-
In the system shown in the figure,
r (t) = sin ωt
The steady-state response c(t) will exhibit a resonance peak at a frequency of—
-
- 4 rad/sec
- 2√2 rad/sec
- 2 rad/sec
- √ 2 rad/sec
- 4 rad/sec
Correct Option: B
From given figure
| T.F. = | = | ||
| R(s) | 1 + 16/(s + 4)s.1 |
or
| = | ||
| R(s) | s( s + 4) + 16 |
| = | ||
| R(jω) | 16 - ω2 + 4jω |
| = | tan-1 |  |  | ||
| √(16 - ω2) + (4ω)2 | 16 - ω2 |
Let,
F(ω) = (16 – ω2)2 + (4ω)2 = ω4 + 256 – 32ω2 + 16ω2
for resonance to occur,
| F(ω) = 0 | |
| dω |
or,
| (ω4 + 256 - 16ω2) = 0 | |
| dω |
4ω3 – 32ω = 0
or
ω2 = 8
ω = 2√2rad/sec
