Find the phasor representation of an equation

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SUMMARY

The discussion focuses on finding the phasor representations of various time functions, specifically voltage and current equations. The phasors for the given functions are derived using Euler's formula, resulting in expressions such as 3e^{-j\frac{\pi}{3}} for v(t) and -2e^{j\frac{3\pi}{4}} for i(t). The conversion process for phasors, particularly for function (d), is clarified, demonstrating how to express -2e^{j\frac{3\pi}{4}} as 2e^{-j\frac{\pi}{4}}. The participants confirm the correctness of the other phasor conversions as well.

PREREQUISITES
  • Understanding of phasors and their significance in electrical engineering.
  • Familiarity with Euler's formula: e^{ix} = cos(x) + isin(x).
  • Knowledge of sinusoidal functions and their representations in the time domain.
  • Basic skills in complex number manipulation.
NEXT STEPS
  • Study the application of Euler's formula in electrical engineering contexts.
  • Learn about phasor analysis in AC circuit theory.
  • Explore the conversion between time domain and frequency domain representations.
  • Investigate the implications of phase shifts in sinusoidal functions.
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing AC signals and their phasor representations.

VinnyCee
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Homework Statement



Find the phasors of the following time functions:

(a) [tex]v(t)\,=\,3\,cos\left(\omega\,t\,-\,\frac{\pi}{3}\right)[/tex]

(b) [tex]v(t)\,=\,12\,sin\left(\omega\,t\,+\,\frac{\pi}{4}\right)[/tex]

(c) [tex]i(x,\,t)\,=\,2\,e^{-3\,x}\,sin\left(\omega\,t\,+\,\frac{\pi}{6}\right)[/tex]

(d) [tex]i(t)\,=\,-2\,cos\left(\omega\,t\,+\,\frac{3\pi}{4}\right)[/tex]

(e) [tex]i(t)\,=\,4\,sin\left(\omega\,t\,+\,\frac{\pi}{3}\right)\,+\,3\,cos\left(\omega\,t\,-\,\frac{\pi}{6}\right)[/tex]

Homework Equations



A short list of conversions from a larger table in the book. These are conversions from time domain sinusoidal functions on the left to cosine-reference phasor functions on the right.

[tex]A\,cos\left(\omega\,t\,+\,\phi_0\right)\,\,\iff\,\,A\,e^{j\,\phi_0}[/tex]

[tex]A\,sin\left(\omega\,t\,+\,\phi_0\right)\,\,\iff\,\,A\,e^{j\left(\phi_0\,-\,\frac{\pi}{2}\right)}[/tex]

The Attempt at a Solution



(a) [tex]3\,e^{-\frac{\pi}{3}\,j}[/tex]

(b) [tex]12\,e^{j\,\left(\frac{\pi}{4}\,-\,\frac{\pi}{2}\right)}\,=\,12\,e^{-\frac{\pi}{4}\,j}[/tex]

(c) [tex]2\,e^{-3\,x}\,e^{j\,\left(\frac{\pi}{6}\,-\,\frac{\pi}{2}\right)}\,=\,2\,e^{-3\,x}\,e^{-\frac{\pi}{3}\,j}\,=\,2\,e^{-3\,x\,-\,\frac{\pi}{3}\,j}[/tex]

(d) [tex]-2\,e^{\frac{3\pi}{4}\,j}[/tex]

(e) [tex]4\,e^{j\,\left(\frac{\pi}{3}\,-\,\frac{\pi}{2}\right)}\,+\,3\,e^{j\,\left(-\frac{\pi}{6}\right)}\,=\,7\,e^{j\,\left(-\frac{\pi}{6}\right)}[/tex]I have a question especially with (d), the answer is given as...

[tex]-2\,e^{j\,\left(\frac{3\pi}{4}\right)}\,=\,-2\,e^{j\,\left(\frac{\pi}{4}\right)}\,=\,2\,e^{-j\,\left(\frac{\pi}{4}\right)}[/tex]

I don't understand how they did the last two conversions in the given answer! Can someone please explain, and say whether the others are correct as well?

Thanks:smile:
 
Last edited:
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what you have to know is just euler formula: exp(ix)=cos(x)+isin(x).
 

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