Questions: 4. The effective voltage supplied by a sine wave is the constant value that would be supplied by a DC power source. Which voltage represents the "effective" voltage of a sinusoidal signal with no DC offset? What if a DC is applied? 5. What two values were used to calculate the Vrms? Which one gives the correct value for both the signal without the DC offset and with the DC offset? Why?

Solution Steps

The effective voltage of a sinusoidal signal is commonly referred to as the root mean square (RMS) voltage. For a pure sinusoidal waveform with no DC offset, the RMS voltage is calculated as:

\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \]

This value represents the equivalent DC voltage that would deliver the same power to a load as the AC signal.

When a DC offset is applied to a sinusoidal signal, the effective voltage (RMS) is calculated by considering both the AC and DC components. The formula for the RMS voltage in this case is:

\[ V_{\text{rms}} = \sqrt{V_{\text{dc}}^2 + \left(\frac{V_{\text{peak}}}{\sqrt{2}}\right)^2} \]

Here, \(V_{\text{dc}}\) is the DC offset voltage, and \(V_{\text{peak}}\) is the peak voltage of the AC component.

The two values used to calculate the RMS voltage are the peak voltage (\(V_{\text{peak}}\)) and the DC offset voltage (\(V_{\text{dc}}\)).

• For a signal without a DC offset, the correct value is \(V_{\text{peak}}/\sqrt{2}\).
• For a signal with a DC offset, the correct value is \(\sqrt{V_{\text{dc}}^2 + (V_{\text{peak}}/\sqrt{2})^2}\).

The reason the latter formula is correct for both cases is that it accounts for the presence of a DC component, which affects the total power delivered by the signal.

Final Answer