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In Frequency Hopping Spread Spectrum, if you knew that the bandwidth B =125MHz and Bss is 500MHZ, then we could say that the length of the k-bit pattern created by the PN equals bits. Note: Please add your answer using numbers ONLY with No spaces. Answer: 2

In Frequency Hopping Spread Spectrum, if you knew that the bandwidth B =125MHz and Bss is 500MHZ, then we could say that the length of the k-bit pattern created by the PN equals bits. Note: Please add your answer using numbers ONLY with No spaces. Answer: 2

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
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**Educational Website Content:** --- **Frequency Hopping Spread Spectrum** --- In Frequency Hopping Spread Spectrum, if you knew that the bandwidth \( B = 125 \text{MHz} \) and \( B_{ss} \) is \( 500 \text{MHz} \), then we could say that the length of the k-bit pattern created by the PN equals ____ bits. --- **Note:** Please add your answer using numbers ONLY with no spaces. --- **Answer:** 2 [The text provided above is accompanied by a field where the answer "2" is indicated with a red arrow suggesting that it is the correct response.] --- Explanation: The k-bit pattern is crucial in frequency hopping systems for spreading the signal over a wider bandwidth. The bandwidth \( B \) typically is the bandwidth of individual frequency channels, while \( B_{ss} \) is the spread spectrum bandwidth. The length of the k-bit pattern can be determined by dividing \( B_{ss} \) by \( B \), which in this case is \( \frac{500 \text{ MHz}}{125 \text{ MHz}} = 4 \). Since the length starts from 0, the final length of k-bit pattern required is 2 bits. **Visual Explanation:** Unfortunately, the original image doesn't include any graphs or diagrams to explain further. It only contains plain text and a numerical answer. --- This content can be used to help understand the essentials of frequency hopping in communication systems, specifically how bandwidth components relate to k-bit patterns generated by pseudorandom sequences.
Transcribed Image Text:**Educational Website Content:** --- **Frequency Hopping Spread Spectrum** --- In Frequency Hopping Spread Spectrum, if you knew that the bandwidth \( B = 125 \text{MHz} \) and \( B_{ss} \) is \( 500 \text{MHz} \), then we could say that the length of the k-bit pattern created by the PN equals ____ bits. --- **Note:** Please add your answer using numbers ONLY with no spaces. --- **Answer:** 2 [The text provided above is accompanied by a field where the answer "2" is indicated with a red arrow suggesting that it is the correct response.] --- Explanation: The k-bit pattern is crucial in frequency hopping systems for spreading the signal over a wider bandwidth. The bandwidth \( B \) typically is the bandwidth of individual frequency channels, while \( B_{ss} \) is the spread spectrum bandwidth. The length of the k-bit pattern can be determined by dividing \( B_{ss} \) by \( B \), which in this case is \( \frac{500 \text{ MHz}}{125 \text{ MHz}} = 4 \). Since the length starts from 0, the final length of k-bit pattern required is 2 bits. **Visual Explanation:** Unfortunately, the original image doesn't include any graphs or diagrams to explain further. It only contains plain text and a numerical answer. --- This content can be used to help understand the essentials of frequency hopping in communication systems, specifically how bandwidth components relate to k-bit patterns generated by pseudorandom sequences.
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