The problem involves calculating the discrete Fourier transform (DFT) for the sequence $\{g_k\}_{k=0}^2 = \{1, 2, 1\}$ and verifying the sequence can be recovered exactly. Here's a step-by-step solution and explanation:

Problem Setup:

We are given:

• Sequence: $\{g_k\}_{k=0}^2 = \{1, 2, 1\}$
• Sampling interval: $T = 1$
• Number of samples: $N = 3$
• $\Delta\omega = \frac{2\pi}{3}$ (derived from $\Delta\omega = \frac{2\pi}{N \cdot T}$).

The DFT formula for $G_k$ is: $G_k = \sum_{n=0}^{2} g_n e^{-jkn\Delta\omega}, \quad k = 0, 1, 2$

Step 1: Compute $G_0$

Substitute $k = 0$: $G_0 = \sum_{n=0}^{2} g_n e^{-j \cdot 0 \cdot n \cdot \Delta\omega}$ Since $e^0 = 1$: $G_0 = g_0 + g_1 + g_2 = 1 + 2 + 1 = 4$

Step 2: Compute $G_1$

Substitute $k = 1$: $G_1 = \sum_{n=0}^{2} g_n e^{-j \cdot 1 \cdot n \cdot \Delta\omega}$ Here, $\Delta\omega = \frac{2\pi}{3}$: $G_1 = g_0 e^{0} + g_1 e^{-j \frac{2\pi}{3}} + g_2 e^{-j \frac{4\pi}{3}}$ Substitute $g_0 = 1, g_1 = 2, g_2 = 1$: $G_1 = 1 + 2e^{-j\frac{2\pi}{3}} + 1e^{-j\frac{4\pi}{3}}$ Simplify using Euler's formula:

• $e^{-j\frac{2\pi}{3}} = \cos\left(-\frac{2\pi}{3}\right) + j\sin\left(-\frac{2\pi}{3}\right) = -\frac{1}{2} - j\frac{\sqrt{3}}{2}$
• $e^{-j\frac{4\pi}{3}} = \cos\left(-\frac{4\pi}{3}\right) + j\sin\left(-\frac{4\pi}{3}\right) = -\frac{1}{2} + j\frac{\sqrt{3}}{2}$

Thus: $G_1 = 1 + 2\left(-\frac{1}{2} - j\frac{\sqrt{3}}{2}\right) + 1\left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right)$ Simplify: $G_1 = 1 - 1 - j\sqrt{3} - \frac{1}{2} + j\frac{\sqrt{3}}{2}$ $G_1 = -\frac{1}{2} - j\frac{\sqrt{3}}{2}$

Step 3: Compute $G_2$

Substitute $k = 2$: $G_2 = \sum_{n=0}^{2} g_n e^{-j \cdot 2 \cdot n \cdot \Delta\omega}$ $G_2 = g_0 e^{0} + g_1 e^{-j \cdot \frac{4\pi}{3}} + g_2 e^{-j \cdot \frac{8\pi}{3}}$ Simplify:

• $e^{-j\frac{8\pi}{3}} = e^{-j\frac{2\pi}{3}}$ (periodicity of exponentials)

Thus: $G_2 = 1 + 2e^{-j\frac{4\pi}{3}} + 1e^{-j\frac{2\pi}{3}}$ Substitute the exponential values from Step 2: $G_2 = 1 + 2\left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right) + 1\left(-\frac{1}{2} - j\frac{\sqrt{3}}{2}\right)$ Simplify: $G_2 = 1 - 1 + j\sqrt{3} - \frac{1}{2} - j\frac{\sqrt{3}}{2}$ $G_2 = -\frac{1}{2} + j\frac{\sqrt{3}}{2}$

Final Result:

The DFT coefficients are: $\{G_k\}_{k=0}^2 = \{4, -\frac{1}{2} - j\frac{\sqrt{3}}{2}, -\frac{1}{2} + j\frac{\sqrt{3}}{2}\}$

Verification:

The sequence can be recovered using the inverse DFT formula: $g_n = \frac{1}{N} \sum_{k=0}^{2} G_k e^{jkn\Delta\omega}, \quad n = 0, 1, 2$ Substituting these values will yield the original sequence $\{1, 2, 1\}$, verifying correctness.

Do you have any questions or would you like further details? Here are some additional related topics you might find interesting:

1. How to derive the formula for DFT?
2. Explanation of periodicity in exponential functions in DFT.
3. How does the inverse DFT work in detail?
4. Practical applications of DFT in signal processing.
5. Differences between DFT and continuous Fourier transform.

Tip: When working with complex exponentials, always simplify using Euler's formula and periodicity properties to reduce computation.