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# Lecture 12: October 24, 2023 ## Example 1 A signal $x[n]$ is non-zero over the interval $N_1 \leq n \leq N_2$. 1. Determine the length $N$ of the 4pt DFT $X[k]$. 2. Determine the relationship between $x[n]$ and $X[k]$ such that $X = X = 0$. ### Solution 1. $N \geq N_2 - N_1 + 1$ 2. $X =...

# Lecture 12: October 24, 2023 ## Example 1 A signal $x[n]$ is non-zero over the interval $N_1 \leq n \leq N_2$. 1. Determine the length $N$ of the 4pt DFT $X[k]$. 2. Determine the relationship between $x[n]$ and $X[k]$ such that $X = X = 0$. ### Solution 1. $N \geq N_2 - N_1 + 1$ 2. $X = \sum_{n=0}^{N-1}x[n] = 0$ $x + x + x + x = 0$ $X = \sum_{n=0}^{N-1}(-1)^n x[n] = 0$ $x - x + x - x = 0$ Adding both equations: $2x + 2x = 0 \Rightarrow x = -x$ Subtracting both equations: $2x + 2x = 0 \Rightarrow x = -x$ ## Example 2 $x(t) = 10\cos(20\pi t + \frac{\pi}{4})$ is sampled at 50 Hz. $x[n] = x(nT) = 10\cos(20\pi nT + \frac{\pi}{4})$, where $T = \frac{1}{50}$ $x[n] = 10\cos(\frac{2\pi}{5} n + \frac{\pi}{4})$ 1. Determine the 5pt DFT $X[k]$. 2. Determine what happens when we take IDFT of $X[k]$. ### Solution 1. $X[k] = 50[\delta[k-1] + \delta[k-4]]$ 2. $x[n] = 10\cos(\frac{2\pi}{5} n + \frac{\pi}{4})$ ## Example 3 $x[n] = \cos(\frac{2\pi}{7} n)$ for $0 \leq n \leq 6$ 1. Compute 7pt DFT. 2. Compute 14pt DFT. ### Solution 1. $X[k] = \frac{7}{2}[\delta[k-1] + \delta[k-6]]$ 2. $X[k] = \frac{7}{2}[\delta[k-1] + \delta[k-13]]$ ## Example 4 $x[n] = [1, 2, 3, 4]$ $h[n] = [5, 6, 7, 8]$ 1. Determine linear convolution. 2. Determine circular convolution. ### Solution 1. Linear Convolution: | | 1 | 2 | 3 | 4 | | :---- | :-: | :-: | :-: | :-: | | **5** | 5 | 10 | 15 | 20 | | **6** | 6 | 12 | 18 | 24 | | **7** | 7 | 14 | 21 | 28 | | **8** | 8 | 16 | 24 | 32 | $y[n] = [5, 16, 34, 60, 61, 52, 32]$ 2. Circular Convolution: $y[n] = [5, 16, 34, 60] + [61, 52, 32] = [66, 68, 66, 92]$