Here are the steps to solve this PLL 4046 design example problem:
1) Let fmin = 8 kHz. Then 8 kHz = 1/(R2(C1+32pF)). Solving for R2 gives R2 = 20 kΩ.
Let fmax = 12 kHz. Then 12 kHz = fmin + 1/(R1(C1+32pF)). Using fmin= 8 kHz, this gives R1 = 10 kΩ.
Using R1 = 10 kΩ and R2 = 20 kΩ, the equation for fmax gives C1 = 100 nF.
2) KO = (fmax - fmin)/VDD = (12 kHz -