Lab 2 - Tanggapan Waktu Rangkaian Orde Dua (1).pdf

𝑅𝐿𝐶

𝑎2

𝑑 2 𝑥(𝑡) 𝑑𝑥(𝑡) + 𝑎1 + 𝑎0 𝑥(𝑡) = 𝑓(𝑡) 2 𝑑𝑡 𝑑𝑡

𝑥(𝑡)

𝑓(𝑡) 𝑎2 𝑎1 𝑎0 𝑑 2 𝑥(𝑡) 𝑑𝑥(𝑡) + 2𝛼 + 𝜔02 𝑥(𝑡) = 𝑓(𝑡) 2 𝑑𝑡 𝑑𝑡

𝛼

𝜔0 𝑥(𝑡)

𝑥𝑛 (𝑡) 𝑥(𝑡) = 𝑥𝑛 (𝑡) + 𝑥𝑓 (𝑡)

𝑥𝑓 (𝑡)

𝑅𝐿𝐶

𝑅𝐿𝐶

𝑓(𝑡) = 0

𝑥𝑛 (𝑡) 𝑑2 𝑥𝑛 (𝑡) 𝑑𝑥𝑛 (𝑡) 𝑎2 + 𝑎 + 𝑎0 𝑥𝑛 (𝑡) = 0 1 𝑑𝑡 2 𝑑𝑡

𝑥𝑛 (𝑡) 𝑥𝑛 (𝑡) 𝑥𝑛 (𝑡) = 𝐴𝑒 𝑠𝑡 𝐴

𝑠 𝑎2 𝐴𝑠 2 𝑒 𝑠𝑡 + 𝑎1 𝐴𝑠𝑒 𝑠𝑡 + 𝑎0 𝐴𝑒 𝑠𝑡 = 0 𝑎2 𝑠 2 + 𝑎1 𝑠 + 𝑎0 = 0

𝑠1 =

−𝑎1 + √𝑎12 − 4𝑎2 𝑎0 2𝑎2

−𝑎1 − √𝑎12 − 4𝑎2 𝑎0 𝑠2 = 2𝑎2

𝑠1

𝑠2

𝑠1 ≠ 𝑠2

𝑥𝑛 (𝑡) = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2 𝑡

𝐹 𝑥𝑓 𝑑 2 𝑥𝑓 𝑑𝑥𝑓 𝑎2 + 𝑎1 + 𝑎0 𝑥𝑓 = 𝐹 2 𝑑𝑡 𝑑𝑡 𝑹𝑳𝑪

𝑅𝐿𝐶 𝑡>0 𝐿

𝑑𝑖 + 𝑅𝑖 + 𝑣 = 𝑉 𝑑𝑡 i L

R

vs = Vu(t)

𝑅𝐿𝐶

𝑖 = 𝐶𝑑𝑣/𝑑𝑡 𝑑2 𝑣 𝑑𝑣 𝐿𝐶 2 + 𝑅𝐶 +𝑣 =𝑉 𝑑𝑡 𝑑𝑡

+ C

-

v

𝑑 2 𝑣 𝑅 𝑑𝑣 1 𝑉 + + 𝑣 = 𝑑𝑡 2 𝐿 𝑑𝑡 𝐿𝐶 𝐿𝐶 𝑑2 𝑣 𝑅 𝑑𝑣 1 + + 𝑣=0 2 𝑑𝑡 𝐿 𝑑𝑡 𝐿𝐶 𝑠 𝑛 = 𝑑𝑛 /𝑑𝑡 𝑛

𝑠2 +

𝑅 1 𝑠+ =0 𝐿 𝐿𝐶 𝑅𝐿𝐶

𝛼= 𝜔02 =

𝑅 2𝐿

1 1 → 𝜔0 = √ 𝐿𝐶 𝐿𝐶

𝑠1 = −𝛼 + √𝛼 2 − 𝜔02 𝑅 𝑅 2 1 √ =− + ( ) − 2𝐿 2𝐿 𝐿𝐶 𝑠2 = −𝛼 − √𝛼 2 − 𝜔02 𝑅 𝑅 2 1 √ =− − ( ) − 2𝐿 2𝐿 𝐿𝐶 𝜔𝑑 𝜔𝑑 = √𝜔02 − 𝛼 2 𝜔0 > 𝛼 𝑠1 = −𝛼 + 𝑗𝜔𝑑

𝑠2 = −𝛼 − 𝑗𝜔𝑑 𝜔02

𝑅𝐿𝐶 𝑡>0

𝛼2

𝑅𝐿𝐶

𝑉

𝑥𝑓 = 𝑉

𝑅𝐿𝐶

𝛼 2 > 𝜔02 𝛼 2 = 𝜔02 𝛼 2 < 𝜔02

𝑅 2

1

(2𝐿) > 𝐿𝐶 𝑅 2

𝑠1 ≠ 𝑠2 )

1

(2𝐿) = 𝐿𝐶 𝑅 2

𝑠1 = 𝑠2 )

1

(2𝐿) < 𝐿𝐶

𝑠1 = 𝑠2∗ )

𝑣(𝑡) = 𝑉 + 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2 𝑡 𝑣(𝑡) = 𝑉 + (𝐴1 + 𝐴2 𝑡)𝑒 −𝛼𝑡 𝑣(𝑡) = 𝑉 + (𝐴1 cos 𝜔𝑑 𝑡 + 𝐴2 sin 𝜔𝑑 𝑡)𝑒 −𝛼𝑡 𝑅𝐿𝐶

𝑅𝐿𝐶

𝐴1

𝐴2

𝑣(0)

𝑑𝑣(0)/𝑑𝑡 𝑣 𝑣

𝑅𝐿𝐶

𝑅𝐿𝐶 𝑅 𝐿

𝑅 𝐿

𝐶

𝐶

𝑅 𝐿 𝑅 Ω

𝐶 𝐶 𝜇

𝑅𝐿𝐶

𝑣

𝐿