12. DNTI AND THE 2EET:Double Null Triple Injection and the Two Extra Element
Theorem
A short and easy way to find the poles and zeros of a circuitcontaining two reactances
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1c LQ Q
1c LQ Q
c LQ Q
mr
LR
tCEiα
Ei
36
5p10k
v SA v−120=β
S BR R4.6k
dC500p 1
62 36S B
B Lvm R RS B m
R RA dBR R r +
≡ = ⇒+ + β
α1dpR
2dpR
1CE with and t dC CUse the 2EET to add and t dC C
( ) Sv
and are already absorbedS BR Rinto a Thevenin equivalent1
Eiβ+
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mr
LR
tCEiα
Ei
36
5p10k
v SA v−120=β
S BR R4.6k
1dpR
For alone, results are already known:tC
( ) Sv(1 )
62S B m
S B m L
R R rm
R R r R+
≡ =β
1
62 36S B
B Lvm R RS B m
R RA dBR R r +
≡ = ⇒+ + β
α
21 / 36Z
mnR r α=∞ = = Ω 21 620Z
LdR mR k=∞ = =
21
1z Z
t nC Rω =∞≡
211
1p Z
t dC Rω =∞≡
1
/ 2880/ 2
51
1 136
1 1z
p
s sMHz
v vm s skHz
A A dBπ
ωπ
ω
− −= =
+ +1Eiβ+
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For alone:dC
mr
LR
tCEiα
Ei
36
5p10k
v SA v−120=β
S BR R4.6k
dC500p 2dpR
( ) Sv
1Eiβ+
(1 )62S B m
S B m L
R R rm
R R r R+
≡ =β
12 0ZnR
=∞ = ( )12 1 2.2Z
S B mdR R R r kβ=∞ = + =1
62 36S B
B Lvm R RS B m
R RA dBR R r +
≡ = ⇒+ + β
α2
/ 2140
1 1361 1
p
v vm s skHz
A A dB πω
= =+ +
122
1p Z
d dC Rω =∞≡
These results are obtainable by inspection;no algebra is required.
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36
mr
LR
tCEiα
Ei
36
5p10k
Ei1+β
v SA v−120=β
S BR R4.6k
dC500p
1dpR
2dpR( ) Sv
1
1
1z
p
s
v vm sA A ω
ω
−=
+1Eiβ+
2
11
p
v vm sA Aω
=+
(1 )62S B m
S B m L
R R rm
R R r R+
≡ =β
With the corner frequencies for and separately now known, t dC Conly and remain to be found.n dK K
Since does not contribute a zero, is irrelevant, andd nC K
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For and :t dC C
mr
LR
tCEiα
Ei
36
5p10k
v SA v−120=β
S BR R4.6k
dC500p
1dpR
2dpR( ) Sv
1
1
1z
p
s
v vm sA A ω
ω
−=
+
1Eiβ+
2
11
p
v vm sA Aω
=+
(1 )62S B m
S B m L
R R rm
R R r R+
≡ =β
1 2 1 2
1
1z
p p p p
s
v vm s s s sd
A AK
ω
ω ω ω ω
−=
+ + +
where
2 1
2 1
0 01 2
1 2
1 1 or (1 )
Z ZS B m Ld dL
d dZ ZL S B md d
R R R R r RRK KmR m mR R rR R β
= =
=∞ =∞≡ = = ≡ = =+
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38(Redundancy check)
For and :t dC C
mr
LR
tCEiα
Ei
36
5p10k
Ei1+β
v SA v−120=β
S BR R4.6k
dC500p
1dpR
2dpR( ) Sv
1
1
1z
p
s
v vm sA A ω
ω
−=
+1Eiβ+
2
11
p
v vm sA Aω
=+
1dK m=
(1 )62S B m
S B m L
R R rm
R R r R+
≡ =β
1 2 1 221 1 1
1
1z
p p p p
s
v vm
m
A As s
ω
ω ω ω ω
−=
⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
The of the quadratic isQ
1 2
1 2
11 2
1 1 1 2
1p p
p p
m p p
p pQ
mω ω
ω ω
ω ω
ω ω= =
++
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0.5In general, , and since m 1 the quadratic always has real roots.Qm
= ≥
For and :t dC C
mr
LR
tCEiα
Ei
36
5p10k
Ei1+β
v SA v−120=β
S BR R4.6k
dC500p
1dpR
2dpR( ) Sv
1
1
1z
p
s
v vm sA A ω
ω
−=
+1Eiβ+
2
11
p
v vm sA Aω
=+
(1 )62S B m
S B m L
R R rm
R R r R+
≡ =β1
dK m=
1 2 1 221 1 1
1
1z
p p p p
s
v vm
m
A As s
ω
ω ω ω ω
−=
⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Here, 62 so the real root approximation is extremely good, and theresult can be written
m =
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40( )1 2 1 2
1
1 1
z
p p p p
s
v vms s
m
A A ω
ω ω ω ω+
−=
⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
For and :t dC C
mr
LR
tCEiα
Ei
36
5p10k
v SA v−120=β
S BR R4.6k
dC500p
1dpR
2dpR( ) Sv
1Eiβ+ (1 )
62S B m
S B m L
R R rm
R R r R+
≡ =β
( )1 2 1 2
1
1 1
z
p p p p
s
v vms s
m
A A ω
ω ω ω ω+
−=
⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( )/ 2
880/ 2 / 2
1237
136
1 1
sMHz
s sMHzkHz
dBπ
π π
−=
+ +
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37kHz
51kHz140kHz 12MHz
880MHz
36dB
Interaction between and t dC Ccauses ʺpole splitting.ʺ
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