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Huang-Jen Chiu

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Compensator Design Example<br />

G<br />

m<br />

=<br />

G m + G s<br />

0.<br />

5(<br />

V<br />

sp<br />

3<br />

− 1)<br />

=<br />

0.<br />

5 × ( 11 − 1)<br />

3<br />

= + 4.<br />

5dB<br />

2.<br />

5<br />

= 4. 5 + 20 log( ) = 4.<br />

5 − 6 = −1.<br />

5dB<br />

5<br />

R = R × 100(<br />

40dB)<br />

= 1k<br />

× 100 = 100kΩ<br />

2<br />

1

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