To solve this problem, let's analyze the given information:

The problem requires us to find:

1. Period (T) of the square wave signal.
2. Frequency (f) of the signal.
3. Duty cycle (in %) of the signal based on the high-level duration.

Given Information

1. The high-level time $t_g = 0.01 \, \text{ms}$.
2. The period time $t_p = 0.5 \, \text{ms}$.

Solution Steps

1. Period (T): The period $T$ of the square wave signal is given by $T = t_p$. $T = 0.5 \, \text{ms}$

2. Frequency (f): The frequency $f$ is the reciprocal of the period $T$: $f = \frac{1}{T} = \frac{1}{0.5 \, \text{ms}} = \frac{1}{0.0005 \, \text{s}} = 2000 \, \text{Hz}$

3. Duty Cycle: Duty cycle (D) is calculated as the ratio of the high-level time to the period, multiplied by 100 to express it as a percentage: $D = \left( \frac{t_g}{T} \right) \times 100 = \left( \frac{0.01 \, \text{ms}}{0.5 \, \text{ms}} \right) \times 100$ Simplifying this: $D = \frac{0.01}{0.5} \times 100 = 2\%$

Summary of Results:

• Period (T): $0.5 \, \text{ms}$
• Frequency (f): $2000 \, \text{Hz}$
• Duty Cycle (D): $2\%$

Would you like more details on any step, or have any questions?

Here are some related questions for further practice:

1. How would the duty cycle change if $t_g$ increased to 0.05 ms?
2. What would be the frequency if $T$ were doubled to 1 ms?
3. How does the duty cycle affect the average output voltage of a PWM signal?
4. If $t_p$ was halved, what would be the new frequency and duty cycle?
5. Why is the duty cycle significant in pulse-width modulation (PWM) applications?

Tip: The duty cycle directly impacts the energy delivered by a pulse signal, which is crucial in power control applications.