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The transfer function has a log-magnitude plot as shown below. Under the assumption that a minimum phase function, its transfer function is—
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1 1 + s
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10 1 + s
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10 s + 10
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√2 s + 0.1
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Correct Option: D
From figure transfer function is of the form
| = (one pole) | |
| (1 + sT) |
as there is 20 dB/dec at
log ω = – 1
or
ω = 1/10
T = 1/ω = 10
Now, log |G| = 1
or
| G | = 10
or
| = 10 | |
| (10ω)2 + 1 |
or
| = 10 | |
| √(10×1/10)2 + 1 |
or
| = 10 | |
| √2 |
k = 10√2
Now,
| T.F. = | |
| s·10 + 1 |
or
| T.F. = | |
| s + 0.1 |
