Thread: Unit step function

1. Unit step function

Q1)f(t) = { 10e^-t 0<t<1 }
-10^e-t t>1

Express the function in terms of the Heaviside unit step function and hence obtain its Laplace transform.

Q2) f(t) = 2 [u(t) –u(t – 4)]+ t u(t – 4)
= 2u(t) – 2 u(t – 4)+ t u(t – 4)
= 2u(t)+ ( t – 2 ) u(t – 4) <----- how to derive to this step, it was addition and now ts multiplication. explaination and guide pls.

Also does anyone know any good sites for Laplace transform read-up(not wiki) and exercises to practise. Unit step fuction sites too.. my notes explaination is not clear enough for me or i am just stupid.

2. Originally Posted by mathsnerd
Q1)f(t) = { 10e^-t 0<t<1 }
-10^e-t t>1

Express the function in terms of the Heaviside unit step function and hence obtain its Laplace transform.
f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

so:

Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]

Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

Q2) f(t) = 2 [u(t) –u(t – 4)]+ t u(t – 4)
= 2u(t) – 2 u(t – 4)+ t u(t – 4)
= 2u(t)+ ( t – 2 ) u(t – 4) <----- how to derive to this step, it was addition and now ts multiplication. explaination and guide pls.
You have:

f(t) = 2u(t) – 2 u(t – 4)+ t u(t – 4)

now you take out the common factor u(t-4) from the last two terms on the right:

f(t) = 2u(t) + u(t – 4)(-2+ t)

rearrange and you have your mysterious line.

RonL

3. f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

so:

Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]

Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

thanks for the help...i apply what u taught me in Q2 and i got this

f(t) = 10 e^{-t} - (10e^{-t} - 10e^{-t}) u(t-1)

how do i get your workings in bold....care to help me explain further... thanks Ron.

btw...the answer is 10/(s+1) - 20 e^-{s+1}/(s+1) .... you got it almost right or my book has the wrong answer.

4. Originally Posted by mathsnerd
f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,
u(t-1) is zero up to t=1 and 1 from there on, but we want a step that is
1 up to t=1 and -1 from there on so that f(t) = 10 e^{-t} up to t=0
and is -10 e^{-t} from there onwards.

So we wamt the step height to be -2 ( +1 to -1) and we need it to be offset
by 1, so we use:

s(t-1) = 2[1/2 - u(t-1)]

which is a step function, which is 1 up to t=1 and -1 from there on.

so:

Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]
We have:

f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

so multiplying this out:

f(t) = 10 e^{-t} - 20 e^{-t} u(t-1), t>0,

Now tale LT:

Lf = L[10 e^{-t} - 20 e^{-t} u(t-1)],

but the LT is a linear transform, so:

Lf = 10 L[e^{-t}] - 20 L[e^{-t} u(t-1)],

Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

thanks for the help...i apply what u taught me in Q2 and i got this

f(t) = 10 e^{-t} - (10e^{-t} - 10e^{-t}) u(t-1)

how do i get your workings in bold....care to help me explain further... thanks Ron.

btw...the answer is 10/(s+1) - 20 e^-{s+1}/(s+1) .... you got it almost right or my book has the wrong answer.