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clean up notation and LaTeX.
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robert bristow-johnson
robert bristow-johnson
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$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $$ X[k] = \frac{1}{N} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $$ x[n] = \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $$ X[k] = \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $$ x[n] = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{j \frac{2\pi}{N} n k} $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{-j \frac{2\pi}{N} n k} $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION 1

$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 2

$ X[k] = \frac{1}{N} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 3

$ X[k] = \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $

NOTATION 4

$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{-j \frac{2\pi}{N} n k} $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

Removed bolded fonts and shouting in the end.
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jojeck ♦
jojeck ♦
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$X$ represents sample in frequency domain and $x$ represents samples in time domain.$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION - 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

PLEASE OBSERVE THE SCALING FACTOR $\frac{1}{N}$ AND CHANGE OF NEGATIVE SIGN $-$ OVER THE EXPONENT TERM ${e^{\left( \cdot \right)j\frac{{2\pi }}{N}nk} }$. WHY IS EVERY NOTATION FOR DFT IS VALID? Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION - 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

PLEASE OBSERVE THE SCALING FACTOR $\frac{1}{N}$ AND CHANGE OF NEGATIVE SIGN $-$ OVER THE EXPONENT TERM ${e^{\left( \cdot \right)j\frac{{2\pi }}{N}nk} }$. WHY IS EVERY NOTATION FOR DFT IS VALID?

$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

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Vineel Kumar Veludandi
Vineel Kumar Veludandi
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please give the reason why every notation for DFT is valid?

$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION - 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION - 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

PLEASE OBSERVE THE SCALING FACTOR $\frac{1}{N}$ AND CHANGE OF NEGATIVE SIGN $-$ OVER THE EXPONENT TERM ${e^{\left( \cdot \right)j\frac{{2\pi }}{N}nk} }$. WHY IS EVERY NOTATION FOR DFT IS VALID?