a,b)Find the phasor form of the following E-field, the polarization type, and plot the vector on the X-y axis (z-0) for one period a) E(z,t) = xZcos(wt kz) y4cos(wt kz) b) E(z,t) = R2cos(wt + kz) - 94sin(wt + kz)

Answers

Answer 1

a) The phasor form of E-field = xˆ 2cos(wt - kz) - yˆ 4cos(wt - kz) is 2√5 L(-63.43).

Polarization type is linearly polarised.

b) The phasor form of E-field = ?

E(z,t) = xˆ 2cos(wt + kz) - yˆ 4sin(wt + kz) is E(z,t) =2√5 L(-143.43). polarization type is elliptical.

We have given that

E⃗(z,t) =2xˆcos(wt - kz) - 4 yˆcos(wt - kz)

|r⃗ | = √((2cos(wt - kz))² + (- 4cos(wt - kz))²)

= √(4cos²(wt - kz) + 16cos²(wt - kz))

= √(20cos²(wt - kz))

= 2√5 cos(wt - kz)

φ = 4cos(wt - kz)/(2cos(wt - kz))

= tan⁻¹ 1(-2) = -63.43°

So, E(z,t) = 2√5 L(-63.43)

b) E(z,t) = xˆ 2cos(wt + kz) - yˆ 4sin(wt + kz)

= xˆ 2cos(wt + kz) - yˆ 4cos(wt + kz -π/2)

|r| = √((2cos(wt + kz))² + (- 4sin(wt -+kz))²)

= √20

phi = tan⁻¹(- 4cos(wt - kz)/(2cos(wt - kz)) - π/2

= tan⁻¹( -2) + π/2 = - 143.43

so, E(z,t) =2√5 L(-143.43)

E(z,t) = xˆ 2cos(wt - kz) - yˆ 4cos(wt - kz)

if wt = 0°

E(z,t) = xˆ 2cos(0 - kz) - yˆ 4cos(0- kz)

= 2xˆ - 4yˆ

wt = 30° , E(z,t) = xˆ 2cos( - kz) - yˆ 4cos(0- kz)

E(z,t) = √3/2(2) xˆ - √3/2(4) yˆ

= √3xˆ - 2√3yˆ

for wt = 45°

E(z,t) = √2 xˆ - 2√2 yˆ

for wt = 90° =>E(z,t) = 0

for wt = 60° , E(z,t) = xˆ - 2 yˆ as seen in above graph that is linear polarised

For part (b) ,

E(z,t) = 2 xˆcos(wt - kz) - 4yˆ sin(wt + kz)

From graph elliptical polarised,

for wt = 0 , E(0,t) = 2 xˆ

for wt = 90° , E(0,t) = -4yˆ

for wt =180° , E(0,t) = -2 xˆ

for wt = 270° , E(0,t) = -4yˆ

Hence, we get the answer of both parts.

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