
    Mar 22, 2014 at 13:43
  • \$\begingroup\$ No, for this circuit there is no L, just a R and a C and sinusoidal excitation. \$\endgroup\$
    – Ghosal_C
    Mar 22, 2014 at 13:45
  • 1
    \$\begingroup\$ Clearly you have tried to solve this on your own, and your work is incredibly neat for a student. You certainly do not look foolish. But as Dave Tweed said, this isn't a good place for the kind of tutorial you need. I would suggest that you look for circuit analysis books at your school's library. Even pretty old books would cover this material. \$\endgroup\$
    – Joe Hass
    Mar 22, 2014 at 14:01

2 Answers 2


That real exponential term represents the transient response of the system. Generally, when doing steady-state sinusoidal analysis, you can simply ignore the transient response altogether, since the real part of the exponent is usually a negative value times t (time), which goes to zero as t → ∞. If not, it means the system is unstable to begin with.

  • \$\begingroup\$ Okay. Thank you for your answer. Is there any substantial literature on transient responses during sinusoidal excitation, because where I am studying from they simply forgot to tell me anything whatsoever about it and assumed that I'd learn by making a fool out of myself on a public forum.. \$\endgroup\$
    – Ghosal_C
    Mar 22, 2014 at 13:30
  • \$\begingroup\$ No, it isn't foolish to ask for information. However, it is a very broad subject that will be difficult to handle in the Q&A format used on this site. Any decent college level textbook on circuit analysis should cover it in detail. I learned this stuff back in college myself, and I don't really have any online references that I can point you to. Perhaps someone else can help. \$\endgroup\$
    – Dave Tweed
    Mar 22, 2014 at 13:39
  • \$\begingroup\$ I'm sorry to have sounded crass, I was just a little pissed at the whole scenario. I was going on and on about the exponential term the whole day and I got a genuine answer about around 5-6 hours of head banging. Thank You very much for your help anyways, I really appreciate it. \$\endgroup\$
    – Ghosal_C
    Mar 22, 2014 at 13:43

The transient part is needed because before and at time 0 the capacitor voltage is 0V. At times tn = n / omega is should also be 0 volt had you only the steady state in your description. But it is not 0 - the transient part describes the details of this. Imagine the phase fi of the sin(w t + fi) had not been 0 or pi it would have an abrupt voltage step at time 0. You need at transient portion to describe how this affects the response.

But seeing your derivation I have one even more important piece of general advice to give: Always derive your equation using symbols as long as possible! So in your case stick with R1, C1, V1 and f labels in the equation and try to solve them analytically. Then insert numerical values into the final equation. Now of course it is not always possible to do this due to the complexity of the equation. But doing so:

  1. gives additional insight into your system having the parameters - like answer the questions - how does the different values affect the solution.
  2. aids in checking that the equations are reasonable as you can verify that the units match up - unit checking is a very power full tool to validate any equations with.

To give an example for point 1) from your particular derivation:

A reasonable question to ask after solving this equation is - why is the transient term x exp(-10t)? Where does the -10 come from? You can only tell by looking back through your derivations to figure that out.

But had you solved using the symbols you would find the term being x exp(-t / (R1 * C1)) immediately giving insight into how R1 and C1 affect the transient's time response. Plus - you can see that the units are correct as the unit of RC is seconds so the argument for the exponential is sec / sec or unitless as it must be.

To illustrate point 2) from your example: Suppose you made a mistake and ended up with an equation for transient of form x exp(-t / (R1 * R1 * C1)) you would immediately be able to spot that this cannot be correct just by looking at the units. Your unitless numbers removes this insight.

I cannot overstress how powerful this is in general for checking equations of physical properties.

So all that said - technically speaking your equation is incorrect because your numbers are unitless. Had you sticked in a proper way to derivation using only numbers, your transient term should read x exp(-10[1 / sec] t) using units (Hz or 1 / sec). Personally I would use 1 / sec here.

  • \$\begingroup\$ helarsen - Hi, FYI when you write on this site, it is parsed as Markdown. That means * and / (among other characters) may be interpreted as commands (e.g. to enable/disable italics). That is why some of the * don't appear in your answer. This is not uncommon when people write formulas here. Solutions include: (a) leave a space before & after Markdown operator characters, or (b) "Escape" Markdown operators in formulas, where needed, by prefixing them with "\" to prevent the parser from trying to interpret them; or (c) use LaTeX (via Mathjax syntax with \$ delimiters) for formulas. \$\endgroup\$
    – SamGibson
    Feb 13 at 23:29
  • \$\begingroup\$ thanks SamGibson for the heads up and detailed suggestions- Yes I did struggle a bit with this. For the moment I have adopted solution (a) into the post and to me it looks readable, considering the simplicity of the equations, so the more powerful LaTeX must be for me to learn and use another day \$\endgroup\$
    – helarsen
    Feb 14 at 10:08

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