The bode plot of the delay transfer function e-3s has the following shape 

Concept:

The transfer function \( G(s) = e^{-Ts} \) represents a pure time delay of T seconds.

When analyzing this in the frequency domain for Bode plots, we substitute \( s = j\omega \):

\( G(j\omega) = e^{-j\omega T} \)

This is a complex exponential which affects only the phase of the frequency response, not the magnitude.

Magnitude:

\( |G(j\omega)| = |e^{-j\omega T}| = 1 \Rightarrow \text{Magnitude (in dB)} = 20\log_{10}(1) = 0~\text{dB} \)

So the magnitude plot is a horizontal straight line at 0 dB.

Phase:

\( \angle G(j\omega) = \text{arg}(e^{-j\omega T}) = -\omega T \)

Phase decreases linearly with frequency. Specifically, for T = 3 seconds:

\( \angle G(j\omega) = -3\omega \) radians

In degrees: \( \angle G(j\omega) = -3\omega \cdot \frac{180}{\pi} \angle G(j\omega) = -3\omega \cdot \frac{180}{\pi} \)

This is a straight line with a negative slope in the phase plot.

Given:

Transfer function: e-3s (pure delay of 3 seconds)

Conclusion:

• Magnitude plot: Constant at 0 dB (flat line)
• Phase plot: Straight line with a negative slope (due to the increasing delay with frequency)