Explanation Given data: Refer to Figure 16.75 in the textbook. Formula used: Write an expression to calculate the value of step input. u ( t ) = { 0 t < 0 1 t > 0 Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of capacitance. Calculation: The given circuit is drawn as shown in Figure 1. For a DC circuit, at steady state condition when time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. The value of current source i s ( t ) calculated as follows: Substitute − 1 for t in i s ( t ) . i s ( t ) = 2 [ 1 − u ( ( − 1 ) ) ] A = 2 [ 1 + ( 0 ) ] A { ∵ u ( t ) = 0 for t < 0 } = 2 A Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the current through inductor is same as the value of current source. i L ( 0 − ) = i s ( t ) = 2 A The resistor ( R 2 ) and the capacitor are connected in parallel. For the parallel connection, the voltage is same. Therefore, the voltage across the capacitor is calculated as follows: v C ( 0 − ) = ( 12 Ω ) ( 2 A ) = 24 V The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 2 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 24 V For time t > 0 , Substitute 1 for t to find i s ( t ) . i s ( t ) = 2 [ 1 − u ( 1 ) ] A = 2 [ 1 − 1 ] A { ∵ u ( t ) = 1 for t > 0 } = 0 A Since the value of current source is zero, it is open circuited and the Figure 1 is reduced as shown in Figure 3. Substitute 8 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 8 Ω Substitute 12 Ω for R 2 in equation (1) to find Z R 2 . Z R 2 = 12 Ω Substitute 2 H for L in equation (2) to find Z L . Z L = s ( 2 H ) = 2 s Substitute 1 18 F for C in equation (3) to find Z C . Z C = 1 s ( 1 18 F ) = 18 s Using element transformation methods with initial conditions convert the Figure 3 into s -domain. Apply Kirchhoff’s voltage law for Figure 4. − v ( 0 ) s + 18 s I ( s ) + 8 I ( s ) + 12 I ( s ) + 2 s I ( s ) − ( 2 s ) i ( 0 ) s = 0 I ( s ) ( 18 s + 8 + 12 + 2 s ) = v ( 0 ) s + ( 2 s ) i ( 0 ) s I ( s ) ( 18 s + 20 + 2 s ) = v ( 0 ) s + ( 2 s ) i ( 0 ) s Substitute 24 for v ( 0 ) , and 2 for i ( 0 ) in above equation

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 52P

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Find the expression of current i(t) and the voltage v(t) for all t>0.

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Chapter 16, Problem 51P

Chapter 16, Problem 53P

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A plane wave propagating in the +z direction in medium 1 is normally incident to medium 2 located at the z=0 plane as below. Both mediums are general, characterized by ( ε i, Mi, Ơi ). tot = [ ει μη σ] [ε, μη σε ] Ex Ex tot E₁₂ (z) = Ee Ex z=0 From conservation of energy: P₁AV'(z=0) + Piav'(z=0) = P2av²(z=0). Using the above show for lossless media that: ( 1 - ||²) = (1/M2 )|T|² .

A plane wave propagating in the +z direction in medium 1 is normally incident to medium 2 located at the z=0 plane as below. Both mediums are general, characterized by ( ε i, Hi, σ¡ ). [ ει μη σ] Ex [ ει μη ση ] Ex tot E₁₂ (z) = E'₁e¹² -122 E(z) = Ee+ E₁₁₁² E₁x z=0 1. Specify the electric field reflection coefficient г and transmission coefficient T: E ΓΔ E E TA EL 2. Show that T=1+г. Can the transmitted electric field amplitude in region 2 be LARGER than the incident electric field amplitude? 3. Determine expressions for P₁AV'(z), PIAV'(z) and P2AV'(z) (note the sign for the reflected power direction should be (-z).