How does a zero in transfer function affect the bandwidth of a system

Question

Suppose we have a system with the closed-loop gain:

\$\displaystyle G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n+\omega_n^2}\$

Which is the default two-pole system closed-loop gain. Let's also assume that bandwith \$\omega_0\$ is defined with \$|G(j\omega_0)| = \frac{1}{\sqrt{2}}|G(j0)|\$ (standard 3dB definition).

Through substitution of these two expressions we end up with:

\$ \omega_0 = \omega_n\sqrt{1-2\zeta^2+\sqrt{4\zeta^4 - 4\zeta^2 + 2}}\$.

-> My question is: If we add a finite zero to the transfer function, such as:

\$\displaystyle G(s) = \frac{(s+z)\omega_n^2}{s^2 + 2\zeta\omega_n+\omega_n^2}\$

what will happen to the bandwidth? Will it change? I know that the finite zero only affects the system in transient phase, and in stable phase (when t \$\rightarrow \infty\$) the zero doesn't affect it. However, I'm not sure about the bandwidth.

Of course, I could substitute G(s) into the bandwidth expression, but I won't get a normal result.

Help :)

Answer 1

What you have initially described is a 2nd order low pass filter then you've made a bit of a complication of things because the bandwidth is \$\omega_n\$ i.e. \$\omega_n\$ is the 3dB point when \$\zeta\$ is not causing the filter to peak i.e. has a value of \$\frac{1}{\sqrt2}\$: -

OK you may be trying to derive a general expression but that doesn't help much visualize the problem so for now I'm assuming \$\zeta\$ = \$\frac{1}{\sqrt2}\$.

Anyway, moving on and ignoring your expression for \$\omega_0\$, the bandwidth of the low-pass filter is from dc to \$\omega_n\$. Then you've modified the 2nd order low-pass filter expression with a 1st order high pass filter (\$s +z\$).

When s is very low, the low pass filter is unaffected other than having a "gain factor" of z. If z is unity then the low pass filter response is unaffected until jw approaches z in magnitude - this then marks a lower 3dB point and the net response climbs with increasing frequency until the original \$\omega_n\$ is reached then, because the low pass filter is a second order, the response starts to fall again.

What you have proposed is a band pass filter with finite gain at dc.

I'm not sure about the bandwidth

If the high pass filter (\$s +z\$) comes into play at significantly lower frequencies than \$\omega_n\$ then the net bandwidth reduces from \$\omega_n\$ to \$\omega_n - z\$.

That's how I see it anyway.