Let's solve **Question 1** in **2 steps** with explanations as requested: ### **Question 1 Statement** The voltage from a square wave generator is of the form shown and has a period of 8 units. Find the Fourier series for function: \[ v(t) = \begin{cases} 0, & -4 < t < 0 \\ 10, & 0 < t < 4 \end{cases} \] with period \( 8 \). --- ## **Step 1: Identify the Fourier Series Components** The general Fourier series for a function of period \( 2L \) (here \( L = 4 \), so period is 8) is: \[ f(t) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi t}{L}\right) + b_n \sin\left(\frac{n\pi t}{L}\right) \right] \] Where: - \( a_0 = \frac{1}{2L} \int_{-L}^{L} f(t) dt \) - \( a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) dt \) - \( b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) dt \) **Explanation:** This step sets up the form of the Fourier series and the expressions for the coefficients. --- ## **Step 2: Calculate the Coefficients** #### 2.1. Compute \( a_0 \): \[ a_0 = \frac{1}{8} \left( \int_{-4}^{0} 0 \, dt + \int_{0}^{4} 10 \, dt \right) = \frac{1}{8} (0 + [10t]_{0}^{4}) = \frac{1}{8} (10 \times 4 - 10 \times 0) = \frac{40}{8} = 5 \] --- #### 2.2. Compute \( a_n \): \[ a_n = \frac{1}{4} \left( \int_{-4}^{0} 0 \cdot \cos\left(\frac{n\pi t}{4}\right)dt + \int_{0}^{4} 10 \cos\left(\frac{n\pi t}{4}\right)dt \right) \] \[ = \frac{10}{4} \int_{0}^{4} \cos\left(\frac{n\pi t}{4}\right)dt \] Let \( u = \frac{n\pi t}{4} \), \( dt = \frac{4}{n\pi} du \), so when \( t=0, u=0 \); when \( t=4, u=n\pi \): \[ = \frac{10}{4} \cdot \frac{4}{n\pi} \int_{0}^{n\pi} \cos u \, du = \frac{10}{n\pi} [\sin u]_{0}^{n\pi} = \frac{10}{n\pi} (\sin n\pi - \sin 0) = 0 \] So, **\( a_n = 0 \) for all \( n \)**. --- #### 2.3. Compute \( b_n \): \[ b_n = \frac{1}{4} \left( \int_{-4}^{0} 0 \cdot \sin\left(\frac{n\pi t}{4}\right)dt + \int_{0}^{4} 10 \sin\left(\frac{n\pi t}{4}\right)dt \right) \] \[ = \frac{10}{4} \int_{0}^{4} \sin\left(\frac{n\pi t}{4}\right)dt \] Let \( u = \frac{n\pi t}{4} \), \( dt = \frac{4}{n\pi} du \): \[ = \frac{10}{4} \cdot \frac{4}{n\pi} \int_{0}^{n\pi} \sin u \, du = \frac{10}{n\pi} [-\cos u]_{0}^{n\pi} = \frac{10}{n\pi} [-\cos(n\pi) + \cos(0)] = \frac{10}{n\pi} [1 - (-1)^n] \] - If \( n \) is even: \( [1 - 1] = 0 \) - If \( n \) is odd: \( [1 - (-1)] = 2 \) So, \[ b_n = \begin{cases} 0, & n \text{ even} \\ \frac{20}{n\pi}, & n \text{ odd} \end{cases} \] --- ## **Final Answer: Fourier Series** \[ v(t) = 5 + \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{20}{n\pi} \sin\left(\frac{n\pi t}{4}\right) \] --- ### **Summary of Steps** - **Step 1:** Set up the Fourier series form and coefficient formulas. - **Step 2:** Calculate each coefficient (\( a_0, a_n, b_n \)), plug them into the series. **Final Answer:** \[ \boxed{ v(t) = 5 + \sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{20}{n\pi} \sin\left(\frac{n\pi t}{4}\right) } \]

Let's proceed step-by-step to solve the first question, which involves finding the Fourier series of a periodic square wave voltage. --- ### **Step 1: Understanding the problem and setting up the Fourier series** **Given:** - The function \( v(t) \) is periodic with period \( T = 8 \). - Over one period: - \( v(t) = 0 \) for \( -4 < t < 0 \), - \( v(t) = 10 \) for \( 0 < t < 4 \). **Approach:** - Recognize that the function is piecewise constant, forming a square wave. - Fourier series for a function with period \( T = 8 \) can be expressed as: \[ v(t) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left( \frac{2\pi n t}{T} \right) + b_n \sin\left( \frac{2\pi n t}{T} \right) \right] \] - The coefficients are given by: \[ a_0 = \frac{1}{T} \int_{-T/2}^{T/2} v(t) \, dt \] \[ a_n = \frac{2}{T} \int_{-T/2}^{T/2} v(t) \cos\left( \frac{2\pi n t}{T} \right) dt \] \[ b_n = \frac{2}{T} \int_{-T/2}^{T/2} v(t) \sin\left( \frac{2\pi n t}{T} \right) dt \] --- ### **Step 2: Calculating the Fourier coefficients** **Note:** - Since \( v(t) \) is defined over \( -4 < t < 4 \), it's convenient to integrate over this interval. - The function is odd-symmetric about \( t=0 \) (since it is 0 for \( -4 < t < 0 \) and 10 for \( 0 < t < 4 \)), so some coefficients will simplify. --- ### **Final step: Computing the coefficients** **(a) Compute \( a_0 \):** \[ a_0 = \frac{1}{8} \left( \int_{-4}^{0} 0 \, dt + \int_{0}^{4} 10 \, dt \right) = \frac{1}{8} (0 + 10 \times 4) = \frac{40}{8} = 5 \] **(Explanation:** The average value over one period is 5 volts.) --- **(b) Compute \( a_n \):** \[ a_n = \frac{2}{8} \int_{-4}^{4} v(t) \cos\left( \frac{\pi n t}{4} \right) dt \] But since \( v(t) \) is non-zero only from 0 to 4: \[ a_n = \frac{1}{4} \int_{0}^{4} 10 \cos\left( \frac{\pi n t}{4} \right) dt \] Calculate: \[ a_n = \frac{10}{4} \int_{0}^{4} \cos\left( \frac{\pi n t}{4} \right) dt \] Let \( u = \frac{\pi n t}{4} \Rightarrow du = \frac{\pi n}{4} dt \Rightarrow dt = \frac{4}{\pi n} du \) When \( t=0 \), \( u=0 \); when \( t=4 \), \( u= \pi n \) So: \[ a_n = \frac{10}{4} \times \frac{4}{\pi n} \int_{0}^{\pi n} \cos u \, du = \frac{10}{\pi n} [ \sin u ]_{0}^{\pi n } \] \[ a_n = \frac{10}{\pi n} ( \sin \pi n - 0 ) \] Since \( \sin \pi n = 0 \) for all integers \( n \), **we conclude:** \[ a_n = 0 \quad \text{for all } n \] --- **(c) Compute \( b_n \):** \[ b_n = \frac{2}{8} \int_{-4}^{4} v(t) \sin\left( \frac{\pi n t}{4} \right) dt \] Again, \( v(t) = 0 \) for \( -4 < t < 0 \), \( v(t) = 10 \) for \( 0 < t < 4 \): \[ b_n = \frac{1}{4} \int_{0}^{4} 10 \sin\left( \frac{\pi n t}{4} \right) dt \] Calculate: \[ b_n = \frac{10}{4} \int_{0}^{4} \sin\left( \frac{\pi n t}{4} \right) dt \] Using the substitution \( u = \frac{\pi n t}{4} \), same as above: \[ b_n = \frac{10}{4} \times \frac{4}{\pi n} \int_{0}^{\pi n} \sin u \, du = \frac{10}{\pi n} [ - \cos u ]_{0}^{\pi n} \] \[ b_n = \frac{10}{\pi n} ( - \cos \pi n + \cos 0 ) = \frac{10}{\pi n} ( - (-1)^n + 1 ) \] \[ b_n = \frac{10}{\pi n} ( 1 - (-1)^n ) \] - For **even \( n \)**: \( (-1)^n = 1 \), so \( b_n = 0 \). - For **odd \( n \)**: \( (-1)^n = -1 \), so: \[ b_n = \frac{10}{\pi n} (1 - (-1)) = \frac{10}{\pi n} \times 2 = \frac{20}{\pi n} \] --- ### **Final Fourier Series:** Putting all together: \[ \boxed{ v(t) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos \left( \frac{\pi n t}{4} \right) + b_n \sin \left( \frac{\pi n t}{4} \right) \right] } \] Since \( a_n=0 \) for all \( n \), \[ v(t) = 5 + \sum_{n=1,3,5,\ldots}^{\infty} \frac{20}{\pi n} \sin \left( \frac{\pi n t}{4} \right) \] --- ### **Final answer:** \[ \boxed{ \textbf{Fourier series of the square wave voltage:} \quad v(t) = 5 + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{20}{\pi n} \sin \left( \frac{\pi n t}{4} \right) } \] --- **Summary:** - The average (DC) component is 5V. - Only sine terms with odd harmonics appear. - The coefficients decay as \( 1/n \). Let me know if you'd like me to proceed with additional questions!

The Fourier series for the given periodic square wave voltage is: \[ \boxed{ v(t) = 5 + \sum_{\substack{n=1,\,3,\,5,\ldots}}^{\infty} \frac{20}{\pi n} \sin \left( \frac{\pi n t}{4} \right) } \] This series includes a constant term of 5 (the average voltage) and sine terms at odd harmonics, with coefficients decreasing proportionally to \( 1/n \).