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).
Show 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.
Step 1: Work out the nominal count.
The clock period is $T_{clk} = 0.1\ \mu\text{s}$, and the input period is close to $T = 1/(100\text{ kHz}) = 10\ \mu\text{s}$, since the input runs near $100$ kHz. Counting clock ticks over one input period should give roughly
\[ N = \frac{T}{T_{clk}} = \frac{10\ \mu\text{s}}{0.1\ \mu\text{s}} = 100 \text{ counts} \]
Step 2: Consider the count landing one tick early or late.
Because the counter gate is not synchronised to the clock, the recorded count can come out as $99$ or $101$ instead of exactly $100$, purely from where the gate edges fall between clock ticks. This $\pm 1$ count uncertainty is the whole source of error here.
Step 3: Turn each boundary count into a frequency.
For a count of $99$: measured period $= 99 \times 0.1\ \mu\text{s} = 9.9\ \mu\text{s}$, so
\[ f = \frac{1}{9.9\ \mu\text{s}} \approx 101.01 \text{ kHz} \]
an error of about $+1.01$ kHz.
For a count of $101$: measured period $= 101 \times 0.1\ \mu\text{s} = 10.1\ \mu\text{s}$, so
\[ f = \frac{1}{10.1\ \mu\text{s}} \approx 99.01 \text{ kHz} \]
an error of about $-0.99$ kHz.
Final Answer:
Both boundary counts give an error whose size rounds to $1$ kHz, so the maximum error in the measured frequency is $1$ kHz.
\[ \boxed{\Delta f \approx 1 \text{ kHz}} \]
