27 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?

This question was previously asked in
UPSC IES Electrical 2022 Prelims Official Paper
View all UPSC IES Papers >
  1. 24.55 MHz
  2. 27.91 MHz
  3. 28.45 MHz
  4. 29.65 MHz

Answer (Detailed Solution Below)

Option 2 : 27.91 MHz
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.4 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

अवधारणा:

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

Latest UPSC IES Updates

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.

More Classification of Receivers Questions

More Receivers Questions

Get Free Access Now
Hot Links: teen patti 100 bonus teen patti royal yono teen patti teen patti lucky teen patti - 3patti cards game