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RG-59 type line has an open circuit impedance of 150L25 degree ohm, and a short circuit impedance of 37.5<-35 degree ohm, ...
Given: Z O C = 150 ∠ 25 ∘ Ω Z_{OC}=150\angle25^\circ\ \Omega Z S C = 37.5 ∠ ( − 35 ∘ ) Ω Z_{SC}=37.5\angle(-35^\circ)\ \Omega Use: Z 0 = Z O C Z S C Z_0=\sqrt{Z_{OC}Z_{SC}} Step 1: Multiply Magnitudes: 150 × 37.5 = 5625 150\times37.5=5625 Angles: 25 ∘ + ( − 35 ∘ ) = − 10 ∘ 25^\circ+(-35^\circ)=-10^\circ Therefore, Z O C Z S C = 5625 ∠ ( − 10 ∘ ) Z_{OC}Z_{SC}=5625\angle(-10^\circ) Step 2: Take square root Z 0 = 5625 ∠ − 10 ∘ 2 Z_0=\sqrt{5625}\angle\frac{-10^\circ}{2} Z 0 = 75 ∠ ( − 5 ∘ ) Ω \boxed{Z_0=75\angle(-5^\circ)\ \Omega} Answer: Z 0 = 75 ∠ − 5 ∘ Ω \boxed{Z_0=75\angle-5^\circ\ \Omega} Option B