計劃進修PhD/Research討論區(18) 如切如磋,如琢如磨

1001 回覆
7 Like 0 Dislike
2017-05-23 00:52:27






2017-05-23 01:00:13







2017-05-23 01:08:01
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula
2017-05-23 01:17:46






2017-05-23 01:24:11







2017-05-23 01:25:46







2017-05-23 01:26:37








2017-05-23 01:29:23








2017-05-23 01:35:13









2017-05-23 01:37:43
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula


ok thank you!
2017-05-23 01:37:58








2017-05-23 01:41:14









2017-05-23 01:42:53
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula


ok thank you!

2017-05-23 01:54:13
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula


ok thank you!


2017-05-23 01:56:46









2017-05-23 01:57:45
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula


ok thank you!



2017-05-23 01:58:14








2017-05-23 01:58:52
問下數: 有條function f(t), example你當係 exp(-at), a>0
let c>0
我要Laplace transform f 的c translation: L(f(t+c))
因為f(t+c)比f(t)係shift 左去左邊而Laplace係用t>=0的information
所以我cutoff 本來的f: f(t)*u(t-c), (u =unit step)
又因為我用左unit step, so sum少左, 所以係s domain scales up 返 exp(s*c)
Therefore:
L(f(t+c)) = exp(s*c)*L(f(t)*u(t-c))
是咪有咁的formula


ok thank you!




2017-05-23 01:59:11








2017-05-23 02:00:18









2017-05-23 02:01:08








2017-05-23 02:02:45

2017-05-23 02:03:41









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