Total field: PEC

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Presentation transcript:

Total field: PEC 𝐸 𝑚||𝑥 𝑖 = 𝐸 𝑚|| 𝑖 cos 𝜃 𝑖 𝛽 𝑥 = 𝛽 𝑖 sin 𝜃 𝑖 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 cos 𝜔𝑡− 𝛽 𝑥 𝑥− 𝜋 2 𝐸 𝑦 = 2 𝐸 𝑚⊥ 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥+ 𝜋 2 = 2 𝐸 𝑚⊥ 𝑖 sin 𝛽 𝑧 𝑧 cos 𝜔𝑡− 𝛽 𝑥 𝑥− 𝜋 2 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 cos 𝜔𝑡− 𝛽 𝑥 𝑥 𝐸 𝑚||𝑥 𝑖 = 𝐸 𝑚|| 𝑖 cos 𝜃 𝑖 𝛽 𝑥 = 𝛽 𝑖 sin 𝜃 𝑖 𝐸 𝑚||𝑧 𝑖 = 𝐸 𝑚|| 𝑖 sin 𝜃 𝑖 𝛽 𝑧 = 𝛽 𝑖 cos 𝜃 𝑖

Reflection by PEC: Total Fields Parallel Polarization 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝐸 𝑥 =0 sin 𝛽 𝑧 𝑧 =0 cos 𝛽 𝑧 𝑧 =0 𝐸 𝑧 =0 𝜆 𝑧 /2 sin 𝛽 𝑧 𝑧 =0 𝐸 𝑥 =0 𝜆 𝑧 /2 𝐸 𝑧 =0 cos 𝛽 𝑧 𝑧 =0 sin 𝛽 𝑧 𝑧 =0 𝐸 𝑥 =0 𝜆 𝑧 =2𝜋/ 𝛽 𝑧

Reflection by PEC: Total Fields Parallel Polarization 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝑧=− 𝜆 𝑧 sin 𝛽 𝑧 𝑧 =0 cos 𝛽 𝑧 𝑧 =0 𝑧=−3 𝜆 𝑧 /4 𝜆 𝑧 /2 sin 𝛽 𝑧 𝑧 =0 𝑧=− 𝜆 𝑧 /2 𝜆 𝑧 /2 𝑧=− 𝜆 𝑧 /4 cos 𝛽 𝑧 𝑧 =0 sin 𝛽 𝑧 𝑧 =0 𝑧=0 𝜆 𝑧 =2𝜋/ 𝛽 𝑧

Reflection by PEC: Total Fields Effect of incident angle 𝜃 𝑖 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝜆 𝑧 =2𝜋/ 𝛽 𝑧 𝛽 𝑧 = 𝛽 𝑖 cos 𝜃 𝑖 𝛽 𝑖 𝜆 𝑧 = 2𝜋 𝛽 𝑖 cos 𝜃 𝑖 = 𝜆 𝑖 cos 𝜃 𝑖 𝜆 𝑧 ≥ 𝜆 𝑖 Wavefront 𝜆 𝑧 /2 𝜃 𝑖 𝜃 𝑖 =0→ 𝜆 𝑧 = 𝜆 𝑖 𝜆 𝑖 /2 𝜃 𝑖 =90°→ 𝜆 𝑧 =∞ 𝜆 𝑥 /2

Reflection by PEC: Total Fields Effect of incident angle 𝜃 𝑖 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝜆 𝑥 =2𝜋/ 𝛽 𝑥 𝛽 𝑥 = 𝛽 𝑖 sin 𝜃 𝑖 𝛽 𝑖 𝜆 𝑥 = 2𝜋 𝛽 𝑖 sin 𝜃 𝑖 = 𝜆 𝑖 sin 𝜃 𝑖 𝜆 𝑥 ≥ 𝜆 𝑖 Wavefront 𝜆 𝑧 /2 𝜃 𝑖 𝜃 𝑖 =0→ 𝜆 𝑥 =∞ 𝜆 𝑖 /2 𝜃 𝑖 =90°→ 𝜆 𝑥 = 𝜆 𝑖 𝜆 𝑥 /2

Reflection by PEC: Total Fields Effect of incident angle 𝜃 𝑖 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝛽 𝑖 2 = 𝛽 𝑥 2 + 𝛽 𝑧 2 𝛽 𝑧 = 𝛽 𝑖 cos 𝜃 𝑖 𝛽 𝑥 = 𝛽 𝑖 sin 𝜃 𝑖 𝜆 𝑧 = 𝜆 𝑖 cos 𝜃 𝑖 𝜆 𝑥 = 𝜆 𝑖 sin 𝜃 𝑖 𝛽 𝑥 2 = 𝛽 𝑖 2 − 𝛽 𝑧 2 𝛽 𝑥 = 𝛽 𝑖 2 − 𝛽 𝑧 2 𝛽 𝑖 2 = 𝛽 𝑥 2 + 𝛽 𝑧 2 1 𝜆 𝑥 = 1 𝜆 𝑖 2 − 1 𝜆 𝑧 2 1 𝜆 𝑖 2 = 1 𝜆 𝑥 2 + 1 𝜆 𝑧 2

Reflection by PEC: Total Fields Boundary conditions 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝑛 × 𝐸 =0 𝑛 =− 𝑎 𝑧 𝐸 𝑥 =0 𝑧=0 b.c. sin 𝛽 𝑧 𝑧 =0 b.c.

Reflection by PEC: Total Fields Boundary conditions 𝐸 𝑥 = 2 𝐸 𝑚||𝑥 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗( 𝛽 𝑥 𝑥+ 𝜋 2 ) 𝐸 𝑧 = −2 𝐸 𝑚||𝑧 𝑖 cos 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥 𝑛 × 𝐸 =0 𝑛 =− 𝑎 𝑧 𝐸 𝑥 =0 𝑧=0 sin 𝛽 𝑧 𝑧 =0

Reflection by PEC: Total Fields Perpendicular Polarization 𝐸 𝑦 = 2 𝐸 𝑚⊥ 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥+ 𝜋 2 𝑛 × 𝐸 =0 𝑛 =− 𝑎 𝑧 𝐸 𝑦 =0 𝑧=0 sin 𝛽 𝑧 𝑧 =0

Reflection by PEC: Total Fields Perpendicular Polarization 𝐸 𝑦 = 2 𝐸 𝑚⊥ 𝑖 sin 𝛽 𝑧 𝑧 𝑒 −𝑗 𝛽 𝑥 𝑥+ 𝜋 2 𝑛 × 𝐸 =0 𝑛 =− 𝑎 𝑧 𝐸 𝑦 =0 𝑧=0 sin 𝛽 𝑧 𝑧 =0 We have solved the two-plate waveguide problem.