Synthesis procedure for microwave resonant circuits.
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1 Synthesis procedure for microwave resonant circuits. The design equations of a series ideal circuit can be derived if it is known a value for the reflection coefficient at a frequency value note equal at the resonant frequency for the circuit. At resonant frequency value in fact it is visible only the purely resistive value for the resonator impedance. Now we suppose to synthesize a circuit with series resonance by knowing a value for the reflection coefficient Γ measured at the frequency ω with characteristic impedance Z. It is easy to derive the expression of the impedance. This value for definition is normalized at the characteristic impedance Z value, compared to the reflection coefficient is measured. + Γ z = r ± jx = Γ We have now the impedance value that represent the circuit once the impedance value is multiplied for Z. Z = Z ( r ± jx) = ± jx The sign of reactive part tell us if the circuit behaviour is inductive or. We know that a reactive positive part indicates an inductive behaviour whereas a negative a capacitive behaviour. An inductive imaginary part implies a synthesis procedure that impose a frequency of Γ greater compared to the resonant frequency of the circuit. This for the fact that the reflection coefficient of a series resonant circuit parameterized in frequency moves it self in clockwise sense as the frequency value increase. If the we measure the Γ in the negative Smith chart part then the frequency will be smaller compared to the resonant frequency.. In every case the loss can be modelled through a series resistance. The synthesis of imaginary part is a little bit more complex; in fact it cannot be done without introduce the resonant frequency and the unloaded quality factor (ω r e il Q U ). Therefore the following two equation system must be solved. ω = X ω ω = () r ω r Q = A Numerical example of the synthesis procedure If we wont to synthesize a resonant circuit with frequency ω r =5GHz with a series kind resonant circuit. This circuit would show at the frequency of ω=3.5ghz the following reflection coefficient Γ =.88 = Γ = 48 Γ
2 3 Since the reflection coefficient measured to the ω=3.5ghz, frequency and since the resonant frequency is ω r =5GHz; we have Γ π Γ π Γ = Γ cos + j sin 8 8 =.7463 j.4663 From which we derive Z + Γ = Z = j 4.74 Γ And then = Ω We must obtain now the pair value that satisfy the constraint on the Γ at the measured frequency and the resonant frequency as well. Then it is necessary to fix the distance between the measured frequency and the resonant frequency. A meter of this distance is the unloaded quality factor Q U. It is helpful to note that a high value of Q U means a great displacement of Γ for small frequency values. = ( π 5 ) 4.74 = π pf Then we obtain ( 5 ) ( π 5 ) = 63pH π π 3.5 Q U 63 π 5 = = Since exist an infinite number of couples that can satisfy the constraint in resonant terms, but for a given value and by fixed the technology the value is fixed as well. We can simulate the system with ADS and verify then the synthesis procedure.
3 4 S-PAAMETES S_Param SP Start=. GHz Stop=. GHz Step=. GHz =3.457 Ohm =63 ph = =.6 pf Term Term Num= Z=5 Ohm Eqn A=[-::.::] Eqn Eye=(-abs(A)-j*A)/(abs(A)++j*(A)) 5.GHz S(,)=.87 / impedance = Z * (.6 - j.) 5.4GHz S(,)=.87 / 73.7 impedance = Z * (.6 + j.5) 4.5GHz S(,)=.87 / -7.3 impedance = Z * (.6 - j.84) Eqn QU=indep()/(indep()-indep()) freq QU <invalid S(,) m m m 5.GHz S(,)=.87 / impedance = Z * (.6 - j.) m 3.5GHz S(,)=.88 / impedance = Z * (.6 - j.87) Eye S(,) freq (.GHz to.ghz) freq (.GHz to.ghz) indep(eye) (. to 4.) Figure
4 5 4 m6 5.GHz imag(z(,))=-.77 Eqn dx=diff(imag(z(,))).e-7 imag(z(,)) m freq, GHz dx 8.E-8 6.E-8 4.E-8.E freq, GHz.3.5 mag(y(,)) freq, GHz Figure At least it is possible to check the existence of complex poles pair, and use the simulator to check how the system answer to a step impulse. In the time domain the frequency of the oscillator is the same of resonance of the system namely 5GHz. In our example we have derived that = 3.457Ω = 63pH =.6 pf 4 = 3.8 < Therefore the condition is satisfied as was appear by the quality factor Q U.
5 6 TANSIENT Tran Tran StopTime= nsec MaxTimeStep=. psec Vout =3.457 Ohm =63 ph = =.6 pf Vin VtStep S Vlow= V t Vhigh= V Delay= nsec ise= psec Vout, mv m6 time= 46.37psec m7 time= 46.4psec Vout=54.3mV Vout=88.4mV m m time, nsec Figure 3 f = = = 5GHz T The parallel resonant circuit. For the parallel resonant circuit the procedure is the same that I ve already used for series. The resonant reflection coefficient trend parameterized with frequency is displaced ever in clockwise sense but the parallel resonant frequency is indicated near to the zero value of phase for the reflection coefficient (remember that the phase assumes 8 of value for the series resonant frequency) therefore in different resonant frequency, an inductive behaviour indicate a frequency smallest compared to the resonant, and a capacitive behaviour a greater frequency.
6 7 Im( Γ) ω e( Γ) ω r ω Figure 4 A numerical example We consider the parallel circuit of the previously example.. ω r =5GHz At the ω=3.5ghz the circuit would show Γ =.88 = Γ = 48 Γ Γ Y = Y =.6 j.66 + Γ The design equation in this case are G =. 6[ S] ω = B ω (6) ωr = Q = ωr = = 6. 5Ω.6 The imaginary part can be founded by solving the system
7 8 = ( π 5 ).66 = π ph ( 5 ) ( π 5 ) =. 8 pf π Q = = U ω r π 3.5 S-PAAMETES S_Param SP Start=. GHz Stop=. GHz Step=. GHz =6.5 Ohm =35 ph = =.8 pf Term Term Num= Z=5 Ohm Figure 5 EqnA= [-::.::] Eqn Eye=(-abs(A)-j*A)/(abs(A)++j*(A)) Eqn QU=(indep())/(indep()-indep()) freq QU <invalid> S(,) m m 3.5GHz S(,)=.88 / impedance = Z * (.6 + j.85) m 5.6GHz S(,)=. / impedance = Z * (.4 - j.33) m Eye S(,) 4.57GHz S(,)=.4 /.7 impedance = Z * (.6 + j.65) 5.GHz S(,)=. /.867 impedance = Z * (.5 + j.) 5.47GHz S(,)=.43 / -.4 impedance = Z * (.66 - j.65) freq (.GHz to.ghz) freq (.GHz to.ghz) indep(eye) (. to 4.) Figure 6 To improve the Quality factor of this circuit t is necessary to reduce the resonant loss namely to model the parallel resistance with a high value resistance.
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