Signals&circuits
RESONANT CIRCUITS
Page 1
RESONANT CIRCUITS
Series resonant. Let’s investigate well-known resonance phenomena. Such effect can be
observed in series RLC circuit (figure 1a). The input impedance of the circuit is
I
R
L
1 ⎞
⎛
Z in ( jω) = R + j ⎜ ωL −
⎟
ωC ⎠
⎝
C
(1)
At some angular frequency ω 0 the reactive term will
equal zero and impedance will be purely resistive. This
condition is known as resonance and ω0 or f0 is the
series resonant frequency. This frequency may be
expressed in terms of the circuit parameters by equating
the reactive term of previous equation (1) to zero:
V
Fig.1. Series RLC circuit
ω0 L −
1
= 0 ; ω0 =
ω0 C
1
LC
;
f0 =
1
2π LC
From Eq. 1 we can see that Z in ( jω) will exhibit minimum impedance equal to R ohms.
So properties of resonant circuit depends on circuit parameters R, L, C. These
parameters are called primary parameters. It is possible to introduce additional so-called
secondary parameters ant to simplify circuit analysis.
The reactive term of the input impedance can be rewritten as follows:
X = ωL −
1
C
1
= ωL
−
ωC
C ωC
L
L
=
L ⎛ ω ω0 ⎞
⎜
⎟
−
C ⎜⎝ ω0 ω ⎟⎠
(2)
Value
L
ρ=
C
= ω0L =
1
ω 0C
(3)
may be named as circuit characteristic impedance or wave impedance. As we can see
characteristic impedance is equal to circuit inductive or capacitive impedances at resonant
frequency.
From Eqs. (1) - (3) we can see that
⎛ ω ω0
Z in ( jω ) = R + jρ ⎜⎜
−
⎝ ω0 ω
⎞
⎟⎟ =
⎠
⎡
ρ ⎛ ω ω 0 ⎞⎤
⎟⎟⎥
R ⎢1 + j ⎜⎜
−
R
ω
ω
0
⎠⎦
⎝
⎣
(4)
Ratio characteristic impedance/circuit resistance usually is named as resonant circuit quality
factor Q:
Q=
ρ
R
=
ω0 L
R
=
1
ω 0CR
(5)
As you remember from previous discussion, the Q of coil is ωL / R at general frequency and,
at the resonant frequency Q0 , is ω 0 L / R (the same it is possible to say about capacitor quality
Signals&circuits
RESONANT CIRCUITS
Page 2
factor, which is equal to 1 / ω 0 CR ). So particular value of the Q factor of the coil or capacitor
at resonant frequency, Q0 , becomes the Q factor of the circuit at the resonance. With
inspection of this value input impedance can be rewritten:
⎡
⎛ ω ω0 ⎞⎤
⎟⎥ ,
−
Z in ( jω) = R ⎢1 + jQ⎜⎜
ω ⎟⎠⎥⎦
⎢⎣
⎝ ω0
or
Z in ( jξ ) = R(1 + jξ ) ,
(6)
⎛ ω ω0 ⎞
⎟⎟ - general mistuning. Let’s introduce additional parameter - absolute
−
here ξ = Q⎜⎜
⎝ ω0 ω ⎠
mistuning Δω = ω − ω0 . It will be convenient latter to use this type of mistuning to simplify
analysis also. When general frequency ω ≈ ω0 ( Δω ≤ ω0 )
ω0
ω ω0 ω0 + Δω
Δω
−
=
−
= 1+
−
ω0 ω
ω0
ω0 + Δω
ω0
1
Δω
≅2
Δω
ω0
1+
ω0
Here Δω / ω0 - relative mistuning
Rearranging the expression for ξ with respect to equation above
ξ ≅ 2Q
Δω
ω0
(7)
Thus, use of the secondary parameters allows to us simplify expression of input
impedance and frequency responses in series
RLC circuit. So, from Eq. (6) follows
R, ohms
Z (ω) = R 1 + ξ 2 ; ϕ Z (ω ) = tan −1 ξ .
(8)
(a)
ϕ,
rad
ω/ω 0
Examples of typical input impedance modulus
and argument dependence on frequency are
demonstrated in Fig.2.
Current in the series RLC circuit
o
o
I in
(b)
o
Vin
V
1
=
= in
Z in ( jω ) R 1 + jξ
(9)
ω/ω
0
Fig.2. Series RLC circuit impedance modulus (a) and
argument (b) dependence on frequency – continuous
curves – results of approximate calculation, - dotted
curves – precise calculation
As we can see, current has maximum at
resonant frequency ( ξ = 0 ). Such current
dependence can be named as resonant curve.
Voltage transfer function can be given by
Signals&circuits
RESONANT CIRCUITS
0
H ( jω ) =
Page 3
0
VC
0
V in
I
1
1
= in 0 =
.
jωC V
jωCR(1 + ξ )
(10)
in
When circuit's quality factor is high (Q >> 1) and general frequency ω ≈ ω 0 ,
expression of voltage transfer function can be rewritten:
H ( jξ ) = −
jQ
.
1 + jξ
(11)
Thus, from the analysis of Eqs. (9,11) follows, that frequency responses of current
resonant curve and voltage transfer function in the series RLC circuit have the same form and
differs in constant multiplier only. Magnitude and phase responses of the series circuit
H (ξ ) =
Q
1 + ξ2
; ϕ(ξ ) = −
π
− tan −1 ξ .
2
(12)
In addition to this it is necessary to notice, that RLC circuit output voltage amplitude at
resonant frequency ( ξ = 0 ) is Q times greater than input voltage amplitude. Thus, resonant in
such circuit is called as voltage resonant and would be very convenient to select the cleared
frequency from adjacent frequencies. For this circuit is obliged to have selective properties
which can be expressed through circuit bandwidth. The method used to determine bandwidth
is based upon the following considerations. At resonance, the power dissipated in circuit is
maximum since the current is maximum. There will then be two frequencies, one as each side
of ω 0 , where the power dissipated is one-half the power of resonance. These frequencies are
designated as the upper ω2 and lower ω1 half-power frequencies. Let us now proceed to
locate these half – power frequencies on resonant curve. At first it is necessary to mention,
that when we are speaking about power dissipation, we are thinking about real power, which
is dissipated, in resistive elements. At ω 0
P0 = I 02 R .
Therefore, at half-power frequencies ω1 and ω2
Pb = I b2 R =
I
1
1
P0 = I 02 R and I = 0 ≅ 0.707 I 0 .
2
2
2
Now we might be desirable to determine the bandwidth of the resonant circuit by
inspection of secondary parameters instead of by direct calculations of the actual circuit.
Because of the same form of current curve and voltage transfer function this tells us
that at boundary frequency H (ω b ) = H (ω 0 ) / 2 and
Q
1 + ξ b2
=
Q
2
Thus, the general mistuning at boundary frequencies ξ b = 1 . Substituting this value into Eq.
(7) we can write
Signals&circuits
RESONANT CIRCUITS
Δω b =
ω0
2Q
=
Page 4
R
=α
2L
Parameter α may be called as limiting mistuning. So, resonant circuit half-power bandwidth
BW = ω 2 − ω 1 = 2α =
H
ω0
Q
.
(13)
From this equation follows very important
another definition of the resonant circuit quality
factor Q:
(a)
ϕ
ω/ω
Q=
0
, rad
ω0
BW
.
(14)
Typical magnitude responses of RLC series
resonant circuit are shown in the Fig. 3 as
example. Fig. 3. Fr. Responses
(b)
Now let’s consider that happens when
resonant circuit is loaded (Fig. 4). Parallel
connection of the capacitor and resistor can be
transformed into series connection (see previous
ω/ω 0
section). When quality of the capacitor is high
(QC≥10), equivalent capacitor has the same
Fig. 3. Series RLC circuit magnitude (a) and phase
capacitance. Resistance of the transformed
(b) frequency responses: – continuous curves –
results of approximate calculation, - dotted curves
equivalent R E = R L (1 + 1 / QC2 ) . Thus, load
– precise calculation
resistance increases circuit resistance. The same
could be observed when it is necessary to evaluate source internal resistance. Therefore, total
circuit resistance
RT = R + Re + RS = R(1 + β )
Here
β=
Re + RS
R
Quality factor of loaded resonant circuit will be decreased. Such proposition follows from
rearranged expression of the quality factor of the loaded RLC series circuit:
QL =
ρ
RT
=
ρ
R (1 + β )
=
Q
1+ β
(15)
Bandwidth of the series RLC circuit BWL becomes wider:
R
L
RS
R
C
VS
(a)
RL
L
RE
RS
VS
CE
(b)
Fig. 4. Loaded RLC-series circuit (a) and its equivalent (b)
RL
Signals&circuits
RESONANT CIRCUITS
I
BW L =
Q
(1 + β ) = BW (1 + β )
(16)
Parallel resonant circuit
L
V
ω0
Page 5
C
RLC circuit, which is build on coil and capacitor connected in
parallel, is called parallel resonant circuit (Fig. 3). Resistor R in
this figure represents coil loss resistance. In some textbooks,
especially older, such circuit is called as anti-resonant circuit.
Input impedance of such circuit can be given by
R
Fig. 5. Parallel RLC resonant
circuit
Here I& = I& L + I&C ,
I&L =
V&
Z& in (ω ) =
I&
V&
; I&C = jωCV&
R + jωL
Thus,
Z in ( jω ) =
1
1
jω C +
R + jωL
=
R + jωL
jωC
1
1 ⎞
⎛
R + j ⎜ ωL −
⎟
ωC ⎠
⎝
.
(17)
As we can see, Z in ( jω) will exhibit minimum impedance at some angular frequency. This
property is known as resonance also.
If coil quality factor is high (Q>>1), then ωL>>R
Z in ( jω ) =
ρ2
1 ⎞
⎛
R + j ⎜ ωL −
⎟
ωC ⎠
⎝
.
(18)
Thus, parallel resonant frequency ω0 or f0 may be expressed in terms of the circuit parameters
by equating the reactive term of previous expression to zero. So, as for series resonant circuit,
ω0 L −
1
= 0 ; ω0 =
ω0 C
1
LC
;
f0 =
1
2π LC
With respect to previous equations for the series circuit, parallel RLC circuit input impedance
expression can be rearranged and given by
Z in ( jω ) =
ρ2
R0 e
R (1 − jξ )
= 0e
1 + jξ
1+ξ 2
(19)
= ρQ – input resistance of the parallel resonant circuit at resonant frequency.
R
Real part of the input impedance is given by
Here R0 e =
Signals&circuits
RESONANT CIRCUITS
Rin =
Page 6
R0 e
1 + jξ 2
(20)
and imaginary part -
X in = −ξ
R0e
= − Rinξ .
1+ξ 2
(21)
Modulus and argument of the impedance are:
Z in (ω ) =
R0e
1+ξ
2
;
ϕ Z (ω ) = tan −1 (−ξ ) .
Current transfer function in the parallel resonant circuit can be found using following
expression:
H I ( jω ) =
I&C
jQ
= jωCZ in (ω ) ≅
.
&I
1 + jξ
in
(22)
As we can see, current has maximum at resonant frequency ( ξ = 0 ). Such type of frequency
response is called parallel RLC circuit resonant curve. As you remember, series RLC circuit
input current (this current is running through capacitor)
dependence on frequency was shown by expression
I
I&IN = V&IN /[ R (1 + jξ )] . So, parallel circuit resonant curve differs
from series circuit resonant curve in constant multiplier only.
Current through capacitor at resonant frequency ω 0 is Q times
L
RS
C
greater then input current. So why parallel RLC circuit is called
V
R
current resonant circuit.
VS
Source resistance influence in to the parallel circuit
parameters. Now let’s try to investigate properties of the
parallel circuit with real source (Fig. 6). Such circuit impedance
is very large, so its parameters depend on input or output circuit
Fig. 6. Parallel RLC circuit with
voltage source
parameters. Let’s consider influence of the source resistance.
Current through this circuit
I& =
V&S
RS + Z in
and voltage across resonant circuit can be given by
V& = I& ⋅ Z in = V&S
At the resonant
Z in
RS + Z in
(23)
Signals&circuits
RESONANT CIRCUITS
V&0 = V&S
Page 7
R0 e
.
RS + R0 e
Thus, parallel RLC circuit frequency response
Z in
&
R S + R0 e
R + Z in
V
= S
=
H V ( jω ) =
.
R0 e
R0 e R S
V&0
R0 e +
R S + R0 e
Z in
(24)
Substituting Z in = R0e (1 + jξ ) in to this equation we can write
H V ( jω ) =
R0 e + R S
=
R0 e + RS (1 + jξ )
Also we can substitute, that ξ = 2Q
H V ( jω ) =
1
⎛
⎞
ξ
⎟⎟
1 + j ⎜⎜
⎝ 1 + R0 e R S ⎠
.
(25)
Δω
:
ω0
1
⎛
Δω ⎞
Q
⎟⎟
⋅
1 + j⎜⎜ 2
+
1
ω
R
R
0e
0 ⎠
S
⎝
=
1
⎛
Δω ⎞
⎟⎟
1 + j⎜⎜ 2Qe
ω
0 ⎠
⎝
.
(26)
From this equation follows, that quality factor of the RLC parallel circuit terminated by
source resistance is less then quality factor of the circuit without this resistance ( RS = 0 ):
Qe =
R
C
Re
L
L
(a)
(b)
Fig. 7. Parallel RLC circuit (a) and its
equivalent (b)
C
Q
.
R0 e
1+
RS
(27)
From this equation also follows, that such circuit
R
bandwidth must be 1 + 0e times wider.
RS
Equivalent scheme of the parallel resonant
circuit. Input impedance for the circuit, shown in
the Fig. 7, can be given by
Z in =
LC
.
R + jωL + 1 jωC
(28)
This equation is obtained supposing that inductive reactance is more great than coils
resistance - ωL >> R .
Input admittance
Signals&circuits
RESONANT CIRCUITS
Yin =
Page 8
R + jωL + 1 jωC
1
R
1
.
=
= 2 + jωC +
Z in
LC
jωL
ρ
(29)
Circuit, shown in the Fig. 7b, has the same admittance as circuit shown in the Fig 7a.. Thus,
this circuit is equivalent to the parallel resonant circuit.
So, circuit branch resistance, which represents coils losses, may be transformed into
separate branch. These properties also follows from the circuit duality principles.
Taped resonant circuits. Impedance transformation. A problem, which frequently arises,
is that coupling a generator or a source to a load where load resistance is not equal to what we
want the generator or source to see. In some cases it is necessary to match internal and load
impedances. As you know, such requirement for internal and load resistances equality arises
from maximum power transfer condition. Usually this situation arises up where load is
parallel RLC resonant circuit with very large input resistance at resonant frequency. In some
cases it is necessary to match devices with various output and input impedances. For example,
final r-f amplifier in a transmitter may require a 5-kohm load for optimum operation. The
transmission line, which is to feed, may only be a 300-ohm line. The problem then is how the
300-ohm line can be matched to the final amplifier to avoid reflections. It should be
understood that impedance matching does not necessarily always mean maximum for power
transfer. Thus some means of stepping the impedance downward must be effected.
Convenient schemes are to tape capacitor or coil in the RLC parallel resonant network.
(Fig. 8 a,b). Such networks could be transformed into form, similar to transformers (Fig. 8
c,d). Let’s try to determine voltage transformation in the network with capacitance division.
Voltages across terminals 1-2 and 3-4 can be given by V&12 = I&2 X C 2 , V&34 = I&1 X L . If circuit
quality factor Q >> 1 , then I&1 = I&2 and V& = I& X . Let’s state, that circuit is be resistive
34
2
L
between the 1-2 terminals: X C 2 = X L − X C1 . Therefore,
V& = I& ( X + X ) . Thus, voltage gain can be given by
34
2
C1
X L = X C1 + X C 2
and
C2
V&34 X C1 + X C 2 C1 + C 2
=
=
.
X C2
C1
V&12
2
Rs
Vs
Rs
C1
R
C
C2
I
I
2
L
1
3
1
a
b
Rs
4
Rs
2
L2
L2
L
C
2
4
1
Vs
L
1
Vs
1
4
R
Vs
L
2
C1
I2
I1
3
c
1
d
Fig. 8. Parallel RLC circuit with taped capacitor (a) and
inductor (c) is acting as impedance transformer (b,d)
C
3
The impedance transformation can be
readily determinated from the
following considerations. The power
put into the network at the 1-2
terminals is
V122
P12 =
.
Z12
If Q is large, the network losses are
negligible, and an equal amount of
power is made available at the 3-4
terminals. Remember that pure
reactive elements cannot absorb
power. Therefore
V342
P34 =
.
Z 34
But P34=P12 and, therefore
Signals&circuits
RESONANT CIRCUITS
Page 9
2
V342 Z 34 ⎛ C1 + C 2 ⎞
⎟ .
=
=⎜
V122 Z12 ⎜⎝ C1 ⎟⎠
If we shall denote ratio
C1
C1 + C2
= pC
(30)
- capacitor-taping coefficient, taped network input voltage and impedance could be expressed
through output voltage and impedance in terms of taping coefficient:
Z12 = pC Z 34 .
V12 = pCV34 ,
2
(31)
Essentially, the same action may be obtained by dividing up the inductor. (Fig. 8 c,d). Again
assuming a high-Q circuit and L1 and L2 are not magnetically coupled we can write:
V&12 = I&2 X C = I&1 X C and V&34 = I&1 X L1 . Since the circuit is resonant at the 1-2 terminals, it also
resonant at 3-4 terminals and X = X + X . Thus, V& = I& ( X + X ) and
C
L1
L2
34
1
L1
L2
V&34
X + X L2
L + L2
= L1
= 1
.
V&12
X L2
L 21
The impedance ratio will again be the square of voltage ratio, if losses are negligible:
2
Z 34 ⎛ L1 + L2 ⎞
⎟ .
=⎜
Z12 ⎜⎝ L2 ⎟⎠
If we shall introduce coil taping coefficient
pL =
L2
,
L1 + L2
(32)
taped network input voltage and impedance could be expressed through output voltage and
impedance in terms of taping coefficient also:
V12 = p LV34 , Z12 = pL Z 34 .
2
(33)
Thus, at its resonant frequency a taped resonant circuit exhibits many characteristics
associated with a transformer. For example, the driving voltage may be stepped up or down.
The impedance levels correspondingly can be stepped also.
RLC parallel circuits could be called as first type resonant parallel circuits. Taped
resonant parallel circuits are called as second type (taped is coil) or third type (taped is
capacitor) resonant circuits.
Such circuits have two resonants - parallel resonant, which can be observed when total
network reactance is equal to zero, and series resonant, which arises up when reactance
between terminals 3-4 is equal to zero.
Parallel resonant in the circuit, shown in the Fig. 8 a, becomes at frequency
Signals&circuits
RESONANT CIRCUITS
1
ω 0 PC =
C ⋅C
L⋅ 1 2
C1 + C 2
Page 10
.
(34)
Circuit that is shown on the Fig. 8c, d has parallel resonant frequency
1
ω 0 PL =
.
C ( L1 + L2 )
R, ohms
(35)
R, ohms
a
c
ωs
H
ω/ω
H
0
b
ωs
ω/ω 0
d
ω/ω
ω/ω
0
0
Fig.9. Typical input impedance (a,c) and magnitude (b,d) frequency responses of the second-type (a,b) and thirdtype (c,d) parallel RLC taped resonant circuits (dotted curves shows responses of the simple parallel resonant
circuit – p=1, continuous curves – responses of the taped circuits with p=0.5)
Series resonant frequencies can be calculated using formulas
ω 0 SC =
1
;
LC2
ω 0 SL =
1
L2C
.
(36)
Typical magnitude responses of such circuits are shown in the Fig. 9.
It is necessary to note that such type of circuit can be used in the frequency band near parallel
resonant frequency. Here properties of magnitude responses are the same as of the first type
resonant circuits.
Signals&circuits
RESONANT CIRCUITS
Page 11
Improving band-pass characteristics with stagger-tuned circuits. Resonant circuits are
acting as simple band-pass networks. The
magnitude response of these circuits appears to
H( ω )
have different degrees of sharpness. It depends
Hmax
on circuit quality (Fig. 10). So, circuit Q
1
measures this sharpness or circuit selectivity. A
higher Q yields a more selective circuit in
2
passing
or
amplifying
the
frequency
components of a signal. However, in trying to
3
approximate the ideal band-pass magnitude
characteristics, all the simple circuits fail to the
same extent. To see this, we replot the curves in
4
figure 10 with the independed variable general
mistuning ξ, which is the ratio of the deviation
ω/ω
0
from the center frequency to half-bandwidth:
Fig.10. Normalized magnitude responses
Δϖ ω − ω 0
of the resonant circuits: 1 – Q=5; 2 –
ξ ≅ 2Q
=
.
Q=10; 3 – Q=15; 4 – Q=20
ω0
0.5BW
Let introduce normalized voltage gain Hn, which
can be given by ratio H (ξ )
. As long as
H max
circuit Q is high, say, Q>>10, the normalized gain
in the neighborhood of the center frequency is
H( ω )
Hmax
1
H n (ξ ) =
1
1+ξ 2
.
A plot of such gain is shown as curve 1 in the
Fig.11. Curve 2 in this figure shows ideal
3
magnitude response of the band-pass filter. So,
band-pass characteristics of the resonant circuit are
ω/ω
0
rather bad in comparison with ideal characteristics.
A simple scheme for obtaining the better
band-pass characteristics is to use a two-stage
amplifier in which the stages are not tuned to the
Fig. 11. Normalized magnitude response
same frequency, but are staggered at frequencies
plots: 1 – simple resonant circuit; 2 –
above and below desired center frequency of the
ideal response of the pass-band filter; 3 –
compete amplifier (Fig.12). The overall gain of the
stager-tuned circuit
amplifier is the product of the gains at each stage.
The magnitude of the resulting response curve can take on a variety of shapes, depending on
the bandwidth of each stage and the
Ideal
difference between the two center
I-st RLC
2-nd RLC
RS
Amplifier
frequencies. Such idea is illustrated
with resultative plots of complete
R1 L C
C2
V1 R 2
1
1
VS
magnitude responses in the Fig. 13.
V2
V1
L
2
Below are presented results needed
for designing an improved bandpass circuit.
1. Each stage should have a
Fig. 12. A stagered-tuned network
bandwidth equal to overall BW,
divided by 2 .
2
B
Signals&circuits
RESONANT CIRCUITS
Page 12
2. The center (resonant) frequencies of the two stages should be
1 BW
ωn ±
.
2 2
Normalized magnitude response, obtained with use of stagertuned circuits, is shown in the Fig.11 as curve 3. This curve is
really better than curve 1. Here, “better” means, that within the
pass-band the magnitude of the gain is more constant, and outside
the pass-band unwanted frequency components are attenuated
more sharply.
Tuned coupled resonant circuits. Another simple scheme for
obtaining the better band-pass characteristics is to use coupledcircuits. Circuits could be coupled using inductive (Fig. 14a), conductive or
autotransformative (Fig. 14b), intrisinc capacitive (Fig. 14c) or exterior capacitive (Fig. 14d)
relations.
Equivalent circuits of the coupled networks.
Analysis of these networks could be
simplified using equivalent circuit (Fig. 15).
Such circuit could be build in analogy to
equivalent circuit of the transformer (look at
networks shown in the Fig. 14a,b). We can see
also, that network, shown in the Fig. 14d, is Ttype and impedances of the equivalent circuit
represent its joint impedances. It is necessary
to note also, that circuits, which are shown in
the Fig. 14 c,d have similar properties because
of
they are related by pair of Δ − Y
equivalent transforms.
At first let’s define impedances in the
equivalent circuit. Impedances ZI and Z2 here
represents resonant circuit self-impedances.
For the circuits, which are shown in the Fig.
14 a, b, these impedances are defined by
1
,
Z 1 = R1 + jω ( L1 − M ) − j
ωC1
H( ω )
(a)
(b)
1
,
ωC 2
Impedance Z12 represents coupling element
and can be denoted as mutual impedance. For
the circuit with inductive coupling
Z 2 = R 2 + jω ( L 2 − M ) − j
Z12 = − jωM
(c)
ω/ω
0
Fig. 13. Possible response curve shapes from
staggered tuning: zero-separation of center
frequency (a), wide separation 9b) and
critical separation for flat-top pass-band (c)
(here dashed curves – magnitude responses of
the separate circuits, continuous curve –
resultative response of staggered network)
In the circuit with conductive-transformative
coupling
Z12 = jωL12
Self-impedances in the last two circuits,
shown in the fig. 14 c, d, can be defined as
follows:
Signals&circuits
C1
M
R1
RESONANT CIRCUITS
R2
C12
R1
L1
L2
C1
C2
R1
L1
2
2
Z 2 = R1 + jωL2 −
1
,
j ωC 2
Mutual impedance in these circuits
(c)
L2
L
R2
R1
1
C1
Z 12 = − j
C2 R
2
L
L
1
,
jωC1
C
(a)
C1
Z 1 = R1 + jωL1 −
R2
L
L1
Page 13
2
12
Now let’s to write mesh equations
for equivalent circuit of the
coupled networks:
C12
C2
(b)
(d)
Fig. 14. Tuned circuits with coupling: a-inductive, bconductive- autotrasformative, c- intrisinc capacitive and
c- exterior capacitive
Here Z 01 = Z 1 + Z 12 ,
1
.
ωC12
Z 01 I&1 − Z12 I&2 = V&1 ,
− Z I& + Z I& = 0.
12 1
02 2
Z 02 = Z 2 + Z 12 .
From these equation follows, that input current phasor could be given by
I&1 =
V&1 Z 02
V&1
=
.
Z 01 Z 02 − Z 122
Z 01 − Z 122 Z 02
This equation can be rearranged as follows:
I&1 =
V&1
,
Z 01 + ΔZ 1
Here ΔZ 1 = − Z 122 Z 02 . This impedance shows influence of output resonant circuit on first,
input, circuit and can be denoted as carried in impedance. Therefore, circuit input impedance
Z IN = Z 01 + ΔZ 1 .
Phasor current in the secondary mesh, which is right-hand side of the circuit in the Fig.15,
I&2 =
Z1
V1
Z
I1
12
I
Z2
2
Fig. 15. Equivalent circuit of the
coupled networks
V&1 ⋅ Z 12 Z 01
V&1 Z 12
=
Z 01 Z 02 − Z 122
Z 02 − Z 122 Z 01
If we shall denote − Z 122 Z 11 = ΔZ 2 , then output
current expression can be rearranged:
V& ⋅ Z Z
I&2 = 1 12 01 .
Z 02 + ΔZ 2
Here carried-in impedance ΔZ 2 shows influence of
Signals&circuits
RESONANT CIRCUITS
Page 14
primary input circuit also. We can see also, that current in the secondary mesh. is created by
voltage controlled voltage source V&1 ⋅ Z 12 Z 01 . Its value depends on mutual impedance Z 01 .
Thus, analysis of these expressions leads to the conclusion, that coupled resonant circuit could
be represented by another equivalent circuit,
Z 01
shown in the Fig16.
Z 02
From this circuit and previous equations follows,
that input voltage in the second circuit is Z C Z 11
Δ Z2
times less then circuit input voltage. We can see
Δ Z1
V
also, that coupled network input and output can
Z12
be taken under consideration as separate circuits.
V1
Z 01
Input impedance. Let’s consider coupled
network, which is shown in the Fig. 14a. This
circuit is often used as input circuit in various
Fig.16. Equivalent circuit of the coupledradio receivers or as output circuit in the
networks
amplifiers. Input impedance of this circuit may
be equated evaluating, that in carried-in impedance Z 12 = jX 12 :
Z IN = Z 01 +
X 122
.
Z 22
Impedances
Z 01 = Z 1 + Z 12 = R1 + jω ( L1 − M ) +
Z 02 = Z 2 + Z 12 = R2 + jω ( L2 − M ) +
1
1
+ jωM = R1 + jωL1 +
,
jωC1
jωC1
1
1
+ j ωM = R 2 + j ω L 2 +
.
j ωC 2
j ωC 2
These equations we can rearrange in the same manner as for separate resonant RLC-circuits:
Z 01 = R1 (1 + jξ 1 ),
Here ξn = 2Q n
Qn =
ω 0n L n
Rn
Z 02 = R2 (1 + jξ 2 ) .
1
ω − ω 0n
– resonant frequencies,
– general mistuning, ω 0 n =
ω 0n
L n ⋅C n
– quality factors, n=1 or 2.
So
Z in = R1 (1 + jξ1 ) +
= R1 + jR1ξ1 +
= R1 +
X 122
X 2 (1 − jξ 2 )
= R1 (1 + jξ1 ) + 12
=
R2 (1 + jξ 2 )
R2 (1 + ξ 2 )
X 122
X 122 ξ 2
−
=
j
R2 (1 + ξ 2 )
R2 (1 + ξ 2 )
X 122
+
R2 (1 + ξ 2 )
⎛
X 122 ξ 2 ⎞
⎟ = Rin + jX in .
j ⎜⎜ R1ξ1 − j
R2 (1 + ξ 2 ) ⎟⎠
⎝
Here Rin - input resistance and X in - input reactance of the coupled networks. Thus, from the
equation above follows, that
Signals&circuits
R in = R1 +
X
in
RESONANT CIRCUITS
Page 15
⎛
⎞
X 122
X 122
=
R
+
1
;
1⎜
2
2 ⎟
R 2 (1 + ξ )
⎝ R1R 2 (1 + ξ2 ) ⎠
= R1ξ1 +
⎛
⎞
X 122ξ2
X 122ξ2
=
R
−
ξ
.
1⎜ 1
2
2 ⎟
R 2 (1 + ξ2 )
R1R 2 (1 + ξ2 ) ⎠
⎝
If we shall denote
X 12
= A , then input resistance and input reactance can be rearranged
R 1R 2
and simplified:
⎛
A 2ξ2 ⎞
X in = R1 ⎜ ξ1 −
⎟.
1 + ξ22 ⎠
⎝
⎛
A2 ⎞
R in = R1 ⎜1 +
⎟;
⎝ 1 + ξ22 ⎠
Previous formula for input impedance becomes
⎛
⎛
A 2ξ 22 ⎞
A2 ⎞
⎟
⎜
⎟.
ξ
+
−
Z in = R1 ⎜⎜1 +
jR
1⎜ 1
2 ⎟
1 + ξ 22 ⎟⎠
⎝ 1+ ξ2 ⎠
⎝
Thus, input impedance depends on first circuit resistance, its general mistuning and has
additional component which depends on general mistuning of the second circuit and
coefficient A. Value A is called coupling power. If X 12 = ωM then
A=
ωM
R1 R2
.
When both circuits have the same resonant frequency, ω 01 = ω 02 = ω 0 , and general frequency
ω ≈ ω0
A2 =
ω 2M 2
R1 R2
≅
ω 0 L1ω 0 L2 M 2
R1 R2
L1 L2
=
ρ1 ρ 2
R1 R2
k 2.
M
– coupling coefficient.
L 1L 2
So, from equation above follows, that coupling power depends on coupling coefficient and
coupled circuits quality factors: A = k Q1Q 2 . When both circuits are the same, then
A = kQ .
Thus, as we can see from equations above, coupling power can be expressed in terms
of coupling coefficient k. Such expression is more convenient than expression in terms of
mutual or coupling impedances. In addition coupling coefficient k can be expressed as
geometrical average of coefficients of voltage transfer from first circuit to second k1 and from
second to first k2:
Here k =
k = k 1k 2 .
In the coupled networks where coupling is realized using inductive connections (Fig 14a)
Signals&circuits
RESONANT CIRCUITS
k1 =
M
,
L1
k2 =
M
,
L2
k=
Page 16
M
;
L1L2
When coupling is build using autotransformative-conductive connections (Fig.14b)
k1 =
L12
,
L1 + L12
L12
,
L 2 + L12
k2 =
k =
L12
;
( L1 + L12 )( L 2 + L12 )
For the networks with interior-capacitive coupling (Fig.14c)
k1 =
C1
,
C 1 + C 12
k2 =
C2
,
C 2 + C 12
k =
C 1C 2
;
(C 1 + C 12 )(C 2 + C 12 )
C2
,
C 2 + C 12
k =
C 1C 2
;
(C 1 + C 12 )(C 2 + C 12 )
And with exterior-capacitive coupling
k1 =
C1
,
C 1 + C 12
k2 =
Thus, then both circuits are build on components with the same nominals, coupling
coefficient can be expressed in terms of mutual reactance and networks characteristic
X 12
impedance: k ≅
.
ρ1 ⋅ ρ 2
Resonant in the circuit becomes when input reactance X in (ξ ) = 0 . When both coupled
networks are build on the same components,
⎛
A2 ⎞
X in (ξ ) = R1ξ1 ⎜1 −
⎟ = 0.
⎝ 1+ ξ2 ⎠
Thus, input impedance would be equal to zero at three general mistuning ξ meanings:
ξ ′ = 0;
ξ ′′ = A 2 − 1;
ξ ′′′ = - A 2 − 1.
In addition, two of them are possible only if A > 1 . When A ≤ 1 , network has only one
resonant frequency ω 0 . When A > 1 , coupled network has two additional resonant
frequencies:
⎛
⎝
1
2Q
⎞
ω
A 2 − 1⎟ = ω 0 + 0
⎠
2
k2−
1
,
Q2
⎛
⎝
1
2Q
⎞
ω
A 2 − 1⎟ = ω 0 − 0
⎠
2
k2−
1
.
Q2
ω 0′′ = ω 0 ⎜1 +
ω 0′′′= ω 0 ⎜1 −
Such conclusion is illustrated by curves presented the Fig.17. Here are shown input resistance
and reactance dependence on frequency. Indeed, when A>1, presented curves have three
extremums. Transition from 1 extreme to 3 becomes when A=1. Thus, coupling power
meaning A = 1 could be called as critical or boundary. Critical coupling coefficient k C = 1Q
corresponds to this coupling power.
Signals&circuits
RESONANT CIRCUITS
Page 17
Frequency responses. Let’s analyze coupled network with inductive coupling (Fig. 14a).
Voltage transfer function for this network can be given
by
R in , ohms
H ( jω ) =
X , ohms
V&2
I&2
Z 12
=
=
.
V&1
jωC 2V&11
jωC 2 Z 01 Z 02 − Z 122
(
When both networks are built on the components with
the same nominals – we can write:
C1 = C 2 = C , Z1 = Z 2 = Z = R(1 + jξ ), Z12 = jωM .
Therefore,
(a)
in
M
H ( jω ) =
ξ
Fig. 17. Coupled networks input
resistance (a) and input reactance
(b) dependence on general
mistuning and coupling power:
A=0.5 – dotted curves, A=1 –
continuous curves and A=1.5 –
dashed curves
2
⎡
⎛ ωM ⎞ ⎤
R 2 C ⎢1 + j 2ξ − ξ 2 + ⎜
⎟ ⎥
⎝ R ⎠ ⎥⎦
⎢⎣
ωM
.
1
1
≅
= Q , we
R
ωRC ω 0 RC
can rewrite voltage transfer function as follows:
Evaluating, that
(b)
)
=A and
H ( jω ) =
AQ
.
1 − ξ + A 2 + j 2ξ
2
Thus, frequency responses of the coupled networks
H (ω ) =
AQ
(1 + A
−ξ
)
;
+ 4ξ
- 2ξ
ϕ (ω ) = arctan
.
1 − ξ 2 + A2
2
2 2
2
Maximum meanings of magnitude response K (ω ) are the same as the functions
(
f (ξ ) = 1 + A 2 − ξ 2
)
2
+ 4ξ 2
extremes. Thus, finding derivatives from this function and equating them to zero, yields:
ξ1 = 0, ξ2 = A 2 − 1, ξ3 = − A 2 − 1.
As we can see, extremes can be observed at the same frequencies as resonant frequencies
determinated at input impedance analysis. Highest possible magnitude response meanings
depends on network quality factor and coupling power. When ξ = 0
H (ω 0 ) =
AQ
.
1 + A2
Signals&circuits
RESONANT CIRCUITS
Page 18
From this expression follows, that output voltage extreme could be reached when A = 1.
Highest possible value of the magnitude response
H max (ω 0 ) =
Q
.
2
Meanings of the magnitude response at frequencies ξ 2,3 = ± A 2 − 1 may be obtained
substituting these frequencies into the expression of the magnitude response:
H (ξ 2 ) = H (ξ 3 ) =
AQ
(1 + A
2
)
(
2
)
− A2 + 1 + 4 A2 − 1
=
Q
.
2
Thus, when A > 1, K (ξ 2 ) = K (ξ 3 ) = K max (ξ 1 ) .
Graphs of the magnitude responses of coupled networks are shown in the figure 18 as
example. Thus, as it is possible to see from these
graphs, pass-band of the coupled networks becomes
H( ω )
wider when coupling power coefficient becomes
Hmax
greater than 1. Unfortunately, hollow appears on the
top of the magnitude response. Some distortions of
the frequency responses are allowed. Let’s find
possible coupling power coefficient values with
(a)
respect to allowed distortions. At the beginning it is
necessary to equate magnitude response with respect
to
its
meaning
at
resonant
frequency
ξ = 0 ) of the coupled network. Such
(ω = ω0 ;
ξ
ϕ ,rad
proposition arises from comparing level of
magnitude response at boundary frequencies with
magnitude level at hollow bottom:
(b)
[1 + A
ξ
Fig.18. Magnitude (a) and phase (b)
responses of the coupled networks at
various coupling power: A=0.5 –
dotted curves; A=1 – continuous lines;
A=1.5 – dashed lines
H (ω ) =
AQ
2
− (ξ
)]
2
* 2
+ 4(ξ * )
=
2
AQ
.
1 + A2
Of course, first solution is ξ * = 0 . Another two
solutions are:
ξ * = ± 2 ( A 2 − 1) .
Assume that allowed hollow depth reaches
0.707Hmax. Thus,
A′ Q
Q
1
H max (ω 0 ) =
=
2
1 + ( A′ )
2
2 2
and allowed A ′ = 2 ± 1 . Thus, A1′ ≅ 2.41, A2′ ≅ 0.41. It is necessary to note once more,
that magnitude response has only one extreme when A1′ < 1 . Thus, coupling power it is
possible to increase until A=2.41. Now let’s consider, how much it is possible to expand passband of the coupled network in comparance with single resonant network. As you remember,
boundary frequencies of the single separate resonant contour are related with general
Signals&circuits
RESONANT CIRCUITS
Page 19
mistuning ξB=± 1 and bandwidth of the
resonant contour BW = 2ξ B . Pass-band of
the coupled network, when its coupling
power is critical (A=1), can be calculated
from the following conditions:
H( ω )
B
Hmax
0.707
H (ξ B ) =
ξ
Fig.19. Magnitude responses of various resonant
networks: separate single resonant contour –
dashed curve (BW=2ξ); coupled network with
critical coupling power – continuous curve (A=1,
BW=2.8ξ); coupled network with largest as
possible coupling power (A=2.41, BW=6.2ξ)
Q
ξ +4
4
B
=
Q
2 2
.
Thus, boundary general mistuning of the
coupled networks with critical coupling
power ξ B = ± 2 and pass-band is 1.4
times wider than pass-band of the
separate resonant network.
Pass-band of the coupled network, when
its coupling power is large as possible
(A=2.41), can be calculated from the
following conditions:
B
H (ξ B ) ≅
2.41Q
(6.81 − ξ ) + 4ξ
2 2
B
2
B
=
Q
2 2
.
Solution of this equation shows, that boundary general mistuning of the coupled network with
allowed largest coupling power ξ B ≅ ±3.1 and pass-band of such network is 2.2 times wider
than pass-band of the coupled network with critical coupling power and 3.1 times wider than
pass-bad of the separate resonant contour. These conclusions are illustrated by normalized
magnitude responses of coupled networks and resonant contour (Fig. 19).
Signals&circuits
RESONANT CIRCUITS
Page 20
Coupled networks tuning. Coupled networks usually are tuned with the aim to get required
frequency response or pass band. Networks could be tuned changing parameters of their
components or coupling impedance. There are some cases.
Current through second network is highest possible. Then reactance of the first network must
be equal to 0. This condition can be achieved only when self-reactance of the network is equal
to carried-in reactance:
X 01 = X 02
X 122
.
Z 022
This condition leads to I partial resonant.
Changing coupling impedance we can find maximal current through second circuit. This
optimal coupling impedance is obtained when
X 12 opt = Z 22
R1
.
R2
Current through second circuit
I 2 max =
V1
2 R1 R2
.
This is complicated resonant. Then we are searching optimal coupling impedance, we must
tune I circuit also, because carried impedance depends on coupling impedance.
Then we are tuning second circuit, we obtain second partial resonant. Its condition
X
22
= X 11
X 122
.
Z 112
Complicated resonant will be at optimal coupling
X 12 opt = Z 11
R2
R1
and maximal current through second circuit
I 2 max =
V1
2 R1 R2
is the same as obtained tuning I circuit.
Then we tuned I and II circuits – X 11 = X
optimal
X 12 opt = R1 ⋅ R 2 .
Then resonant is full, maximal current
22
= 0, Z 11 = R 1 , Z 22 = R 2 and coupling is
Signals&circuits
I 2 max =
V1
2 R1 R2
.
Current through I circuit is
I 1 opt =
V1
.
2R1
RESONANT CIRCUITS
Page 21