Frequency Counter Error Analysis

The problem asks for the maximum error in a frequency measurement performed by a digital counter. We are given the counter's time resolution and the approximate frequency of the input signal.

Understanding Measurement Parameters

• Input Signal Frequency: $f_{in} \approx 100\ \text{kHz} = 100 \times 10^3\ \text{Hz}$.
• Counter Resolution: The counter increments every $0.1\ \mu\text{s}$. This value represents the smallest time interval the counter can resolve, thus defining the accuracy of any time measurement it performs. Let this be $\Delta t = 0.1\ \mu\text{s} = 0.1 \times 10^{-6}\ \text{s}$.
• Measurement Method: The counter measures the period ($T$) of the input signal. The frequency is then calculated as $f = 1/T$.

Calculating Period and Error

First, calculate the period ($T$) of the input signal:

$ T = \frac{1}{f_{in}} = \frac{1}{100 \times 10^3\ \text{Hz}} = 10 \times 10^{-6}\ \text{s} = 10\ \mu\text{s} $

The counter measures this period $T$. The maximum error in the period measurement ($\Delta T$) is equal to the counter's resolution:

$ \Delta T = \pm \Delta t = \pm 0.1\ \mu\text{s} = \pm 0.1 \times 10^{-6}\ \text{s} $

Determining Frequency Error

The error in the measured frequency ($\Delta f$) is related to the error in the measured period ($\Delta T$). We can approximate this relationship using calculus. If $f = 1/T$, then the change in $f$ ($\Delta f$) due to a change in $T$ ($\Delta T$) is given by:

$ \Delta f \approx \left| \frac{df}{dT} \right| \Delta T $

Calculate the derivative of $f$ with respect to $T$:

$ \frac{df}{dT} = \frac{d}{dT} (T^{-1}) = -T^{-2} = -\frac{1}{T^2} $

Now, substitute the values into the error approximation:

$ \Delta f \approx \frac{1}{T^2} \Delta T $

$ \Delta f \approx \frac{1}{(10 \times 10^{-6}\ \text{s})^2} \times (0.1 \times 10^{-6}\ \text{s}) $

$ \Delta f \approx \frac{1}{(10^{-5}\ \text{s})^2} \times (10^{-7}\ \text{s}) $

$ \Delta f \approx \frac{1}{10^{-10}\ \text{s}^2} \times 10^{-7}\ \text{s} $

$ \Delta f \approx 10^{10}\ \text{s}^{-2} \times 10^{-7}\ \text{s} = 10^3\ \text{s}^{-1} = 1000\ \text{Hz} $

Convert the error to kHz:

$ \Delta f \approx 1\ \text{kHz} $

The maximum error in the measured frequency is approximately 1 kHz.