Exercise 9.14 An harmonic oscillator?

(a)
(b)


An answer:

H1() = 1 + 2jωLR jωLR H2() = 1 + jωRC jωRC H3() = R1 R
(c)


An answer:

90H 1() 0 90H 2() 0 H3() = ±180
(d)


An answer:

The loop gain is Aloop() = 1+2jωLR jωLR 1+jωRC jωRC R1 R

It can be shown mathematically that Aloop can be N 360 only at ω = 0. This is not considered as a valid oscillation frequency.

(e)


An answer:

H1() = 1 + 2jωLR jωLR H2() = 1 + jωRC jωRC H3() = 1 jωRC
(f)


An answer:

90H 1() 0 90H 2() 0 H3() = 90
(g)


An answer:

The loop gain is Aloop() = Abuf 1+2jωLR jωLR 1+jωRC jωRC 1 jωRC

The denominator is imaginary. If the numerator can be set to an imaginary value for a non-zero finite ω then this circuit can be dimensioned to oscillate harmonically. Working things out a little leads to:

Aloop() = Abuf 1 + (2LR + RC) + 2()2LC ()3LC2R ωosc = 1 2LC Abuf = RC 2RC + 4LR