Op-Amp Transient Response and Voltage Saturation (Electrical Engineering) Solution

1. Domain and Vertical Asymptote Analysis

1. Name the Governing Formulas: The Rational Function Domain Restriction Rule and Vertical Asymptote Existence Criterion: (These principles are formalized in Vertical Asymptotes.)
2. Find Singularities: Set the denominator to zero:
   $$t^2 + 4t + 3 = 0 \implies (t + 1)(t + 3) = 0 \implies t = -1 \quad \text{and} \quad t = -3$$
   Algebraic Domain across $\mathbb{R}$:
   $$(-\infty, -3) \cup (-3, -1) \cup (-1, \infty)$$
3. Physical Domain Verification: The amplifier circuit is switched on at $t = 0$. On the operational physical domain $[0, \infty)$, both singular points ($t = -1\text{ ms}$ and $t = -3\text{ ms}$) lie strictly in negative time prior to energization. Thus, no vertical asymptotes or discontinuities exist during circuit operation; $V(t)$ is smooth and continuous for all $t \ge 0$.

2. Initial Voltage Condition ($y$-Intercept)

Evaluate $V(t)$ at energization time $t = 0$:

3. Horizontal Asymptote & Steady-State Saturation

1. Name the Governing Formula: The Horizontal Asymptote Degree Dominance Rule for Balanced Rational Functions:
   $$\lim_{t \to \infty} \frac{a_n t^n + \dots}{b_m t^m + \dots} = \frac{a_n}{b_m} \quad (\text{when } n = m)$$
   (This theorem is detailed in Horizontal Asymptotes.)
2. Compute Limit as $t \to \infty$:
   $$\lim_{t \to \infty} V(t) = \lim_{t \to \infty} \frac{15t^2}{t^2 + 4t + 3} = \lim_{t \to \infty} \frac{15}{1 + \frac{4}{t} + \frac{3}{t^2}} = \frac{15}{1 + 0 + 0} = 15\text{ Volts}$$
3. Engineering Interpretation: The horizontal asymptote is $V = 15\text{ V}$. In physical circuit hardware, this corresponds to the positive DC power supply voltage rail ($+V_{CC} = +15\text{ V}$), setting the upper physical ceiling for the operational amplifier output.

4. Monotonicity via First Derivative Sign Analysis

1. Name the Governing Formula: The Quotient Rule for Differentiation:
   $$\left[\frac{u}{v}\right]' = \frac{u'v - uv'}{v^2}$$
2. Differentiate $V(t)$: Here $u(t) = 15t^2 \implies u'(t) = 30t$, and $v(t) = t^2 + 4t + 3 \implies v'(t) = 2t + 4$.
   $$V'(t) = \frac{(30t)(t^2 + 4t + 3) - (15t^2)(2t + 4)}{(t^2 + 4t + 3)^2}$$
   Expand the numerator:
   $$\text{Num} = (30t^3 + 120t^2 + 90t) - (30t^3 + 60t^2) = 60t^2 + 90t = 30t(2t + 3)$$
   Thus:
   $$V'(t) = \frac{30t(2t + 3)}{(t^2 + 4t + 3)^2}$$
3. Sign Analysis on $[0, \infty)$:
   • At $t = 0$: $V'(0) = 0$ (the response starts with zero initial slew rate).
   • For all $t > 0$: $30t > 0$, $2t + 3 > 0$, and the denominator squared is strictly positive.
   • Therefore, $V'(t) > 0$ for all $t > 0$. By the Monotonicity Theorem, the output voltage is strictly increasing, guaranteeing monotonic transient charging with zero parasitic ringing or overshoot.

5. Amplifier 90% Settling Time

Target settling threshold: