# Step response of third-order continuous-time transfer function

I have a transfer function of the form:

$$H(s) = \frac{b\omega_n s^2 + a\omega_n^2 s + \omega_n^3}{s^3 + b\omega_n s^2 + a\omega_n^2 s + \omega_n^3}$$

If it matters, $$a=b=2$$. Is anyone aware of a simple explicit form of the step response for this transfer function (i.e. something directly in terms of $$\omega_n$$, $$a$$, and $$b$$)?

• For those particular values of a and b, the denominator is quite easy to factor (poles at (s/w)= -1, and +/- 120 degrees). You can do partial fractions and solve H(s).s. – Juancho Sep 8 at 16:28

It does matter that $$a=b=2$$, because this gives a relatively nice looking solution. As mentioned in a comment, you should split up the given transfer function $$H(s)$$. But first, let's introduce a normalized variable $$p$$:

$$p=\frac{s}{\omega_n}\tag{1}$$

Now we can write the given transfer function as

$$\hat{H}(p)=\frac{2p^2+2p+1}{p^3+2p^2+2p+1}\tag{2}$$

The denominator can be factored as

$$p^3+2p^2+2p+1=(p+1)(p^2+p+1)\tag{3}$$

Partial fraction expansion of $$\hat{H}(p)$$ gives

\begin{align}\hat{H}(p)&=\frac{1}{p+1}+\frac{p}{p^2+p+1}\\&=\frac{1}{p+1}+\frac{p}{(p+\frac12)^2+\frac34}\tag{4}\end{align}

The Laplace transform of the step response equals the transfer function divided by $$p$$:

\begin{align}\hat{A}(p)=\frac{\hat{H}}{p}&=\frac{1}{p(p+1)}+\frac{1}{(p+\frac12)^2+\frac34}\\&=\frac{1}{p}-\frac{1}{p+1}+\frac{1}{(p+\frac12)^2+\frac34}\tag{5}\end{align}

The inverse Laplace transform of $$(5)$$ is (table)

\begin{align}\hat{a}(t)&=u(t)-e^{-t}u(t)+\frac{2}{\sqrt{3}}e^{-t/2}\sin(\sqrt{3}t/2)u(t)\\&=\left[1-e^{-t}+\frac{2}{\sqrt{3}}e^{-t/2}\sin(\sqrt{3}t/2)\right]u(t)\tag{6}\end{align}

Finally, denormalization according to $$(1)$$ taking into account that $$\hat{A}(p)=\omega_nA(s)$$ (because of division by $$p$$) gives the desired step response:

\begin{align}a(t)&=\hat{a}(\omega_nt)\\&= \left[1-e^{-\omega_nt}+\frac{2}{\sqrt{3}}e^{-\omega_nt/2}\sin(\sqrt{3}\omega_nt/2)\right]u(t)\tag{7}\end{align}

• This appears correct aside from an extra $\omega_n$ scale factor in front of the final expression. – rhz Sep 10 at 1:20
• I think I understand why you wrote (7) with the scaling (since it seems to follow from the properties of Laplace Transforms). However, clearly the system should settle at 1, not $\omega_n$. – rhz Sep 10 at 1:35
• @rhz: You're right, I've corrected it. The factor $\omega_n$ is already taken care of by division by $p$ in $(5)$. – Matt L. Sep 10 at 7:33
• Why is the extra factor accounted for by the 1/p associated with the step? – rhz Sep 10 at 15:49
• @rhz: Because $1/p=\omega_n/s$, so the necessary multiplication by $\omega_n$ happens by multiplying by $1/p$. What is left is the scaling of $t$ to obtain the denormalized result. – Matt L. Sep 10 at 16:08