DocTron-Formula / asset /test_jsons /paragraph-level.json
zhongyufeng02
Track binary files with Git LFS
c0be972
[
{
"qid": "part-00025-8bdd594e-bf2f-4dee-8277-9c33e8468d28-c000_4210_multi",
"messages": [
{
"content": "<image> Analyze the image. Extract and output only the LaTeX formulas present in the image, in LaTeX code format. Ignore inline formulas, all other text, and do not include any explanations.",
"role": "user"
},
{
"content": "\\begin{align}\n&{\\text{Start with defining $2N$-Point DFT of $x_{2N}[n]\\ldots$}}\\\\ X_{2N}[k] &= \\sum_{n=0}^{2N-1} {x_{2N}[n] e^{-j\\frac{2\\pi}{2N}nk}} \\\\ \\\\ &{\\text{first split the sum and then substitude $n=n-N$ in the $2^{nd}$}}\\\\ &= \\sum_{n=0}^{N-1} {x_{2N}[n] e^{-j\\frac{2\\pi}{N}n(k/2)}} + \\sum_{n=N}^{2N-1} {x_{2N}[n] e^{-j\\frac{2\\pi}{2N}nk}} \\\\ \\\\ &{\\text{recognise $x_{2N}[n]=x_N[n]$, $x_{2N}[n+N]=x_N[n]$ }}\\\\ &= \\sum_{n=0}^{N-1} {x_{2N}[n] e^{-j\\frac{2\\pi}{N}n(k/2)}} + \\sum_{n=0}^{N-1} {x_{2N}[n+N] e^{-j\\frac{2\\pi}{2N}(n+N)k}} \\\\ \\\\ &{\\text{Expand the 2nd sum's multiplier factor}}\\\\ &= \\sum_{n=0}^{N-1} {x_N[n] e^{-j\\frac{2\\pi}{N}n(k/2)}} + \\sum_{n=0}^{N-1} {x_N[n] e^{-j\\frac{2\\pi}{N}(n+N)(k/2)}} \\\\ \\\\ &{\\text{Replace $e^{-2j\\frac\\pi N}$ with $W_N$ for simplicity }} \\\\ &= \\sum_{n=0}^{N-1} {x_N[n] W_N^{nk/2}} + W_N^{Nk/2}\\sum_{n=0}^{N-1} {x_N[n] W_N^{nk/2}} \\\\ \\\\ &{\\text{Recognise the sums as $N$-Point DFT $X[k]$ of $x[n]$, at $\\frac k2$}}\\\\ &= X_N[k/2] + e^{-j\\pi k} X_N[k/2] \\\\ \\\\ &= X_N[k/2] \\cdot \\big( 1 + (-1)^k \\big) \\\\ \\\\ &= \\begin{cases} 2 X_N[k/2] , &\\scriptstyle{\\text{k=0,2,4,...,2N-2 }}\\\\ 0 , &\\scriptstyle{\\text{k=1,3,5,...,2N-1 }}\\\\ \\end{cases}\n\\end{align}\n$$\nX_{2N}[k] = 0 ~~,~~ \\text{for}~~ k=1,3,5...,2N-1\n$$",
"role": "assistant"
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],
"images": [
"./asset/images/question_part-00025-8bdd594e-bf2f-4dee-8277-9c33e8468d28-c000_4210_multi.png"
],
"model_answer": "\\begin{align}\n\\text{Start with defining 2N-Point DFT of } x_{2N}[n] \\ldots \\\\ X_{2N}[k] &= \\sum_{n=0}^{2N-1} x_{2N}[n] e^{-j \\frac{2\\pi}{2N} nk} \\\\ \\\\ &\\text{first split the sum and then substitute } n = n-N \\text{ in the } 2^{nd} \\\\ &= \\sum_{n=0}^{N-1} x_{2N}[n] e^{-j \\frac{2\\pi}{2N} n (k/2)} + \\sum_{n=N}^{2N-1} x_{2N}[n] e^{-j \\frac{2\\pi}{2N} nk} \\\\ \\\\ &\\text{recognise } x_{2N}[n] = x_N[n], \\ x_{2N}[n+N] = x_N[n] \\\\ &= \\sum_{n=0}^{N-1} x_{2N}[n] e^{-j \\frac{2\\pi}{2N} n (k/2)} + \\sum_{n=0}^{N-1} x_{2N}[n+N] e^{-j \\frac{2\\pi}{2N} (n+N) k} \\\\ \\\\ &\\text{Expand the 2nd sum's multiplier factor} \\\\ &= \\sum_{n=0}^{N-1} x_N[n] e^{-j \\frac{2\\pi}{N} n (k/2)} + \\sum_{n=0}^{N-1} x_N[n] e^{-j \\frac{2\\pi}{N} (n+N) (k/2)} \\\\ \\\\ &\\text{Replace } e^{-2j\\frac{\\pi}{N}} \\text{ with } W_N \\text{ for simplicity} \\\\ &= \\sum_{n=0}^{N-1} x_N[n] W_N^{nk/2} + W_N^{Nk/2} \\sum_{n=0}^{N-1} x_N[n] W_N^{nk/2} \\\\ \\\\ &\\text{Recognise the sums as N-Point DFT } X[k] \\text{ of } x[n] \\text{, at } \\frac{k}{2} \\\\ &= X_N[k/2] + e^{-j \\pi k} X_N[k/2] \\\\ \\\\ &= X_N[k/2] \\cdot \\big( 1 + (-1)^k \\big) \\\\ \\\\ &= \\begin{cases} 2 X_N[k/2], & \\text{k = 0,2,4,...2N2} \\\\ 0, & \\text{k = 1,3,5,...2N-1} \\end{cases} \\\\\n\\end{align}\n$$\nX_{2N}[k] = 0 ~~,~~ \\text{for}~~ k = 1,3,5...,2N-1\n$$",
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]