Editing Dual-modulus prescaler (section)

==The problem==
A frequency synthesizer produces an output frequency, f<sub>o</sub>, which, divided by the modulus, N, is the reference frequency, f<sub>r</sub>:

<math> \frac{f_o}{N} = f_r \Rightarrow f_o = Nf_r</math>

The modulus is generally restricted to integer values, as the comparator will match when the waveform is in phase. Typically, the possible frequency multiples will be the channels for which the radio equipment is designed, so f<sub>r</sub> will usually be equal to the [[channel spacing]]. For example, on narrow-band [[radiotelephone]]s, a channel spacing of 12.5&nbsp;kHz is typical.

Suppose that the programmable divider, using N, is only able to operate at a maximum clock frequency of 10&nbsp;MHz, but the output f<sub>o</sub> is required to be in the hundreds of MHz range. Interposing a fixed prescaler that can operate at this frequency range with a division ratio M of, say, 40 drops the output frequency into the operating range of the programmable divider. However, a factor of 40 has been introduced into the equation, so the output frequency is now:

<math>f_o = 40Nf_r</math>

If f<sub>r</sub> remains at 12.5&nbsp;kHz, only every 40th channel can be obtained. Alternatively, if f<sub>r</sub> is reduced by a factor of 40 to compensate; it becomes 312.5&nbsp;Hz, which is much too low to give good filtering and lock performance characteristics. It also means that programming the divider becomes more complex, as the modulus needs to be verified so that only those that give true channels are used, not every 1/40th of a channel that is available.