Automated Audio Amplifier Loop Gain (Electrical & Computer Engineering) Solution

In professional audio electronics design, an automated gain control circuit stabilizes output volume using an active feedback amplifier loop. The loop amplification gain factor $G$ (dimensionless ratio) depends on an input control bias voltage $v$ (in Volts, $\text{V}$) and a digital potentiometer feedback resistance $r$ (in kilo-ohms, $\text{k}\Omega$, for $r > 0$):

1. Compute the first-order partial derivatives $G_v$ and $G_r$.
2. Compute the second-order partial derivatives ($G_{vv}, G_{rr}, G_{vr}, G_{rv}$).
3. Verify Clairaut's Theorem for the mixed partials $G_{vr}$ and $G_{rv}$.
4. Construct the $2 \times 2$ Hessian matrix $\mathbf{H}_G(v, r)$ and evaluate it at the reference bias operating point $(v, r) = (2\text{ V}, e\text{ k}\Omega)$.
5. Calculate the Hessian determinant $D = \det(\mathbf{H}_G)$ at $(2, e)$ and determine whether the gain surface exhibits a local extremum or a saddle behavior at this operating point.

Step-by-Step Solution:

1. Compute First-Order Partial Derivatives: State the First-Order Partial Derivative Formulas (see Partial Derivatives):

   $$G_v = \frac{\partial}{\partial v}[v^3 \ln(r)] = 3v^2 \ln(r)$$
   $$G_r = \frac{\partial}{\partial r}[v^3 \ln(r)] = v^3 \left(\frac{1}{r}\right) = \frac{v^3}{r}$$

2. Compute Second-Order Partial Derivatives: State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives):

   • Pure Partials:
     $$G_{vv} = \frac{\partial}{\partial v}[3v^2 \ln(r)] = 6v \ln(r)$$
     $$G_{rr} = \frac{\partial}{\partial r}\left[v^3 r^{-1}\right] = v^3 (-r^{-2}) = -\frac{v^3}{r^2}$$
   • Mixed Partials:
     $$G_{vr} = \frac{\partial}{\partial r}[3v^2 \ln(r)] = 3v^2 \left(\frac{1}{r}\right) = \frac{3v^2}{r}$$
     $$G_{rv} = \frac{\partial}{\partial v}\left[\frac{v^3}{r}\right] = \frac{3v^2}{r}$$

3. Verify Clairaut's Theorem:

   $$G_{vr} = G_{rv} = \frac{3v^2}{r}$$
   For all $r > 0$, the function $G(v, r)$ is infinitely differentiable ($C^\infty$), confirming Clairaut's Theorem.

4. Construct and Evaluate Hessian Matrix at $(v, r) = (2, e)$:

   $$\mathbf{H}_G(v, r) = \begin{bmatrix} G_{vv} & G_{vr} \\ G_{rv} & G_{rr} \end{bmatrix} = \begin{bmatrix} 6v \ln(r) & \frac{3v^2}{r} \\ \frac{3v^2}{r} & -\frac{v^3}{r^2} \end{bmatrix}$$
   • Since $\ln(e) = 1$:
     • $G_{vv}(2, e) = 6(2)(1) = 12$
     • $G_{rr}(2, e) = -\frac{2^3}{e^2} = -\frac{8}{e^2}$
     • $G_{vr}(2, e) = \frac{3(2^2)}{e} = \frac{12}{e}$
       $$\mathbf{H}_G(2, e) = \begin{bmatrix} 12 & \frac{12}{e} \\ \frac{12}{e} & -\frac{8}{e^2} \end{bmatrix}$$

5. Compute Hessian Determinant and Classify Geometry:

   $$D = \det(\mathbf{H}_G) = G_{vv} G_{rr} - (G_{vr})^2 = (12)\left(-\frac{8}{e^2}\right) - \left(\frac{12}{e}\right)^2$$
   $$D = -\frac{96}{e^2} - \frac{144}{e^2} = -\frac{240}{e^2} \approx -\frac{240}{7.389} \approx -32.48$$
   Because $D < 0$, the gain surface possesses negative Gaussian curvature, confirming that the operating point $(2, e)$ is on a hyperbolic saddle geometry. In feedback control loop design, negative Hessian determinants indicate parameter trade-offs: increasing voltage boosts gain while increasing resistance suppresses gain, requiring active closed-loop stabilization.