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

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

By definition;

We can integrate this by parts;

Then plugging into the IBP formula:

Gives us:

Hence:

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

Over the interval

# 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))) #

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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