# Bode plot of second-order system

I was reading about second-order systems in Control systems engineering by Norman Nise and I am confused by the idea that the phase at high frequencies of a second-order system is equal to 180° (check picture attached) and for the inverse of the same transfer function, the phase at high frequencies will be equal to -180°.

In fact, if we specify the interval of study of our angles as (-180,180], we should have 180° for both cases, since for both cases the transfer function will be equivalent to a negative number (-ω2 or -1/ω2 for me is the same as -2 and -1/2 values in phase).

What is wrong with my reasoning?

• No - it is not correct that "the phase at high frequencies will be equal to -180°." For a second-order sysytem, the phase will APPROACH the -180deg line for rising frequencies but will never be "equal to -180 deg".
– LvW
Feb 21 at 8:54
• yes but that's not the question Feb 21 at 15:54
• Correct - this was not the question, but a correction to an incorrect statement.
– LvW
Feb 21 at 17:30
• Do you have any idea about my question? Feb 21 at 18:01
• In your post - G(s) is the denominator of a 2nd-order transfer function, correct? And you are asking about the properties of 1/G(s) ? Why? Because such a transfer function cannot exist in reality.
– LvW
Feb 22 at 8:42

if we specify the interval of study of our angles as (-180,180] ...

Why would you? Transfer functions are basically complex numbers: $$\Z=A+jB=|Z| \angle{\tan(B/A)}\$$. Think about the polar representation again.

since for both cases the transfer function will be equivalent to a negative number

No.

• For lower frequencies: $$\s=j \omega_1\$$ where $$\\omega_1 << \omega_n\$$ (i.e. way lower) therefore $$\\omega_1/\omega_n<<1 \rightarrow\omega_1/\omega_n\approx0\$$, so: $$G(s)=\omega_n^2\Big(\frac{s^2}{\omega_n^2}+2\zeta \frac{s}{\omega_n}+1\Big)\\ \Rightarrow G(j\omega_1)\approx\omega_n^2((j0)^2+2\zeta\ 0+1)=\omega_n^2$$

So at lower frequencies the transfer function is an always-positive constant:

$$G\approx\omega_n^2=\omega_n^2+j0=\omega_n^2 \angle{0°}$$

• Likewise, for higher frequencies: $$\s=j \omega_2\$$ where $$\\omega_2 >> \omega_n\$$ (i.e. way higher) therefore $$\\omega_n/\omega_2<<1 \rightarrow\omega_n/\omega_2\approx0\$$, so: $$G(s)=s^2\Big(1+2\zeta \frac{\omega_n}{s}+\frac{\omega_n^2}{s^2}\Big) \\ \Rightarrow G(j\omega_2)\approx(j\omega_2)^2(1+2\zeta\ 0+ 0)=(j\omega_2)^2=-\omega_2^2$$

So at higher frequencies the transfer function is an always-negative constant:

$$G\approx -\omega_n^2=-\omega_n^2+j0=\omega_n^2 \angle{180°}$$

• Thank you for your reply but I am talking about 1/G and for high frequencies we say it will be equal to - 180 Feb 21 at 11:06