Explanation Given data: Refer to given circuit and waveform in the textbook. Formula used: 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 capacitor in s -domain. Z C = 1 s C (2) Here, C is the value of capacitance. Write a general expression of convolution integral. y ( t ) = ∫ − ∞ ∞ h ( λ ) x ( t − λ ) d λ (3) Here, h ( λ ) is the unit impulse function, and x ( t − λ ) is the shifted input. Calculation: The given circuit is redrawn as shown in Figure 1. Use equation (1) to find Z R 1 . Z R 1 = 5 k Ω = 5 × 10 3 Ω { ∵ 1 k = 10 3 } = 5000 Use equation (1) to find Z R 2 . Z R 2 = 20 k Ω = 20 × 10 3 Ω { ∵ 1 k = 10 3 } = 20000 Substitute 0.4 μ F for C in equation (2) to find Z C . Z C = 1 s ( 0.4 μ F ) = 1 s ( 0.4 × 10 − 6 F ) = 2.5 × 10 6 s Convert the Figure 1 into s -domain as shown in Figure 2. Use source transformation to convert the current source into voltage source. The voltage source is calculated as follows: V = 5000 I g Now, the Figure 2 is reduced as shown in Figure 3. Apply voltage division rule in Figure 3 to find V o . V o = ( 20000 5000 + 20000 + 2.5 × 10 6 s ) 5000 I g = ( 20000 25000 s + 2.5 × 10 6 s ) 5000 I g = ( 20000 s 25000 ( s + 100 ) ) 5000 I g = 4000 s I g s + 100 Simplify the above equation to find H ( s ) = V o I g H ( s ) = V o I g = 4000 ( s + 100 − 100 ) s + 100 = 4000 ( s + 100 ) − 4 × 10 5 s + 100 = 4000 ( s + 100 ) s + 100 − 4 × 10 5 s + 100 = 4000 − 4 × 10 5 s + 100 Apply inverse Laplace transform for above equation to find h ( t ) . h ( t ) = 4000 δ ( t ) − 4 × 10 5 e − 100 t u ( t ) { ∵ ℒ − 1 ( 1 s + a ) = e − a t u ( t ) ℒ − 1 ( 1 ) = δ ( t ) } Substitute λ for t in above equation to find h ( λ ) . h ( λ ) = 4000 δ ( λ ) − 4 × 10 5 e − 100 t u ( λ ) (4) The given waveform is drawn as shown in Figure 4. From Figure 4, i g ( t ) = { 10 mA 0 < t < 4 ms − 20 mA 4 ms < t < 6 ms 0 t > 6 ms The waveform of i g ( t − λ ) for time 0 < t < 4 ms is drawn as shown in Figure 5. From Figure 5, i g ( t − λ ) = { 10 mA 0 < λ < t for 0 < t < 4 ms For time 0 < t < 4 ms , δ ( λ ) = { 1 0 < λ < t , 0 < t < 4 ms 0 otherwise Refer to Figure 1, the input is i g and output is v o . Therefore, the equation (3) becomes as follows: v o ( t ) = ∫ − ∞ ∞ h ( λ ) i g ( t − λ ) d λ (5) Substitute 4000 δ ( λ ) − 4 × 10 5 e − 100 t u ( λ ) for h ( λ ) , and value of i g ( t − λ ) as mentioned above in equation (5). v o ( t ) = ∫ 0 t ( 10 mA ) 4000 δ ( λ ) − 4 × 10 5 e − 100 t u ( λ ) d λ = ∫ 0 t ( 10 × 10 − 3 ) ( 4000 δ ( λ ) − 4 × 10 5 e − 100 t u ( λ ) ) d λ = ∫ 0 t 40 δ ( λ ) − 4000 e − 100 t u ( λ ) d λ = 40 − 4000 [ − e − 100 t 100 ] 0 t Simplify the above equation as follows: v o ( t ) = 40 − 40 [ − e − 100 t + e − 100 ( 0 ) ] = 40 − 40 [ − e − 100 t + 1 ] = 40 + 40 e − 100 t − 40 = 40 e − 100 t 0 < t < 4 ms The waveform of i g ( t − λ ) for time 4 ms < t < 6 ms is drawn as shown in Figure 6

Electric Circuits Plus Mastering Engineering with Pearson eText 2.0 - Access Card Package (11th Edition) (What's New in Engineering)

11th Edition

ISBN: 9780134814117

Author: NILSSON, James W., Riedel, Susan

Publisher: PEARSON

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Chapter 13, Problem 76P

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Electric Circuits Plus Mastering Engineering with Pearson eText 2.0 - Access Card Package (11th Edition) (What's New in Engineering)

Ch. 13.2 - The parallel circuit in Example 13.1 is placed in...Ch. 13.3 - Prob. 2AP Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.3 - The dc current and dc voltage sources are applied...Ch. 13.3 - Prob. 6AP Ch. 13.3 - Using the results from Example 13.7 for the...Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.4 - Derive the numerical expression for the transfer...Ch. 13.5 - Find (a) the unit step and (b) the unit impulse...Ch. 13.5 - The unit impulse response of a circuit is υo(t) =...

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