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If the incoming signal to the mixer is 3 600 kHz and the first IF is 9 MHz, at which one of the following frequencies would the local oscillator (LO) operate?
5 400 kHz
The IF is the difference between LO and RF, or LO = RF + IF (if LO is higher) or LO = IF - RF (if LO is lower, less common for up-conversion). If IF = |LO - RF|, then LO = RF +/- IF. For up-conversion to 9MHz from 3.6MHz, LO = 9MHz - 3.6MHz = 5.4MHz is not standard. If IF=9MHz and RF=3.6MHz, common practice is LO = RF+IF = 3.6+9 = 12.6MHz or LO = IF-RF = 9-3.6 = 5.4MHz (if IF>RF). Here, 5400kHz (5.4MHz) + 3600kHz (3.6MHz) = 9000kHz (9MHz). So LO is 5.4 MHz and RF is 3.6 MHz for a 9 MHz IF sum, or if IF is difference, |5.4-3.6| = 1.8 not 9. Or |LO-3.6|=9, so LO = 12.6 or LO = -5.4 (not phys). If LO=5.4MHz, and RF is higher, RF=LO+IF = 5.4+9 = 14.4. The option 5400 kHz for LO implies IF = |RF - LO| = |3600 - 5400| = |-1800| = 1.8 MHz, not 9 MHz. OR, if RF = 5400 - 3600 = 1800. Question implies LO is 5.4MHz and RF is 3.6MHz for a sum IF of 9MHz or a difference IF of 1.8MHz. If LO = RF + IF for high-side injection: LO = 3.6MHz + 9MHz = 12.6MHz. If LO = RF - IF for low-side injection: LO = 3.6MHz - 9MHz (not possible for positive freq). If IF = LO - RF, 9MHz = LO - 3.6MHz => LO = 12.6MHz. If IF = RF - LO, 9MHz = 3.6MHz - LO => LO = 3.6 - 9 = -5.4MHz (not possible). The answer 5400 kHz (5.4 MHz) suggests an error in the question or common interpretation. However, if the mixer output selected is LO+RF, then 5.4 MHz + 3.6 MHz = 9 MHz. This is unusual for a first IF but possible. Or, it means the LO is 5.4 MHz and the RF is actually different to produce 9 MHz IF.
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