SME
Communication Systems · Problem type
Superheterodyne LO and image
Use this for “A radio with a 455 kHz IF is tuned to 1000 kHz. Find the LO and image frequencies.”
Diagram
fLO = fc + fIF = 1,000 kHz + 455 kHz = 1,455 kHz. fimage = fc + 2 × fIF = 1,000 kHz + 2 × 455 kHz = 1,910 kHz.
The mixer takes both the station and the image to |f − fLO| = 455 kHz: only the RF stage, tuned to the station, can reject the image.
This is an example. Type your own values to start a new problem.
fLO (kHz)
=
fc (kHz)
+
fIF (kHz)
fimage (kHz)
=
fc (kHz)
+
2
×
fIF (kHz)
Formulas
fLO = fc + fIF
1,455 = 1,000 + 455
fimage = fc + 2 × fIF
1,910 = 1,000 + 2 × 455
Assumptions
•
High-side injection: the local oscillator sits above the station, fLO = fc + fIF.
•
The mixer passes the difference |f − fLO|, so any signal fIF from the LO lands in the IF.
•
The image is the other such signal, fIF above the LO; only the RF stage can reject it.
Step-by-step
WE KNOW
fc = 1,000 kHz, fIF = 455 kHz
FIND
local oscillator frequency (fLO), image frequency (fimage)
Step 1 · Find local oscillator frequency (fLO)
Use fLO = fc + fIF
The oscillator runs one IF above the station, so the mixer’s difference is the IF.
fLO = 1,000 + 455
fLO = 1,455 kHz
Step 2 · Find image frequency (fimage)
Use fimage = fc + 2 × fIF
The image sits one IF above the oscillator, a mirror of the station one IF below it: two IFs above the station.
fimage = 1,000 + 2 × 455
fimage = 1,000 + 910
fimage = 1,910 kHz
Check
Put the numbers back in. Both sides should match.
1,455 = 1,000 + 455 ✓
1,910 = 1,000 + 2 × 455 ✓
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