(a) To determine Find the output voltage v o using convolution integral. Explanation Given data: Refer to given circuit in the textbook. The value of input voltage is, v i ( t ) = 5 [ u ( t ) − u ( t − 0.5 ) ] V (1) Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (2) 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 (3) 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 (4) Here, C is the value of capacitance. Write a general expression of convolution integral. y ( t ) = ∫ − ∞ ∞ h ( λ ) x ( t − λ ) d λ (5) Here, h ( λ ) is the unit impulse function, and x ( t − λ ) is the shifted input. Calculation: From the given data the circuit is redrawn as shown in Figure 1. Substitute 2 Ω for R in equation (2) to find Z R . Z R = 2 Substitute 100 mH for L in equation (3) to find Z L . Z L = s ( 100 mH ) = s ( 100 × 10 − 3 H ) = 0.1 s Substitute 100 mF for C in equation (3) to find Z C . Z C = 1 s ( 100 mF ) = 1 s ( 100 × 10 − 3 F ) { ∵ 1 m = 10 − 3 } = 10 s Convert the Figure 1 into s -domain as shown in Figure 2. Use voltage division rule in Figure 2 to find V o . V o = ( ( 10 s ) 2 + 0.1 s + 10 s ) V i = ( ( 10 s ) 0.1 s 2 + 2 s + 10 s ) V i = ( 10 s ) ( s 0.1 ( s 2 + 20 s + 100 ) ) V i = ( 100 s 2 + 20 s + 100 ) V i Rearrange the above equation to find V o V i . V o V i = 100 s 2 + 20 s + 100 On solving the quadratic equation in denominator of above equation it becomes as follows: V o V i = 100 ( s + 10 ) 2 Write a general expression to calculate transfer function. H ( s ) = V o V i Substitute 100 ( s + 10 ) 2 for V o V i in above equation to find H ( s ) . H ( s ) = 100 ( s + 10 ) 2 Apply inverse Laplace transform for above equation to find h ( t ) . h ( t ) = 100 t e − 10 t u ( t ) { ∵ ℒ − 1 ( 1 ( s + a ) n ) = t n − 1 ( n − 1 ) ! e − a t u ( t ) } Substitute λ for t in above equation to find h ( λ ) . h ( λ ) = 100 λ e − 10 λ u ( λ ) Refer to Figure 1, the input is v i and output is v o . Therefore, the equation (5) becomes as follows: v o ( t ) = ∫ − ∞ ∞ h ( λ ) v i ( t − λ ) d λ (6) Substitute t − λ for t in equation (1) to find v i ( t − λ ) . v i ( t ) = 5 [ u ( t − λ ) − u ( t − λ − 0.5 ) ] V (7) The waveform for equation (1) with time λ is drawn as shown in Figure 3. The waveform for inverted function of equation (1) is drawn as shown in Figure 4. The waveform for equation (7) is drawn as shown in Figure 5. Substitute 100 e − 10 λ u ( λ ) for h ( λ ) , and equation (6) in equation (5) to find v o ( t ) . v o ( t ) = ∫ − ∞ ∞ [ 100 t e − 10 λ u ( λ ) ] [ 5 u ( t − λ ) − 5 u ( t − λ − 0.5 ) ] d λ v o ( t ) = ∫ − ∞ ∞ [ [ 500 t e − 10 λ u ( λ ) ] u ( t − λ ) ] − [ [ 500 t e − 10 λ u ( λ ) ] u ( t − λ − 0 (b) To determine Sketch the output voltage for time 0 < t < 1 s .

ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

11th Edition

ISBN: 9780134894300

Author: NILSSON

Publisher: PEARSON

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Chapter 13, Problem 70P

(a)

To determine

Find the output voltage vo using convolution integral.

(b)

To determine

Sketch the output voltage for time 0<t<1 s.

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Chapter 13, Problem 69P

Chapter 13, Problem 71P

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Chapter 13 Solutions

ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

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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