double null triple injection and the two extra element

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

v.0.1 3/07 http://www.RDMiddlebrook.com 12. DNTI & the 2EET

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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

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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.


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

= =

=∞ =∞≡ = = ≡ = =+


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= =

++


39max

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 =


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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double null triple injection and the two extra element

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