9 Appendix 3: Derivation of Crossover Frequency

In the Poles and Zeros section, we established the analytical expressions and locations of all dominant poles and zeros in the system—namely the injection zero, LC double pole, and high-frequency zero. Building on that understanding, we now move to derive the crossover frequency FC using a logarithmic (Bode plot) approach, by tracking how the gain evolves across frequency.

Figure 9-1. Derivation of Crossover Frequency
We start from the low-frequency gain A0, which comes from the DC loop gain:
Equation 9-1. 
A 0 = 20 log 10 1 R 1 C INJ

From here, we move across frequency step-by-step, following the slopes introduced by poles and zeros:

From F0 to F1:

A single pole dominates and the slope becomes –20 dB/dec

At F1, a zero is introduced:
Equation 9-2. 
F 1 = 1 2 πR INJ C INJ :

A zero is introduced by the ripple injection network, cancelling the earlier pole and the slope becomes 0 dB/dec

At F2, LC double pole is introduced:

Equation 9-3. 
F 2 = 1 2 π LC OUT :

The LC double pole appears and the slope becomes –40 dB/dec

At F3, a zero is introduced:
Equation 9-4. 
F 3 = 1 2 πR 1 C FF :

This high-frequency zero reduces slope to –20 dB/dec

Using the standard logarithmic gain relation:
Equation 9-5. 
Gain 2 = Gain 1 + ( slope ) · log ( F 2 F 1 )

Now we translate the slope behavior into exact gain equations:

From A0 to A1:
Equation 9-6. 
A 1 = A 0 - 20 log 10 ( F 1 F 0 ) = 20 log 10 ( R INJ R 1 2 C INJ )
From A1 to A2: Flat region (0 dB/dec), so:
Equation 9-7. 
A 2 = A 1
From A2 to A3:
Equation 9-8. 
A 3 = A 2 - 40 log 10 ( F 3 F 2 )
From A3 to crossover FC: At crossover, gain = 0 dB:
Equation 9-9. 
0 = A 3 - 20 log 10 ( F C F 3 )

This equation becomes the key to solving for FC.

We now substitute all expressions back and solve them systematically:

Starting from:
Equation 9-10. 
0 = 20 log 10 ( R INJ R 1 2   C INJ ) - 40 log 10 ( F 3 F 2 ) - 20 log 10 ( F C F 3 )
Rearranging:
Equation 9-11. 
log 10 ( F C ) = log 10 ( F 3 ) + log 10 ( R INJ R 1 2   C INJ ) - 2 log 10 ( F 3 F 2 )
Exponentiating:
Equation 9-12. 
F C = F 3 · R INJ R 1 2   C INJ · F 2 F 3 2
Substituting the known pole-zero frequencies:
Equation 9-13. 
F 2 = 1 2 π LC OUT ,   F 3 = 1 2 πR 1 C FF
After simplification, we arrive at the final compact expression:
Equation 9-14. 
F C = R INJ · C FF 2 πLC OUT