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20 10 (t) 0 -10 -20 m 10 20 30 -20 -10 0 Time t (msec) 40

20 10 (t) 0 -10 -20 m 10 20 30 -20 -10 0 Time t (msec) 40

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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### Understanding a Sinusoidal Wave Plot #### Question: The figure given below is a plot of a sinusoidal wave. From the plot, determine values for the amplitude (A), phase (ϕ), and frequency (ω₀) needed in the representation: \[ x(t) = A \cos (\omega_0 t + \phi) \] Provide the answer as numerical values, including the units where applicable. #### Detailed Analysis of the Plot: **Graph Description:** - **X-axis (Time \(t\)):** The time is measured in milliseconds (msec) ranging from -20 to 40. - **Y-axis (\(x(t)\)):** The values range from -20 to 20. - The plotted wave completes one full cycle from approximately 0 to 20 msec on the time axis. #### Solution Approach: 1. **Amplitude (A):** The amplitude of this wave is the maximum value of \(x(t)\) from the equilibrium position (y=0). The observed amplitude is \(A = 20\). 2. **Frequency (ω₀):** Determine the period \(T\) of one complete cycle on the time axis. The wave goes from 0 to 20 msec for one full cycle. Hence, the period \(T = 20\) msec. To find the angular frequency, use the relationship: \[ \omega_0 = \frac{2\pi}{T} \] Substituting \(T = 20\) msec: \[ \omega_0 = \frac{2\pi}{20 \times 10^{-3}} = \frac{2\pi}{0.02} \approx 314.16 \text{ rad/sec} \] 3. **Phase (ϕ):** The phase shift can be determined by finding the horizontal shift of the wave from the standard cosine function. The graph shows the wave crossing the vertical axis (t=0) at the origin and reaching its positive peak value at t= ~2 msec, indicating no significant phase shift from a standard cosine function. Therefore, the phase shift \( \phi = 0 \) radians. ### Conclusion: Thus, the values are: - **Amplitude (A):** 20 - **Angular Frequency (ω₀):** 314.16 rad/sec - **Phase (ϕ):**
Transcribed Image Text:### Understanding a Sinusoidal Wave Plot #### Question: The figure given below is a plot of a sinusoidal wave. From the plot, determine values for the amplitude (A), phase (ϕ), and frequency (ω₀) needed in the representation: \[ x(t) = A \cos (\omega_0 t + \phi) \] Provide the answer as numerical values, including the units where applicable. #### Detailed Analysis of the Plot: **Graph Description:** - **X-axis (Time \(t\)):** The time is measured in milliseconds (msec) ranging from -20 to 40. - **Y-axis (\(x(t)\)):** The values range from -20 to 20. - The plotted wave completes one full cycle from approximately 0 to 20 msec on the time axis. #### Solution Approach: 1. **Amplitude (A):** The amplitude of this wave is the maximum value of \(x(t)\) from the equilibrium position (y=0). The observed amplitude is \(A = 20\). 2. **Frequency (ω₀):** Determine the period \(T\) of one complete cycle on the time axis. The wave goes from 0 to 20 msec for one full cycle. Hence, the period \(T = 20\) msec. To find the angular frequency, use the relationship: \[ \omega_0 = \frac{2\pi}{T} \] Substituting \(T = 20\) msec: \[ \omega_0 = \frac{2\pi}{20 \times 10^{-3}} = \frac{2\pi}{0.02} \approx 314.16 \text{ rad/sec} \] 3. **Phase (ϕ):** The phase shift can be determined by finding the horizontal shift of the wave from the standard cosine function. The graph shows the wave crossing the vertical axis (t=0) at the origin and reaching its positive peak value at t= ~2 msec, indicating no significant phase shift from a standard cosine function. Therefore, the phase shift \( \phi = 0 \) radians. ### Conclusion: Thus, the values are: - **Amplitude (A):** 20 - **Angular Frequency (ω₀):** 314.16 rad/sec - **Phase (ϕ):**
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