Let's break down the problem step by step: We have a **trapezoidal wave** with a **90° flat portion** (per half-cycle), and **amplitude 1** of **100 A**. **Assumptions & Notation:** - "Amplitude 1" is the height of the flat portion, i.e., the maximum value is 100 A. - The wave is symmetric about the time axis and periodic (like a modified square wave with sloped edges). - The flat top lasts for 90° (π/2 radians) per half-cycle. - The rest of the half-cycle (total 180° per half cycle) is the sloped portion (rising/falling edges), so each edge is (180° - 90°)/2 = 45° (π/4 radians). - The period is \( T \), and the angular frequency is \( \omega_0 = 2\pi/T \). Let's proceed with the calculations. --- ## 1. **Waveform Construction** Let the period \( T \) correspond to \( 2\pi \) radians. - **Flat portion**: from \( \theta = \frac{\pi}{4} \) to \( \theta = \frac{3\pi}{4} \), value = 100 A - **Rising edge**: from \( \theta = 0 \) to \( \theta = \frac{\pi}{4} \), rises from 0 to 100 A - **Falling edge**: from \( \theta = \frac{3\pi}{4} \) to \( \theta = \pi \), falls from 100 A to 0 We can define the waveform for \( 0 \leq \theta \leq \pi \) (half cycle), and then it is odd or even extended as needed. - For \( 0 \leq \theta < \frac{\pi}{4} \): \( i(\theta) = 100 \cdot \frac{4}{\pi} \theta \) - For \( \frac{\pi}{4} \leq \theta < \frac{3\pi}{4} \): \( i(\theta) = 100 \) - For \( \frac{3\pi}{4} \leq \theta < \pi \): \( i(\theta) = 100 \cdot \left[1 - \frac{4}{\pi} (\theta - \frac{3\pi}{4}) \right] \) --- ## 2. **(a) Crest Factor (CF)** **Crest Factor** is: \[ CF = \frac{I_{\text{peak}}}{I_{\text{rms}}} \] Here, \( I_{\text{peak}} = 100 \) A. We need to calculate \( I_{\text{rms}} \): \[ I_{\text{rms}} = \sqrt{\frac{1}{2\pi} \int_0^{2\pi} i^2(\theta) d\theta} \] But due to symmetry, it's easier to compute over \( 0 \) to \( \pi \), then double it. \[ I_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^{2\pi} i^2(\theta) d\theta = \frac{1}{\pi} \int_0^{\pi} i^2(\theta) d\theta \] Now, split the interval: 1. \( 0 \to \frac{\pi}{4} \): \( i(\theta) = 100 \frac{4}{\pi} \theta \) 2. \( \frac{\pi}{4} \to \frac{3\pi}{4} \): \( i(\theta) = 100 \) 3. \( \frac{3\pi}{4} \to \pi \): \( i(\theta) = 100 \left[1 - \frac{4}{\pi} (\theta - \frac{3\pi}{4}) \right] \) Let's compute each segment. --- ### **First Segment: \( 0 \to \frac{\pi}{4} \)** \[ i(\theta) = 100 \cdot \frac{4}{\pi} \theta \] \[ i^2(\theta) = 100^2 \cdot \left(\frac{4}{\pi}\right)^2 \theta^2 = 10{,}000 \cdot \frac{16}{\pi^2} \theta^2 \] \[ \int_0^{\frac{\pi}{4}} i^2(\theta) d\theta = 10{,}000 \cdot \frac{16}{\pi^2} \int_0^{\frac{\pi}{4}} \theta^2 d\theta \] \[ \int \theta^2 d\theta = \frac{\theta^3}{3} \] So: \[ \int_0^{\frac{\pi}{4}} \theta^2 d\theta = \frac{1}{3} \left(\frac{\pi}{4}\right)^3 = \frac{1}{3} \cdot \frac{\pi^3}{64} = \frac{\pi^3}{192} \] \[ \int_0^{\frac{\pi}{4}} i^2(\theta) d\theta = 10{,}000 \cdot \frac{16}{\pi^2} \cdot \frac{\pi^3}{192} = 10{,}000 \cdot \frac{16 \pi^3}{192 \pi^2} \] \[ = 10{,}000 \cdot \frac{16 \pi}{192} = 10{,}000 \cdot \frac{\pi}{12} \approx 10{,}000 \cdot 0.2618 = 2,618 \] --- ### **Second Segment: \( \frac{\pi}{4} \to \frac{3\pi}{4} \)** \[ i(\theta) = 100 \implies i^2(\theta) = 10,000 \] \[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 10,000 d\theta = 10,000 \cdot \left( \frac{3\pi}{4} - \frac{\pi}{4} \right) = 10,000 \cdot \frac{2\pi}{4} = 10,000 \cdot \frac{\pi}{2} \approx 10,000 \cdot 1.5708 = 15,708 \] --- ### **Third Segment: \( \frac{3\pi}{4} \to \pi \)** Let \( x = \theta - \frac{3\pi}{4} \), so as \( \theta \) goes from \( \frac{3\pi}{4} \) to \( \pi \), \( x \) goes from 0 to \( \pi - \frac{3\pi}{4} = \frac{\pi}{4} \). \[ i(\theta) = 100 \left[1 - \frac{4}{\pi} x \right] \] \[ i^2(\theta) = 100^2 \left(1 - \frac{4}{\pi} x \right)^2 = 10,000 \left(1 - \frac{8}{\pi} x + \frac{16}{\pi^2} x^2 \right) \] Integrate from \( x=0 \) to \( x=\frac{\pi}{4} \): \[ \int_0^{\frac{\pi}{4}} 10,000 \left(1 - \frac{8}{\pi} x + \frac{16}{\pi^2} x^2 \right) dx \] \[ = 10,000 \left[ x - \frac{8}{\pi} \frac{x^2}{2} + \frac{16}{\pi^2} \frac{x^3}{3} \right]_0^{\frac{\pi}{4}} \] Plug \( x = \frac{\pi}{4} \): - \( x = \frac{\pi}{4} \) - \( x^2 = \frac{\pi^2}{16} \) - \( x^3 = \frac{\pi^3}{64} \) So: \[ = 10,000 \left[ \frac{\pi}{4} - \frac{8}{\pi} \cdot \frac{1}{2} \cdot \frac{\pi^2}{16} + \frac{16}{\pi^2} \cdot \frac{1}{3} \cdot \frac{\pi^3}{64} \right] \] \[ = 10,000 \left[ \frac{\pi}{4} - \frac{8}{\pi} \cdot \frac{\pi^2}{32} + \frac{16}{\pi^2} \cdot \frac{\pi^3}{192} \right] \] \[ = 10,000 \left[ \frac{\pi}{4} - \frac{8 \pi^2}{32 \pi} + \frac{16 \pi^3}{192 \pi^2} \right] \] \[ = 10,000 \left[ \frac{\pi}{4} - \frac{\pi}{4} + \frac{\pi}{12} \right] \] \[ = 10,000 \left[ 0 + \frac{\pi}{12} \right] = 10,000 \cdot \frac{\pi}{12} \approx 2,618 \] --- ### **Sum of All Segments** Total integral over \( 0 \to \pi \): \[ S = 2,618 + 15,708 + 2,618 = 20,944 \] So, \[ I_{\text{rms}}^2 = \frac{1}{\pi} \cdot 20,944 \] \[ I_{\text{rms}} = \sqrt{ \frac{20,944}{\pi} } = \sqrt{ 6,670.1 } \approx 81.7 \text{ A} \] --- ### **Crest Factor** \[ CF = \frac{I_{\text{peak}}}{I_{\text{rms}}} = \frac{100}{81.7} \approx \boxed{1.22} \] --- ## 3. **(b) Distortion Factor (DF)** **Distortion Factor (DF)** is the ratio of the **rms value of the fundamental** to the **total rms value**: \[ DF = \frac{I_{1, \text{rms}}}{I_{\text{rms}}} \] We need to compute the rms value of the fundamental. The rms value of the fundamental is \( \frac{I_1}{\sqrt{2}} \), where \( I_1 \) is the amplitude of the **first harmonic (fundamental) component**. Let's compute the amplitude of the fundamental. ### **Fourier Series Coefficient for Fundamental** The general formula for the \( n \)-th harmonic amplitude (\( a_n \)) for a periodic function \( f(\theta) \) over \( 0 \) to \( 2\pi \): \[ a_n = \frac{2}{2\pi} \int_0^{2\pi} f(\theta) \sin(n\theta) d\theta \] \[ a_n = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \sin(n\theta) d\theta \] For odd symmetry, only sine terms exist. For \( n = 1 \): \[ a_1 = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \sin(\theta) d\theta \] But due to symmetry, double the half-cycle: \[ a_1 = \frac{2}{\pi} \int_0^{\pi} f(\theta) \sin(\theta) d\theta \] We'll compute this integral in three segments. --- #### **First Segment: \( 0 \to \frac{\pi}{4} \)** \( f(\theta) = 100 \frac{4}{\pi} \theta \) \[ I_1^{(1)} = \frac{2}{\pi} \int_0^{\frac{\pi}{4}} 100 \frac{4}{\pi} \theta \sin(\theta) d\theta = \frac{2}{\pi} \cdot 100 \cdot \frac{4}{\pi} \int_0^{\frac{\pi}{4}} \theta \sin(\theta) d\theta = \frac{800}{\pi^2} \int_0^{\frac{\pi}{4}} \theta \sin(\theta) d\theta \] Integral: \[ \int \theta \sin(\theta) d\theta = -\theta \cos(\theta) + \int \cos(\theta) d\theta = -\theta \cos(\theta) + \sin(\theta) \] So, \[ \int_0^{a} \theta \sin(\theta) d\theta = [-\theta \cos(\theta) + \sin(\theta)]_0^a \] \[ = [-a \cos(a) + \sin(a)] - [0 - 0] = -a \cos(a) + \sin(a) \]