Question
Download Solution PDFPLL आवृत्ति संश्लेषक में, यदि 500 MHz आवृत्ति वाला क्रिस्टल ऑसिलेटर 2 से विभाजित नेटवर्क से गुजरता है, तो PLL में इनपुट आवृत्ति _______ MHz है।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
आवृत्ति संश्लेषक के रूप में उपयोग किए जाने वाले PLL (फेज लॉक लूप) में, एक क्रिस्टल ऑसिलेटर एक उच्च स्थिर आवृत्ति प्रदान करता है। यदि इस आवृत्ति को N से विभाजित नेटवर्क से गुजारा जाता है, तो आउटपुट आवृत्ति
दिया गया है:
क्रिस्टल ऑसिलेटर आवृत्ति,
2 से विभाजित नेटवर्क: N =2
गणना:
PLL में इनपुट आवृत्ति =
अंतिम उत्तर:
2) 250 MHz
Last updated on Jun 7, 2025
-> RRB JE CBT 2 answer key 2025 for June 4 exam has been released at the official website.
-> Check Your Marks via RRB JE CBT 2 Rank Calculator 2025
-> RRB JE CBT 2 admit card 2025 has been released.
-> RRB JE CBT 2 city intimation slip 2025 for June 4 exam has been released at the official website.
-> RRB JE CBT 2 Cancelled Shift Exam 2025 will be conducted on June 4, 2025 in offline mode.
-> RRB JE CBT 2 Exam Analysis 2025 is Out, Candidates analysis their exam according to Shift 1 and 2 Questions and Answers.
-> The RRB JE Notification 2024 was released for 7951 vacancies for various posts of Junior Engineer, Depot Material Superintendent, Chemical & Metallurgical Assistant, Chemical Supervisor (Research) and Metallurgical Supervisor (Research).
-> The selection process includes CBT 1, CBT 2, and Document Verification & Medical Test.
-> The candidates who will be selected will get an approximate salary range between Rs. 13,500 to Rs. 38,425.
-> Attempt RRB JE Free Current Affairs Mock Test here
-> Enhance your preparation with the RRB JE Previous Year Papers.