Let's solve the problem of finding the sequence $x(n)$ whose Discrete Fourier Transform (DFT) is given as $\{20, -4+4j, -4, -4-4j\}$ using the matrix method.

Step-by-Step Solution using Matrix Method

The DFT and inverse DFT can be represented using matrix multiplication. Given the DFT values $X(k) = \{20, -4+4j, -4, -4-4j\}$, we can find the time-domain sequence $x(n)$ by using the inverse DFT matrix.

1. Formulate the Inverse DFT Matrix

For a sequence of length $N = 4$, the inverse DFT matrix $W^{-1}$ is given by:

1 & 1 & 1 & 1 \\ 1 & e^{j\frac{2\pi}{4}} & e^{j\frac{4\pi}{4}} & e^{j\frac{6\pi}{4}} \\ 1 & e^{j\frac{4\pi}{4}} & e^{j\frac{8\pi}{4}} & e^{j\frac{12\pi}{4}} \\ 1 & e^{j\frac{6\pi}{4}} & e^{j\frac{12\pi}{4}} & e^{j\frac{18\pi}{4}} \end{bmatrix}.$$ We simplify the exponents: $$W^{-1} = \frac{1}{4} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & j & -1 & -j \\ 1 & -1 & 1 & -1 \\ 1 & -j & -1 & j \end{bmatrix}.$$ #### 2. Multiply the Inverse DFT Matrix with the DFT Values To find the sequence $$x(n)$$, multiply $$W^{-1}$$ by the column vector of the DFT values $$X(k)$$: $$x = W^{-1} \cdot X,$$ where $$X = \begin{bmatrix} 20 \\ -4 + 4j \\ -4 \\ -4 - 4j \end{bmatrix}.$$ Perform the matrix multiplication: $$x = \frac{1}{4} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & j & -1 & -j \\ 1 & -1 & 1 & -1 \\ 1 & -j & -1 & j \end{bmatrix} \begin{bmatrix} 20 \\ -4 + 4j \\ -4 \\ -4 - 4j \end{bmatrix}.$$ #### 3. Compute Each Element of $$x(n)$$ Let's compute each element $$x(n)$$ for $$n = 0, 1, 2, 3$$. - **For $$x(0)$$:** $$x(0) = \frac{1}{4} (20 + (-4 + 4j) + (-4) + (-4 - 4j)).$$ Combine the terms: $$x(0) = \frac{1}{4} (20 - 4 + 4j - 4 - 4 - 4j).$$ Simplify: $$x(0) = \frac{1}{4} (8) = 2.$$ - **For $$x(1)$$:** $$x(1) = \frac{1}{4} (20 + j(-4 + 4j) - 4 - j(-4 - 4j)).$$ Compute each multiplication: $$x(1) = \frac{1}{4} \left( 20 + (-4j - 4) + (-4) + (4j - 4) \right).$$ Combine like terms: $$x(1) = \frac{1}{4} (20 - 4 - 4 - 4j + 4j - 4) = \frac{1}{4} (16) = 4.$$ - **For $$x(2)$$:** $$x(2) = \frac{1}{4} (20 + (-1)(-4 + 4j) + (-4) + (-1)(-4 - 4j)).$$ Compute: $$x(2) = \frac{1}{4} (20 + (4 - 4j) - 4 + (4 + 4j)).$$ Combine like terms: $$x(2) = \frac{1}{4} (20 + 4 - 4 - 4j + 4 + 4j) = \frac{1}{4} (24) = 6.$$ - **For $$x(3)$$:** $$x(3) = \frac{1}{4} (20 + (-j)(-4 + 4j) + (-4) + j(-4 - 4j)).$$ Compute: $$x(3) = \frac{1}{4} (20 + (4j + 4) - 4 + (-4j - 4)).$$ Combine like terms: $$x(3) = \frac{1}{4} (20 + 4 - 4 - 4j + 4j - 4) = \frac{1}{4} (16) = 4.$$ ### Final Sequence The final sequence $$x(n)$$ is: $$x(n) = \{2, 4, 6, 4\}.$$ Would you like a more detailed explanation, or do you have any questions? Here are some related questions to consider: 1. What are the properties of the DFT matrix? 2. How does the matrix method compare to the direct computation of the IDFT? 3. Can we use the matrix method for sequences longer than 4 elements? 4. What are some practical applications of the inverse DFT in signal processing? 5. How does the Fast Fourier Transform (FFT) algorithm improve upon the DFT matrix method? **Tip:** The matrix method for calculating the DFT and IDFT is straightforward but can be computationally intensive for large sequences, which is why the FFT algorithm is preferred for practical applications.