# Laplace Transforms/Differential Equations using MATLAB

For the following differential equation:

where constant B = 7 and C = 0, I need to find the roots/zeros, do a partial fraction expansion, an inverse Laplace transform and plot the function f(t). I then need to plot the poles and zeros of the differential equation.

I'm stuck on finding the laplace transform/transfer function of the differential equation. This is what I've tried:

Help appreciated, thanks.

Well, we have that:

$$\text{y}'''\left(t\right)+6\cdot\text{y}''\left(t\right)+11\cdot\text{y}'\left(t\right)+3\cdot\text{y}\left(t\right)=\text{B}\cdot\text{u}''\left(t\right)+\text{C}\cdot\text{u}'\left(t\right)+\text{u}\left(t\right)$$

Assuming that all the initial conditons are equal to $0$, we get for Laplace transform:

$$\text{s}^3\cdot\text{Y}\left(\text{s}\right)+6\cdot\text{s}^2\cdot\text{Y}\left(\text{s}\right)+11\cdot\text{s}\cdot\text{Y}\left(\text{s}\right)+3\cdot\text{Y}\left(\text{s}\right)=\text{B}\cdot\text{s}^2\cdot\text{U}\left(\text{s}\right)+\text{C}\cdot\text{s}\cdot\text{U}\left(\text{s}\right)+\text{U}\left(\text{s}\right)$$

Simplifying it bit:

$$\text{Y}\left(\text{s}\right)\cdot\left(\text{s}^3+6\cdot\text{s}^2+11\cdot\text{s}+3\right)=\text{U}\left(\text{s}\right)\cdot\left(\text{B}\cdot\text{s}^2+\text{C}\cdot\text{s}+1\right)$$

So, we also get:

$$\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}=\frac{\text{B}\cdot\text{s}^2+\text{C}\cdot\text{s}+1}{\text{s}^3+6\cdot\text{s}^2+11\cdot\text{s}+3}$$

So, for the poles and zeros:

• Zeros: $$\text{B}\cdot\text{s}^2+\text{C}\cdot\text{s}+1=0\space\Longleftrightarrow\space\text{s}=\frac{-\text{C}\pm\sqrt{\text{C}^2-4\cdot\text{B}}}{2\cdot\text{B}}$$
• Poles: $$\text{s}^3+6\cdot\text{s}^2+11\cdot\text{s}+3=0\space\Longleftrightarrow\space\text{s}\approx-0.3283\space\space\space\wedge\space\space\space\text{s}\approx-2.83585\pm1.04687i$$
• Many thanks for your reply Jan Eerland. I have a few questions.. When you take the laplace of both sides you end up with the transfer function Y(s)/U(s) = H(s)? Since C is 0, the zeroes quadratic will be s^2 + 1 = 0 so s = +-1? – somers Mar 20 '17 at 13:35
• Once we have the roots for numerator and denominator of Y(s)/U(s) we can do partial fraction expansion? How do you work out the poles and end up with complex numbers? – somers Mar 20 '17 at 13:35
• @stevesy You're welcome, I'm glad that I could help! Well when $\text{C}=0$, we get for the zeros: $$\text{s}=\frac{-0\pm\sqrt{0^2-4\cdot\text{B}}}{2*\text{B}}=\pm\frac{i}{\sqrt{\text{B}}}$$ – Jan Eerland Mar 20 '17 at 13:38
• You get (sqrt -4)/2 = +-1? I don't understand why you have i/(sqrt B). – somers Mar 20 '17 at 14:13
• @stevesy Huh, that is so so so so wrong, the square root of a negative number becomes an imaginairy number!! – Jan Eerland Mar 20 '17 at 14:34