Question
Download Solution PDF27 MHz के RF वाहक और 455 kHz की IF केंद्र आवृत्ति वाले उच्च-पक्ष अंतः क्षेपण का उपयोग करने वाले सिटीजन बैंड ग्राही के लिए, प्रतिबिंब आवृत्ति क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
प्रतिबिंब आवृत्ति: वह संकेत जो व्यक्तिकरण का कारण बनता है उसे 'प्रतिबिंब आवृत्ति' कहा जाता है।
प्रतिबिंब आवृत्ति और मध्यवर्ती आवृत्ति निम्नानुसार संबंधित हैं:
\(f_{si}=f_s+2 \space IF\)
fs = वाहक आवृत्ति
fsi = प्रतिबिंब आवृत्ति
IF = मध्यवर्ती आवृत्ति
गणना:
दिया गया है, fs = 27 MHz
IF = 455 kHz = 0.455 MHz
\(f_{si}=27 + 2(0.455)\)
fsi = 27.91 MHz
Last updated on Jul 2, 2025
-> ESE Mains 2025 exam date has been released. As per the schedule, UPSC IES Mains exam 2025 will be conducted on August 10.
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.