Question

A first order system with a time constant of 1 minute is subjected to frequency response analysis. At an input frequency of 1 radian / minute, the phase shift is:

• A. \(45^\circ\)
• B. \(-90^\circ\)
• C. \(-180^\circ\)
• D. \(-45^\circ\)

Hint:

For any first-order system, when the operating frequency is exactly equal to the reciprocal of the time constant (\( \omega = \frac{1}{\tau} \)), the argument inside the inverse tangent reduces to 1, making the phase shift consistently equal to \(-45^\circ\).

• At \(\omega \rightarrow 0\), \(\phi \rightarrow 0^\circ\)

• At \(\omega \rightarrow \infty\), \(\phi \rightarrow -90^\circ\)

Answer 1

The Correct Option is D

Concept: Frequency response analysis evaluates how a system reacts to stable sinusoidal inputs. For a given input frequency \( \omega \), the output of a linear system is also a sinusoid of the same frequency, but shifted in phase by an angle \( \phi \).

Step 1: Finding the standard expression for a first-order phase shift.
The standard transfer function of a first-order process is represented in the Laplace domain as: \[ G(s) = \frac{K}{\tau s + 1} \] Where \(\tau\) is the process time constant. Substituting \( s = j\omega \) to perform the frequency domain mapping gives: \[ G(j\omega) = \frac{K}{j\omega\tau + 1} \] The phase angle \(\phi\) is calculated as the argument of the complex transfer function: \[ \phi = \angle G(j\omega) = \angle K - \angle(1 + j\omega\tau) \] Since \(K\) is a real positive constant scalar, its angle is \(0^\circ\). Thus: \[ \phi = 0 - \tan^{-1}(\omega\tau) = -\tan^{-1}(\omega\tau) \]

Step 2: Substituting the given numerical values.
The problem statement provides the following specific operational values:

• Time constant, \( \tau = 1 \text{ minute} \)

• Input frequency, \( \omega = 1 \text{ radian/minute} \)
Plugging these metrics directly into our phase shift equation: \[ \phi = -\tan^{-1}(1 \times 1) = -\tan^{-1}(1) \] Since \(\tan(45^\circ) = 1\), the inverse tangent of 1 is \(45^\circ\) (or \(\frac{\pi}{4}\) radians): \[ \phi = -45^\circ \] The negative sign indicates a phase lag, meaning the output sine wave trails behind the input sine wave by \(45^\circ\).