ECE 0101: Linear Circuits and Systems

Homework Assignment #6

1. Determine v(t) for the circuit of Fig. 1 when L = 1 H and vs = 0 for t ≥ 0. The initial
conditions are v(0) = 6 V and dv/dt(0) = – 3000 V/s.

Fig. 1

Answer: v(t) = – 2e–100t + 8e–400t V

Solution:

dv ( 0 )
v ( 0 ) = 6, = −3000
dt

v ( v −vs )  L 
Using operators, the node equation is: Csv + + = 0 or  LCs 2 + s + 1 v = v
R sL  R  s

1 1
So the characteristic equation is: s 2 + s+ = 0
RC LC
 s1,2 = − 250  2502 − 40, 000 = − 100, − 400

So v ( t ) = Ae −100t + Be −400t
v (0) = 6 = A + B
dv ( 0 )  A = −2
= − 3000 = − 100 A − 400 B 
dt  B = 8
 v ( t ) = − 2e −100t + 8e −400t t>0
2. Find vc(t) for t > 0 for the circuit shown in Fig. 2.

Fig. 2

Answer: vc(t) = (3 + 6000t)e–2000t V

Solution:

t 0

di c dv
+ v c = 0, i c = 10−5 c
KVL a : 100i c + .025
dt dt
2
d vc dv
 2
+ 4000 c + 4 106 v c = 0
dt dt
s + 4000 s + 4 10 = 0  s = −2000, − 2000  vc ( t ) = A1e −2000t + A2te −2000t
2 6

t = 0− (Steady − State)
dvc ( 0+ )
iL = ic ( 0 −
)=0 = ic ( 0 +
)  = 0
dt
vc ( 0− ) = 3 V = vc ( 0 + )

so vc ( 0+ ) = 3 = A1
dvc ( 0+ )
= 0 = −2000 A1 + A2  A2 = 6000
dt
vc ( t ) = ( 3 + 6000t ) e−2000t V
3. Find vc(t) for t > 0 for the circuit of Fig. 3. Assume steady-state conditions exist at t =
0–.

Fig. 3

Answer: vc(t) = – 8te–2t V

Solution:

t0 KCL at v c : − vc dt + vc + 1

t
( )
dvc
4 dt
=0

d 2 vc dv
 + 4 c + 4vc = 0
dt dt

s 2 + 4s + 4 = 0, s = −2, − 2  vc ( t ) = A1 e−2t + A2 t e−2t

t = 0− (Steady-State) vc ( 0− ) = 0 = vc ( 0+ ) & iL ( 0− ) =
20 V
= 2 A = iL ( 0 + )
10 
Since vc ( 0+ ) = 0 then ic ( 0+ ) = −iL ( 0+ ) = −2 A
dvc ( 0+ ) ic ( 0+ )
 = = −8 V
dt 1 S
4
So vc ( 0+ ) = 0 = A1
dvc ( 0+ )
= − 8 = A2
dt
 vc ( t ) = − 8te −2t V
4. The input to the circuit shown in Fig. 4 is the voltage of the voltage source, vs. The
output is the capacitor voltage v(t), and is given by the following differential
equation:
R1v s d2 R1 R 2 d 1
= v (t ) + v (t ) + v (t )
LC ( R1 + R 2 ) dt L ( R1 + R 2 ) dt LC
If vs= 1 V, R1= 1 Ω, R2= 2 Ω, L= 3 H, and C= 1 F, find v(t), for t > 0.

Fig. 4

Solution:
2 1
Natural response c/ch equation: 𝑠 2 + 𝑠 + = 0
9 3

1 √26
𝑠1,2 = − ± 𝑗 → under damped response
9 9

𝑣𝑐 (𝑡 ) = (𝑣𝑐 (0+ ) − 𝑣𝑐 (∞))𝑣𝑐 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 + 𝑣𝑐 (∞)

−𝑡 √26 √26
𝑣𝑐 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 = 𝑒 9 [𝐾1 cos ( 𝑡) + 𝐾2 𝑠𝑖𝑛( 𝑡)]
9 9
𝑣𝑐 (0− ) = 𝑣𝑐 (0+ ) = 0

1
𝑣𝑐 (∞) = 𝑉
3
Using initial conditions:
1 1
𝑣𝑐 (0) = 0 = − 𝐾1 → 𝐾1 = 1
3 3
𝑑𝑣𝑐 (𝑡) 𝑖𝑐 (𝑡) 𝑑𝑣𝑐 (0) 𝑖𝑐 (0) 1 √26 1
= → = =0= − ( 27 𝐾2 ) = → 𝐾2 =
𝑑𝑡 𝐶 𝑑𝑡 𝐶 27 √26

1 −𝑡 √26 1 √26
𝑣𝑐 (𝑡 ) = [ 1 − 𝑒 9 [cos ( 𝑡) + 𝑠𝑖𝑛 ( 𝑡)]] 𝑉, 𝑓𝑜𝑟 𝑡 ≥ 0
3 9 √26 9

5. The circuit shown in Fig. 5, the capacitor voltage v(t) is represented by the following
second-order differential equation:
𝑑 2 𝑣(𝑡) 𝑅1 𝑅2 𝑑𝑣(𝑡) 1 𝑣𝑠
+ + 𝑣 ( 𝑡 ) =
𝑑𝑡 2 𝐿 (𝑅1 + 𝑅2 (1 + 𝑏)) 𝑑𝑡 𝐿𝐶 𝐿𝐶

If vs= 1 V, b= 1, R1= 10 Ω, R2= 15 Ω, L= 1.5 H, and C= 1 F, find v(t), for t > 0.

Fig. 5

Solution:
5 2
Natural response c/ch equation: 𝑠 2 + 𝑠 + = 0
2 3

𝑠1 = −0.3035 and 𝑠2 = −2.196 → Over damped response

𝑣𝑐 (𝑡 ) = (𝑣𝑐 (0+ ) − 𝑣𝑐 (∞))𝑣𝑐 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 + 𝑣𝑐 (∞)

𝑣𝑐 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 = 𝐾1 𝑒 −0.3035𝑡 + 𝐾2 𝑒 −2.196𝑡

𝑣𝑐 (0− ) = 𝑣𝑐 (0+ ) = 0
 v s − v oc 
ia = 
R1  v s R 2 (1 + b )
  v oc =
v oc  R1 + R 2 (1 + b )
ia + bia =
R 2 

𝑣𝑜𝑐 = 𝑣𝑐 (∞) = 0.75 𝑉

𝑣𝑐 (𝑡 ) = 0.75(1 − [𝐾1 𝑒 −0.3035𝑡 + 𝐾2 𝑒 −2.196𝑡 ])

Using initial conditions to get 𝐾1 and 𝐾2 :

𝑣𝑐 (0) = 0 = 0.75(1 − [𝐾1 + 𝐾2 ] → 𝐾1 = 1 − 𝐾2

𝑑𝑣𝑐 (𝑡) 𝑖𝑐 (𝑡) 𝑑𝑣𝑐 (0) 𝑖𝑐 (0)
= → = =0
𝑑𝑡 𝐶 𝑑𝑡 𝐶

𝑑𝑣𝑐 (𝑡)
= 0.75([0.3035 𝐾1 𝑒 −0.3035𝑡 + 2.196 𝐾2 𝑒 −2.196𝑡 ]
𝑑𝑡

𝑑𝑣𝑐 (0)
= 0 = 0.75([0.3035 𝐾1 + 2.196 𝐾2 ] → 𝐾2 = −0.1604
𝑑𝑡

Hence, 𝐾1 = 1.1604

Therefore

𝑣𝑐 (𝑡 ) = 0.75 (1 − 1.1604𝑒 −0.3035𝑡 + 0.1604 𝑒 −2.196𝑡 ) 𝑉, 𝑓𝑜𝑟 𝑡 ≥ 0