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Give step-by-step solution with explanation and final answer:Problem 5 : : Obtain the Fourier representation for the following signals a) x(n) ={.-10,203, B 0,1,0310203,..} _ 3 b) ym ={..321232123212,..}

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Give step-by-step solution with explanation and final answer:Uploaded ImageProblem 5 : : Obtain the Fourier representation for the following signals a) x(n) ={.-10,203, B 0,1,0310203,..} _ 3 b) ym ={..321232123212,..}

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Asked by: Abhishek
Created at: 2025-09-21 14:29:33

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Let's solve each part step by step. --- ## **a) \( x(n) = \{ ..., 1, 0, 2, 0, 3, \underline{1}, 0, 1, 0, 3, 1, 0, 2, 0, 3, ... \} \)** ### **Step 1: Identify the Period \( N \)** Let's find the period by checking the repeating pattern: - Sequence: \( 1, 0, 2, 0, 3 \) - Sequence after next 5: \( 1, 0, 2, 0, 3 \) So, **period \( N = 5 \)**. ### **Step 2: Write One Period** Let's write one period starting from \( n = 0 \): - \( x(0) = 1 \) - \( x(1) = 0 \) - \( x(2) = 2 \) - \( x(3) = 0 \) - \( x(4) = 3 \) ### **Step 3: Discrete-Time Fourier Series (DTFS) Coefficients** For a periodic sequence with period \( N \): \[ X(k) = \frac{1}{N} \sum_{n=0}^{N-1} x(n) e^{-j \frac{2\pi}{N} k n} \] where \( k = 0, 1, ..., N-1 \). #### **Compute \( X(k) \):** \[ X(k) = \frac{1}{5} \left[ x(0) + x(1)e^{-j\frac{2\pi}{5}k} + x(2)e^{-j\frac{4\pi}{5}k} + x(3)e^{-j\frac{6\pi}{5}k} + x(4)e^{-j\frac{8\pi}{5}k} \right] \] Substitute values: - \( x(0) = 1 \) - \( x(1) = 0 \) - \( x(2) = 2 \) - \( x(3) = 0 \) - \( x(4) = 3 \) \[ X(k) = \frac{1}{5} \left[ 1 + 0 + 2e^{-j\frac{4\pi}{5}k} + 0 + 3e^{-j\frac{8\pi}{5}k} \right] \] \[ X(k) = \frac{1}{5} \left[ 1 + 2e^{-j\frac{4\pi}{5}k} + 3e^{-j\frac{8\pi}{5}k} \right] \] --- ## **b) \( y(n) = \{ ..., 3, 2, 1, 2, \underline{3}, 2, 1, 2, 3, 2, 1, 2, ... \} \)** ### **Step 1: Identify the Period \( N \)** Let's check the repeating pattern: - \( 3, 2, 1, 2 \) - Next: \( 3, 2, 1, 2 \) - So, **period \( N = 4 \)** ### **Step 2: Write One Period** Starting from \( n = 0 \): - \( y(0) = 3 \) - \( y(1) = 2 \) - \( y(2) = 1 \) - \( y(3) = 2 \) ### **Step 3: DTFS Coefficients** \[ Y(k) = \frac{1}{4} \sum_{n=0}^{3} y(n) e^{-j \frac{2\pi}{4} k n} \] Substitute values: \[ Y(k) = \frac{1}{4} \left[ 3 + 2e^{-j\frac{2\pi}{4}k} + 1e^{-j\frac{4\pi}{4}k} + 2e^{-j\frac{6\pi}{4}k} \right] \] \[ Y(k) = \frac{1}{4} \left[ 3 + 2e^{-j\frac{\pi}{2}k} + e^{-j\pi k} + 2e^{-j\frac{3\pi}{2}k} \right] \] --- ## **Final Answers** ### **a)** \[ \boxed{ X(k) = \frac{1}{5} \left[ 1 + 2e^{-j\frac{4\pi}{5}k} + 3e^{-j\frac{8\pi}{5}k} \right], \quad k = 0,1,2,3,4 } \] ### **b)** \[ \boxed{ Y(k) = \frac{1}{4} \left[ 3 + 2e^{-j\frac{\pi}{2}k} + e^{-j\pi k} + 2e^{-j\frac{3\pi}{2}k} \right], \quad k = 0,1,2,3 } \] If you need the explicit numerical values for each \( k \), let me know!

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