Question

Form factor of a sine wave is

• A. 0.75
• B. 0.65
• C. 0.50
• D. 1.11

Hint:

The form factor is a useful metric for identifying waveform distortions. For instance: - A pure sine wave has a form factor of 1.11. - A square wave has an RMS value equal to its peak and average values, yielding a form factor of exactly 1.00 (the lowest possible value for an AC signal). - A triangular/sawtooth wave has a form factor of $\frac{2}{\sqrt{3}} \approx 1.15$.

Answer 1

The Correct Option is D

Concept: The Form Factor ($k_f$) of an alternating current (AC) or periodic waveform is a dimensionless ratio that characterizes its overall shape. It is defined as the ratio of the Root Mean Square (RMS) value of the waveform to its mathematical average (mean) value over a half-cycle: \[ \text{Form Factor } (k_f) = \frac{\text{RMS Value}}{\text{Average Value}} \] Let us derive this value step-by-step for a standard pure sinusoidal wave defined by $v(t) = V_m \sin(\theta)$, where $V_m$ is the peak amplitude and $\theta = \omega t$.

Step 1: Calculate the Root Mean Square (RMS) Value ($V_{\text{rms}}$) The RMS value represents the effective DC-equivalent heating value of the alternating wave. It is computed over a full period $2\pi$: \[ V_{\text{rms}} = \sqrt{\frac{1}{2\pi} \int_{0}^{2\pi} v^2(\theta) \, d\theta} = \sqrt{\frac{1}{2\pi} \int_{0}^{2\pi} V_m^2 \sin^2(\theta) \, d\theta} \] Using the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$: \[ V_{\text{rms}} = \sqrt{\frac{V_m^2}{4\pi} \int_{0}^{2\pi} (1 - \cos(2\theta)) \, d\theta} = \sqrt{\frac{V_m^2}{4\pi} \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{2\pi}} \] \[ V_{\text{rms}} = \sqrt{\frac{V_m^2}{4\pi} \cdot (2\pi - 0)} = \sqrt{\frac{V_m^2}{2}} = \frac{V_m}{\sqrt{2}} \approx 0.707 \, V_m \]

Step 2: Calculate the Average Value ($V_{\text{avg}}$) over a half-cycle For a symmetrical AC wave, the average value over a full cycle is zero. Therefore, the average value is calculated over a positive half-cycle from $0$ to $\pi$: \[ V_{\text{avg}} = \frac{1}{\pi} \int_{0}^{\pi} V_m \sin(\theta) \, d\theta = \frac{V_m}{\pi} \left[ -\cos(\theta) \right]_{0}^{\pi} \] \[ V_{\text{avg}} = \frac{V_m}{\pi} \left( -\cos(\pi) - (-\cos(0)) \right) = \frac{V_m}{\pi} (1 + 1) = \frac{2V_m}{\pi} \approx 0.637 \, V_m \]

Step 3: Compute the Form Factor ratio Now, substitute the derived expressions for $V_{\text{rms}}$ and $V_{\text{avg}}$ into the form factor definition: \[ k_f = \frac{V_{\text{rms}}}{V_{\text{avg}}} = \frac{\frac{V_m}{\sqrt{2}}}{\frac{2V_m}{\pi}} = \frac{\pi}{2\sqrt{2}} \] Performing the numerical evaluation using the values $\pi \approx 3.14159$ and $\sqrt{2} \approx 1.41421$: \[ k_f = \frac{3.14159}{2 \times 1.41421} = \frac{3.14159}{2.82842} \approx 1.1107 \] Rounding to two decimal places gives a value of 1.11, which corresponds to Option (4).