A system function has a pole at s=0 and a zero at s= -1. The constant multiplier is unity. For an excitation cos(t), What is the steady-state response?

Explanation:

Analysis of the Given System Function:

The system function provided has a pole at \( s = 0 \) and a zero at \( s = -1 \). The constant multiplier is unity. This means the transfer function \( H(s) \) of the system can be represented as:

\[ H(s) = \frac{s + 1}{s} \]

This transfer function describes the relationship between the input and output signals in the Laplace domain. Now, the excitation signal is given as \( \cos(t) \), which can be expressed in the Laplace domain as:

\[ \cos(t) \xrightarrow{\mathcal{L}} \frac{s}{s^2 + 1} \]

Determining the Steady-State Response:

The steady-state response is obtained by multiplying the Laplace transform of the input signal with the transfer function \( H(s) \). Therefore, the output in the Laplace domain is:

\[ Y(s) = H(s) \cdot \text{Input}(s) = \frac{s + 1}{s} \cdot \frac{s}{s^2 + 1} \]

Now, simplifying \( Y(s) \):

\[ Y(s) = \frac{(s + 1)}{s} \cdot \frac{s}{s^2 + 1} = \frac{s + 1}{s^2 + 1} \]

Partial Fraction Expansion:

To simplify further and determine the steady-state response in the time domain, we use partial fraction expansion to rewrite \( Y(s) \):

\[ Y(s) = \frac{s}{s^2 + 1} + \frac{1}{s^2 + 1} \]

The terms above correspond to the Laplace transforms of sinusoidal functions. Using the inverse Laplace transform, we get:

\[ y(t) = \mathcal{L}^{-1}\left(\frac{s}{s^2 + 1}\right) + \mathcal{L}^{-1}\left(\frac{1}{s^2 + 1}\right) \]

From Laplace transform tables:

\[ \mathcal{L}^{-1}\left(\frac{s}{s^2 + 1}\right) = \cos(t), \quad \mathcal{L}^{-1}\left(\frac{1}{s^2 + 1}\right) = \sin(t) \]

Thus, the output in the time domain is:

\[ y(t) = \cos(t) + \sin(t) \]

Final Expression for Steady-State Response:

The expression \( \cos(t) + \sin(t) \) can be written in a single sinusoidal form using the trigonometric identity:

\[ \cos(t) + \sin(t) = \sqrt{2} \sin(t + 45^\circ) \]

Therefore, the steady-state response is:

\[ y(t) = \sqrt{2} \sin(t + 45^\circ) \]

Correct Answer:

The correct option is:

Option 1: \( \sqrt{2} \sin(t + 45^\circ) \)

Additional Information

Analysis of Other Options:

Let’s evaluate why the other options are incorrect:

Option 2: \( \sqrt{2} \sin(t - 45^\circ) \)

This option represents a sinusoidal function shifted by \( -45^\circ \). However, the analysis above clearly shows that the phase shift caused by the system is \( +45^\circ \), not \( -45^\circ \). Hence, this option is incorrect.

Option 3: \( \sin(t - 45^\circ) \)

This option assumes a magnitude of 1 for the steady-state response. However, the magnitude of the response is \( \sqrt{2} \), as derived earlier. Additionally, the phase shift is \( +45^\circ \), not \( -45^\circ \). Therefore, this option is incorrect.

Option 4: \( \sin(t) \)

This option suggests no phase shift or amplitude scaling, which does not align with the transfer function analysis. The system introduces both a magnitude change and a phase shift, resulting in \( \sqrt{2} \sin(t + 45^\circ) \). Hence, this option is incorrect.

Conclusion:

The steady-state response analysis confirms that the correct answer is \( \sqrt{2} \sin(t + 45^\circ) \). The other options fail to account for either the phase shift or the amplitude scaling introduced by the system. Understanding the transfer function and performing proper Laplace domain analysis are crucial for solving such problems accurately.