(a) To determine Find the unit step response of the circuit. Explanation Given data: Refer to given circuit 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 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 redrawn as shown in Figure 1. Substitute 2 Ω for R in equation (1) to find Z R . Z R = 2 Substitute 1 H for L in equation (2) to find Z L . Z L = s ( 1 H ) = s Substitute 0.1 F for C in equation (3) to find Z C . Z C = 1 s ( 0.1 F ) = 10 s Convert the Figure 1 into s -domain as shown in Figure 2. Refer to Figure 2, the capacitor is connected in parallel with the series combination of resistor and inductor. The equivalent impedance is calculated as follows: Z eq = ( 2 + s ) ∥ 10 s = ( 2 + s ) ( 10 s ) 2 + s + 10 s = 20 s + 10 ( s 2 + 2 s + 10 s ) = ( 20 + 10 s s ) ( s 2 + 2 s + 10 s ) Z eq = 10 ( s + 2 ) s 2 + 2 s + 10 The value of voltage is calculated as follows: V o = Z eq I g Rearrange the above equation to find the transfer function V o I g . V o I g = Z eq Substitute 10 ( s + 2 ) s 2 + 2 s + 10 for Z eq in above equation to find V o I g . H ( s ) = V o I g = 10 ( s + 2 ) s 2 + 2 s + 10 (4) From equation (4), the characteristic equation of denominator is written as follows: s 2 + 2 s + 10 = 0 Compare the above equation with the quadratic equation ( a s 2 + b s + c = 0 ) . a = 1 b = 2 c = 10 Write an expression to solve quadratic equation. s 1 , 2 = − b ± b 2 − 4 a c 2 Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. Substitute 1 for a , 2 for b , and 10 for c in above equation to find s 1 , 2 . s 1 , 2 = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 10 ) 2 = − 2 ± 4 − 40 2 = − 2 ± − 36 2 = − 2 ± j 6 2 Simplify the equation. s 1 , 2 = − 2 ± j 6 2 , − 2 ± j 6 2 = − 1 + j 3 , − 1 − j 3 Therefore, the roots of characteristic equations are real and imaginary. s 1 = − 1 + j 3 s 2 = − 1 − j 3 Now, the equation (4) is written as follows: H ( s ) = V o I g = 10 ( s + 2 ) ( s − ( − 1 + j 3 ) ) ( s − ( − 1 − j 3 ) ) H ( s ) = V o I g = 10 ( s + 2 ) ( s + 1 − j 3 ) ( s + 1 + j 3 ) (5) For a step input, i g ( t ) = u ( t ) Apply Laplace transform for above equation to find I g . I g = 1 s { ∵ L ( u ( t ) ) = 1 s } Rearrange equation (5) to find V o . V o = ( 10 ( s + 2 ) ( s + 1 − j 3 ) ( s + 1 + j 3 ) ) I g (6) Substitute 1 s for I g in equation (6) to find V o . V o = ( 10 ( s + 2 ) ( s + 1 − j 3 ) ( s + 1 + j 3 ) ) ( 1 s ) = 10 ( s + 2 ) s ( s + 1 − j 3 ) ( s + 1 + j 3 ) Take partial fraction for above equation. V o = 10 ( s + 2 ) s ( s + 1 − j 3 ) ( s + 1 + j 3 ) = A s + B ( s + 1 − j 3 ) + C ( s + 1 + j 3 ) (7) The equation (7) can also be written as follows: 10 ( s + 2 ) s ( s + 1 − j 3 ) ( s + 1 + j 3 ) = A ( s + 1 − j 3 ) ( s + 1 + j 3 ) + B s ( s + 1 + j 3 ) + C s ( s + 1 − j 3 ) s ( s + 1 − j 3 ) ( s + 1 + j 3 ) Simplify the above equation as follows: 10 ( s + 2 ) = A ( s + 1 − j 3 ) ( s + 1 + j 3 ) + B s ( s + 1 + j 3 ) + C s ( s + 1 − j 3 ) (8) Substitute 0 for s in equation (8) to find A . 10 ( 0 + 2 ) = A ( 0 + 1 − j 3 ) ( 0 + 1 + j 3 ) + B ( 0 ) ( 0 + 1 + j 3 ) + C ( 0 ) ( 0 + 1 − j 3 ) 20 = A ( 1 − j 3 ) ( 1 + j 3 ) + 0 + 0 10 A = 20 Simplify the above equation to find A (b) To determine Find the unit impulse response of the circuit.

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 13.5, Problem 10AP

(a)

To determine

Find the unit step response of the circuit.

(b)

To determine

Find the unit impulse response of the circuit.

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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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