# 5. Laplace Transform of a Periodic Function *f*(*t*)

If function *f*(*t*) is:

Periodicwith periodp> 0, so thatf(t+p) =f(t), and

f_{1}(t) isone period(i.e. one cycle) of the function, written using Unit Step functions,

then

`Lap{f(t)}= Lap{f_1(t)}xx 1/(1-e^(-sp))`

**NOTE:** In English, the formula says:

The Laplace Transform of the periodic function

f(t) with periodp, equals the Laplace Transform ofone cycleof the function, divided by `(1-e^(-sp))`.

### Examples

Find the Laplace transforms of the periodic functions shown below:

(a)

Graph of periodic unit ramp function.

Answer

From the graph, we see that the first period is given by:

`f_1(t)=t*[u(t)-u(t-1)]` and that the period is `p=2`.

`Lap{f_1(t)}`

`= Lap{t*[(u(t)-u(t-1)]}`

`= Lap{t*u(t)}- Lap{t*u(t-1)}`

Now

`t*u(t-1)` `=(t-1)*u(t-1)+u(t-1)`

So

`Lap{t*u(t)}- Lap{t*u(t-1)}`

`= Lap{t*u(t)}-` ` Lap{(t-1)*u(t-1)+u(t-1)}`

`= Lap{t*u(t)}- ` `Lap{(t-1)*u(t-1)}-` ` Lap{u(t-1)}`

`=1/s^2-e^(-s)/s^2-e^(-s)/s`

`=(1-e^(-s)-se^(-s))/(s^2)`

Hence, the Laplace transform of the periodic function, *f*(*t*) is given by:

`Lap{f(t)}` `=((1-e^(-s)-se^(-s))/s^2)xx1/(1-e^(-2s))`

`=(1-e^(-s)-se^(-s))/(s^2(1-e^(-2s))`

Please support IntMath!

(b) Saw-tooth waveform:

Graph of saw-tooth waveform.

Answer

We can see from the graph that

`f_1(t)=a/bt*[u(t)-u(t-b)]`

and that the period is `p = b`.

So we have

`Lap{f_1(t)}`

`= Lap{a/bt*[u(t)-u(t-b)]}`

`=a/b Lap{t*u(t)-t*u(t-b)}`

(We next subtract, then add a "`b`" term in the middle, to achieve the required form.)

`=a/b Lap{t*u(t)-` `{:(t-b+b)*u(t-b)}`

`=a/b Lap{t*u(t)-` `(t-b)*u(t-b) -` `{: b*u(t-b)}`

(We now find the Laplace Transform of the individual pieces.)

`=a/b[ Lap{t*u(t)}-` ` Lap{(t-b)*u(t-b)}-` `{: Lap{b*u(t-b)}{:]`

`=a/b(1/s^2-(e^(-bs))/s^2-(be^(-bs))/s)`

`=(a(1-e^(-bs)-bse^(-bs)))/(bs^2)`

So the Laplace Transform of the periodic function is given by

`Lap{f(t)}` `=(a(1-e^(-bs)-bse^(-bs))) / (bs^2(1-e^(-bs))`

(c) Full-wave rectification of sin *t*:

Graph of `f(t)=sin t*{u(t)-u(t-pi)}`, with period `pi`.

Answer

Here,

`f_1(t)=sin t*{u(t)-u(t-pi)}`

and the period, `p=pi`.

`Lap{f_1(t)}`

`= Lap{sin\ t*(u(t)-u(t-pi))}`

`= Lap{sin t*u(t)}+` ` Lap{sin(t-pi)*u(t-pi)}`

`=1/(s^2+1)+(e^(-pis))/(s^2+1)`

`=(1+e^(-pis))/(s^2+1)`

So the Laplace Transform of the periodic function is given by:

`Lap{f(t)}=(1+e^(-pis))/((s^2+1)(1-e^(-pis))`

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!