2260
F 08
Homework #2 prob 3 solution
EX:
+
v1
–
C
L2
+
vg –
R1
L1
i2
i1
R2
At t = 0, vg(t) switches instantly from –vo to vo.
a)
Write the state-variable equations for the circuit in terms of the state vector:
 i1 
r  
x = i2 
v1
b)
Evaluate the state vector at t = 0+.
€
SOL'N: a) We put first derivatives of state-variables on the left side of our equations,
and we write expressions involving only state-variables, component
values, and source values on the right side of our equations.
As an intermediate step, we write the derivatives in terms of nonderivatives. This is made possible by the basic equations for L's and C's:
diL v L
=
dt
L
dvC iC
=
dt
C
€
€
€
Also, we use the value of vg for t > 0, since the equations apply for t > 0.
Thus, we have vg = vo.
di1 v L1 v o − v1
=
=
dt
L1
L1
di2 v L2 (v o − v1 ) − i2 R2
=
=
dt
L2
L2
dv1 iC i1 + i2 + (v o − v1 ) /R1
=
=
dt
C
C
€
b) We find the initial conditions by finding state-variable values at t = 0–.
€ Because they will not change instantly, they will have the same value at
time t = 0+.
At time t = 0–, the L's acts like wires, and the C acts like an open circuit.
This means the currents below the C on the right side will be zero:
i1 (0− ) = 0
i2 (0− ) = 0
€ Since L1 is a wire, we can find the voltage on the C from a voltage loop
through vg, C, and L1:
€
v1 (0− ) = v g = −v o
Our values at t = 0+ are the same as at t = 0–:
€
€
 i (0+ )   0 
1
 
r + 

x (0 ) = i2 (0+ ) =  0 
v (0+ ) −v 
o
1
