When would you use a Laplace transform to figure something out about a circuit? Does the circuit have to have capacitors?


The LaPlace transform isn't really used in DC analysis of circuits. Typically, you would just use your standard "Circuits 101" tools to analyze the DC properties of a circuit. These are:

  • Kirchhoff's Current Law (Nodal Analysis)
  • Kirchhoff's Voltage Law (Loop Analysis)
  • Thevenin's (or Norton's) Theorem

In DC, capacitors are treated as open circuits (hence why they are also known "DC blockers") and inductors are treated as close circuits (or dead shorts).

LaPlace transforms are used in both frequency and transient analysis. They replace the more complex method of solving differential equations, or using convolution to determine the response of a circuit to an input.

Transient analysis builds upon the basic DC analysis, but determines what the circuit will do in response to a transient (a step input, an impulse, a ramp, etc).

Frequency analysis then builds upon the transient analysis to determine the response of a circuit to periodic signals (sine wave, square wave, triangle waves, as an example).

LaPlace analysis is frequently used in

  • Filter design (creating a circuit with specific frequency responses)
  • Controls (forcing a system or circuit to react with particular transient and frequency responses).
  • Higher level analysis of most electrical engineer (power, communications, electromagnetics,etc).
  • \$\begingroup\$ So if you put a capacitor in a DC circuit, it will act as an open circuit, and nothing will work? \$\endgroup\$ – 200ok404notfound Mar 23 '11 at 6:01
  • \$\begingroup\$ Assuming that the circuit is ONLY DC, and there are no transients (switches closing or opening, signal inputs, etc). DC analysis is best though of as an analogy to statics in mechanical engineering. Nothing in your circuit is moving or changing, and it is at an equilibrium. AC analysis is when you add some sort of dynamic input to your system. Once you add a dynamic to the system, then you will have you use LaPlace analysis to determine what the circuit will do over time. \$\endgroup\$ – mjcarroll Mar 23 '11 at 6:07
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    \$\begingroup\$ What if the voltage source comes from an AC adaptor and it's going into a voltage regulator with capacitors connected to it? Why would a voltage regulator need capacitors if the voltage is DC? \$\endgroup\$ – 200ok404notfound Mar 23 '11 at 6:11
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    \$\begingroup\$ I would rather say that correct functioning of the real circuit (with noise and transients) relies on there being a capacitor. :) Also, without any capacitors a real voltage regulator may oscillate or do some other strange things. \$\endgroup\$ – jpc Mar 23 '11 at 12:21
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    \$\begingroup\$ @200ok404notfound I am confused by your comment about how the correct function relies on their being noise. This makes no sense. In an ideal world when your voltage line is pure DC the capacitors will act as if they just weren't there. It will still function though. \$\endgroup\$ – Kellenjb Mar 23 '11 at 13:57

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