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3. A 400 m length of coax runs the full length of the UMD campus (MWAH to Labovitz School of Business) with a velocity factor of 0.92 and is terminated with an open. How much time elapses for a pulse at the source end to travel to the open and reflect back to the source?

3. A 400 m length of coax runs the full length of the UMD campus (MWAH to Labovitz School of Business) with a velocity factor of 0.92 and is terminated with an open. How much time elapses for a pulse at the source end to travel to the open and reflect back to the source?

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**Question 3: Coaxial Cable Pulse Travel Time Calculation** A coaxial cable, 400 meters in length, runs the entire length of the UMD campus (from MWAH to the Labovitz School of Business). The cable has a velocity factor of 0.92 and is terminated with an open. The task is to determine how much time elapses for a pulse at the source end to travel to the open termination and reflect back to the source. **Solution Approach:** 1. **Understand the Context:** - A 400-meter coaxial cable is involved. - It has a velocity factor of 0.92 (indicating the speed of signal travel compared to the speed of light in vacuum). - The cable is terminated with an open, meaning no signal absorption at the end. 2. **Calculate the Effective Signal Speed:** - The speed of light in a vacuum is approximately \(3 \times 10^8 \) meters/second. - The effective speed of the signal through the cable is given by: \[ \text{Effective Speed} = \text{Velocity Factor} \times \text{Speed of Light} = 0.92 \times 3 \times 10^8 \text{ m/s} \] 3. **Determine the Total Distance Traveled:** - Since the pulse must travel to the end of the cable and back, the total distance is \( 2 \times 400 \text{ m} = 800 \text{ m} \). 4. **Calculate the Time Elapsed:** - Using the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] - Substitute the values to get: \[ \text{Time} = \frac{800 \text{ m}}{0.92 \times 3 \times 10^8 \text{ m/s}} \] - Simplify to find the time in seconds. By completing these calculations, the precise time taken for the pulse to travel to the open end and back to the source can be determined.
Transcribed Image Text:**Question 3: Coaxial Cable Pulse Travel Time Calculation** A coaxial cable, 400 meters in length, runs the entire length of the UMD campus (from MWAH to the Labovitz School of Business). The cable has a velocity factor of 0.92 and is terminated with an open. The task is to determine how much time elapses for a pulse at the source end to travel to the open termination and reflect back to the source. **Solution Approach:** 1. **Understand the Context:** - A 400-meter coaxial cable is involved. - It has a velocity factor of 0.92 (indicating the speed of signal travel compared to the speed of light in vacuum). - The cable is terminated with an open, meaning no signal absorption at the end. 2. **Calculate the Effective Signal Speed:** - The speed of light in a vacuum is approximately \(3 \times 10^8 \) meters/second. - The effective speed of the signal through the cable is given by: \[ \text{Effective Speed} = \text{Velocity Factor} \times \text{Speed of Light} = 0.92 \times 3 \times 10^8 \text{ m/s} \] 3. **Determine the Total Distance Traveled:** - Since the pulse must travel to the end of the cable and back, the total distance is \( 2 \times 400 \text{ m} = 800 \text{ m} \). 4. **Calculate the Time Elapsed:** - Using the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] - Substitute the values to get: \[ \text{Time} = \frac{800 \text{ m}}{0.92 \times 3 \times 10^8 \text{ m/s}} \] - Simplify to find the time in seconds. By completing these calculations, the precise time taken for the pulse to travel to the open end and back to the source can be determined.
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