Homework Solution Chapter 1

# Homework Solution Chapter 1 - 21e Pole at 6 zero at 6...

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Homework Solution Chapter 1 Note that c transforms to 1/sc, L transforms to Ls, a constant transforms to 1/s and cos(t) transforms to s/ (s 2 +1).

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1.7a. Gain of the ideal inverting opamp : A v = - =- / ZfZi 1 scR = 10s This circuit acts as an integrator.
b. A v = - = +( ) ZfZi 5k 10K||1sc 2k = . ( + )( + ) 2 5 s 300 s 100 This is a lag circuit: one pole, one zero. P<z. 21c. double zero at +-2i, single zero at -5, Quad pole at +-2i and double pole at -6. Asymptotically unstable, BIBO marginally stable.

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Unformatted text preview: 21e. Pole at +6, zero at +6 cancel. Other poles are on LHP. Asymptotically unstable, BIBO stable. 23. T(s) = + ( - ) + + ( - ) ks 4s s 2 1 ks 4s s 2 = ( + ) + -+ k s 4 s2 k 2s 4k For k=1, K=2 and k=20 find the roots of the denaminator: K=1 eq. - + = s2 s 4 0 real part is positive RHP unstable K=2 eq. + = s2 4 0 roots on imaginary axis marginally stable K=20 eq. + + = s2 18s 4 0 roots on LHP stable. 24b. 26b. 26c....
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Homework Solution Chapter 1 - 21e Pole at 6 zero at 6...

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