(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 .

11th Edition

ISBN: 8220106795262

Author: Riedel

Publisher: YUZU

See similar textbooks

Videos

Question

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13, Problem 69P

Chapter 13, Problem 71P

Students have asked these similar questions

A. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and Manchester code. Data sequence= (00110100) B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of 50 dB. If the message is a single tone with fm-5 kHz. Determine: 1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio. 2) Minimum required system bandwidth.

Find Io using Mesh analysis

FM station of 100 MHz carrier frequency modulated by a 20 kHz sinusoid with an amplitude of 10 volt, so that the peak frequency deviation is 25 kHz determine: 1) The BW of the FM signal. 2) The approximated BW if the modulating signal amplitude is increased to 50 volt. 3) The approximated BW if the modulating signal frequency is increased by 70%. 4) The amplitude of the modulating signal if the BW is 65 kHz.

Chapter 13 Solutions

EBK ELECTRIC CIRCUITS

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) =...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.