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142 2. Tutorial
In[33]:= voutsimp3 = Together[
V$RL /. First[Solve[approxceamp3, V$RL]]]
Out[33]=
C1 C2 R1 R2 RL RC RE s 2 RC Rpi$Q1 s 2 gm$Q1 RC Ro$Q1 Rpi$Q1 s 2
V1 R1 RC RE R2 RC RE R1 RE Ro$Q1 R2 RE Ro$Q1
R1 RC Rpi$Q1 R2 RC Rpi$Q1 R1 Ro$Q1 Rpi$Q1 R2 Ro$Q1 Rpi$Q1
gm$Q1 R1 RE Ro$Q1 Rpi$Q1 gm$Q1 R2 RE Ro$Q1 Rpi$Q1
C1 R1 R2 RC RE s C2 R1 RC RE RL s C2 R2 RC RE RL s
C1 R1 R2 RE Ro$Q1 s Cbc$Q1 R1 R2 RE Ro$Q1 s C2 R1 RC RE Ro$Q1 s
C2 R2 RC RE Ro$Q1 s C2 R1 RE RL Ro$Q1 s 22
C2 Cbc$Q1 R1 R2 RC RE Ro$Q1 s 2 C1 C2 R1 R2 RE RL Ro$Q1 s 2
C2 Cbc$Q1 R1 R2 RE RL Ro$Q1 s 2 C2 Cbe$Q1 R1 R2 RC RE Rpi$Q1 s 2
C1 C2 R1 R2 RC RL Rpi$Q1 s 2 C2 Cbe$Q1 R1 R2 RC RL Rpi$Q1 s 2
C2 Cbe$Q1 R1 R2 RE RL Rpi$Q1 s 2 C1 C2 R1 R2 RC Ro$Q1 Rpi$Q1 s 2
C2 Cbc$Q1 R1 R2 RC Ro$Q1 Rpi$Q1 s 2 C2 Cbe$Q1 R1 R2 RC
Ro$Q1 Rpi$Q1 s 2 C1 C2 gm$Q1 R1 R2 RC RE Ro$Q1 Rpi$Q1 s 2
C2 Cbc$Q1 gm$Q1 R1 R2 RC RE Ro$Q1 Rpi$Q1 s 2 C1 C2 R1 R2 RL
Ro$Q1 Rpi$Q1 s 2 C2 Cbc$Q1 R1 R2 RL Ro$Q1 Rpi$Q1 s 2
C2 Cbe$Q1 R1 R2 RL Ro$Q1 Rpi$Q1 s 2 C2 Cbc$Q1 gm$Q1 R1 R2 RC RL
Ro$Q1 Rpi$Q1 s 2 C1 C2 gm$Q1 R1 R2 RE RL Ro$Q1 Rpi$Q1 s 2
C2 Cbc$Q1 gm$Q1 R1 R2 RE RL Ro$Q1 Rpi$Q1 s 2
As compared to the result of our previous equation-based simplification this transfer function is quite
lengthy again. Still, approximating the matrix has reduced the degree of the solution. Therefore,
we can apply solution-based approximation to compute a simplified transfer function which is more
compact than our SAG-only solution and valid over a larger frequency range than our first SBG
result.
In[34]:= voutSBGSAG = ApproximateTransferFunction[voutsimp3,
s, dpceamp, 0.1]
Out[34]=
C1 C2 gm$Q1 R1 R2 RC RL Ro$Q1 Rpi$Q1 s 2 V1
gm$Q1 R1 RE Ro$Q1 Rpi$Q1 gm$Q1 R2 RE Ro$Q1 Rpi$Q1
C1 gm$Q1 R1 R2 RE Ro$Q1 Rpi$Q1 C2 gm$Q1 R1 RE RL Ro$Q1 Rpi$Q1
C2 gm$Q1 R2 RE RL Ro$Q1 Rpi$Q1
s C1 C2 gm$Q1 R1 R2 RE RL Ro$Q1 Rpi$Q1 s 2
The common factor which appears in the coefficients of the numerator and the denominator of this
transfer function can be cancelled by algebraic simplification:
In[35]:= voutsimp3s = Simplify[voutSBGSAG]
C1 C2 R1 R2 RC RL s
Out[35]=
2 V1
RE R1 R2 C1 R1 R2 s 1 C2 RL s
The above result is very useful because it allows for reading off approximate symbolic expressions
for the poles. To calculate these explicitly we must solve the denominator of the transfer function
for s:
In[36]:= poles = Solve[Denominator[voutsimp3s] == 0, s]
R1 R2
Out[36]= s
C1 R1 R2 , s 1
C2 RL
From our SAG-only approximation (voutsimp) alone we cannot determine the poles so easily because
the denominator is a polynomial of degree . On the other hand, the SBG-only approximation is