Fall 1998 Exam 1
1. 6V
2. a) Vo = R1*i1 + R2*(i1-i2)
      R2*(i2-i1) + R3*i2 + R4*(i2-i3) = 0
      R4*(i3-i2) + R5*i3 + L*di3/dt = 0
   b) -Io + V1/R1 + (V1-V2)/R2 = 0
      (V2-V1)/R2 + C*dV2/dt + (V2-V3)/R3 = 0
      (V3-V2)/R3 - alpha*(V1-V2) = 0
3. a) (d^2)*iR2/dt^2 + 2*diR2/dt +2iR2 = 0
   b) iR2(O+) = 0
      diR2/dt(0+) = 4
   c) There is no particular solution.
   d) A = 2^1.5
      alpha = 4
      w = 1
      theta = -135 degrees
4. V = te^(-2*t)

Fall 1998 Exam 2
1. a) [vi = 10*e^(-j*pi/4)]  60*j*C*(v1-vi) - j*v1/(60*L) + (v1-v2)/R = 0
                             (v2-v1)/R + j*60*C*v2 - j*(v2-v3)/(60*L) = 0
                             (v3-v2)/(j*60*L) - alpha*(vi-v1) = 0
   b) f = 1/(2*pi*(L*C)^0.5)
      Z = R
   c) v(t) = 1.28*cos(120*pi*t + 5*pi/6)
2. Graph points are given and they are connected with line segments: 
   (x1,y1) (x2,y2) ... -> (0,0) (2,2) (4,0) is a triangle for example.
   a) 1st graph: (0,1) (1,1) (2,-1) (3,-1)
      2nd graph: (0,0) (1,2) (2,2) (3,0)
      3rd graph: (0,0) (2,2) (3,2) (x,2) for all x
   b) h(t) = [4*t*e^(2*t)]*u(t)
3. a) A*cos(3*t) + B*sin(3*t)
   b) C*e^[(-1+3*j)*t] + D*e^[(-1-3*j)*t]
   c) C*e^[(-1+3*j)*t] + D*e^[(-1-3*j)*t]
   d) 0
   e) C*e^[(-1+3*j)*t] + D*e^[(-1-3*j)*t]
   f) A*cos(3*t) + B*sin(3*t)
4. a) 69.025
   b) 0

Fall 1999 Exam 2
1. a) L = 2,8
   b) ig = 0.25*cos(500*t) or 0.1*cos(500*t) A
2. a) The system is time invariant as a shift in the integrated input
      will shift the output by the same amount.
      The system has memory as the integral depends on past values of s.
      The system is noncausal as y(t) depends on values of f(s) such 
      that t < s < t + to.
   b) y(t) = integral[x(t0)*h(t-t0)] from -infinity to t and y(2t) = 
      integral[x(t0)*h(2*t-t0)] from -infinity to 2*t.  Then, 2*x(2*t) 
      convoluted with h(2*t) is integral[2*x(2*t0)*h(2*t-2*t0] 
      integrated from -infinity to 2*t.  Setting k = 2*t0 (dk = 2*dt0)
      yields integral[x(k)*h(2*t-k)] integrated from -infinity to 2*t.
      This is the same form as the second equation above, so the equality
      holds.
   c) Graph points are given and they are connected with line segments: 
      (x1,y1) (x2,y2) ... -> (0,0) (2,2) (4,2) (6,0) is a trapezoid for 
      example. 
      (0,1) (3,1) (6,2) (9,2)
   d) 1st graph: (0,0) (1,1) (t,t) for all t
      2nd graph: (-e/2,0) (0,0.5) (e/2,1) (t,1) for all t
      3rd graph: u(t)
   e) h(t) = [2*e^(-t/6)]*u(t)
3. a) -2*e^(-2*t) + 3*e^(-t)
   b) A*e^(-2*t) + B*e^(-t) + C*[e^(-3*t)]*cos(3*t+D)
   c) e^(-2*t) - 2*e^(-t) + 1
   d) [e^(-2*t) - 2*e^(-t) + 1]*u(t) - 2*[e^(-2*t+2) - 2*e^(-t+1) + 1]
       *u(t-1) + [e^(-2*t+4) - 2*e^(-t+2) + 1]*u(t-2)
4. y(t) = t^2 for t > 0