Empfohlen
Empfohlen
Weitere ähnliche Inhalte
Ähnlich wie Clase 11 - Mat IV - Transformada de Laplace (2).pdf
Ähnlich wie Clase 11 - Mat IV - Transformada de Laplace (2).pdf (20)
Kürzlich hochgeladen
Kürzlich hochgeladen (20)
Clase 11 - Mat IV - Transformada de Laplace (2).pdf
- 1. 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨𝑷𝑳𝑨𝑪𝑬 𝑳−𝟏 𝒂𝒇 𝒔 ± 𝒃𝒈(𝒔) = 𝒂𝑳−𝟏 𝒇(𝒔) ± 𝒃𝑳−𝟏 𝒈(𝒔) = 𝒂𝑭 𝒕 ± 𝒃𝑮 𝒕 𝑺𝒆𝒂 𝑭 𝒕 𝒖𝒏𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒄𝒐𝒏𝒕í𝒏𝒖𝒂 𝒑𝒐𝒓 𝒕𝒓𝒂𝒎𝒐𝒔 𝒚 𝒅𝒆 𝒐𝒓𝒅𝒆𝒏 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒄𝒊𝒂𝒍 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳 𝑭 𝒕 = 𝒇 𝒔 , 𝒅𝒐𝒏𝒅𝒆 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 𝒔𝒆 𝒍𝒍𝒂𝒎𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒅𝒂 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒆 𝑳𝒂𝒑𝒍𝒂𝒄𝒆. 𝑺𝒊 𝒒𝒖𝒆𝒓𝒆𝒎𝒐𝒔 𝒉𝒂𝒍𝒍𝒂𝒓 𝑭 𝒕 𝒄𝒖𝒂𝒏𝒅𝒐 𝒇 𝒔 = 𝟑 𝒔 − 𝟒 , 𝒆𝒏 𝒆𝒇𝒆𝒄𝒕𝒐 𝑳−𝟏 𝒇(𝒔) = 𝟑. 𝑳−𝟏 𝒇(𝒔) = 𝟑𝒆𝟒𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑭 𝒕 = 𝟑𝒆𝟒𝒕 𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝑷𝑹𝑶𝑷𝑰𝑬𝑫𝑨𝑫𝑬𝑺 𝑫𝑬 𝑳𝑨 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨𝑷𝑳𝑨𝑪𝑬 𝟏. 𝑷𝒓𝒐𝒑𝒊𝒆𝒅𝒂𝒅 𝒅𝒆 𝑳𝒊𝒏𝒆𝒂𝒍𝒊𝒅𝒂𝒅 𝑷𝒐𝒓 𝒆𝒋𝒆𝒎𝒑𝒍𝒐: 𝑯𝒂𝒍𝒍𝒂𝒓 𝑭 𝒕 , 𝒔𝒊 𝒇 𝒔 = 𝟏 𝒔𝟐 − 𝟏 𝒔𝟐 + 𝟒 + 𝟏 𝒔 − 𝟒 𝑷𝒐𝒓 𝒆𝒋𝒆𝒎𝒑𝒍𝒐: 𝑯𝒂𝒍𝒍𝒂𝒓 𝑭 𝒕 , 𝒄𝒖𝒂𝒏𝒅𝒐 𝒇 𝒔 = 𝒔 − 𝟒 (𝒔 − 𝟒)𝟐+𝟗 𝟐. 𝑷𝒓𝒊𝒎𝒆𝒓𝒂 𝒑𝒓𝒐𝒑𝒊𝒆𝒅𝒂𝒅 𝒅𝒆 𝑻𝒓𝒂𝒔𝒍𝒂𝒄𝒊ó𝒏 𝑺𝒊 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝒇(𝒔 − 𝒂) = 𝒆𝒂𝒕𝑭 𝒕
- 2. 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨 𝑫𝑬𝑹𝑰𝑽𝑨𝑫𝑨 𝑺𝒊 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 න 𝟎 ∞ 𝒇 𝒖 𝒅𝒖 = 𝑭(𝒕) 𝒕 𝑺𝒊 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝒇 𝒏 (𝒔) = −𝟏 𝒏 𝒕𝒏 𝑭(𝒕) 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨𝑺 𝑰𝑵𝑻𝑬𝑮𝑹𝑨𝑳𝑬𝑺 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑷𝒐𝒓 𝒆𝒋𝒆𝒎𝒑𝒍𝒐: 𝑯𝒂𝒍𝒍𝒂𝒓 𝑳−𝟏 𝒔 (𝒔 − 𝟏)𝟓 𝑳−𝟏 𝟏 (𝒔 − 𝟏)𝟓 = 𝒆−𝒕 𝑳−𝟏 𝟏 𝒔𝟓 = 𝒆−𝒕 𝒕𝟒 𝟐𝟒 , 𝒅𝒆 𝒅𝒐𝒏𝒅𝒆: 𝑳−𝟏 𝒔 (𝒔 − 𝟏)𝟓 = 𝒕𝟑 𝒆−𝒕 𝟔 − 𝒕𝟒 𝒆−𝒕 𝟐𝟒 = 𝒆−𝒕 𝟐𝟒 𝟒𝒕𝟑 − 𝒕𝟒 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨 𝑴𝑼𝑳𝑻𝑰𝑷𝑳𝑰𝑪𝑨𝑪𝑰𝑶𝑵 𝑷𝑶𝑹 𝑺 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑺𝒊 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 , 𝒚 𝑭 𝟎 = 𝟎, 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝒔𝒇(𝒔) = 𝑭′ 𝒕
- 3. 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑫𝑬 𝑳𝑨 𝑫𝑰𝑽𝑰𝑺𝑰𝑶𝑵 𝑷𝑶𝑹 𝑺 𝑻𝒆𝒏𝒆𝒎𝒐𝒔 𝒒𝒖𝒆: 𝑳−𝟏 𝑳𝒏(𝟏 + 𝟏 𝒔𝟐 ) = 𝑳−𝟏 𝑳𝒏 𝒔𝟐 + 𝟏 − 𝑳𝒏𝒔𝟐 𝑺𝒊 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝒇(𝒔) 𝒔 = න 𝟎 ∞ 𝒇 𝒖 𝒅𝒖 𝑷𝒐𝒓 𝒆𝒋𝒆𝒎𝒑𝒍𝒐: 𝑯𝒂𝒍𝒍𝒂𝒓 𝑳−𝟏 𝟏 𝒔 𝑳𝒏(𝟏 + 𝟏 𝒔𝟐 ) 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝑳−𝟏 𝟏 𝒔 𝑳𝒏(𝟏 + 𝟏 𝒔𝟐 ) = න 𝟎 ∞ 𝟐(𝟏 − 𝒄𝒐𝒔𝒖) 𝒖 𝒅𝒖 𝒍𝒖𝒆𝒈𝒐 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: 𝑳−𝟏 𝑳𝒏(𝟏 + 𝟏 𝒔𝟐 ) = 𝟐(𝟏 − 𝒄𝒐𝒔𝒕) 𝒕 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝑳−𝟏 𝑳𝒏 𝒔𝟐 + 𝟏 − 𝑳𝒏𝒔𝟐 = 𝟏 𝒕 𝑳−𝟏 𝟐𝒔 𝒔𝟐 + 𝟏 − 𝟐 𝒔 = − 𝟏 𝒕 𝟐𝒄𝒐𝒔𝒕 − 𝟐
- 4. 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑷𝑶𝑹 𝑬𝑳 𝑴𝑬𝑻𝑶𝑫𝑶 𝑫𝑬 𝑭𝑹𝑨𝑪𝑪𝑰𝑶𝑵𝑬𝑺 𝑷𝑨𝑹𝑪𝑰𝑨𝑳𝑬𝑺 𝑪𝒐𝒎𝒐 𝟏𝟏𝒔𝟐 − 𝟐𝒔 + 𝟓 (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) 𝒆𝒔 𝒓𝒂𝒄𝒊𝒐𝒏𝒂𝒍 𝒑𝒓𝒐𝒑𝒊𝒂 𝟏𝟏𝒔𝟐 − 𝟐𝒔 + 𝟓 (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) = 𝑨 𝒔 − 𝟐 + 𝑩 𝟐𝒔 − 𝟏 + 𝑪 𝒔 + 𝟏 = 𝑨 𝟐𝒔 − 𝟏 𝒔 + 𝟏 + 𝑩 𝒔 − 𝟐 𝒔 + 𝟏 + 𝑪(𝒔 − 𝟐)(𝟐𝒔 − 𝟏) (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) 𝑬𝒔𝒕𝒆 𝒎é𝒕𝒐𝒅𝒐 𝒔𝒆 𝒂𝒑𝒍𝒊𝒄𝒂 𝒂 𝒍𝒂𝒔 𝒇𝒖𝒏𝒄𝒊𝒐𝒏𝒆𝒔 𝒓𝒂𝒄𝒊𝒐𝒏𝒂𝒍𝒆𝒔 𝒑𝒓𝒐𝒑𝒊𝒂𝒔 𝒅𝒆 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝑷𝒏(𝒔) 𝑸𝒎(𝒔) , 𝒎 > 𝒏 𝑯𝒂𝒍𝒍𝒂𝒓 𝑳−𝟏 𝟏𝟏𝒔𝟐 − 𝟐𝒔 + 𝟓 (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) 𝑺𝒊 𝒔 + 𝟏 = 𝟎 ; 𝒔 = −𝟏 ; 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑪 = 𝟐 𝑺𝒊 𝒔 − 𝟐 = 𝟎 ; 𝒔 = 𝟐 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑨 = 𝟓 ; 𝑺𝒊 𝟐𝒔 − 𝟏 = 𝟎 ; 𝒔 = 𝟏 𝟐 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑩 = −𝟑 𝑳−𝟏 𝟏𝟏𝒔𝟐 − 𝟐𝒔 + 𝟓 (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) = 𝑳−𝟏 𝟓 𝒔 − 𝟐 − 𝟑 𝟐𝒔 − 𝟏 + 𝟐 𝒔 + 𝟏 = 𝑳−𝟏 𝟓 𝟏 𝒔 − 𝟐 − 𝟑 𝟐 𝟏 (𝒔 − 𝟏 𝟐) + 𝟐 𝟏 𝒔 + 𝟏 𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝟏. 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏
- 5. 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑭𝑶𝑹𝑴𝑼𝑳𝑨 𝑫𝑬𝑳 𝑫𝑬𝑺𝑨𝑹𝑹𝑶𝑳𝑳𝑶 𝑫𝑬 𝑯𝑬𝑨𝑽𝑰𝑺𝑰𝑫𝑬 𝑸 𝒔 = 𝒔 − 𝟐 𝒔 + 𝟏 𝒔 + 𝟑 = 𝒔𝟑 + 𝟐𝒔𝟐 − 𝟓𝒔 − 𝟔 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑸′ 𝒔 = 𝟑𝒔𝟐 + 𝟒𝒔 − 𝟓 𝑺𝒆𝒂𝒏 𝑷 𝒔 𝒚 𝑸 𝒔 𝒑𝒐𝒍𝒊𝒏𝒐𝒎𝒊𝒐𝒔, 𝒅𝒐𝒏𝒅𝒆 𝒆𝒍 𝒈𝒓𝒂𝒅𝒐 𝒅𝒆 𝑷 𝒔 𝒆𝒔 𝒎𝒆𝒏𝒐𝒓 𝒒𝒖𝒆 𝒆𝒍 𝒈𝒓𝒂𝒅𝒐 𝑸 𝒔 . 𝑺𝒊 𝑸 𝒔 𝒕𝒊𝒆𝒏𝒆 𝒓𝒂í𝒄𝒆𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒕𝒆𝒔 𝒂𝟏, 𝒂𝟐, ⋯ , 𝒂𝒏 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝑷 𝒔 𝑸 𝒔 = 𝒌=𝟏 𝒏 𝑷(𝒂𝒌) 𝑸′(𝒂𝒌) 𝒆𝒂𝒌𝒕 𝑯𝒂𝒍𝒍𝒂𝒓 𝑳−𝟏 𝟏𝟗𝒔 + 𝟑𝟕 (𝒔 − 𝟐)(𝒔 + 𝟏)(𝒔 + 𝟑) 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 E𝒋𝒆𝒎𝒑𝒍𝒐 𝑳−𝟏 𝟏𝟏𝒔𝟐 − 𝟐𝒔 + 𝟓 (𝒔 − 𝟐)(𝟐𝒔 − 𝟏)(𝒔 + 𝟏) = 𝟓𝒆𝟐𝒕 − 𝟑 𝟐 𝒆 𝒕 𝟐 + 𝟐𝒆−𝒕
- 6. 𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑳−𝟏 𝟏𝟗𝒔 + 𝟑𝟕 (𝒔 − 𝟐)(𝒔 + 𝟏)(𝒔 + 𝟑) = 𝑷(𝟐) 𝑸′(𝟐) . 𝒆𝟐𝒕 + 𝑷(−𝟏) 𝑸′(−𝟏) 𝒆−𝒕 + 𝑷(−𝟑) 𝑸′(−𝟑) 𝒆−𝟑𝒕 = 𝟕𝟓 𝟏𝟓 𝒆𝟐𝒕 + 𝟏𝟖 −𝟔 𝒆−𝒕 + −𝟐𝟎 𝟏𝟎 𝒆−𝟑𝒕 𝑷𝒐𝒓 𝒕𝒂𝒏𝒕𝒐 𝑳−𝟏 𝟏𝟗𝒔 + 𝟑𝟕 (𝒔 − 𝟐)(𝒔 + 𝟏)(𝒔 + 𝟑) = 𝟓𝒆𝟐𝒕 − 𝟑𝒆−𝒕 − 𝟐𝒆−𝟑𝒕 𝑪𝒐𝒎𝒐 𝑷 𝒔 = 𝟏𝟗𝒔 + 𝟑𝟕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝑷 𝟐 = 𝟕𝟓 , 𝑷 −𝟏 = 𝟏𝟖 , 𝑷 −𝟑 = −𝟐𝟎 𝒍𝒖𝒆𝒈𝒐 𝑸′ 𝟐 = 𝟏𝟓 ; 𝑸′ −𝟏 = −𝟔 , 𝑸′ −𝟑 = 𝟏𝟎
- 7. 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑳𝑨 𝑪𝑶𝑵𝑽𝑶𝑳𝑼𝑪𝑰𝑶𝑵 𝑷𝑨𝑹𝑨 𝑳𝑨 𝑻𝑹𝑨𝑵𝑺𝑭𝑶𝑹𝑴𝑨𝑫𝑨 𝑰𝑵𝑽𝑬𝑹𝑺𝑨 𝑺𝒆𝒂𝒏 𝑳−𝟏 𝒇(𝒔) = 𝑭 𝒕 𝒚 𝑳−𝟏 𝒈(𝒔) = 𝑮 𝒕 , 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝑳−𝟏 𝒇 𝒔 . 𝒈(𝒔) = න 𝟎 𝒕 𝑭 𝒖 𝑮 𝒕 − 𝒖 𝒅𝒖 = 𝑭 ∗ 𝑮 , 𝒅𝒐𝒏𝒅𝒆 𝑭 ∗ 𝑮 𝒆𝒔 𝒍𝒂 𝒄𝒐𝒏𝒗𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒅𝒆 𝑭 𝒚 𝑮 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓: 𝑳−𝟏 𝟏 (𝒔 − 𝟏)(𝒔 + 𝟒) 𝑯𝒂𝒄𝒆𝒎𝒐𝒔 𝒇 𝒔 = 𝟏 𝒔 − 𝟏 , 𝒚 𝒈 𝒔 = 𝟏 𝒔 + 𝟒 𝒅𝒆 𝒅𝒐𝒏𝒅𝒆 𝑳−𝟏 𝒇 𝒔 = 𝒆𝒕 = 𝑭 𝒕 : 𝑳−𝟏 𝒈 𝒔 = 𝒆−𝟒𝒕 = 𝑮 𝒕 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝑷𝒐𝒓 𝒆𝒍 𝒕𝒆𝒐𝒓𝒆𝒎𝒂 𝒅𝒆 𝒄𝒐𝒏𝒗𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒔𝒆 𝒕𝒊𝒆𝒏𝒆 ∶ 𝑳−𝟏 𝟏 (𝒔 − 𝟏)(𝒔 + 𝟒) = න 𝟎 𝒕 𝑭 𝒖 𝑮 𝒕 − 𝒖 𝒅𝒖 = න 𝟎 𝒕 𝒆𝒖 𝒆−𝟒(𝒕−𝒖) 𝒅𝒖 = 𝒆−𝟒𝒕 න 𝟎 𝒕 𝒆𝟓𝒖 𝒅𝒖 = 𝒆𝒕 𝟓 − 𝒆−𝟒𝒕 𝟓
- 8. 𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝟐 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝐿−1 𝑠2 (𝑠2 + 4)2 𝑳−𝟏 𝒔𝟐 (𝒔𝟐 + 𝟒)𝟐 = 𝑳−𝟏 𝒔 𝒔𝟐 + 𝟒 . 𝒔 𝒔𝟐 + 𝟒 𝒚 𝒉𝒂𝒄𝒊𝒆𝒏𝒅𝒐 𝒇 𝒔 = 𝒔 𝒔𝟐 + 𝟒 , 𝒈 𝒔 = 𝒔 𝒔𝟐 + 𝟒 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐: 𝑳−𝟏 𝒇(𝒔) = 𝒄𝒐𝒔 𝟐𝒕 = 𝑭 𝒕 𝒚 𝑳−𝟏 𝒈(𝒔) = 𝒄𝒐𝒔 𝟐𝒕 = 𝑮(𝒕) 𝑳−𝟏 𝒔𝟐 (𝒔𝟐 + 𝟒)𝟐 = න 𝟎 𝒕 𝑭 𝒖 𝑮 𝒕 − 𝒖 𝒅𝒖 = න 𝟎 𝒕 𝒄𝒐𝒔𝟐𝒖𝒄𝒐𝒔 𝟐𝒕 − 𝟐𝒖 𝒅𝒖 = 𝒄𝒐𝒔𝟐𝒕 𝒕 𝟐 + 𝒔𝒆𝒏𝟒𝒕 𝟖 + 𝒔𝒆𝒏𝟐𝒕 𝒔𝒆𝒏𝟐 𝟐𝒕 𝟒 = 𝒕𝒄𝒐𝒔𝟐𝒕 𝟐 + 𝒔𝒆𝒏𝟐𝒕 𝟒 = 𝟎 𝒕 𝒄𝒐𝒔𝟐𝒖 𝒄𝒐𝒔𝟐𝒕𝒄𝒐𝒔𝟐𝒖 + 𝒔𝒆𝒏𝟐𝒕𝒔𝒆𝒏𝟐𝒖 𝒅 𝒖 = 𝒄𝒐𝒔𝟐𝒕 𝟎 𝒕 𝒄𝒐𝒔𝟐 𝟐𝒖𝒅𝒖 + 𝒔𝒆𝒏𝟐𝒕 𝟎 𝒕 𝒔𝒆𝒏𝟐𝒖𝒄𝒐𝒔𝟐𝒖𝒅𝒖 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐: 𝑳−𝟏 𝒔𝟐 (𝒔𝟐 + 𝟒)𝟐 = 𝒕𝒄𝒐𝒔𝟐𝒕 𝟐 + 𝒔𝒆𝒏𝟐𝒕 𝟒 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏
- 9. 𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶𝑺 2. 𝐿−1 6𝑠 𝑠2 + 2𝑠 − 6 3. 𝐿−1 2𝑠2 + 5𝑠 − 4 𝑠3 + 𝑠2 − 2𝑠 1. 𝐿−1 2𝑠 + 3 𝑠2 + 9 4. 𝐿−1 𝑠 + 1 9𝑠2 + 6𝑠 + 5 5. 𝐿−1 𝑠2 − 3 (𝑠 − 2)(𝑠 − 3)(𝑠2 + 2𝑠 + 5) 6. 𝐿−1 1 (𝑠 + 1)(𝑠2 + 1) 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒅𝒂 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒆 𝑳𝒂𝒑𝒍𝒂𝒄𝒆