so I have encountered a hitch in my understanding of analyzing steady state AC circuits. I have looked everywhere for the answer to my question and I can find nothing. Usually when we analyze ss AC circ. we convert everything into the "Phasor domain." Now I have heard people call it Phasor domain and Frequency domain. Well I used the Fourier transform to derive the inductive and capacitive reactances just by simply taking the transform of:

\$I = C\frac{dv}{dt}\$ and \$V = L\frac{di}{dt}\$.

Okay so that makes sense we want to analyze the circuit in the frequency domain so we use the Fourier transform to do that. But what I never see justified is how you can take a time domain source such as cos(w0t+phi) and suddenly just write it as a Phasor such as He^(jphi), where H is the magnitude. I always see the argument that you can write cosine as the real part of a complex exponential and since the frequency does not change you can just ignore it and then you get a phasor. That is not very rigorous and it bothers me. I feel like I should be able to take a Fourier transform of the source and get a phasor from that when I plug in the condition that w=w0. So I found the Fourier Transform of a phase shifted cosine and when plugging in the above condition I get pie^(jphi). Here is the derivation here,https://imgur.com/a/QNU8lt4. In the image they used a minus phi but if you use a plus phi then you get pie^(jphi). The problem is this almost works but there is a factor of PI in there which as we know when we write the phasor form of a cosine there is no pi in the mag unless we put one there on purpose. Does anyone know where my thinking is going a miss? This has been driving me insane. Also sorry for the lack of TEX use, I am not familiar and its late and I needed to get this question out there. Thank you!

  • 1
    \$\begingroup\$ Check out how to use TEX if you're interested. \$\endgroup\$ – Dmitry Grigoryev Sep 19 at 6:48
  • \$\begingroup\$ Example: \$\pi\$ is "pi" with "\$" before and after. \$\endgroup\$ – Andy aka Sep 19 at 7:01
  • \$\begingroup\$ To determine frequency response of an LCR circuit, it's normal to replace \$\small L\$ by \$\small j \omega L\$, and \$\small C\$ by \$\frac{1}{j\omega C}\$, where \$\small \omega\$ is frequency in rad/s. This method is derived from the Laplace transform analysis where a sinusoid is the forcing function and transient terms are assumed to have dissipated, leaving the steady state sinusoidal terms only (i.e. the frequency response). This does not lack rigour, it just means that we don't need to go through the same routine analysis process from first principles every time. \$\endgroup\$ – Chu Sep 19 at 9:20
  • \$\begingroup\$ @Chu I am not asking about reactances in the frequency domain. I can get those from Fourier transform. What I am asking is can you show me a derivation of when you take the Fourier transform of your forcing function say, cos(wot+phi), and you set the condition that w=wo that you get a phasor representation of your signal in the frequency domain with a just a mag. and phase. \$\endgroup\$ – EECE Sep 19 at 17:55
  • \$\begingroup\$ I wouldn't use the Fourier transform for this task. The better approach is the Laplace transform, then transform to the frequency domain via \$s\rightarrow j\omega \$, which gives the steady-state frequency response \$\endgroup\$ – Chu Sep 19 at 19:04

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