1. Problem (ii): Given the time $T = 2$ ms, find the frequency $f$ of the signal. Frequency and time period are related by the formula: $$f = \frac{1}{T}$$ Since $T = 2$ ms = $2 \times 10^{-3}$ s, $$f = \frac{1}{2 \times 10^{-3}} = 500 \text{ Hz}$$ So, the frequency of the signal is 500 Hz. 2. Problem (iii): Given vertical scale = 1 V/div and waveform spans 3 divisions, find the peak voltage $V_p$. Peak voltage is calculated by multiplying the vertical scale by the number of divisions: $$V_p = 1 \text{ V/div} \times 3 \text{ div} = 3 \text{ V}$$ So, the peak voltage is 3 V. 3. Problem (c)(i): What is a multimeter? A multimeter is an electronic measuring instrument that combines several measurement functions in one unit, typically measuring voltage, current, and resistance. Advantages of a multimeter: - It is versatile and can measure multiple electrical quantities. - Portable and easy to use for troubleshooting. Disadvantages of a multimeter: - Limited accuracy compared to specialized instruments. - Can be damaged if used improperly or on high voltage/current. 4. Problem (c)(ii): List two limitations of a multimeter. - Cannot measure very high frequencies accurately. - Limited to measuring basic electrical parameters; cannot measure complex signals like waveform shapes or phase angles.