Adding two cosine waves, result in the standard A \cdot \cos(\omega t+B) form y(t)=\cos

Laney Spears

Laney Spears

Answered question

2022-01-25

Adding two cosine waves, result in the standard Acos(ωt+B) form
y(t)=cos(40t)0.3cos(40t16)
This combines two waves of the same frequency but different phases, and the objective here is to add them and present the result in the simplified form, as a single real amplitude multiplied by a single cosine. I have tried complex analysis, via Re[ei40t(10.3ei16)], but I still can't get a result out of it.

Answer & Explanation

lorugb

lorugb

Beginner2022-01-26Added 13 answers

Re[ej40t[10.3ej16]] now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, [10.3ej16]=1.280.0863j=1.282ej0.0673
since wave is nothing but = Re part of [phasor  ×ejωt]
so, your wave is 1.282cos(ωt0.0673) in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put ω=40

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