我沒有放棄
:^(
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J痕叔叔
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
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
:^(
J痕叔叔
ok thank you!
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!
:^(
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Adam_Smith
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!
:^(
Michaelzaki
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!
:^(
J痕叔叔
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!
:^(
Michaelzaki
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!
:^(
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