Parametric Gain Curve With a Required Minimum
A bench engineer is calibrating a voltage-gain stage whose gain is modeled by G(v) = (v² + a v + 25)/(v + 1) for input v > 0, with v in volts and G dimensionless. The trim parameter a is fixed during manufacture. The specification requires the gain to reach a local minimum at v = 3.0 V; the engineer must determine a, verify that this stationary point is the global minimum on v > 0, and then find the input range for which the gain stays at or below 20. A colleague argues that because the expression contains a square, any positive a will give a minimum at 3.0 V and that the gain can be kept below 20 for all larger inputs. The bench supply limits v to 12 V, so the answer must be checked within that interval.
1. Determine the value of a that places a stationary point at v = 3.0 V, and establish the corresponding minimum gain.
Read the worked answer
Differentiate using the quotient rule:
G′(v) = [(2v + a)(v + 1) − (v² + a v + 25)]/(v + 1)² = [v² + 2v + a − 25]/(v + 1)².
At a stationary point the numerator is zero. Setting v = 3 gives 9 + 6 + a − 25 = 0, so a = 10. Then
G(v) = (v² + 10v + 25)/(v + 1), and G(3) = (9 + 30 + 25)/4 = 16.
The derivative numerator becomes v² + 2v − 15 = (v + 5)(v − 3). For 0 < v < 3 it is negative, and for v > 3 it is positive, so G decreases then increases and v = 3 gives the global minimum on v > 0. The minimum gain is 16, not 0; the square in the expression does not by itself guarantee either the location or the value of the minimum.
2. Solve G(v) ≤ 20 for v > 0 and test the colleague’s claim that the gain remains below 20 for all larger inputs.
Read the worked answer
Because v + 1 > 0, multiply without reversing the inequality:
v² + 10v + 25 ≤ 20(v + 1) v² − 10v + 5 ≤ 0.
The roots are v = [10 ± √(100 − 20)]/2 = 5 ± 2√5, approximately 0.528 and 9.472. The quadratic is negative between the roots, so the gain is at most 20 on the interval [5 − 2√5, 5 + 2√5], or about [0.53 V, 9.47 V].
The colleague is wrong about “all larger inputs”: at v = 12 V, G = (144 + 120 + 25)/13 = 289/13 ≈ 22.2, which exceeds 20. The gain eventually behaves like v + 9, so it grows without bound. The valid range within the bench limit is about 0.53 V to 9.47 V.
3. If the input is restricted to v ≥ 6.0 V, where does the minimum gain occur and what is its value?
Read the worked answer
For v > 3, G′(v) = (v + 5)(v − 3)/(v + 1)² is positive, so G is strictly increasing on [6, 12]. Therefore the smallest gain on this restricted interval occurs at the left endpoint v = 6.0 V:
G(6) = (36 + 60 + 25)/(6 + 1) = 121/7 ≈ 17.29.
The minimum is about 17.3 at 6.0 V, not at 3.0 V, because 3.0 V is outside the restricted interval. The maximum on the restricted interval would occur at v = 12 V and equals about 22.2.