Image frequency calculation in a superheterodyne receiver A superheterodyne receiver is tuned such that its mixer sees a desired RF of 555 kHz, while the local oscillator (LO) input is at 1010 kHz and the IF is 455 kHz. What is the image frequency for this tuning configuration?

Introduction / Context:
Image frequency rejection is a crucial aspect of superhet design. Signals at the image will also convert to IF and can cause double spotting unless sufficiently attenuated by the RF front end.

Given Data / Assumptions:

• IF = 455 kHz, LO = 1010 kHz.
• Desired RF = 555 kHz since 1010 − 555 = 455.
• High-side injection (LO above RF).

Concept / Approach:
For high-side injection, the image frequency is located at f_image = f_LO + IF (since both f_LO − f_RF and f_image − f_LO can equal IF depending on side). Equivalently, f_image = f_RF + 2IF.

Step-by-Step Solution:
Compute f_RF from LO and IF: f_RF = f_LO − IF = 1010 − 455 = 555 kHz.Image for high-side LO: f_image = f_RF + 2IF.Calculate: f_image = 555 + 2455 = 555 + 910 = 1465 kHz.Thus the image is 1465 kHz.

Verification / Alternative check:
Using the alternate relation f_image = f_LO + IF = 1010 + 455 = 1465 kHz gives the same result.

Why Other Options Are Wrong:
555 kHz: The desired signal, not the image.

1010 kHz: The LO frequency.

1920 kHz: Unrelated arithmetic combination.

910 kHz: Equals 2IF; not the image frequency here.

Common Pitfalls:
Confusing image formula for high-side and low-side injection; always check whether LO is above or below the RF.

Final Answer:
1465 kHz