From 03d0923a014fcce8aa6d818de64b31b8bbbc6713 Mon Sep 17 00:00:00 2001
From: Lovino Maxime <maximelovino@gmail.com>
Date: Thu, 30 Mar 2017 10:18:29 +0200
Subject: [PATCH] added missing i in TFD
---
cours.tex | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/cours.tex b/cours.tex
index 2340155..3caac50 100644
--- a/cours.tex
+++ b/cours.tex
@@ -2830,9 +2830,9 @@ Montrons à présent que la transformée inverse discrète de la transformée de
discrète donne bien la suite de départ
\begin{align}
f[n]&=\frac{1}{N}\sum_{k=0}^{N-1} \fh[k] e^{\frac{2\pi i k n}{N}},\nonumber\\
- &=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f[m] e^{-\frac{2\pi k m}{N}} e^{\frac{2\pi i k n}{N}},\nonumber\\
- &=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f[m] e^{\frac{2\pi k (n-m)}{N}},\nonumber\\
- &=\frac{1}{N}\sum_{m=0}^{N-1} f[m] \sum_{k=0}^{N-1} e^{\frac{2\pi k (n-m)}{N}},\nonumber\\
+ &=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f[m] e^{-\frac{2\pi i k m}{N}} e^{\frac{2\pi i k n}{N}},\nonumber\\
+ &=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f[m] e^{\frac{2\pi i k (n-m)}{N}},\nonumber\\
+ &=\frac{1}{N}\sum_{m=0}^{N-1} f[m] \sum_{k=0}^{N-1} e^{\frac{2\pi i k (n-m)}{N}},\nonumber\\
&=\frac{1}{N}\sum_{m=0}^{N-1} f[m] N \delta_{nm},\nonumber\\
&=f[n].
\end{align}
--
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