Unit step function/Heaviside function.

Printable View

April 22nd 2008, 01:33 AM
Danshader

Unit step function/Heaviside function.

well i have 2 questions.

1) Consider the function given below:

f(t) = 0, 0 < t < 2
sin(t) t > 2

Find the Laplace transform of the function,
Well what i found was:

L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
= integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
infinity for sin(t).exp(-st)
= 0 + integration from 2 to infinity for sin(t).exp(-st)

if i were to use integration by parts will end up with:
= const + integration from 2 to infinity for cos (t).exp(-st)/s

which brings me back to my initial problem of integrating the trigonometry.

So my question now is how do i solve this unending integration?

2) Doing some exercise in a maths book i found another similar question:

f(t) = sin(t), 0 <= t <= p
0 t >= pi

and the goven answer was:
[ 1+ exp(-s.pi) ]/ (s^2 +1)

so what i did for this question was:

L{f(t)} = L{ sin(t) [H(t-0) - H(t-pi)] + 0.H(t-pi)}
= L{ sin(t).H(t)} - L{sin(t).H(t-pi)}
= 1/(s^2 +1) - [exp(-s.pi)] / [(s^2 +1)]
= [ 1+ exp(-s.pi) ]/ (s^2 +1)

the formula given for the heaviside function was:
L{f(t-a)H(t-a)} =exp(-as).F(s)

In the above solution i made i didn't used the general form whereby i was suppose to use L{sin(t-pi).H(t-pi)} instead of L{sin(t).H(t-pi)}. still i get the same answer.

then i look at other examples like:
L{exp(-t).H(t-6)}
the answer was exp[-6(s+1)]/(s+1)

if i were to do the same thing:
a = 6
L{exp(-t)} = 1/(s+1)
since it is a shifting function the solution for the heaviside will be exp[6(s+1)]
combining both i will get exp[-6(s+1)]/(s^2 +1)

So my question is it correct in what i am doing where i straight away perform Laplace transform on the function before the heaviside without converting (taking the last example) exp(-t) to exp[-(t-6)] before doing the calculations? my lecturer taught me to convert it but i am not sure about the way i am solving it now

well as a matter of fact i dont really understand how heaviside works yet so this questions could have already been solved or part of the theory itself o.o"

All help is appreciated.
April 22nd 2008, 06:46 AM
mr fantastic

Quote:

Originally Posted by Danshader

well i have 2 questions.

1) Consider the function given below:

f(t) = 0, 0 < t < 2
sin(t) t > 2

Find the Laplace transform of the function,
Well what i found was:

L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
= integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
infinity for sin(t).exp(-st)
= 0 + integration from 2 to infinity for sin(t).exp(-st)

if i were to use integration by parts will end up with:
= const + integration from 2 to infinity for cos (t).exp(-st)/s

which brings me back to my initial problem of integrating the trigonometry.

So my question now is how do i solve this unending integration?

2) Doing some exercise in a maths book i found another similar question:

f(t) = sin(t), 0 <= t <= p
0 t >= pi

and the goven answer was:
[ 1+ exp(-s.pi) ]/ (s^2 +1)

so what i did for this question was:

L{f(t)} = L{ sin(t) [H(t-0) - H(t-pi)] + 0.H(t-pi)}
= L{ sin(t).H(t)} - L{sin(t).H(t-pi)}
= 1/(s^2 +1) - [exp(-s.pi)] / [(s^2 +1)]
= [ 1+ exp(-s.pi) ]/ (s^2 +1)

the formula given for the heaviside function was:
L{f(t-a)H(t-a)} =exp(-as).F(s)

In the above solution i made i didn't used the general form whereby i was suppose to use L{sin(t-pi).H(t-pi)} instead of L{sin(t).H(t-pi)}. still i get the same answer.

then i look at other examples like:
L{exp(-t).H(t-6)}
the answer was exp[-6(s+1)]/(s+1)

if i were to do the same thing:
a = 6
L{exp(-t)} = 1/(s+1)
since it is a shifting function the solution for the heaviside will be exp[6(s+1)]
combining both i will get exp[-6(s+1)]/(s^2 +1)

So my question is it correct in what i am doing where i straight away perform Laplace transform on the function before the heaviside without converting (taking the last example) exp(-t) to exp[-(t-6)] before doing the calculations? my lecturer taught me to convert it but i am not sure about the way i am solving it now

well as a matter of fact i dont really understand how heaviside works yet so this questions could have already been solved or part of the theory itself o.o"

All help is appreciated.

I have no time now but will look at it later if no-one else has.
April 22nd 2008, 09:32 AM
Peritus

$ \int\limits_2^\infty {\sin (t)e^{ - st} dt = \left. { - \frac{{\sin (t)}} {s}e^{ - st} } \right|} _2^\infty + \int\limits_0^\infty {\frac{{\cos (t)}} {s}e^{ - st} dt = } $

$= \frac{{\sin 2}} {s}e^{ - 2s} - \left. {\frac{{\cos (t)}} {{s^2 }}e^{ - st} } \right|_2^\infty - \int\limits_0^\infty {\frac{{\sin (t)}} {{s^2 }}e^{ - st} dt = }$

$= \frac{{\sin 2}} {s}e^{ - 2s} + \frac{{\cos 2}} {{s^2 }}e^{ - 2s} - \frac{1} {{s^2 }}\int\limits_0^\infty {\sin (t)e^{ - st} dt = }$

and now the "tricky" part:

$ \Leftrightarrow \left( {1 + \frac{1} {{s^2 }}} \right)\int\limits_2^\infty {\sin (t)e^{ - st} dt = } \left( {\frac{{\sin 2}} {s} + \frac{{\cos 2}} {{s^2 }}} \right)e^{ - 2s} $
$ \Leftrightarrow \int\limits_2^\infty {\sin (t)e^{ - st} dt = } \frac{{s\left( {\sin (2)} \right) + \cos (2)}} {{s^2 + 1}}e^{ - 2s}$

-------------------------------------------------------------------------------
2. can be done the same way

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~

1. can be solved in a much simpler way if we use the well known Laplace transforms:

$ \begin{gathered} L\left\{ {\sin (\omega t)u(t)} \right\} = \frac{\omega } {{s^2 + \omega ^2 }} \hfill \\ L\left\{ {\cos (\omega t)u(t)} \right\} = \frac{s} {{s^2 + \omega ^2 }} \hfill \\ \end{gathered} $

$ \int\limits_2^\infty {\sin (t)e^{ - st} dt} = \int\limits_0^\infty {\sin (t)u(t - 2)e^{ - st} dt} $

which is equivalent to:

$ \int\limits_0^\infty {\sin (t + 2)u(t)e^{ - s(t + 2)} dt} $

and now for a little trigonometry:

$ \begin{gathered} e^{ - 2s} \int\limits_0^\infty {\left[ {\sin (t)\cos (2) + \sin (2)\cos (t)} \right]} e^{ - st} dt = \hfill \\ \hfill \\ \cos (2)e^{ - 2s} L\left\{ {\sin (t)} \right\} + \sin (2)e^{ - 2s} L\left\{ {\cos (t)} \right\} = ... \hfill \\ \end{gathered} $
April 22nd 2008, 10:30 AM
Isomorphism

Quote:

Originally Posted by Danshader

well i have 2 questions.

1) Consider the function given below:

f(t) = 0, 0 < t < 2
sin(t) t > 2

Find the Laplace transform of the function,
Well what i found was:

L[f(t)] = L { 0.[H(t-0) - H(t-2)] + sin(t).H(t-2)}
= integration from 0 to 2 for 0.exp^(-st) + integration from 2 to
infinity for sin(t).exp(-st)
= 0 + integration from 2 to infinity for sin(t).exp(-st)

if i were to use integration by parts will end up with:
= const + integration from 2 to infinity for cos (t).exp(-st)/s

which brings me back to my initial problem of integrating the trigonometry.

So my question now is how do i solve this unending integration?

Dont guess it is unending with just one try. As Peritus showed, with a couple of try you can get it in terms of sin again. This is a useful trick.

However you can try doing this from a different angle. Use the definition $\sin t = \frac{e^{jt} - e^{-jt}}{2j}$
$\int_2^{\infty} \sin t \, e^{-st} \, dt = \frac1{2j} \int_2^{\infty} e^{jt-st} - e^{-jt-st}\, dt$

$\frac1{2j} \left(\int_2^{\infty} e^{jt-st} \, dt - \int_2^{\infty} e^{-jt-st} \, dt\right) = \frac1{2j} \left(\frac1{j-s}\,e^{jt-st} + \frac1{j+s}\,e^{-jt-st} \right)$ evaluated between 2 and infinity.

This is where ROC hops in. To make this transform meaningful we have to choose Re(s) > 0.

Now because of this $t \rightarrow \infty \Rightarrow e^{jt-st} \rightarrow 0$ . Similar story happens with the other integral. Now evaluate and do some manipulations.
April 23rd 2008, 08:57 AM
Danshader

xD ahhh that make sense. I also found that same method in a book ^-^ thanks for your time
April 23rd 2008, 08:34 PM
Isomorphism

Quote:

Originally Posted by Danshader

xD ahhh that make sense. I also found that same method in a book ^-^ thanks for your time

Just curious :)
Which method are you referring to? Peritus' or mine?