Question

Find the Laplace Transform `f(t)=t/T` Over the interval `[0,T]`?

Answer 1

`ℒ \ {f(t)} = 1/(Ts^2 )\ \ \` provided `s>0`

Explanation:

By definition;

`ℒ \ {f(t)} = int_(0)^(oo) e^(-st) \ f(t) \ dt`

And so if we have `f(t)=t/T`, for some constant `T`, then using the definition we have;

`ℒ \ {f(t)} = int_(0)^(oo) e^(-st) \ t/T \ dt`
`" " = 1/T \ int_(0)^(oo) e^(-st) \ t \ dt`

We can integrate this by parts;

Let `{ (u,=t, => (du)/dt,=1), ((dv)/dt,=e^(-st), => v,=-1/se^(-st) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dt) \ dt = (u)(v) - int \ (v)((du)/dt) \ dt`

Gives us:

`ℒ \ {f(t)} = int_(0)^(oo) e^(-st) \ t/T \ dt`
`" " = 1/T { [(t)(-1/se^(-st))]_0^oo - int_0^oo \ (-1/se^(-st))(1) \ dt }`
`" " = 1/T { [-t/se^(-st)]_0^oo + 1/s int_0^oo \ e^(-st) \ dt }`
`" " = 1/T { 0 + [-e^(-st)/s^2]_0^oo }`
`" " = 1/T 1/s^2 \ \ \` provided `s>0`

Hence:

`ℒ \ {f(t)} = 1/(Ts^2 )\ \ \` provided `s>0`

Answer 2

`ℒ \ {f(t)} = (1-e^(-Ts) - Tse^(-Ts))/(Ts^2(1-e^(-Ts)))`

Explanation:

Over the interval `[0,T]` then `f(t)=t/T` (a straight line passing through the origin with gradient `1/T`). We can extend the function to `x in RR` (forming a single wave of the saw-wave function) using the Heavyside Unit step function:

`u(t) = { (0,t lt 0), (1, t gt 1) :}`

So that for a single period:

`f_1(t) = t/T( u(t) - u(t-T) )`

Then;

`ℒ \ {f_1(t)} = ℒ \ {t/T( u(t) - u(t-T) )}`
`" " = 1/T \ ℒ \ {t( u(t) - u(t-T) )}`
`" " = 1/T \ ℒ \ {tu(t) - tu(t-T) }`

We can then manipulate the function as follows;

`ℒ \ {f_1(t)} = 1/T \ ℒ \ {tu(t) - (t-T+T)u(t-T) }`
`" " = 1/T \ ℒ \ {tu(t) - (t-T)u(t-T) - (T)u(t-T)}`

We then take the Laplace Transform of the individual pieces, I won't derive from first principles but instead just use lookup tables and quote the result, as follows:

# {: ( f(t), ℒ \ [f(t)] ), ( tu(t), 1/s^2), ( (t-T)u(t-T), e^(-Ts)/(s^2) ), ( Tu(t-T), (Te^(-Ts))/s) :} #

And so we get:

`ℒ \ {f_1(t)} = 1/T \ {1/s^2 - e^(-Ts)/(s^2) - (Te^(-Ts))/s}`
`" " = 1/T \ { (1-e^(-Ts) - Tse^(-Ts))/(s^2) }`
`" " = (1-e^(-Ts) - Tse^(-Ts))/(Ts^2)`

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

`ℒ \ {f(t)} = (1-e^(-Ts) - Tse^(-Ts))/(Ts^2(1-e^(-Ts)))`