Automated Audio AGC Amplifier Loop Stability (Electrical & Computer Engineering) Solution

An audio electronics hardware engineer in the UNLV Entertainment Engineering & Design program is tuning an automated gain control (AGC) feedback pre-amplifier circuit for live stage microphone feeds. The small-signal total harmonic distortion energy $E(v, r)$ (in microjoules, $\mu\text{J}$) across the feedback loop depends on the input reference bias voltage $v$ (in Volts) and the digital feedback resistance parameter $r$ (in kilo-ohms).

Circuit network analysis models the distortion energy surface over the operating domain by:

1. Formulate the first-order partial derivatives $\frac{\partial E}{\partial v}$ and $\frac{\partial E}{\partial r}$ and assemble the gradient vector $\nabla E(v, r)$.
2. Set up and solve the simultaneous linear system $\nabla E(v, r) = \mathbf{0}$ to find all quiescent operating bias points $(v_0, r_0)$.
3. Compute the second-order partial derivatives $E_{vv}$, $E_{rr}$, and the cross-derivative $E_{vr}$.
4. Calculate the Hessian discriminant $D = E_{vv}E_{rr} - (E_{vr})^2$ and apply the Second Derivative Test to classify the bias point.
5. In audio hardware design, determine whether this circuit tuning configuration minimizes signal distortion or risks unstable oscillation. State the minimum distortion energy $E_{\min}$.

Step-by-Step Solution:

1. Formulate the Gradient Vector: State the Two-Variable Gradient Formula (defined in The Gradient Vector):

   $$\nabla E(v, r) = \left\langle \frac{\partial E}{\partial v}, \frac{\partial E}{\partial r} \right\rangle = \frac{\partial E}{\partial v}\hat{i} + \frac{\partial E}{\partial r}\hat{j}$$
   Compute the partial derivatives:
   $$\frac{\partial E}{\partial v} = \frac{\partial}{\partial v}(v^2 - 4vr + 5r^2 - 6v + 4) = 2v - 4r - 6$$
   $$\frac{\partial E}{\partial r} = \frac{\partial}{\partial r}(v^2 - 4vr + 5r^2 - 6v + 4) = -4v + 10r$$
   The evaluated gradient is:
   $$\nabla E(v, r) = \langle 2v - 4r - 6, \, -4v + 10r \rangle = (2v - 4r - 6)\hat{i} + (-4v + 10r)\hat{j}$$

2. Locate the Quiescent Bias Point: State the Critical Point Condition (defined in Critical Points in Multivariable Functions):

   $$\nabla E(v, r) = \mathbf{0} \implies \begin{cases} 2v - 4r - 6 = 0 \\ -4v + 10r = 0 \end{cases}$$
   From the second equation:
   $$10r = 4v \implies v = \frac{10}{4}r = 2.5r$$
   Substitute $v = 2.5r$ into the first equation:
   $$2(2.5r) - 4r - 6 = 0 \implies 5r - 4r = 6 \implies r = 6\text{ k}\Omega$$
   Now calculate $v$:
   $$v = 2.5(6) = 15\text{ V}$$
   The single quiescent critical bias point is $(v_0, r_0) = (15, 6)$.

3. Compute Second-Order Partial Derivatives: State the Second-Order Partial Derivatives (defined in Second-Order Partial Derivatives):

   $$E_{vv} = \frac{\partial}{\partial v}(2v - 4r - 6) = 2$$
   $$E_{rr} = \frac{\partial}{\partial r}(-4v + 10r) = 10$$
   $$E_{vr} = \frac{\partial}{\partial r}(2v - 4r - 6) = -4$$

4. Calculate Hessian Discriminant and Classify: State the Hessian Discriminant Formula (defined in Second Derivative Test for Multivariable Functions):

   $$D = E_{vv}E_{rr} - (E_{vr})^2$$
   Substitute the second derivatives:
   $$D = (2)(10) - (-4)^2 = 20 - 16 = 4$$
   Evaluate criteria:
   • $D = 4 > 0$
   • $E_{vv} = 2 > 0$ Since $D > 0$ and $E_{vv} > 0$, the bias point $(15\text{ V}, 6\text{ k}\Omega)$ is a local minimum.

5. Engineering Evaluation of Stability and Distortion: Because the critical point is a true local minimum with positive curvature in all parameter directions ($D > 0, E_{vv} > 0$), small drift in battery supply voltage $v$ or resistor thermal tolerance $r$ causes distortion to rise symmetrically rather than inducing divergent feedback runaway. The loop is stably tuned.

   Substitute $(15, 6)$ into the energy function:

   $$E(15, 6) = (15)^2 - 4(15)(6) + 5(6)^2 - 6(15) + 4$$
   $$E(15, 6) = 225 - 360 + 5(36) - 90 + 4 = 225 - 360 + 180 - 90 + 4 = -41\ \mu\text{J}\ (\text{relative to reference offset})$$
   The circuit achieves minimal harmonic distortion at $v = 15\text{ V}$ and $r = 6\text{ k}\Omega$.