Encontrar la magnitud de la función de transferencia

Bien, tenemos el siguiente circuito:

^{simular este circuito : esquema creado con CircuitLab}

Al analizar un transistor necesitamos usar las siguientes relaciones :

• $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
• ganancia de transistor$\beta$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$

Usando KCL, podemos escribir:

$$\begin{cases} \text{I}_\text{y}=\text{I}_\text{c}+\text{I}_4\\ \\ \text{I}_\text{t}=\text{I}_3+\text{I}_4\\ \\ \text{I}_\text{x}=\text{I}_2+\text{I}_3\\ \\ \text{I}_8=\text{I}_5+\text{I}_7\\ \\ \text{I}_7=\text{I}_6+\text{I}_\text{y}\\ \\ \text{I}_1=\text{I}_5+\text{I}_6\\ \\ \text{I}_\text{x}=\text{I}_1+\text{I}_2\\ \\ \text{I}_8=\text{I}_\text{t}+\text{I}_\text{c}\\ \\ \beta=\frac{\text{I}_\text{c}}{\text{I}_\text{t}} \end{cases}\tag3$$

Usando KVL, podemos escribir:

$$\begin{cases} \text{I}_\text{x}=\frac{\text{V}_\text{x}-\text{V}_5}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_5}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_5-\text{V}_4}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{y}-\text{V}_4}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_1}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_1-\text{V}_2}{\text{R}_6}\\ \\ \text{I}_7=\frac{\text{V}_2}{\text{R}_7}\\ \\ \text{V}_4-\text{V}_1=\alpha \end{cases}\tag4$$

Ahora, puedes resolver tu problema usando:

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, \[Beta] == Ic/It, 
   I8 == It + Ic, Ix == (Vx - V5)/R1, I2 == (V5)/R2, 
   I3 == (V5 - V4)/R3, I4 == (Vy - V4)/R4, I5 == (V1)/R5, 
   I7 == (V1 - V2)/R6, I7 == V2/R7, V4 - V1 == \[Alpha]}, {Ix, I2, I3,
    It, I1, I4, Ic, I8, I5, I7, I6, Iy, V5, V4, V3, V1, V2}]]

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EDITAR

Asumamos$\alpha=\frac{7}{10}$,$\beta=100$y$\text{V}_\text{x}=10$. En el dominio s (complejo) necesitamos resolver el siguiente sistema de ecuaciones (utilicé Mathematica 12.0 para resolverlo):

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, I8 == It + Ic, 
   100 == Ic/It, Ix == ((10/s) - V5)/600, 
   I2 == V5/(((((s*20*10^(-6)))*((1/(s*2*10^(-9)))))/(((1/(s*2*10^(-9)\
))) + ((s*20*10^(-6)))))), I3 == (V5 - V4)/(1/(s*1*10^(-9))), 
   I4 == ((18/s) - V4)/100, I5 == V1/2000, 
   I7 == (V1 - V2)/((1/(s*100*10^(-9)))), I7 == V2/100, 
   V4 - V1 == (7/10)/s}, {V1, V2, V4, V5, Ic, It, Ix, Iy, I1, I2, I3, 
   I4, I5, I6, I7, I8}]]

Ahora, usé la transformada inversa de Laplace para trazar el voltaje$\text{V}_2\left(t\right)$:

Dónde:

• $$\lim_{t\to0}\text{V}_2\left(t\right)=-\frac{7}{10}\tag5$$
• $$\lim_{t\to\infty}\text{V}_2\left(t\right)=0\tag6$$

El máximo se produce cuando$t\approx1.20098\space\mu\text{s}$y entonces es igual a:

$$\hat{\text{V}}_2\left(1.20098\right)\approx16.6022\space\text{V}\tag7$$