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

If function f(t) is:

Periodic with period p > 0, so that f(t + p) = f(t), and

f1(t) is one 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 period p, equals the Laplace Transform of one cycle of 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.

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))

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(b) Saw-tooth waveform:

Graph of saw-tooth waveform.

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.

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))

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