## 1. Definition

The inverting amplifier assembly has a differential linear amplifier and two resistors

## 2. Simulation

### 2.a. Transfer function

The amplifier is modeled as a linear quadrupole, with an open loop gain of the following form: g (s) = g01 + sw0

We use the Mathematica module:

Get["../simulin/simulineaire.m"];

First, the circuit is defined in the following function:

inverseur[r1_,r2_,g0_,w0_]:=Module[{n,A,B,g}, n=4; A=Table[0,{n},{n}]; B=Table[0,{n}]; A=ajouterResistance[A,1,2,r1]; A=ajouterResistance[A,2,3,r2]; g=g0/(1+s/w0); {A,B}=ajouterSourceTensionSTCT[A,B,3,4,4,2,g]; {A,B}=ajouterMasse[A,B,4]; {A,B}=definirEntree[A,B,1]; Return[{A,B}]; ]

Here is the transfer function

{A,B}=inverseur[r1,r2,g0,w0]; h=transfert[A,B,3]/.s->I*omega

-g0 r2 w0g0 r1 w0+i ω r1+i ω r2+r1 w0+r2 w0

### 2.b. Bode diagram

The open loop gain is fixed at g0 = 105 and the cutoff pulse w0 = 20π (10 Hz). The resistors are chosen so as to obtain a gain of 10 at low frequency:

{A,B}=inverseur[10^3,10^4,10^5,20*N[Pi]]; h=transfert[A,B,3];

bodeGain[h,0,6,0,60]

Let’s increase the gain by a factor of 100:

{A,B}=inverseur[10^3,10^6,10^5,20*N[Pi]]; h=transfert[A,B,3];

bodeGain[h,0,6,0,60]

bodePhase[h,0,6]

There is a reduction in the bandwidth of 2 decades.