Step 1: Read the asymptotic plot.
For frequencies below \(\omega_0\), the magnitude falls at \(-20\) dB/decade as frequency increases. For frequencies above \(\omega_0\), the magnitude is flat at \(0\) dB.
Step 2: Recall what a falling low-frequency asymptote means.
A slope of \(-20\) dB/decade at low frequency is the signature of a pole at the origin, an integrator-type term, because the magnitude of \(1/s\) falls exactly at that rate as frequency increases.
Step 3: Recall what a flat high-frequency asymptote at 0 dB means.
Once the magnitude becomes flat at 0 dB for large \(\omega\), the transfer function must behave like a constant of magnitude 1 at high frequency, which happens when a numerator term and a denominator term of the same order cancel out.
Step 4: Test option (A).
\[
G(s)=\frac{1+\dfrac{s}{\omega_0}}{\dfrac{s}{\omega_0}}
\]
For \(\omega\ll\omega_0\), the numerator is close to 1 and the denominator is close to \(\dfrac{j\omega}{\omega_0}\), so
\[
|G(j\omega)|\approx\frac{\omega_0}{\omega}
\]
which falls at \(-20\) dB/decade as \(\omega\) rises, exactly as the plot shows. For \(\omega\gg\omega_0\), both the numerator and denominator are dominated by the \(\dfrac{s}{\omega_0}\) term, so \(|G(j\omega)|\to1\), giving the flat \(0\) dB region. This matches the plot on both sides of \(\omega_0\).
Step 5: Rule out the other options.
Option (B) is the reciprocal of (A); its magnitude would rise at \(+20\) dB/decade for \(\omega\ll\omega_0\), which is wrong. Options (C) and (D) both contain a factor \(1-\dfrac{s}{\omega_0}\) in place of \(1+\dfrac{s}{\omega_0}\), which places a pole or zero in the right half of the \(s\)-plane. Such a term describes a non-minimum-phase or unstable element, which is not the simplest and standard system that a plain asymptotic magnitude sketch like this one is meant to represent. Option (A) is the well-behaved, minimum-phase choice whose magnitude asymptotes match the plot exactly.
Step 6: Final conclusion.
\[
\boxed{G(s)=\dfrac{1+\dfrac{s}{\omega_0}}{\dfrac{s}{\omega_0}}}
\]
Hence the correct option is (A).
