La fonction de transfert dans le domaine fréquentiel de H(f) est donnée par:
H(f)=15sin⁡(π5fTe)sin⁡(πfTe)e−jπ4fTe{"version":"1.1","math":"H(f)=\frac{1}{5}\frac{\sin(\pi 5fT_e)}{\sin(\pi fT_e)}e^{-j\pi4fT_e}"}
L'expression du spectre de H(f) est:

Question

La fonction de transfert dans le domaine fréquentiel de H(f) est donnée par:

H(f)=15sin⁡(π5fTe)sin⁡(πfTe)e−jπ4fTe{"version":"1.1","math":"H(f)=\frac{1}{5}\frac{\sin(\pi 5fT_e)}{\sin(\pi fT_e)}e^{-j\pi4fT_e}"}

L'expression du spectre de H(f) est:

Answer

1 5 | sin ⁡ ( π 5 f T e ) | | sin ⁡ ( π f T e ) | × cos ⁡ ( 4 π f T e ) {"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5fT_e)|}{|\sin(\pi fT_e)|}	imes\cos(4\pi fT_e)"}

Answer

1 5 | sin ⁡ ( π 5 f T e ) | | sin ⁡ ( π f T e ) | × sin ⁡ ( 4 π f T e ) {"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5fT_e)|}{|\sin(\pi fT_e)|}	imes\sin(4\pi fT_e)"}

Answer

1 5 sin ⁡ ( π 5 f T e ) sin ⁡ ( π f T e ) {"version":"1.1","math":"\frac{1}{5}\frac{\sin(\pi 5fT_e)}{\sin(\pi fT_e)}"}

Answer

1 5 | sin ⁡ ( π 5 f T e ) | | sin ⁡ ( π f T e ) | × e − π 4 f T e {"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5fT_e)|}{|\sin(\pi fT_e)|}	imes e^{-\pi4fT_e}"}

Answer

{"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5f/fe)|}{|\sin(\pi f/fe)|}"} 1 5 | sin ⁡ ( π 5 f T e ) | | sin ⁡ ( π f T e ) | {"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5fT_e)|}{|\sin(\pi fT_e)|}"}

Answer

1 5 | sin ⁡ ( π 5 f T e ) | | sin ⁡ ( π f T e ) | × | cos ⁡ ( 4 π f T e ) − sin ⁡ ( 4 π f T e ) | {"version":"1.1","math":"\frac{1}{5}\frac{|\sin(\pi 5fT_e)|}{|\sin(\pi fT_e)|}	imes|\cos(4\pi fT_e) -\sin(4\pi fT_e)|"}

Crowdly

La fonction de transfert dans le domaine fréquentiel de H(f) est donnée par: H (...

Want instant access to all verified answers on learning.devinci.fr?