# Poles of a second-order closed-loop transfer function

I was reading about the effects of poles and zeroes of a second-order transfer function on its time response and it was said that the real part of the pole of transfer function gives the exponential decay frequency of that system.

What does the imaginary part represent? It would be really helpful if you could explain it in detail.

If you read about the real part then you must be able to read about the imaginary part, as well, along with explanations. What you're asking can take a good part of a book.

Otherwise, to find out how the real- and imaginary parts affect a generic 2nd order biquad, write it this way:

\begin{align} H(s)&=K\dfrac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0} \tag{1} \\ &=K\dfrac{s^2-2\Re{}_z s+\Re{}_z^2+\Im{}_z^2}{s^2-2\Re{}_ps+\Re{}_p^2+\Im{}_p^2} \tag{2} \end{align}

Where $$\\Re\$$ is considered negative for a Hurwitz polynomial. Now you can apply the inverse Laplace transform and you'll get:

\begin{align} h(t)&=K\Big\{\delta(t)+\mathrm{e}^{\Re{}_pt}\big[A\sin(\Im{}_pt)+B\cos(\Im{}_pt)\big]\Big\} \tag{3} \\ A&=\dfrac{\Re{}_p^2-\Im{}_p^2+\Re{}_z^2+\Im{}_z^2-2\Re{}_p\Re{}_z} {\Im{}_p} \\ B&=2(\Re{}_p-\Re{}_z) \end{align}

If you look at the terms, closely, $$\\Im{}_p\$$ (the imaginary part of the pole) appears as the argument for the oscillating terms, $$\\sin\$$ and $$\\cos\$$. This implies an underdamped case, otherwise it would have been an overdamped case, $$\\sinh\$$ and $$\\cosh\$$. If none are present then it's critically damped.

Note that (1) differs from (2) in that $$\K\$$ is different for the two, which is reflected in the $$\\delta(t)\$$ term: for (1) it would have been $$\K\frac{a_2}{b_2}\$$, which means that the $$\K\$$ in (2) is, in fact: $$\K\frac{\Re{}_z^2+\Im{}_z^2}{\Re{}_p^2+\Im{}_p^2}\$$. But, since it's only a matter of scaling, I've considered it as one.

And for the "why" part, it helps if you think of what the real- and imaginary parts of a pole represent:

• $$\s=\sigma\pm j\omega =\Re\pm j\Im\$$ is the general representation of the pole;
• $$\\Re\$$ is on the X axis, the damping, $$\\zeta\$$ ,and the quality factor,$$\Q\$$, are dependent on it, and it appears as the argument for $$\\mathrm{e}^{\Re t}\$$, so it sets the decaying envelope;
• $$\\Im\$$ is on the imaginary axis, where the frequency lies ($$\j\omega\$$), so it sets the oscillation.

The imaginary value of the pole represents the damped frequency of oscillation, Wd.

The higher the imaginary part of the pole is in value, the higher will be the damped frequency of oscillation.

If the poles of a closed loop system have an imaginary component (the poles are not real) then the damping ratio is less than 1, the poles are a complex conjugate pair and when such a system is subject to a step input then the output will undergo several diminishing amplitude oscillations before settling down to a steady state value (assuming that the system is not unstable). The frequency of those oscillations is known as the damped frequency, Wd. This is in contrast to the undamped natural frequency, Wn which is the theoretical frequency which the closed loop system will oscillate at if there is no damping and the system is on the verge of instability.

The value of Wd depends on the values of Wn and zeta, the damping ratio.

If zeta = 0 then Wd = Wn and the poles are on the imaginary axis.

Try this picture from my basic website: - So, the point on the $$\j\omega\$$ axis coincident with the positive pole is $$\\omega_n\sqrt{1-\zeta^2}\$$ and, this is sometimes called the damped frequency.