From acd29ce2c6ce4e0b42223affcbe25d5c4584b7ad Mon Sep 17 00:00:00 2001
From: IliasN <iliascfpt@gmail.com>
Date: Sat, 12 Jun 2021 16:48:42 +0200
Subject: [PATCH] Correction correction exercice 3.2 fourrier
---
exercices/fourier_serie1.md | 8 ++++----
1 file changed, 4 insertions(+), 4 deletions(-)
diff --git a/exercices/fourier_serie1.md b/exercices/fourier_serie1.md
index 1d2df73..031d1a5 100644
--- a/exercices/fourier_serie1.md
+++ b/exercices/fourier_serie1.md
@@ -216,7 +216,7 @@ Corrigé +.#
En utilisant la formule
$$
-f[n]=\sum_{k=0}^{N-1}\hat f[n]e^{2\pi ink/N},
+f[n]=\frac{1}{N}\sum_{k=0}^{N-1}\hat f[n]e^{2\pi ink/N},
$$
on peut calculer la TFD de $\hat f=\{2, -1-i, 0, -1+i\}$ avec $N=4$.
On obtient donc
@@ -225,7 +225,7 @@ f[0]=\hat f[0]+\hat f[1]+\hat f[2]+\hat f[3]=0.
$$
Et ainsi de suite on obtient
\begin{align}
-f[1]&=\hat f[0]+\hat f[1]e^{\pi i/2}+\hat f[2]e^{\pi i}+\hat f[3]e^{3\pi i/2}=2+i(-1-i)+(-i)(-1+i)=4,\\
-\hat f[2]&=f[0]+f[1]e^{\pi i}+f[2]e^{2\pi i}+f[3]e^{3\pi i}=2+(-1)(-1-i)-1(-1+i)=4,\\
-\hat f[3]&=f[0]+f[1]e^{3\pi i/2}+f[2]e^{3\pi i}+f[3]e^{9\pi i/2}=2-i(-1-i)+i(-1+i)=0.
+f[1]&=\frac{1}{4}(\hat f[0]+\hat f[1]e^{\pi i/2}+\hat f[2]e^{\pi i}+\hat f[3]e^{3\pi i/2})=\frac{1}{4}(2+i(-1-i)+(-i)(-1+i))=1,\\
+\hat f[2]&=\frac{1}{4}(f[0]+f[1]e^{\pi i}+f[2]e^{2\pi i}+f[3]e^{3\pi i})=\frac{1}{4}(2+(-1)(-1-i)-1(-1+i))=1,\\
+\hat f[3]&=\frac{1}{4}(f[0]+f[1]e^{3\pi i/2}+f[2]e^{3\pi i}+f[3]e^{9\pi i/2})=\frac{1}{4}(2-i(-1-i)+i(-1+i))=0.
\end{align}
--
GitLab