Problem 2.81 For the circuit of Problem 2.80, generate a bounce diagram for thecurrent and plot its time history at the middle of the line.

Solution: Using the values forΓg andΓL calculated in Problem 2.80, we reversetheir signs when using them to construct a bounce diagram for the current.

I+1 =

V+1

Z0=

15075

= 2 A,

I+2 =

V+2

Z0=

−15075

= −2 A,

I+∞ =

V∞

ZL= 1.33 A.

2A

-0.5A

-0.25A

62.5mA

31.25mA

-7.79mA

4 µs

2 µs

1 µs

3 µs

5 µs

z = 0 z = 200 m

Γ = -Γg =12

Γ = -ΓL = -

t t

t = 0

-3.90mA6 µs

I1 (t) 1

4

Figure P2.81: (a) Bounce diagram forI1(t) in reaction toVg1(t).

-2A

0.5A

0.25A

-62.5mA

-31.25mA

7.79mA

4.4 µs

2.4 µs

1.4 µs

3.4 µs

5.4 µs

z = 0 z = 200 m

Γ = -Γg =12

Γ = -ΓL = -

t t

t = 0.4 µs

3.90mA6.4 µs

I2 (t) 1

4

Figure P2.81: (b) Bounce diagram for currentI2(t) in reaction toVg2(t).

(i) I1(l/2, t) due toVg1(t):

t (µs)

I ( 100, t )1

0.5 1.5 2.5 3.5

2A

1.51.25 1.3125 1.3333

Figure P2.81: (c) I1(l/2, t).

(ii) I2(l/2, t) due toVg2(t):

t (µs)

I ( 100, t )2

0.9 1.9 2.9 3.9

-2A

-1.5-1.25 -1.3125 -1.3333

Figure P2.81: (d) I2(l/2, t).

(iii) Net currentI(l/2, t) = I1(l/2, t)+ I2(l/2, t):

t (µs)

I ( 0, t )

2A

0.9

1.5

0.5

1.9 2.5 2.5

0.5 0.5

-0.5

-0.25

.0625

Figure P2.81: (e) TotalI(l/2, t).