The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, is -180° This system has

The correct option is one LHP pole and one RHP zero at the same frequency.

Analysis of the Bode Plot:
From the provided image:

1. The Bode magnitude plot is constant across all frequencies. This indicates that the system is an all-pass filter, where the gain is constant for all frequencies.
2. The phase starts at $0°$ at low frequencies ($f\to 0$) and asymptotically approaches $-180°$ at high frequencies ($f\to \infty$).

Transfer Function Derivation:
For a system to have a constant magnitude at all frequencies, the poles and zeros must be symmetric with respect to the imaginary axis in the s-plane. Since the system is stable, its poles must lie in the Left Half Plane (LHP).

Let us consider a first-order system with a pole in the Left Half Plane (LHP) at $s=-a$ (where $a>0$) and a zero in the Right Half Plane (RHP) at $s=a$ at the same frequency:

$H\left(s\right)=\frac{a-s}{a+s}$

1. Magnitude Response:
Substituting $s=j\omega$:

$\left|H\right(j\omega \left)\right|=\frac{|a-j\omega | }{|a+j\omega |}=\frac{\sqrt{a^2+{\omega }^2}}{\sqrt{a^2+{\omega }^2}}=1$

This confirms that the magnitude is constant at all frequencies.

2. Phase Response:
The phase of the system is given by:

$\angle H\left(j\omega \right)=\angle (a-j\omega )-\angle (a+j\omega )=-\arctan \left(\frac{\omega }{a}\right)-\arctan \left(\frac{\omega }{a}\right)=-2\arctan \left(\frac{\omega }{a}\right)$

Checking the asymptotic values:
- At low frequencies ($\omega \to 0$):

$\angle H\left(j0\right)=-2\arctan \left(0\right)=0°$

- At high frequencies ($\omega \to \infty$):

$\angle H(j\infty )=-2\arctan (\infty )=-2(90°)=-180°$

This matches the phase response shown in the diagram. Thus, the system has one LHP pole and one RHP zero at the same frequency.