Step response of a first order system

[VinnyCee]

Homework Statement

Find the unit step response of the transfer function...

a) G(s)\,=\,\frac{4}{s\,+\,4}

b) G(s)\,=\,\frac{2}{0.2s\,+\,1}

Homework Equations

General first order step response equation...

C(s)\,=\,R(s)\,G(s)\,=\,\frac{a}{s(s\,+\,a)}, where R(s)\,=\,\frac{1}{s}

then do an inverse Laplace transform...

c(t)\,=\,1\,-\,e^{-at}

The Attempt at a Solution

Part a) is simple enough. I just plugged into formula above and got c(t)\,=\,1\,-\,e^{-4t}

However, part b) is where I am confused. To get the G(s) into the form needed (i.e. ~ \frac{a}{s\,+\,a}), I divided both the numerator and denominator by 0.2...

G(s)\,=\,\frac{2}{0.2s\,+\,1}\,=\,\frac{10}{s\,+\,5}\,=\,2\left(\frac{5}{s\,+\,5}\right)

But now the form is not exactly as needed in the first order system equations. What do I do?

I tried taking out a 2 from the numerator, and got an answer, just not sure if it's right though.

Is this right for part b)...

c(t)\,=\,2\,\left[1\,-\,e^{-5t}\right]

 

[klouchis]

That's right, remember that the Laplace transform is a linear operation. If f(t) has a Laplace transform of F(s), then a*f(t) has a Laplace transform of a*F(s). Assuming "a" is a scalar quantity. The same linearity is true for inverse Laplace transforms.