Signal Cutoff Gain (Electrical & Computer Engineering) Solution

Part 1: Why direct evaluation fails

Direct substitution of the target frequency $f = 4\text{ kHz}$ into the gain equation yields:

G(4) = \frac{4^2 - 16}{4 - 4} = \frac{16 - 16}{0} = \frac{0}{0}

In mathematics, division by zero is undefined, and

\frac{0}{0}

represents an indeterminate form. Physically, this means the filter model cannot compute the voltage gain at that exact transition frequency without using limits.

Part 2: Numerical evaluation near the cutoff frequency

We simplify the gain equation for all frequencies $f \neq 4\text{ kHz}$ by factoring the numerator:

G(f) = \frac{(f - 4)(f + 4)}{f - 4} = f + 4 \quad (\text{for } f \neq 4)

We evaluate the gain at frequencies arbitrarily close to $4\text{ kHz}$ from both below and above:

Approaching from the left ( $f \to 4^-$ ):
- $G(3.9) = 3.9 + 4 = 7.9$
- $G(3.99) = 3.99 + 4 = 7.99$
- $G(3.999) = 3.999 + 4 = 7.999$
Approaching from the right ( $f \to 4^+$ ):
- $G(4.1) = 4.1 + 4 = 8.1$
- $G(4.01) = 4.01 + 4 = 8.01$
- $G(4.001) = 4.001 + 4 = 8.001$

Summary table of gain values:

Frequency $f$ (kHz)	$3.9$	$3.99$	$3.999$	$4$	$4.001$	$4.01$	$4.1$
Gain $G(f)$	$7.9$	$7.99$	$7.999$	?	$8.001$	$8.01$	$8.1$

Conclusion: As the frequency pinches closer to the cutoff $f = 4\text{ kHz}$ from both sides, the gain values approach a stable value of exactly $8$ . Thus, while the gain $G(4)$ is undefined, the limit is well-defined:

\lim_{f \to 4} G(f) = 8