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CHAPTER 4 - Impedance Matching 180 Figure 4.3-10 Schematic corresponding to frequency response in Figure 4.3-9 To work out the frequency response we looked at the voltage across the load resistor (VP2) and the voltage across the generator (VP1), we then calculated the ratio of these voltages and converted it into dB. ( ) ( ) At this point we must specify what we mean by attenuation and gain. In this case, since we have only passive elements present, the gain of the filter is mostly negative and may only reach zero dB as a maximum. In actual facts, the signal is attenuated across the majority of the frequency band of interest. However, attenuation is defined as a positive value which is the ratio of input over output voltage hence we cannot use this term here! If we look at the ratio shown above, it is gain, albeit negative gain, that we are looking at. It is matter of nomenclature but we must avoid confusion! Note how we have used a very high value for our load resistor R2=100 kΩ to simulate a high impedance load. This was done to enable us to assess the impact of the source impedance alone by minimising the effect of a load. If we now increase the source impedance R1 to 1000Ω , we can see how the Q of the circuit changes dramatically as shown in Figure 4.3-11. Our circuit has become a lot more frequency-selective! Neither of the curves shown in Figure 4.3-11 however addresses the impact of a load impedance on the circuit since our R2 at the moment is so large that it does not allow a significant amount of current though it. Figure 4.3-11 Increasing the source impedance, increases the Q ACVS ID=V1 Mag=1 V Ang=0 Deg Offset=0 V DCVal=0 V RES ID=R1 R=50 Ohm IND ID=L1 L=50 nH CAP ID=C1 C=25 pF RES ID=R2 R=1e5 Ohm M_PROBE ID=VP1 M_PROBE ID=VP2 Xo Xn . . . SWPVAR ID=SWP1 VarName="RS" Values={ 50,1000 } UnitType=None RS=50 Gain (dB) Q=1.1 Q=22.4 Conquer Radio Frequency 180 www.cadence.com/go/awr

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