Laplace and pole zero diagram.

[Steve Collins]

The following diff. equation describes the functionality of a system with respect to time. However, it is not known how the system will behave when stimulated. Apply a forward Laplace transform to determine damping ratio and pole zeros. Plot a pole zero diagram and comment on stability.

d²y/dt² + 6dy/dt + 10y = 63dy/dt + 63xPutting like terms on either side of the equation:

d²y/dt² + 6dy/dt - 63dy/dt + 10y = 63x

= d²y/dt² - 57dy/dt + 10y = 63x

Laplace transform:

s²Y(s) - 57sY(s) + 10Y(s) = 63 ... (63 because unit impulse used to stimulate?)

Simplify for Y(s):

Y(s).(s² - 57s + 10) = 63

Solve for Y(s):

Y(s) = 63/(s² - 57s + 10)

Using quadratic formula to find poles:

s= (57 +/- √57² - 4 x 1 x 10)/(2 x 1)

= (57 +/- 56.648)/2 = 28.5 +/- 56.648 (I was expecting a complex number!)

I think that I am nearly there, but I suspect that I have gone wrong.

 

[rude man]

The following diff. equation describes the functionality of a system with respect to time. However, it is not known how the system will behave when stimulated. Apply a forward Laplace transform to determine damping ratio and pole zeros. Plot a pole zero diagram and comment on stability.

d²y/dt² + 6dy/dt + 10y = 63dy/dt + 63x


Putting like terms on either side of the equation:

d²y/dt² + 6dy/dt - 63dy/dt + 10y = 63x

= d²y/dt² - 57dy/dt + 10y = 63x

Laplace transform:

s²Y(s) - 57sY(s) + 10Y(s) = 63 ... (63 because unit impulse used to stimulate?)

Factor out Y(s)F(s) = X(s) from the above transformed equation with X(s) = 1 (unit impulse input).

Then Y(s)/X(s) = 1/F(s) and let H(s) = 1/F(s) so that now Y(s) = X(s)H(s). H(s) is now the system transfer function. Given any X(s) you can now compute Y(s) and consequently y(t).

Note that the magnitude of the input impulse has nothing to do with pole/zero location, nor damping ratio. You only use it if you want to compute Y(s) and y(t).

Note also that with an impulse input, all the poles & zeros of the output Y(s) are due to the system H(s) only.