Page 91 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

Equating real and imaginary parts we get
R IN = R + 3.5
1
0 = jX + j0.5
1

and while R 1 can be any real number, we must have jX = – j0.5 and thus
1
Z = R – j0.5 :
1
1
3.
a. BW = f – f = 1075 – 925 = 150 KHz . Then,
1
2
f = f + BW 2 = 925 + 150 2 = 1000 KHz
e
e
0
1
f
b. The exact value of is the geometric mean of and and thus
f
f
1
0
2
3
f = f ˜ f = 925 + 1075 10 = 997.18 KHz

0
2
1
f 1000 Z C
0
0
c. Q 0P = -------------- = ------------ = 20 3 . Also, Q 0P = ---------- . Then
e
f –
G
f
150
1
2
6
Z C 2Sf C 2S u 10 u 10 – 6 3S
0
0
G = ---------- = --------------- = -------------------------------------- = ------ = 0.94 : – 1
Q 0P Q 0P 20 3 10
e
and
1 1 1
L = ---------- = ------------------- = ------------------------------------------- = 0.025 PH
Z C 4S f C 4S u 10 12 u 10 – 6
2 2
2
0
0
4.
f 500 Z C
0
0
a. Q 0P = --------- = --------- = 25 . Also, Q 0P = ---------- or
G
20
BW
Q ˜ G 25 u 10 – 3 – 9
0P
C = ------------------ = ------------------------------ = 7.96 u 10 F = 7.96 nF
Z 0 2S u 5 u 10 5
1 1 1 – 6
L = ---------- = ------------------- = ------------------------------------------------------------------------ = 12.73 u 10 H = 12.73 PH
Z C 4S f C 4S u 25 u 10 10 u 7.96 u 10 – 9
2
2 2
0
0
I = V Y = V G = 20 u 10 – 3 A = 20 mA
0 0
0
0
b. f = f – BW 2 = 500 – 10 = 490 KHz and f = f + BW 2 = 500 + 10 = 510 KHz
e
e
2
0
1
0
Circuit Analysis II with MATLAB Applications 2-25
Orchard Publications