Home │ Audio
Home Page 
Copyright © 2017 by Wayne Stegall
Updated July 14, 2017. See Document History at end for
details.

(1) 
H_{suminnerloop}(s) = 
s
ω_{hp} 
+  1  + 
ω_{lp}
s 
Figure 1: First schematic. 
Figure 2: First SPICE model. 
* xover 300 and 5kHz v1 vin 0 dc 0 ac 1 sin 1.41421 0 rin vin 0 47k emain emo 0 emvp 0 100k r1 vin emvp 10k rfi emvp eio 10k rfu emvp euo 10k rfd emvp edo 10k eint eio 0 0 eivn 100k ri1 emo eivn 10k ci1 eivn eio 53n euni euo 0 0 euvn 100k ru1 emo euvn 10k ru2 euvn euo 10k ediff edo 0 0 edvn 100k cd1 emo edvn 6.77n rd1 edvn edo 4.7k .end .control ac dec 30 2 200k plot db(eio) db(euo) db(edo) plot db(eio+euo+edo) .endc 
Figure
3:
Bode
plot
shows
two poles each for lowpass and highpass functions.: 
Figure 4: Treble boost reveals two pole anomaly. 
(2) 
H_{sum′}(s) = 
(s + ω_{lp})(s + ω_{hp})
s 
(3) 
H_{sum′}(s) = 
s^{2} + s(ω_{lp}
+ ω_{hp}) + ω_{lp}ω_{hp}
s 
(4) 
H_{sum′}(s) = 
s + (ω_{lp} + ω_{hp}) +  ω_{lp}ω_{hp}
s 
(5) 
H_{sum}(s) = 
H′(s)
ω_{hp} 
= 
s
ω_{hp} 
+  ω_{lp} + ω_{hp}
ω_{hp} 
+ 
ω_{lp}
s 
(6) 
H_{sum}(s) = 
(s + ω_{lp})(s + ω_{hp})
sω_{hp} 
(7) 
H_{int+zero}(s) =  ω_{lp}
s 
+ 
ω_{lp}
ω_{hp} 
= 
(s
+ ω_{hp})ω_{lp}
sω_{hp} 
(8) 
H_{lp}(s) = 

= 
ω_{lp}
s + ω_{lp} 
(9) 
H_{diff+zero}(s) =  s
ω_{hp} 
+ 
ω_{lp}
ω_{hp} 
= 
s
+ ω_{lp}
ω_{hp} 
(10) 
H_{hp}(s) = 

= 
s
s + ω_{hp} 
(11) 
H_{flat}(s) = H_{sum}(s) – H_{int+zero}(s) – H_{int+zero}(s) 
(12) 
H_{flat}(s) =  s
ω_{hp} 
+  ω_{lp} + ω_{hp}
ω_{hp} 
+ 
ω_{lp}
s 
–  ω_{lp}
s 
+ 
ω_{lp}
ω_{hp} 
– 
s
ω_{hp} 
+ 
ω_{lp}
ω_{hp} 
(13) 
H_{flat}(s) = 1 –  ω_{lp}
ω_{hp} 
(14) 
H_{bp}(s) = 

= 
s(ω_{hp} – ω_{lp})
(s + ω_{lp})(s + ω_{hp}) 

Figure 5: Second circuit adds zeros to integrator and differentiator sections. 
(15) 
C_{i1} = 
1
2πf_{lp}R_{i1} 
= 
1
2π × 300Hz × 10kΩ 
= 53.0516nF 
(16) 
R_{i2} = R_{i1} ×  f_{lp}
f_{hp} 
= 10kΩ ×  300Hz
5kHz 
= 600Ω 
(17) 
C_{d1} = 
1
2πf_{hp}R_{d2} 
= 
1
2π × 5kHz × 10kΩ 
= 3.1831nF 
(18) 
R_{d1} = R_{d2} ×  f_{hp}
f_{lp} 
= 10kΩ ×  5kHz
300Hz 
= 166.667kΩ 
(19) 
R_{u2} = R_{u1}  1 –  f_{lp}
f_{hp} 
= 10kΩ  1 –  300Hz
5kHz 
= 9.4kΩ 
Figure 6: Second SPICE deck using calculated values. 
* xover 300 and 5kHz v1 vin 0 dc 0 ac 1 sin 1.41421 0 rin vin 0 47k emain emo 0 emvp 0 100k r1 vin emvp 10k rfi emvp eio 10k rfu emvp euo 10k rfd emvp edo 10k eint eio 0 0 eivn 100k ri1 emo eivn 10k ci1 eivn ri2ci1 53.0516n ri2 ri2ci1 eio 600 euni euo 0 0 euvn 100k ru1 emo euvn 10k ru2 euvn euo 9.4k ediff edo 0 0 edvn 100k cd1 emo edvn 3.183n rd1 emo edvn 166.667k rd2 edvn edo 10k .end .control ac dec 30 2 200k plot db(eio) db(euo) db(edo) plot db(eio+euo+edo) .endc 
Figure
7:
Response
now
lacks
the
unwanted
extra
zeros. 
Figure 8: Inverting Summer 


^{1}Ron Mancini, "Analyzing feedback loops containing secondary amplifiers," 2005, Texas Instruments, ti.com, link.
Document History
July 14, 2017 Created.
July 14, 2017 Corrected some grammar and omissions.
July 14, 2017 Added footnote validating reference referred in
Stabiility Design..