# 1) Inspect symmetry → which coefficients are zero From the plot, the waveform \(i(t)\) is even: \(i(-t) = i(t)\). The positive pulse area equals the magnitude of the negative pulse area, so the average (DC) is zero. Therefore: - All sine coefficients are zero: \(b_n = \) for all \(n\). - The DC term is zero: \(a_ = \). - Only cosine coefficients \(a_n\) (for \(n \ge 1\)) can be nonzero. (Thus, the Fourier series will contain only cosines with zero DC.) --- # 2) Compute \(a_n\) (analytical expression) Using the real Fourier cosine coefficient formula: \[ a_n = \frac{2}{T} \int_{-T/2}^{T/2} i(t) \cos(n \omega_ t) \, dt, \qquad \omega_ = \frac{2\pi}{T}. \] Since \(i(t)\) is even, this simplifies to: \[ a_n = \frac{4}{T} \int_{}^{T/2} i(t) \cos(n \omega_ t) \, dt. \] From the diagram, the values of \(i(t)\) on \([, T/2]\) are: \[ i(t) = \begin{cases} I_{pk}, & \le t < T/6, \\ , & T/6 \le t < T/3, \\ - I_{pk}, & T/3 \le t \le T/2. \end{cases} \] Thus, \[ a_n = \frac{4}{T} \left[ \int_{}^{T/6} I_{pk} \cos(n \omega_ t) \, dt + \int_{T/6}^{T/3} \, dt + \int_{T/3}^{T/2} (-I_{pk}) \cos(n \omega_ t) \, dt \right]. \] Evaluate the two nonzero integrals: \[ \int \cos(\alpha t) \, dt = \frac{\sin(\alpha t)}{\alpha}. \] Let \(\omega_ = 2\pi / T\). Then: \[ \int_{}^{T/6} I_{pk} \cos(n \omega_ t) \, dt = I_{pk} \frac{\sin(n \omega_ T/6)}{n \omega_} = I_{pk} \frac{\sin(n \pi / 3)}{n \omega_}. \] And: \[ \int_{T/3}^{T/2} (-I_{pk}) \cos(n \omega_ t) \, dt = -I_{pk} \frac{\sin(n \omega_ T/2) - \sin(n \omega_ T/3)}{n \omega_} = -I_{pk} \frac{\sin(n \pi) - \sin(2 n \pi / 3)}{n \omega_}. \] Since \(\sin(n \pi) = \): \[ = I_{pk} \frac{\sin(2 n \pi / 3)}{n \omega_}. \] Add the two integrals and multiply by \(\frac{4}{T}\): \[ a_n = \frac{4}{T} \cdot \frac{I_{pk}}{n \omega_} \left[ \sin(n \pi/3) + \sin(2 n \pi/3) \right]. \] Using \(n \omega_ = 2\pi n / T \Rightarrow T(n \omega_) = 2 \pi n\): \[ a_n = \frac{4 I_{pk}}{2 \pi n} \left[ \sin(n \pi/3) + \sin(2 n \pi/3) \right] = \frac{2 I_{pk}}{\pi n} \left[ \sin(n \pi/3) + \sin(2 n \pi/3) \right]. \] Apply the sum-to-product identity: \[ \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \] with \(A = n \pi/3\), \(B = 2 n \pi/3\): \[ \sin(n \pi/3) + \sin(2 n \pi/3) = 2 \sin \left(\frac{n \pi/3 + 2 n \pi/3}{2}\right) \cos \left(\frac{n \pi/3 - 2 n \pi/3}{2}\right) = 2 \sin \left(\frac{3 n \pi/3}{2}\right) \cos \left(-\frac{n \pi/3}{2}\right). \] Simplify: \[ = 2 \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right). \] Thus, \[ a_n = \frac{2 I_{pk}}{\pi n} \times 2 \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right) = \frac{4 I_{pk}}{\pi n} \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right). \] Since \(b_n = \) and \(a_ = \), the Fourier coefficients are: \[ \boxed{ a_n = \frac{4 I_{pk}}{\pi n} \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right), \quad b_n=, \quad a_=. } \] **Remarks:** - \(\sin(n \pi/2) = \) for all even \(n\), so only odd \(n\) produce nonzero \(a_n\). - For odd \(n\), \(\sin(n \pi/2) = \pm 1\), alternating sign depending on \(n \pmod 4\). --- # 3) Polar form \((A_n, \theta_n)\) The amplitude and phase for the \(n\)-th harmonic: \[ A_n = \sqrt{a_n^2 + b_n^2} = |a_n|, \quad \theta_n = \begin{cases} , & a_n > , \\ \pi, & a_n < , \\ \text{undefined (zero amplitude)}, & a_n=. \end{cases} \] Using the closed form for \(a_n\): \[ \boxed{ A_n = \left| \frac{4 I_{pk}}{\pi n} \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right) \right|, \quad \theta_n = \arg \left( \sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi}{6}\right) \right). } \] --- # 4) Final Fourier series (alternate form) Since only cosine terms are present: \[ \boxed{ i(t) = \sum_{n=1}^{\infty} a_n \cos(n \omega_ t), \quad \text{with } a_n \text{ as above and } a_ = b_n = . } \]

# 1) Inspect symmetry → which coefficients are zero From the plot, the waveform \(i(t)\) is even: \(i(-t) = i(t)\). The positive pulse area equals the magnitude of the negative pulse area, so the average (DC) is zero. Therefore: - All sine coefficients are zero: \(b_n = 0\) for all \(n\). - The DC term is zero: \(a_0 = 0\). - Only cosine coefficients \(a_n\) (for \(n \ge 1\)) can be nonzero. (Thus, the Fourier series will contain only cosines with zero DC.) --- # 2) Compute \(a_n\) (analytical expression) Using the real Fourier cosine coefficient formula: \[ a_n = \frac{2}{T} \int_{-T/2}^{T/2} i(t) \cos(n \omega_0 t) \, dt, \qquad \omega_0 = \frac{2\pi}{T}. \] Since \(i(t)\) is even, this simplifies to: \[ a_n = \frac{4}{T} \int_{0}^{T/2} i(t) \cos(n \omega_0 t) \, dt. \] From the diagram, the values of \(i(t)\) on \([0,T/2]\) are: \[ i(t) = \begin{cases} I_{pk}, & 0 \le t < T/6, \\ 0, & T/6 \le t < T/3, \\ -I_{pk}, & T/3 \le t \le T/2. \end{cases} \] Thus, \[ a_n = \frac{4}{T} \left[ \int_{0}^{T/6} I_{pk} \cos(n \omega_0 t) \, dt + \int_{T/6}^{T/3} 0 \, dt + \int_{T/3}^{T/2} (-I_{pk}) \cos(n \omega_0 t) \, dt \right]. \] Evaluate the two nonzero integrals: \[ \int \cos(\alpha t) \, dt = \frac{\sin(\alpha t)}{\alpha}. \] Let \(\omega_0 = \frac{2\pi}{T}\). Then: \[ \int_{0}^{T/6} I_{pk} \cos(n \omega_0 t) \, dt = I_{pk} \frac{\sin\left(n \omega_0 \frac{T}{6}\right)}{n \omega_0} = I_{pk} \frac{\sin\left(n \frac{\pi}{3}\right)}{n \omega_0}. \] And: \[ \int_{T/3}^{T/2} (-I_{pk}) \cos(n \omega_0 t) \, dt = -I_{pk} \frac{\sin\left(n \omega_0 \frac{T}{2}\right) - \sin\left(n \omega_0 \frac{T}{3}\right)}{n \omega_0} = -I_{pk} \frac{\sin(n \pi) - \sin\left(2n \pi/3\right)}{n \omega_0}. \] Since \(\sin(n \pi) = 0\): \[ = I_{pk} \frac{\sin\left(2n \pi/3\right)}{n \omega_0}. \] Add the two integrals and multiply by \(\frac{4}{T}\): \[ a_n = \frac{4}{T} \cdot \frac{I_{pk}}{n \omega_0} \left[ \sin\left(n \frac{\pi}{3}\right) + \sin\left(2n \frac{\pi}{3}\right) \right]. \] Using \(n \omega_0 = n \cdot \frac{2\pi}{T} \Rightarrow T(n \omega_0) = 2\pi n\): \[ a_n = \frac{4 I_{pk}}{2 \pi n} \left[ \sin\left(n \frac{\pi}{3}\right) + \sin\left(2n \frac{\pi}{3}\right) \right] = \frac{2 I_{pk}}{\pi n} \left[ \sin\left(n \frac{\pi}{3}\right) + \sin\left(2n \frac{\pi}{3}\right) \right]. \] Apply the sum-to-product identity: \[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right), \] with \(A = n \frac{\pi}{3}\), \(B = 2n \frac{\pi}{3}\): \[ \sin\left(n \frac{\pi}{3}\right) + \sin\left(2n \frac{\pi}{3}\right) = 2 \sin\left(\frac{3n \pi/3}{2}\right) \cos\left(-\frac{n \pi/3}{2}\right) = 2 \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right). \] Thus, \[ a_n = \frac{2 I_{pk}}{\pi n} \cdot 2 \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right) = \frac{4 I_{pk}}{\pi n} \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right). \] Since \(b_n = 0\) and \(a_0 = 0\), the Fourier coefficients are: \[ \boxed{a_n = \frac{4 I_{pk}}{\pi n} \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right), \quad b_n = 0, \quad a_0 = 0.} \] **Remarks:** - \(\sin\left(\frac{n \pi}{2}\right) = 0\) for every even \(n\), so only odd \(n\) produce nonzero \(a_n\). - For odd \(n\), \(\sin\left(n \pi/2\right) = \pm 1\), alternating sign depending on \(n \mod 4\). --- # 3) Polar form \((A_n, \theta_n)\) The amplitude and phase for the \(n\)-th harmonic: \[ A_n = \sqrt{a_n^2 + b_n^2} = |a_n|, \quad \theta_n = \begin{cases} 0, & a_n > 0, \\ \pi, & a_n < 0, \\ \text{undefined (zero amplitude)}, & a_n = 0. \end{cases} \] Using the closed form for \(a_n\): \[ \boxed{A_n = \left| \frac{4 I_{pk}}{\pi n} \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right) \right|, \quad \theta_n = \arg\left(\sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{6}\right)\right).} \] --- # 4) Final Fourier series (alternate form) Since only cosine terms are present: \[ \boxed{i(t) = \sum_{n=1}^{\infty} a_n \cos(n \omega_0 t), \quad \text{with } a_n \text{ as above and } a_0 = b_n = 0.} \]