In a superheterodyne receiver, if the signal frequency is fs and the intermediate frequency is fi, what is the formula for the image frequency fsi that must be rejected by the RF front end?

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Introduction / Context:The image frequency is a well-known challenge in superheterodyne receivers. It arises because both the desired frequency and another undesired frequency can produce the same intermediate frequency (IF) after mixing. To avoid distortion, image frequency must be rejected by RF filters. Given Data / Assumptions: Superheterodyne receiver with IF = fi. Signal frequency = fs. Local oscillator frequency = fs + fi (high-side injection assumed). Concept / Approach:The image frequency is defined as the frequency that is separated from fs by twice the IF. Mathematically, fsi = fs + 2fi for high-side injection. For low-side injection, fsi = fs − 2fi. Step-by-Step Solution: Local oscillator flo = fs + fi.Mixing: |flo − fs| = fi.For an undesired frequency fsi: |flo − fsi| = fi.So fsi = flo + fi = (fs + fi) + fi = fs + 2fi. Verification / Alternative check: Example: fs = 1200 kHz, fi = 450 kHz ⇒ fsi = 1200 + 900 = 2100 kHz. Why Other Options Are Wrong: fs − 2fi: applies to low-side injection only, not general formula.fs ± fi: corresponds to LO or IF, not image.2fs − fi: unrelated expression. Common Pitfalls: Forgetting the ±2fi relationship; confusing LO with image frequency. Final Answer: fsi = fs + 2fi

In a superheterodyne receiver, if the signal frequency is fs and the intermediate frequency is fi, what is the formula for the image frequency fsi that must be rejected by the RF front end?

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