Question

A digital counter increments its count every \(0.1\) \(\mu\text{s}\). The counter is used in an instrument to measure frequency. The counter is operated for one period of the input signal to the instrument to measure its frequency. The input signal is approximately at \(100\) kHz. The maximum error in the measured frequency is kHz (rounded off to the nearest integer).

Hint:

The count can be off by one clock tick because the counter is not synchronised with the input signal. Use \(\Delta f \approx f^2 T_{clk}\) to size that error.

Answer 1

Correct Answer: 1

Step 1: Identify the measurement method.
The instrument finds the input frequency by counting clock pulses over exactly one period of the input signal. Each clock pulse takes \(T_{clk} = 0.1\) \(\mu\text{s}\), so the count \(N\) recorded over one input period \(T\) is \(N = T/T_{clk}\).

Step 2: State the source of the error.
The counter is not synchronised with the input signal, so the gate can open or close up to one clock tick early or late. This gives a classic \(\pm 1\) count (quantization) error in a period-counting instrument, and it limits how precisely \(T\), and so \(f = 1/T\), can be measured.

Step 3: Convert the count error into a frequency error.
For a small error, the fractional error in the measured period is about the same size as the fractional error contributed by one missing or extra clock pulse:
\[ \frac{\Delta T}{T} \approx \frac{T_{clk}}{T} \]
Since \(f = 1/T\), a small error in \(T\) gives about the same size fractional error in \(f\):
\[ \frac{\Delta f}{f} \approx \frac{T_{clk}}{T} = f\, T_{clk} \]
So the maximum error in the measured frequency is
\[ \Delta f \approx f^2\, T_{clk} \]

Step 4: Substitute the given values.
With \(f = 100\) kHz \(= 10^5\) Hz and \(T_{clk} = 0.1\) \(\mu\text{s} = 10^{-7}\) s:
\[ \Delta f = (10^5)^2 \times 10^{-7} = 10^{10} \times 10^{-7} = 1000 \text{ Hz} = 1 \text{ kHz} \]

Final Answer:
The maximum error in the measured frequency is \(1\) kHz. \[ \boxed{1 \text{ kHz}} \]