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.

1. Analysis of the Bode Plot Image
By inspecting the provided image, we can observe two plots as a function of log-frequency, labeled on the horizontal axis as
$\log \left(f\right)$
- The upper curve, labeled "magnitude", is a flat horizontal line, indicating that the system's magnitude is constant across all frequencies.
- The lower curve, labeled "phase", starts at
$0^{\circ }$
and drops to a high-frequency asymptotic value of
$-{180}^{\circ }$
as frequency increases.

2. System Stability
For a system to be stable, all of its poles must lie in the Left Half-Plane (LHP) of the s-domain. Therefore, a first-order stable system must have its pole at:
$$s=-a(\text{where}a>0)$$

3. All-Pass Nature (Constant Magnitude)
A system that has a constant magnitude response across all frequencies is known as an all-pass system. For the magnitude to remain constant, the transfer function must satisfy:
$$\left|H\right(j\omega \left)\right|=\text{constant}$$
This condition is fulfilled when the zeros of the system are the mirror reflections of its poles across the imaginary axis. Since the pole is at
$s=-a$
(in the LHP), the zero must be located at
$s=+a$
(in the Right Half-Plane, RHP) at the same frequency magnitude $a$.

4. Transfer Function Derivation
Let us construct the transfer function with a pole at
$s=-a$
and a zero at
$s=a$
with a gain adjustment for phase alignment:
$$H\left(s\right)=\frac{a-s}{s+a}$$
Substituting
$s=j\omega$
to obtain the frequency response:
$$H\left(j\omega \right)=\frac{a-j\omega }{a+j\omega }$$

5. Verifying Magnitude and Phase
The magnitude is:
$$\left|H\right(j\omega \left)\right|=\frac{|a-j\omega |}{|a+j\omega |}=\frac{\sqrt{a^2+{\omega }^2}}{\sqrt{a^2+{\omega }^2}}=1$$
This confirms the magnitude is indeed constant at all frequencies.

The phase of the transfer function is given by:
$$\angle H\left(j\omega \right)=\angle (a-j\omega )-\angle (a+j\omega )=-{\tan }^{-1}\left(\frac{\omega }{a}\right)-{\tan }^{-1}\left(\frac{\omega }{a}\right)=-2{\tan }^{-1}\left(\frac{\omega }{a}\right)$$
Now let's compute the phase at the frequency limits:
- At low frequency ($\omega \to 0$):
$$\angle H\left(j0\right)=-2{\tan }^{-1}\left(0\right)=0^{\circ }$$
- At high frequency ($\omega \to \infty$):
$$\underset{\omega \to \infty }{\lim }\angle H\left(j\omega \right)=-2\left({90}^{\circ }\right)=-{180}^{\circ }$$
This perfectly matches the asymptotic phase values shown on the graph, confirming the presence of one stable LHP pole and one RHP zero at the same frequency.