Signals & Systems - Laplace - Oscillatory Component

[Angello90]

Homework Statement

What is that oscillatory component? And is my answer for the following correct?

$$x(t) = u(t)$$
$$H(s) = \frac{R}{R + sL}$$

$$y(t)$$ will contain oscillatory component if
$$R > \frac{1}{L}$$
True or False?

Homework Equations

Basic Laplace Transform:
$$u(t) \longrightarrow \frac{1}{s}$$
$$e^{-at}u(t) \longrightarrow \frac{1}{s+a}$$

The Attempt at a Solution

$$H(s) = \frac{\frac{R}{L}}{s + \frac{R}{L}}$$
$$X(s) = \frac{1}{s}$$
$$Y(s) = \frac{1}{s} * \frac{\frac{R}{L}}{s + \frac{R}{L}}$$
where $$*$$ donates multiplication not convolution.

Doing partial fraction expansion I got:
$$Y(s) = \frac{1}{s} - \frac{1}{s + \frac{R}{L}}$$
which in time domain is:
$$y(t) = u(t) - e^{\frac{-Rt}{L}}u(t)$$

Now, clearly, if:
$$R > \frac{1}{L}$$
that exponential e will converge to 0, and signal would eventually be a DC signal - 1 which is donated by u(t) right? So output y(t) doesn't have an oscillatory component?

 

[viscousflow]

Euler's identity is $$e^{ix} = \cos(x) + i\sin(x)$$ Add this to your list of relevant equations and the answer should be clear.

 

[Angello90]

viscousflow I don't really see how I could use this. I assume I need to change
$$e^{\frac{-Rt}{L}}u(t)$$
But I don't know how, cause I don't have i. Should I take one i outside?
$$e^{\frac{i.i.Rt}{L}}u(t)$$
Than I would end up with:
$$e^{\frac{i.i.Rt}{L}}u(t) = (Cosh(\frac{Rt}{L}) - Sinh(\frac{Rt}{L}))u(t)$$
Am I going in right direction?

 

[viscousflow]

What it boils down to is, for any first order system to be oscillatory, your denominator needs to have a complex number. Therefore, it needs to be of the form $$s + (a+bi)$$

 

[Angello90]

Ok, so should I do the partial fraction expansion as I did and work on:
$$Y(s) = \frac{1}{s} - \frac{1}{s + \frac{R}{L}}$$
or start over? Or... there is no complex component in dominator, therefore there is no oscillatory component?

Sorry for being annoying.

 

[viscousflow]

No, all you need to do is perform the substitution in your second equation under "Relevant equations"

$$e^{-at}u(t) \longrightarrow \frac{1}{s+a}$$

where $$a = b+ci$$

You need to divide top and bottom by R

 

[Angello90]

But isn't this rather wierd? I would get
$$Y(s) = \frac{1}{s} - \frac{\frac{1}{R}}{\frac{s}{R} + \frac{1}{L}}$$
and than I should transform from Laplace to time domain?

 

[viscousflow]

Seems as though you made it a bit complicated. Let me help you out a bit.

$$H(s) = \frac{R}{R + sL}$$

Divide by R

$$Y(s) = \frac{1}{ s(\frac{L}{R}s + 1)}$$ or $$Y(s) =\frac{1}{ s(s + \frac{R}{L})}$$

your roots are
$$s=0; s= -\frac{R}{L}$$

Thus, R can be complex and L can be real.

 

[Angello90]

Ok I found this extract in the book that my department is using for signals and systems, saying that,
Assuming that we have:
$$\frac{A_{1}}{s-\alpha - j\omega_{0}} + \frac{A_{2}}{s-\alpha + j\omega_{0}}$$
can be replaces by:
$$\frac{B_{1}s + B_{2}}{(s-\alpha - j\omega_{0})((s-\alpha + j\omega_{0})}$$
and simplifying:
$$\frac{B_{1}s + B_{2}}{(s-\alpha)^{2} + \omega_{0}^{2}}$$

And there is a transition one can use:
$$e^{at}cos(\omega_{0}t) \longleftrightarrow \frac{(s-\alpha)}{(s-\alpha)^{2} + \omega_{0}^{2}}$$
and
$$e^{at}sin(\omega_{0}t) \longleftrightarrow \frac{\omega_{0}}{(s-\alpha)^{2} + \omega_{0}^{2}}$$

can these transitions be used?

 

[Angello90]

Hold on, so by saying that s=0, s=-R/L, and assuming that R is and imaginary number and L is real, that eqn has a oscillatory component? My lecture never covered that part of the course, yet he examines on this .

Will this hold the fact that R>1/L? Can imaginary number be greater than rational number?