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I've stacked on solving this problem and can't find an answer in the web, so I decided to ask a question here. There are the problem: Preselector of boadcast receiver contains of one oscillation circuit, that has a bandwidth 8 kHz at LW range. Quality factor at SW range is 120. How will the selectivity of the receiver on the image frequency channel be changed if we will switch from LW range($f_{RF}=280 kHz$) to SW range($f_{RF}=12.04 MHz$)?

The answer is: It will decrease for $17,4 dB$.

The selectivity os given by formula $\sigma=10\lg(1+\xi^2)$, where $\xi=Q(\frac{f}{f_0}-\frac{f_0}{f})\approx\frac{2Q\Delta f}{f_0}$. And for the image frequency channel $\Delta f=2f_{IF}$ I'm understand what is going on and how to solve this problem if the $f_{IF}$ if done, because I can get the quality factor on LW range from the bandwidth, but there are no $f_{IF}$ in the problem.

Any ideas will be appreciated!

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  • $\begingroup$ I think you mean $\xi=Q \left (\frac{f_{high}}{f_0}-\frac{f_{low}}{f_0} \right )$, because $f - f = 0$. Inherent in this problem is that the input filter will be adjustable, will have a single LC resonator and will have a constant $Q$ -- these are all pretty optimistic assumptions, and appear to be based on 1940's radio practice. You may want to work out the problem leaving $f_{IF}$ as a variable and see if the answer as a ratio depends strongly on $f_{IF}$. $\endgroup$ – TimWescott Nov 29 '20 at 18:21
  • $\begingroup$ Sorry, it’s a typo in formula. $\endgroup$ – KhaBeleth Nov 29 '20 at 21:06

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