Chapter 6,
Questions:
6.1 6.2 6.3 6.7 6.8 6.9
6.1 Solution:
Vout
Vin
a)
Vout
=
Vin
( jω ) =
R
1
1
=
=
R + jωL 1 + jωL / R 1 + j 2.5 × 10−6 ω
(
1
1 + (ωL / R)
2
(
=
ϕ (ω ) = − arctan 2.5×10−6 ω
1
1 + 6.25 × 10−12 ω 2
)
The plots obtained using Matlab are shown below:
b)
c)
)
d)
6.2 Solution:
First, we find the Thévenin equivalent circuit seen by the capacitor:
RT = 500 500 = 250 Ω
and
500
v
vOC =
v in = in
500 + 500
2
1
jω C
1
1
v out
=
=
=
1+ jω RT C 1+ j (0.05ω )
vOC RT + 1
jω C
a)
v out
=
vOC
1
1 + 0.0025ω 2
v out 1 v out
=
=
2 vOC
v in
1
4 + 0.01ω 2
ϕ (ω ) = − arctan(0.05 ω )
b)
c)
The plots obtained using Matlab are shown below:
d)
6.3 Solution:
First, we find the Thévenin equivalent circuit seen by the capacitor:
RT = 2000 2000 + 1000 = 2000 Ω
and
2000
v
v in = in
vOC =
2000 + 2000
2
1
v out
jω C
1
1
=
=
=
1 + jω RT C 1 + j (0.02ω )
vOC RT + 1
jω C
a)
v out
=
vOC
1
1 + 4 ×10−4 ω 2
v out 1 v out
=
=
v in
2 vOC
0.5
1 + 4 × 10−4 ω 2
ϕ (ω ) = − arctan(0.02 ω )
b)
The plots obtained using Matlab are shown below:
c)
d)
6.7 Solution:
a)
As ω → ∞,
As ω → 0,
Z L → ∞ ⇒ Open ⇒Z → ∞
Z L → 0 ⇒ Short ⇒ Z → R
b)
KVL : − V i + I i Z R + I i Z L = 0
Z ( jω ) = V i = Z L + Z R = jωL + R
Ii
⎛
ωL ⎞
c) Z ( jω ) = R + j ωL = R ⎜ 1+ j
⎟
⎝
R ⎠
R
2000
rad
ω L
= 1000 k
d) c = 1 ⇒ ω c = =
−3
R
L
s
2 ⋅10
e)
⎛
rad ⎞
2 ⋅ 10 −3 ⋅ 10 5 ⎞⎟
⎛
Z ⎜100k
= 2000(1 + j 0.1) = 2.01 kΩ∠5.71°
⎟ = R⎜1 + j
⎜
⎟
2000
s ⎠
⎝
⎝
⎠
⎛
rad ⎞
2 ⋅ 10 -3 ⋅ 10 6 ⎞⎟
⎛
Z ⎜1M
= 2000(1 + j 1) = 2.82 kΩ ∠45.00°
⎟ = R⎜ 1 + j
⎜
⎟
s ⎠
2000
⎝
⎝
⎠
⎛
rad ⎞
2 ⋅ 10 −3 ⋅ 10 7 ⎞⎟
⎛
Z ⎜10M
= 2000(1 + j 10 ) = 20.10 kΩ ∠84.29°
⎟ = R ⎜1 + j
⎜
⎟
s ⎠
2000
⎝
⎝
⎠
Note, in particular, the behavior of the impedance one decade below and one decade above the cutoff
frequency.
6.8 Solution:
As ω → ∞,
Z L → ∞ ⇒ Open, Z C → 0
⇒ Short ⇒Z → R1
As ω → 0,
Z C → ∞ ⇒ Open, Z L → 0
⇒ Short ⇒ Z → R1 + R2
a)
b)
Z[jω ] =
= R1 +
V[jω ]
ZC [ Z R2 + Z L ]
= Z R1 +
= R1 +
I[jω ]
ZC + [ Z R2 + Z L ]
jωC
=
jωC
( R 1 [ 1 - ω 2 LC ] + R 2 ) + j ( ω R 1 R 2 C + ωL ) (− j )
R 2 + j ωL
=
⋅
⇒
2
1 - ω LC + j ω R 2 C
[ 1 - ω 2 LC ] + j ω R 2 C
(− j )
(
ωR C + j (ω LC − 1)
ω (R1R2C + L) + j ω R1LC − R1 − R2
2
⇒ Z [ jω ] =
1
] [ R 2 + jωL ]
jωC
1
+ [ R 2 + jωL ]
jωC
[
2
2
)= R R C + L ⋅
1 2
R2C
1+ j
ω 2 R1LC − R1 − R2
ω (R1R2C + L)
1+ j
ω 2 LC − 1
ωR2C
Both f1[ω] and f2[ω] can be positive or negative, and therefore equal to plus or minus one
c)
depending on the frequency; therefore, both cases must be considered.
ω c [ R1 R 2 C + L ] = ± 1
f 1 [ω c] =
R 1 [ 1 - ω 2c LC ] + R 2
1
+ R2
R
] ω c - R1
= 0
ω 2c ± [ 2 +
C
L
R1
R 1 LC
1
1100
1
rad
R2
+
=
+
= 13.69 k
-9
0.19
s
L
R1C
[ 2300 ] [ 55⋅10 ]
3400
R1 + R 2
=
R 1 LC
[ 2300] [ 0.19 ] [ 55⋅10 -9 ]
ωc = −
rad
= 141.46 M 2
s
1
1
[ ± 13.69 ⋅10 3 ] ±
( [ ± 13.69 ⋅10 3 ]2 - 4[1][ - 141.5 ⋅10 6 ] )1/2
2
2
rad
rad
ω c4 = 20.569 k
s
s
Where only the positive answers are physically valid, i.e., a negative frequency is physically
impossible.
1
ω 2c LC - 1 = ± 1
⇒ ω 2c ± [ R 2 ] ω c = 0
f 2 [ω c] =
L
LC
ω c R2C
1100
rad
1
1
rad
R2
=
= 5.79 k
=
= 95.69 M 2
-9
L
0.19
s
LC
s
[ 0.19 ] [ 55⋅10 ]
1
1
6
2
1/2
= ± 2895 ± 10201
[ ± 5790 ] ±
( [ ± 5790 ] + 4[1][ 95.69 ⋅10 ] )
ωc = 2
2
rad
rad
⇒ ω c2 = 7.31 k
ω c3 = 13.09 k
s
s
Again, the negative roots were rejected because they are physically impossible.
d)
Magnitude and phase response in semilogarithmic frequency plots:
= ± 6.845 ⋅10 3 ± 13.724 ⋅10 3
⇒ ω c1 = 6.879 k
6.9 Solution
Analysis:
a)
As ω → ∞ :
0
Z C → 0 ∠ − 90 ⇒ Short
0
H v → 0 ∠ − 90
VD :
As ω → 0 :
VD :
Hv →
R2
∠ 00
R1 + R 2
ZC Z R2
=
ZC + Z R2
Z eq =
b)
VD :
0
Z C → ∞ ∠ − 90 ⇒ Open
H v [jω ] =
=
V o [jω ]
V i [jω ]
=
[
1
] [ R2 ]
jωC
1
+ R2
jωC
Z eq
Z R 1 + Z eq
R2
+
+
j ω R1 R 2 C
R1
R2
jωC
jωC
=
R2
1 + j ω R2C
R2
1 + jω R 2 C
1 + j ωR 2 C
=
=
R2
1
+ jω R 2 C
+
R1
1 + j ω R2C
1
R2
=
ω
+
R1
R 2 1 + j R1 R 2 C
R1 + R 2
c)
f[ ω c ] =
Ho =
ω c R1 R 2 C = 1
R1 + R 2
ωc =
1900
R2
=
1300 + 1900
R1 + R 2
1300 + 1900
[ 1300 ] [ 1900] [ 0.5182 ⋅10
-6
= 2.5 k
]
rad
s
= 0.5938 = 20 ⋅ Log[0.5938] = - 4.527 dB