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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