If we have a transfer function that shows no peaking in the magnitude bode plot (Starting from a flatline and then rolling off). Does this mean that there is no resonant frequency? Or do we consider the point at which the curve begins to roll off the resonant frequency?
I understand that resonant frequency is the location at which we have the maximum value so I'm assuming that there isn't a resonant frequency in this case but I wanted to be sure.
My answer applies to higher-than-1st-order systems.
There will always be a resonant point even if you can't see it. You need to understand how "poles" work. Take a look at this: -
Even if there doesn't appear to be a resonance in the bode plot there will be a "pole" that is present and this pole represents the resonant frequency even though the "dampening" is causing it not to appear in the bode plot. Here is what a 2nd order low pass filter looks like with varying degrees of Q (where Q = \$\dfrac{1}{2\zeta}\$): -
If you could determine the phase angle where the output shifted by 90 degrees from the input you would find the resonant frequency even if it doesn't appear to have a "peak" in the bode plot.
Peaks in the frequency response can only exist in systems with conjugate complex poles. For an underdamped (\$\zeta<1\$ or \$Q > 0.5\$) second-order system, the peak appears specifically for \$\zeta<1/\sqrt{2}=0.707\$.
$$H(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$
where \$\omega_n\$ is the natural frequency (also called corner frequency when considering assymptotes), the peak
$$M_p=\frac{1}{2\zeta\sqrt{1-\zeta^2}}$$
occurs at resonant frequency
$$\omega_p=\omega_n\sqrt{1-2\zeta^2}$$
Note on figure below: When varying the damping ratio \$\zeta\$, the peak follows a specific curve. In filter theory, that special value for \$\zeta=0.707\$ corresponds to a Butterworth response. The magnitude curve is sais to be maximally flat (no peak). The meaning of \$w_n\$ for the Butterworth response is the same as for the first-order case, that is, \$w_n\$ represents the -3 dB frequency, also called cuttoff frequency. Only in this case. Also, \$w_n=w_p\$, causes an infinite response (undamped system - oscillator).
It is absolutely correct to say that there is no resonant frequency defined if the system is sufficiently underdamped.
Is there a mathematical threshold for damping constant below which the system will just have no resonance? Yes. D = 2^(-0.5) = 0.707 approx. ( D = damping constant)
For a general second order system with transfer function as : Wn^2/{s^2 + 2*sDWn + Wn^2} Wn = undamped natural frequency.
You can compute the resonance frequency Wr by differentiating w.r.t Wn and equating the result to 0. The result will be : Wr = Wn*sqrt{1-2D^2} which can only be real if D > 1/sqrt{2}
Hope I have answered your query to satisfaction.
