CHAPTER 3. FILTER CIRCUITS58->Vin->VoutRLCFigure 3.12: Circuit with components in parallel at the output.3.7Ampliﬁer ModelEnough of ﬁlters. Lets now look at the simple ampliﬁer model in ﬁgure 3.14vout(jω)= A(jω)vin(jω),(3.74)vout(s)= A(s)vin(s).(3.75)Notice that vout is with respect to ground while vin is a voltage diﬀerence. For a typicaloperational ampliﬁer|H|≈ 106 at ω =0. As ω→∞, A(ω) < A(ω = 0) due to internalcapacitance, ie. the ampliﬁer behaves like a low-pass ﬁlter.A(s)= A0Hlow(s),(3.76)where Hlow is the transfer function of a low-pass ﬁlter.3.7.1One-, Two- and Three-Pole Ampliﬁer ModelsThe simplest ampliﬁer hasA1(s)=A01+ s/sa,(3.77)where sa is a real positive number (corner frequency). There is a single pole at−sa on thenegative real axis. This means that the impulse response will decay exponentially withoutringing.If we cascade multiple (three) single-pole ampliﬁers togetherA3(s)=A0(1 + s/sa)(1 + s/sb)(1 + s/sc).(3.78)There will still be no oscillation to an impulse.