Let's analyze the periodic signal step by step to address each part of the problem. ## **Given Signal Description:** waveform is a pulse that: - Has \( I_{pk} \)positive and negative). - centered at \( t = ). - Is active (non-zero) between \( -T/ \) and \( T/6). - Is zero elsewhere. - period of the is \( T). --- ## **a) Fourier coefficients expected to be zero and why** ### **Explanation:** - The signal is **even** because it is symmetric around \( t= \): the positive and negative pulses are mirror images. - The signal contains only **cosine terms** in its Fourier series because it's symmetric about the vertical axis (even function). - The **sine Fourier coefficients** \( b_n \) are **zero** because the signal is even. ### **Fourier Coefficients:** - \( a_ \) (average value): may be non-zero because of the rectangular pulse. - \( a_n \) (cosine coefficients): may be non-zero. - \( b_n \) (sine coefficients): **expected to be zero** because the signal is even. **Answer:** - \( b_n = \) for all \( n \), because the signal is even, and sine functions are odd, resulting in zero Fourier sine coefficients. --- ## **b) Derive analytical expressions for \( a_ \), \( a_n \), and \( b_n \)** ### **Fourier Series general forms:** \[ x(t) = a_ + \sum_{n=1}^{\infty} \left[ a_n \cos \left( \frac{2\pi n t}{T} \right) + b_n \sin \left( \frac{2\pi n t}{T} \right) \right] \] ### **Since the signal is even:** - \( b_n = \). ### **Calculating \( a_ \):** \[ a_ = \frac{1}{T} \int_{-T/2}^{T/2} x(t)\, dt \] Since the pulse is non-zero only between \( -T/6 \) and \( T/6 \): \[ a_ = \frac{1}{T} \int_{-T/6}^{T/6} I_{pk} \, dt + \int_{-T/6}^{T/6} (-I_{pk}) \, dt \] But note the pulse is positive \( I_{pk} \) between \( -T/6 \) and \( T/6 \), and zero elsewhere. \[ a_ = \frac{1}{T} \times \left[ \int_{-T/6}^{T/6} I_{pk} \, dt \right] \] \[ a_ = \frac{1}{T} \times I_{pk} \times \left( \frac{T}{3} \right) = \frac{I_{pk}}{3} \] --- ### **Calculating \( a_n \):** \[ a_n = \frac{2}{T} \int_{}^{T/2} x(t) \cos \left( \frac{2\pi n t}{T} \right) dt \] or equivalently, for a symmetric pulse: \[ a_n = \frac{2}{T} \int_{-T/2}^{T/2} x(t) \cos \left( \frac{2\pi n t}{T} \right) dt \] Since \( x(t) \) is non-zero only in \( -T/6 \) to \( T/6 \): \[ a_n = \frac{2}{T} \int_{-T/6}^{T/6} I_{pk} \cos \left( \frac{2\pi n t}{T} \right) dt \] Because the integrand is even, limits can be doubled: \[ a_n = \frac{4 I_{pk}}{T} \int_{}^{T/6} \cos \left( \frac{2\pi n t}{T} \right) dt \] Calculating the integral: \[ a_n = \frac{4 I_{pk}}{T} \left[ \frac{\sin \left( \frac{2\pi n t}{T} \right)}{\frac{2\pi n}{T}} \right]_^{T/6} \] \[ a_n = \frac{4 I_{pk}}{T} \times \frac{T}{2\pi n} \sin \left( \frac{2\pi n}{T} \times \frac{T}{6} \right) \] \[ a_n = \frac{2 I_{pk}}{\pi n} \sin \left( \frac{\pi n}{3} \right) \] --- ### **Summary:** \[ \boxed{ \begin{aligned} a_ &= \frac{I_{pk}}{3} \\ a_n &= \frac{2 I_{pk}}{\pi n} \sin \left( \frac{\pi n}{3} \right) \\ b_n &= \end{aligned} } \] --- ## **c) Expressions for the Fourier coefficients \( A_n \) and \( \theta_n \) in the polar form** ### **Polar form:** \[ C_n = A_n e^{j \theta_n} = a_n + j b_n \] Since \( b_n = \), the coefficients are purely real: \[ A_n = |a_n + j | = |a_n| \quad \text{and} \quad \theta_n = \quad \text{if } a_n > \] or \[ \theta_n = \pi \quad \text{if } a_n < \] ### **Expressions:** \[ \boxed{ A_n = \left| \frac{2 I_{pk}}{\pi n} \sin \left( \frac{\pi n}{3} \right) \right| } \] \[ \theta_n = \begin{cases} , & \text{if } a_n > \\ \pi, & \text{if } a_n < \end{cases} \] --- ## **Summary:** | **Fourier Coefficients** | **Expression** | |--------------------------|----------------| | \( a_ \) | \( \frac{I_{pk}}{3} \) | | \( a_n \) | \( \frac{2 I_{pk}}{\pi n} \sin \left( \frac{\pi n}{3} \right) \) | | \( b_n \) | (since the signal is even) | | \( A_n \) (polar magnitude) | \( \left| a_n \right| \) | | \( \theta_n \) (phase) | if \( a_n > \), \( \pi \) if \( a_n < \) | --- Let me know if you'd like more detailed calculations for specific \( n \)!