Download The impedance network

The impedance network
A semi-infinite “ladder” netIn
I0
I1
V1
V2
Vn
Vn+1
work is composed by impedances V0
...
...
Z1 = Z1 (ω) and Z2 = Z2 (ω) as
Z1
shown in the figure.
a) Calculate the total impedance
Z2
(Z0 ) of the semi-infinite network. How can a real finite
...
...
network be terminated after N
steps in order that its properties resemble those of the semi-infinite one?
b) Let Vn be the voltage on the la n-th node. Find the relation between Vn and Vn+1 and, from this,
the expression of Vn as a function of the “input” value V0 . Discuss the result for a purely resistive
network (Z1 = R1 , Z2 = R2 ).
Now consider a LC network, i.e. having Z1 = −iωL and Z2 = −1/iωC, with an input signal
V0 (t) = V0 e−iωt .
c) Find the frequency range which allows signal transmission on the network. Show that there exist
a cut-off frequency ωT such that for ω > ωT the signal is damped, and calculate find the damping
factor.
d) Now exchange L and C, i.e. put Z1 = −1/iωC and Z2 = −iωL, and discuss the signal transmission
for such “CL” network.
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Solution
a)
Z0
=
Z1
Z2
Z0
The infinite network is a sequence of identical circuit elements. If
we add another identical element on the input side, the impedance
becomes
Z0′ = Z1 +
Z2 Z0
.
Z2 + Z0
(1)
But the impedance of the infinite ladder cannot change by the addition of a single element, thus
Z0′ = Z0 . From this condition we obtain a second order algebraic equation for Z0 with the solution
2
1/2
Z1
Z1
Z0 =
+
+ Z1 Z2
.
(2)
2
4
The other solution has been discarded because in the case of real, positive impedances (the purely
resistive case below) it would give an unphysical negative value. A finite ladder may be terminated
with an impedance Z0 .
b) Let In the current flowing through the impedance Z1 . The voltage drop across the impedance
is between two nodes is Vn − Vn+1 = In Z1 . On the other hand, Vn = In Z0 holds. Thus, solving for
Vn+1 /Vn ≡ α we obtain
α≡
Vn+1
Z1
=1−
.
Vn
Z0
(3)
Thus, if V0 (t) = V0 e−iωt is the voltage at the n = 0 node, at the n-th node Vn = αn V0 e−iωt . For the
resistive network
2
1/2
R1
R1
+
+ R1 R2
≡ R0
(4)
Z0 =
2
4
is a real number, thus α = 1 − R1 /R0 < 1. The signal is damped along the propagation by a factor
of αn after n nodes.
c) For the LC network
2 2
1/2
iωL i
ω L
iωL
L 2
+
−
(ωT − ω 2 )1/2 − iω ,
(5)
Z0 = −
=
2
4
2 ωC
2
√
where ωT ≡ 2/ LC. Thus
α=
(ωT2 − ω 2 )1/2 + iω
.
(ωT2 − ω 2 )1/2 − iω
(6)
If ω < ωT then Z0 is a real number and α is the ratio of complex conjugate numbers, i.e. a complex
number with unitary modulus:
α = eiφ ,
φ = 2 arctan[ω/(ωT2 − ω 2 )1/2 ] .
2
(7)
Thus Vn = V0 einφ−iωt , i.e. the signal propagates along the network. The above equation also gives
the dispersion relation
ω = ωT | sin(φ/2)| .
(8)
If ω > ωT then Z0 is an imaginary number,
Z0 = ±i(ω 2 − ωT2 )1/2 ,
(9)
±(ω 2 − ωT2 )1/2 + ω
α=
.
±(ω 2 − ωT2 )1/2 − ω
(10)
and α is real:
Taking the root with the minus sign, we have |α| < 1, and the signal is damped. The other root
corresponds to an unphysical situation since it gives |α| > 1, i.e. an amplification of the signal along
the network without an external energy source.
The LC network behaves as a low-pass filter, since signals at frequencies ω > ωT are attenuated after N nodes by a factor |α|N . For
large N the “transmission curve” as a function of the frequency is close
to that of an ideal low-pass filter, for which the transmission is zero for
ω > ωT . The figure shows |α| and |α|4 (dashed) as a function of ω.
d) By proceeding as in point c) we obtain
"
#
1/2
1/2
i
1
1
i
1
i
1
Z0 =
− 2
+ − 2 2+
(−iωL)
+
,
(11)
=
2ωC
4ω C
ωC
2C
ωT2
ω
ω
and
α=
(ωT−2 − ω −2 )1/2 − iω −1
.
(ωT−2 − ω −2 )1/2 + iω −1
(12)
Propagating waves are obtained when |α| = 1 i.e. when ω > ωT . In the opposite case ω < ωT the
signals are damped, i.e. the network acts as a high-pass filter.
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