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Tài liệu Bài giải mạch P14 doc
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Mô tả chi tiết
Chapter 14, Solution 1.
1 j RC
j RC
R 1 j C
R
( )
i
o
+ ω
ω = + ω
ω = = V
V
H
H(ω) = 0
0
1 j
j
+ ω ω
ω ω , where RC
1
ω0 =
2
0
0
1 ( )
H ( )
+ ω ω
ω ω = H ω =
ω
ω − π
φ = ∠ ω = 0
-1 tan
2
H( )
This is a highpass filter. The frequency response is the same as that for P.P.14.1
except that ω0 = 1 RC . Thus, the sketches of H and φ are shown below.
H
ω0 = 1/RC ω
1
0.7071
0
0
90°
φ
ω0 = 1/RC ω
45°
Chapter 14, Solution 2.
= + ω = + ω
ω = 1 j L R
1
R j L
R
H( )
0 1 j
1
+ ω ω , where
L
R
ω0 =
2
0 1 ( )
1
H ( )
+ ω ω = H ω =
ω
ω
φ = ∠ ω = 0
-1 H( ) -tan
The frequency response is identical to the response in Example 14.1 except that
ω0 = R L. Hence the response is shown below.
φ
H
ω0 = R/L ω
0.7071
1
0
ω
ω0 = R/L
-45°
-90°
0°
Chapter 14, Solution 3.
(a) The Thevenin impedance across the second capacitor where V is taken is o
1 sRC
R
Th R R ||1 sC R
+
Z = + = +
R 1 sC 1 sRC
1 sC i
Th i + = + = V
V V
ZTh
sC
1
+
Vo
−
+
− VTh
1 sC (1 sRC)(1 sC )
1 sC
Th
i
Th
Th
o Z
V
V
Z
V
+ + ⋅ = + =
(1 sRC)(1 sRC sRC (1 sRC))
1
(1 sC )(1 sRC)
1
s)
i Th
o
+ + + + = + + = = V Z
V
H(
H(s) = s R C 3sRC 1
1
2 2 2 + +
(b) RC (40 10 )(2 10 ) 80 10 0.08 3 -6 -3 = × × = × =
There are no zeros and the poles are at
= = RC
- 0.383
s1 - 4.787
= = RC
- 2.617
s 2 - 32.712
Chapter 14, Solution 4.
(a) 1 j RC
R
j C
1
R || + ω = ω
R j L(1 j RC)
R
1 j RC
R
j L
1 j RC
R
( )
i
o
+ ω + ω =
+ ω
ω +
+ ω
ω = = V
V
H
H(ω) = - RLC R j L
R
2 ω + + ω
(b) 1 j C(R j L)
j C(R j L)
R j L 1 j C
R j L
( ) + ω + ω
ω + ω = + ω + ω
+ ω
H ω =
H(ω) = 1 LC j RC
- LC j RC
2
2
− ω + ω
ω + ω
Chapter 14, Solution 5.
(a) R j L 1 j C
1 j C
( )
i
o
+ ω + ω
ω
ω = = V
V
H
H(ω) = 1 j RC LC
1
2 + ω − ω
(b) 1 j RC
R
j C
1
R || + ω = ω
R j L(1 j RC)
j L(1 j RC)
j L R (1 j RC)
j L
( )
i
o
+ ω + ω
ω + ω = ω + + ω
ω
ω = = V
V
H
H(ω) = R j L RLC
j L RLC
2
2
+ ω − ω
ω − ω
Chapter 14, Solution 6.
(a) Using current division,
R j L 1 j C
R
( )
i
o
+ ω + ω
ω = = I
I
H
1 j (20)(0.25) (10)(0.25)
j (20)(0.25)
1 j RC LC
j RC
( ) 2 2 + ω − ω
ω = + ω −ω
ω
H ω =
H(ω) = 2 1 j 5 2.5
j 5
+ ω − ω
ω
(b) We apply nodal analysis to the circuit below.
1/jωC
+ − V Io x
0.5 Vx
I R s jωL
j L 1 j C
0.5
R
x x x
s ω + ω
− = +
V V V
I
But 2 (j L 1 j C) j L 1 j C
0.5 x o
x
o → = ω + ω
ω + ω = V I
V
I
j L 1 j C
0.5
R
1
x
s
ω + ω = +
V
I
2(j L 1 j C)
1
R
1
2 (j L 1 j C) o
s
ω + ω = +
I ω + ω
I
1
R
2(j L 1 j C)
o
s +
ω + ω = I
I
j RC 2(1 LC)
j RC
1 2(j L 1 j C) R
1 ( ) 2
s
o
ω + − ω
ω = + ω + ω
ω = = I
I
H
j 2(1 0.25)
j ( ) 2 ω+ − ω
ω
H ω =
H(ω) = 2 2 j 0.5
j
+ ω− ω
ω
Chapter 14, Solution 7.
(a) 0.05 = 20log10 H
2.5 10 log10 H -3 × =
= = × -3 2.5 10 H 10 1.005773
(b) - 6.2 = 20log10H
- 0.31 = log10H
= = -0.31 H 10 0.4898
(c) 104.7 = 20log10 H
5.235 = log10 H
= = 5.235 H 10 5 1.718 × 10
Chapter 14, Solution 8.
(a) H = 0.05
HdB = 20log10 0.05 = - 26.02 , φ = 0°
(b) H = 125
HdB = 20log10 125 = 41.94 , φ = 0°
(c) = ∠ ° + = 4.472 63.43 2 j
j10
H (1)
HdB = 20log10 4.472 = 13.01 , φ = 63.43°
(d) = − = ∠ ° +
+
+ = 3.9 j1.7 4.254 - 23.55 2 j
6
1 j
3
H (1)
HdB = 20log10 4.254 = 12.577 , φ = - 23.55°
Chapter 14, Solution 9.
(1 j )(1 j 10)
1
( ) + ω + ω
H ω =
HdB = -20log10 1+ jω − 20log10 1+ jω/10
-tan ( ) tan ( /10) -1 -1 φ = ω − ω
The magnitude and phase plots are shown below.
HdB
0.1
-40
1 j /10
1
20 log10 + ω
1+ jω
1
20log10
-20
1 10 100 ω
1
1 j
1
arg
φ
-135°
1 j /10 arg + ω
+ ω
-90°
-180°
0.1 1 10 100 ω
-45°
Chapter 14, Solution 10.
ω
ω +
= ω + ω
ω =
5
j 1j 1
10
j (5 j )
50 H(j )
jω
1 20log
ω +
5
j 1
1 20log
20 log1
-40
0.1
10
1 100 ω
20
-20
HdB
40
φ
-135°
1 j / 5
1
arg
+ ω
jω
1
arg
-90°
-180°
0.1 10
1
100 ω
-45°
Chapter 14, Solution 11.
j (1 j 2)
5(1 j 10) ( )
ω + ω
+ ω
H ω =
HdB = 20log10 5 + 20log10 1+ jω 10 − 20log10 jω − 20log10 1+ jω 2
-90 tan 10 tan 2 -1 -1 φ = °+ ω − ω