Question

Phase lag of a first order system is given by _______

• A. \(\tan^{-1}(\omega\tau)\)
• B. \(\tan^{-1}(-\omega\tau)\)
• C. \(\tan^{-1}(2\omega\tau)\)
• D. \(\tan^{-1}(-2\omega\tau)\)

Hint:

A negative sign in the phase angle always represents a phase lag.
At very low frequencies ($\omega \to 0$), the phase lag is $0^\circ$.
At the corner frequency ($\omega = 1/\tau$), the phase lag is exactly $-45^\circ$ or $\tan^{-1}(-1)$.
At very high frequencies ($\omega \to \infty$), the phase lag approaches $-90^\circ$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the mathematical expression of the phase lag (phase angle) of a first-order system as a function of frequency $\omega$ and time constant $\tau$.
Phase lag measures the delay in phase between the input sinusoidal signal and the resulting output sinusoidal response of the system.

Step 2: Key Formula or Approach:
The transfer function of a standard first-order system is given by: \[ G(s) = \frac{1}{\tau s + 1} \] To perform frequency response analysis, we substitute $s = j\omega$ to get the sinusoidal transfer function: \[ G(j\omega) = \frac{1}{1 + j\omega\tau} \] The phase angle $\phi(\omega)$ of a complex transfer function $G(j\omega) = \frac{A(j\omega)}{B(j\omega)}$ is calculated as: \[ \phi(\omega) = \angle A(j\omega) - \angle B(j\omega) \]

Step 3: Detailed Explanation:
Let us apply the phase angle formula to the first-order system:
The numerator of $G(j\omega)$ is $1$, which is a purely real number. Its phase angle is: \[ \angle(1) = 0^\circ = 0 \] The denominator of $G(j\omega)$ is $1 + j\omega\tau$, which is a complex number with real part $1$ and imaginary part $\omega\tau$.
Its phase angle is: \[ \angle(1 + j\omega\tau) = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}(\omega\tau) \] Therefore, the total phase angle $\phi(\omega)$ is: \[ \phi(\omega) = \angle(1) - \angle(1 + j\omega\tau) = 0 - \tan^{-1}(\omega\tau) = -\tan^{-1}(\omega\tau) \] Using the trigonometric property of the arctangent function, $\tan^{-1}(-x) = -\tan^{-1}(x)$, we can write: \[ \phi(\omega) = \tan^{-1}(-\omega\tau) \] This negative phase angle represents a phase lag, meaning the output lags behind the input.

Step 4: Final Answer
Thus, the phase lag of a first-order system is given by $\tan^{-1}(-\omega\tau)$, which corresponds to option (B).