Question

The open loop transfer function of a unity negative feedback system is \[ G(s) = \frac{k}{s(1+sT_1)(1+sT_2)}, \] where $k$, $T_1$ and $T_2$ are positive constants. The phase cross-over frequency, in rad/s, is

• A. $\dfrac{1}{\sqrt{T_1T_2}}$
• B. $\dfrac{1}{T_1T_2}$
• C. $\dfrac{1}{T_1\sqrt{T_2}}$
• D. $\dfrac{1}{T_2\sqrt{T_1}}$

Hint:

For phase crossover frequency, focus on the total phase shift of $-180^\circ$. With two first-order factors, the condition simplifies to $\omega^2 T_1 T_2 = 1$.

Answer 1

The Correct Option is A

Step 1: Phase condition for phase crossover frequency.
Phase crossover frequency $\omega_{pc}$ is the frequency at which the phase angle of $G(j\omega)$ is $-180^\circ$.
Step 2: Phase angle of $G(j\omega)$.
\[ G(j\omega) = \frac{k}{j\omega (1+j\omega T_1)(1+j\omega T_2)} \] Phase contribution: - From $j\omega$: $-90^\circ$ - From $(1+j\omega T_1)$: $-\tan^{-1}(\omega T_1)$ - From $(1+j\omega T_2)$: $-\tan^{-1}(\omega T_2)$
So total phase: \[ \phi = -90^\circ - \tan^{-1}(\omega T_1) - \tan^{-1}(\omega T_2) \] Step 3: Apply phase crossover condition.
At phase crossover: \[ -90^\circ - \tan^{-1}(\omega T_1) - \tan^{-1}(\omega T_2) = -180^\circ \] \[ \Rightarrow \tan^{-1}(\omega T_1) + \tan^{-1}(\omega T_2) = 90^\circ \] Step 4: Simplify using identity.
\[ \tan^{-1}(x) + \tan^{-1}(y) = 90^\circ \;\;\Rightarrow\;\; xy=1 \] So, $\omega^2 T_1 T_2 = 1$. Step 5: Solve for $\omega$.
\[ \omega = \frac{1}{\sqrt{T_1T_2}} \] \[ \boxed{\dfrac{1}{\sqrt{T_1T_2}}} \]