Question

Sketch the bode plots for

G(1) s = 1/(s + 2)
G(2) s = 10/(s + 40)
G(3) s = 8/(s + 5)

Answer 1

Bode plots depict how the magnitude and phase of a system's output respond to different frequencies in a concise graphical representation. In another words, Bode plots illustrate the frequency response of a system. The bode plots can be found on the attachment.

Step-by-step explanation:

For
`\( G(1) = (1)/(s + 2) \)`, its magnitude plot starts high at low frequencies and decreases by 20 dB/decade as frequency increases, representing a first-order system. The phase plot starts at 0° and asymptotically approaches -90° as frequency rises, indicating a phase lag due to the single pole at -2.

`\( G(2) = (10)/(s + 40) \)` exhibits a similar trend with a steeper slope in the magnitude plot (20 dB/decade) due to the higher pole frequency. The phase begins at 0° and descends more rapidly, reaching -90° quicker due to the pole at -40.

For
`\( G(3) = (8)/(s + 5) \)`, it starts at a higher magnitude compared to G(1) due to the gain. Its slope decreases by 20 dB/decade, similar to G(1), but the phase plot starts at 0° and reaches -90° at a different frequency due to the pole at -5.

These trends depict how the systems react to different frequencies, impacting both magnitude and phase.