Superheterodyne broadcast receiver (no RF amplifier) Given a loaded Q of the antenna coupling circuit equal to 100 and an intermediate frequency (IF) of 455 kHz, find the image frequency corresponding to a desired station at 1000 kHz for a standard high-side local oscillator configuration.

Introduction / Context:
Image frequency is a classic topic in superheterodyne receiver design. When mixing a wanted RF signal with a local oscillator (LO), another unwanted frequency (the image) can also convert to the same IF, potentially causing interference. Understanding how to compute the image frequency and why high front-end Q is desirable is fundamental for receiver selectivity and performance.

Given Data / Assumptions:

• Desired station (signal) frequency fs = 1000 kHz.
• Intermediate frequency IF = 455 kHz.
• Typical broadcast receiver with a high-side LO (LO above the signal).
• No RF amplifier stage; antenna coupling circuit has loaded Q = 100 (useful for image rejection discussion but not needed to compute the image frequency itself).

Concept / Approach:

For a high-side LO, the oscillator frequency is fLO = fs + IF. The image frequency fim is the frequency that also satisfies |fLO − fim| = IF, which leads to fim = fs + 2IF. This places the image on the higher-frequency side of the desired station in AM broadcast receivers with high-side injection.

Step-by-Step Solution:

Verification / Alternative check:

Alternatively use fim = fs + 2IF = 1000 kHz + 2*455 kHz = 1910 kHz. Both routes agree, confirming the result.

Why Other Options Are Wrong:

1010 kHz and 545 kHz are near-image distractors but do not satisfy the mixing condition with IF = 455 kHz. 1455 kHz is the LO, not the image. 1365 kHz does not meet the image equation.

Common Pitfalls:

Confusing LO frequency with image frequency; using low-side injection without stating it; or assuming Q changes the image frequency (it affects image rejection, not the image location).

Final Answer:

1910 kHz