Fractional bandwidth (IF BP filter in a duplexing filter)

I have a question regarding the calculation of fractional bandwidth, in an example in Pozars book, "Microwave and RF Design of Wireless Systems", section 10.1 (Receiver Architecture, Duplexing) (page 337 in the edition of year 2000):

The example is as follows:

The AMPS and IS-54 cellular telephone system mobile units transmit over the frequency range of 824-849 MHz, and receive over the frequency range of 869-894 MHz. These two bands are each divided into 832 channels, each 30 kHz wide, to provide full-duplex communication. The first IF frequency of the receiver is 88 MHz. Compare the fractional bandwidth of the IF bandpass filter and a hypothetical tunable bandpass filter used at the RF stage.

In the solution, they conclude that:

a) At the first IF frequency of 88 MHz, the fractional bandwidth of the 50 kHz IF filter is: Δf/f = 50kHz/88MHz=0.05/88=0.06%,
whereas the same passband at the (midband) RF receive frequency of 882 MHz would be Δf/f = 0.05/882=0.006%.

My question is: from where comes the figure "50 kHz"?

Is it based upon the conditions given in the example, that is calculated based upon these? Or is it an assumption? If so, please help me with explaining how you come to the conclusion that Δf=50 kHz.

(I guess they have assumed that the duplex filter is made as a single unit, with a cross-over point at the 3dB level. )

Many thanks in advance!

Best regards,
/A

PS. I include a few images which hopefully gives a clear picture of the problem.

Figure 10.5: Block diagram of a double-conversion superheterodyne receiver
Figure 10.7: Transmit and receive passband responses of a duplexing filter

I think that they are being generous with the bandwidth of the 30 kHz wide signal just to get some numbers to show the magnitude of the problem.

Many thanks for your response.

Could it be that is has to do with an assumption that the duplex filter is made as a single unit, with a cross-over point at the 3dB level (as shown in figure 10.7)?

/A

Have another question regarding the same example:

The receive frequency of 869 MHZ maps to a low-pass prototype (normalised to a low-pass filter with a cut-off frequency of 1 Hz) of:

Δ=(849-824)/836.5=0.03.
f′= (1/0.03)*(869/836.5-836.5/869)=2.54.

The filter design graphs attached show that a filter of order N=4 is required for 50 dB attenuation. Could you please explain why? That is, what considerations are used when the filter design graph (attached) is used?

Many thanks in advance!

/A