Complex exponentials & phasors

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SUMMARY

The discussion focuses on converting the expression x(t) = 2sin(ω0t + 45°) + cos(ω0t) into the form x(t) = Acos(ω0t + φ) using phasor notation. The solution involves expressing the sine function as a cosine function and applying phasor representation, resulting in the final form of 2.798cos(ω0t - 30.3612°). Key steps include converting the sine term to cosine, using rectangular coordinates, and then transforming back to polar form.

PREREQUISITES
  • Understanding of phasor notation and its application in sinusoidal functions
  • Familiarity with trigonometric identities and transformations
  • Knowledge of converting between rectangular and polar coordinates
  • Basic proficiency in handling complex numbers in electrical engineering contexts
NEXT STEPS
  • Study the principles of phasor analysis in electrical engineering
  • Learn about the conversion between rectangular and polar forms of complex numbers
  • Explore trigonometric identities relevant to sinusoidal functions
  • Investigate applications of phasors in AC circuit analysis
USEFUL FOR

Students and professionals in electrical engineering, particularly those focusing on signal processing and AC circuit analysis, will benefit from this discussion.

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Homework Statement


x(t) = 2sin([tex]\omega[/tex]0t + 45o) + cos([tex]\omega[/tex]0t)

Express x(t) in the form x(t) = Acos([tex]\omega[/tex]0t + [tex]\phi[/tex])

The Attempt at a Solution



I don't really know when to begin; I can't find anything about it in the textbook.
 
Last edited:
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well expand out Acos(ω0t+φ) and then equate coefficients.
 
I'm not sure what that = 45 is doing. I'm going to assume you meant + 45

[tex]2sin(\omega_0 t + 45^o) + cos(\omega_0t)[/tex]
our first step in using phasor notation is to define each sinusoid as either a sine or cosine:
[tex]2cos(\omega_0 t +45^o - 90^o) + cos(\omega_0 t)[/tex]
[tex]2cos(\omega_0 t -45^o) + cos(\omega_0 t)[/tex]
we then define what our phasor is at 0 degrees:
[tex]cos(\omega_0t) = (1 \angle 0^o)[/tex]
apply it:
[tex](2 \angle -45^o) + (1 \angle 0^o)[/tex]
break up into rectangular coordinates:
[tex]2cos(-45^o) + 2sin(-45^o)j + 1[/tex]
use common real and imaginary part arithmetic to bring back into polar form:
[tex](2.798 \angle -30.3612)[/tex]
bring out of phasor form:
[tex]2.798cos(\omega_o t - 30.3612^o)[/tex]
 

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