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.
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))`
(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))`
