How do I find the magnitude and phase of the frequency response function for a third order system using transfer functions?

Question

A third order system is described by: $$\frac{{\rm d}^3 y}{{\rm d}t^3} + 2 \frac{{\rm d}^2 y}{{\rm d}t^2} + 6 \frac{{\rm d} y}{{\rm d}t} + 5y = u + 2 \frac{{\rm d} u}{{\rm d}t}.$$ Determine the magnitude and phase of its frequency response function.

I know I have to change the equation into a transfer function but after that is where I get stuck because it doesn't simply very nicely. Any help would be appreciated, thanks in advance!

edit: This is the work I've done Work I've done

Answer 1

Laplace transform:

$$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t\tag1$$

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So, we get:

$$\text{n}_1\cdot\text{y}'''\left(t\right)+\text{n}_2\cdot\text{y}''\left(t\right)+\text{n}_3\cdot\text{y}'\left(t\right)+\text{n}_4\cdot\text{y}\left(t\right)=\text{n}_5\cdot\text{u}\left(t\right)+\text{n}_6\cdot\text{u}'\left(t\right)\tag2$$

Where $\text{n}_\text{a}$ is a constant for every $\text{a}$

Now, when we take the Laplace transform of both sides, we need to know:

1. $$\mathcal{L}_t\left[\text{n}_1\cdot\text{y}'''\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_1\cdot\int_0^\infty\text{y}'''\left(t\right)e^{-\text{s}t}\space\text{d}t=$$ $$\text{n}_1\cdot\left(\text{s}^3\cdot\text{Y}\left(\text{s}\right)-\text{s}^2\cdot\text{y}\left(0\right)-\text{s}\cdot\text{y}'\left(0\right)-\text{y}''\left(0\right)\right)\tag3$$
2. $$\mathcal{L}_t\left[\text{n}_2\cdot\text{y}''\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_2\cdot\int_0^\infty\text{y}''\left(t\right)e^{-\text{s}t}\space\text{d}t=\text{n}_2\cdot\left(\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}'\left(0\right)\right)\tag4$$
3. $$\mathcal{L}_t\left[\text{n}_3\cdot\text{y}'\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_3\cdot\int_0^\infty\text{y}'\left(t\right)e^{-\text{s}t}\space\text{d}t=\text{n}_3\cdot\left(\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)\right)\tag5$$
4. $$\mathcal{L}_t\left[\text{n}_4\cdot\text{y}\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_4\cdot\int_0^\infty\text{y}\left(t\right)e^{-\text{s}t}\space\text{d}t=\text{n}_4\cdot\text{Y}\left(\text{s}\right)\tag6$$
5. $$\mathcal{L}_t\left[\text{n}_5\cdot\text{u}\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_5\cdot\int_0^\infty\text{u}\left(t\right)e^{-\text{s}t}\space\text{d}t=\text{n}_5\cdot\text{U}\left(\text{s}\right)\tag7$$
6. $$\mathcal{L}_t\left[\text{n}_6\cdot\text{u}'\left(t\right)\right]_{\left(\text{s}\right)}:=\text{n}_6\cdot\int_0^\infty\text{u}'\left(t\right)e^{-\text{s}t}\space\text{d}t=\text{n}_6\cdot\left(\text{s}\cdot\text{U}\left(\text{s}\right)-\text{u}\left(0\right)\right)\tag8$$

Assuming that a initial conditions are equal to zero, we get:

$$\text{n}_1\cdot\text{s}^3\cdot\text{Y}\left(\text{s}\right)+\text{n}_2\cdot\text{s}^2\cdot\text{Y}\left(\text{s}\right)+\text{n}_3\cdot\text{s}\cdot\text{Y}\left(\text{s}\right)+\text{n}_4\cdot\text{Y}\left(\text{s}\right)=\text{n}_5\cdot\text{U}\left(\text{s}\right)+\text{n}_6\cdot\text{s}\cdot\text{U}\left(\text{s}\right)\tag9$$

Simplifying it a little bit:

$$\text{Y}\left(\text{s}\right)\cdot\left(\text{n}_1\cdot\text{s}^3+\text{n}_2\cdot\text{s}^2+\text{n}_3\cdot\text{s}+\text{n}_4\right)=\text{U}\left(\text{s}\right)\cdot\left(\text{n}_5+\text{n}_6\cdot\text{s}\right)\tag{10}$$

So, we also know that:

$$\text{H}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}=\frac{\text{n}_5+\text{n}_6\cdot\text{s}}{\text{n}_1\cdot\text{s}^3+\text{n}_2\cdot\text{s}^2+\text{n}_3\cdot\text{s}+\text{n}_4}\tag{11}$$

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For the respons you're looking for, we can use:

$$\text{s}=\text{j}\omega\tag{12}$$

Where $\text{j}^2=-1$

So, we will also get:

$$\text{H}\left(\text{j}\omega\right)=\frac{\text{Y}\left(\text{j}\omega\right)}{\text{U}\left(\text{j}\omega\right)}=\frac{\text{n}_5+\text{n}_6\cdot\left(\text{j}\omega\right)}{\text{n}_1\cdot\left(\text{j}\omega\right)^3+\text{n}_2\cdot\left(\text{j}\omega\right)^2+\text{n}_3\cdot\left(\text{j}\omega\right)+\text{n}_4}\tag{13}$$

Simplying it a little bit, gives us:

$$\text{H}\left(\text{j}\omega\right)=\frac{\text{Y}\left(\text{j}\omega\right)}{\text{U}\left(\text{j}\omega\right)}=\frac{\text{n}_5+\text{n}_6\cdot\omega\cdot\text{j}}{\text{n}_4-\text{n}_2\cdot\omega^2+\text{j}\cdot\omega\cdot\left(\text{n}_3-\text{n}_1\cdot\omega^2\right)}\tag{14}$$

For the magnitude we get:

$$\left|\text{H}\left(\text{j}\omega\right)\right|=\frac{\left|\text{Y}\left(\text{j}\omega\right)\right|}{\left|\text{U}\left(\text{j}\omega\right)\right|}=\frac{\sqrt{\text{n}_5^2+\text{n}_6^2\cdot\omega^2}}{\sqrt{\left(\text{n}_4-\text{n}_2\cdot\omega^2\right)^2+\omega^2\cdot\left(\text{n}_3-\text{n}_1\cdot\omega^2\right)^2}}\tag{15}$$

And for the phase, you can use:

$$\arg\left(\text{H}\left(\text{j}\omega\right)\right)=\arg\left(\frac{\text{Y}\left(\text{j}\omega\right)}{\text{U}\left(\text{j}\omega\right)}\right)=\arg\left(\text{Y}\left(\text{j}\omega\right)\right)-\arg\left(\text{U}\left(\text{j}\omega\right)\right)\tag{16}$$

Now, when $\text{n}_5\space\wedge\space\text{n}_6\cdot\omega\in\mathbb{R}^+$:

$$\arg\left(\text{Y}\left(\text{j}\omega\right)\right)=\arctan\left(\frac{\text{n}_6\cdot\omega}{\text{n}_5}\right)\tag{17}$$