A phasor can be thought of as a vector, representing a complex number, rotating around the polar plane at a certain frequency in radians/second. The projection of such a vector onto the real axis gives a cosine with a certain amplitude and phase. This problem will show the algebra of phasors which would help you with some of the trigonometric identities that are hard to remember.

(a) When you plot y(t) = Asin(Ω₀t) you notice that it is a cosine x(t) = Acos(Ω₀t) shifted in time, i.e.,

how much is this shift Δ_t? Better yet, what is Δ_θ = ΩθΔt or the shift in phase? One thus only need to consider cosine functions with different phase shifts instead of sines and cosines.

(b) From above, the phasor that generates 

The phasor corresponding to the sine y(t) should then be Ae-^jπ/2. Obtain an expression for y(t) similar to the one for x(t) in terms of this phasor.

(c) From the above results, give the phasors corresponding to -x(t) = -Acos(Ω₀t) and -y(t) = -sin(Ω₀t). Plot the phasors that generate cos, sin, -cos and -sin for a given frequency. Do you see now how these functions are connected? How many radians do you need to shift in positive or negative direction to get a sine from a cosine, etc.

(d) Suppose then you have the sum of two sinusoids, for instance z(t) = x(t) + y(t) = Acos(Ω₀t) + Asin(Ω₀t), adding the corresponding phasors for x(t) and y(t) at some time, e.g., t = 0, which is just a sum of two vectors, you should get a vector and the corresponding phasor. For x(t), y(t), obtain their corresponding phasors and then obtain from them the phasor corresponding to z(t) = x(t) + y(t).

(e) Find the phasors corresponding to

Transcribed Image Text:

y(t) = Asin(ot) = A cos(No(t – A;)) = x(t – A;) r(t) = A cos(Qot) is Ae10 so that r(t) = Re[Ae³°e%ot].