4.49
(a) omega_0 = 10 MHz
   alpha = 10^7 s^{-1}
   zeta = 1
(c) v(t) = 0
(d) v(t) = 10^9*t*e^{-10^7*t}

4.35
v(t) = 25cos(10t)-25sin(10t)+25e^{-t/0.1}

6.59
v_o = 0.4975cos(200*pi*t+84.29^o) + 3.536cos(2000*pi*t+45^o)

6.81
(a) H(f) = (1-4pi^2*fLC)/(1-4pi^2*fLC+j2pi*fRC)
(c) H(f) ~= 1
(d) H(f) ~= 1

14.7
(a) -2V
(b) -1V
(c) 3V
(d) 0V
(e) 3V

14.18
A_1 = -4/3
A_2 = -8/3

14.21
(a) i_0 = (v_1-v_2)/R
   Z_0 = infinite
(b) i_0 = -v_in/R_f
   Z_0 = infinite

14.26
i_0 = -(R_1+R_2)/R_2 * i_in
Z_0 = infinite

14.40
(a) A_{CL,ideal} = -10
   A_{CL,actual} = -9.9989
(b) Z_{in,ideal} = 1 kOhm
   Z_{in,actual} = 1.0001 kOhm
(c) Z_{out,ideal} = 0 Ohm
   Z_{out,actual} = 2.75 mOhm

P1
v_o = 12V-2/3*v_a
v_o/v_a = -2/3

P2
v_o/v_i = -2/((1+jw/w1)(1+jw/w2))
w1 = R/L
w2 = 1/(2RC)

P3
v_o/v_i = w^2*L-jwR

P4
v_o = R_2/R_1*(1+R_4/R_3)

P5
V_o = 1.25*V_A-0.75*V_B

P6
It will saturate if A = 1.6*B - 6 or A = 1.6*B + 9
When A=2B this means B = -15 V or B = 22.5 V.