(b) Determine the incident densities Power electric field 1 mv/m. Electric field incident E Incident Power density Pincident Wave Incident ad incident at Region O Мо n = no = √260 छ Powen = Reflection Transmitted Preflected ETransmitted where 1) 1207 Incident Power electric field = at Esuflected 7.5 2 12₂ when. the Power Density IM₂ 8₂ N than reflected Power Density Preflected= Co. Power Denaity P E transmitted Reflected & transmittied the amplitude of incident interface is addumed 1 mu/m = DA 120x ohm. 2 0.230 0.230 mu/m = 00.265x10 W/m² 0.265 Nano w/m² [Reflection Coefficient ] x Ein = 0.230m) 2 x 120 व 7.016 x10" 0-230 Transmitted Powed Density = 1207 √& = 235062 [1-1T1] Eincident x 1 m v/m (Ereflected) 240 W/m² R 2 (E transmitted) ² 21₂ [ Eincident] 24 0.77 mulm [ & traindmitted] 24, (0.77 mulm) 2 x 235.62 1 1 milm 2 -1208 √√2.56 S then, -9 1 x 10 Nano W/m w/m

The compensation techniques are the methods that are employed in the electrical control systems to enhance the performance of the system. The performance are generally affected by the factors like lag in the compensating devices, slow response of the sys…