Q1:
Q2:
1k
470 ohms
2.2k
Parameter
R=470 Ω
R=1.0 kΩ
R=2.2 kΩ
Resonant Frequency
(π0)
799. 8π»π§
799. 8π»π§
799. 8π»π§
Lower Cutoff (π1)
625. 0π»π§
605. 0π»π§
256. 7π»π§
Upper Cutoff (π2)
1037. 0π»π§
1350. 0π»π§
2200. 0π»π§
Bandwidth (π΅π)
412. 0π»π§
745. 0π»π§
1943. 3π»π§
Simulated Quality
Factor (π)
1. 94
1. 07
0. 41
Analytical Quality
Factor (π)
1. 92
0. 91
0. 41
The resonant frequency f0βremains strictly constant at 799.8 Hz regardless of
resistance because it depends only on L (180 mH) and C (220 nF). As R increases
from 470 Ω to 2.2 kΩ, the lower cutoff frequency f1βdecreases (625 Hz→256.7 Hz)
while the upper cutoff frequency f2βincreases (1.037 kHz→2.2 kHz). This expands
the total bandwidth from 412 Hz to 1.943 kHz, causing the Quality Factor Q to drop
from 1.94 down to 0.41, verifying that Q is inversely proportional to resistance R.
Q3:
4.3k
Q4:
Parallel:
For an ideal parallel π
πΏπΆ circuit driven by a current source (or evaluated via input admittance
π(π )), the admittance is:
1
1
π(π ) = π
+ π πΏ + π πΆ
Setting π = πω:
1
(
1
π(πω) = π
+ π ωπΆ − ωπΏ
)
1. Resonant Frequency (ω0)
Resonance occurs when the imaginary part of the admittance (the total susceptance) equals
zero:
$$\omega_0 C - \frac{1}{\omega_0 L} = 0 \implies \omega_0^2 = \frac{1}{LC} \implies \omega_0
= \frac{1}{\sqrt{LC}}$$
5
Given ω0 = 100ππππ/π = 10 πππ/π :
πΏπΆ =
1
2
ω0
=
1
5 2
(10 )
−10 2
= 10
π
2. Bandwidth (β) and Quality Factor (π)
The bandwidth β is defined as the frequency range between the upper and lower half-power (
− 3ππ΅) cutoff frequencies ω1 and ω2:
1
β = ω2 − ω1 = π
πΆ (πππ/π )
Given β = 5ππππ/π = 5000πππ/π :
1
1
−4
π
πΆ = β = 5000 = 2 × 10 π
The Quality Factor π is:
π=
ω0
β
=
100ππππ/π
5ππππ/π
= 20
3. Component Selection
Since π
πΏπΆ circuit parameters leave one degree of freedom, selecting a standard baseline
resistance π
yields the inductance and capacitance. Choosing π
= 1πβ¦:
ββ Capacitance (πΆ):β
$$C = \frac{1}{\beta \cdot R} = \frac{1}{(5000\text{ rad/s})(1000\ \Omega)} = 0.2\
\mu\text{F} \quad (200\text{ nF})$$
ββ Inductance (πΏ):β
$$L = \frac{1}{\omega_0^2 C} = \frac{1}{(10^5)^2 (0.2 \times 10^{-6})} = 0.5\text{ mH}
\quad (500\ \mu\text{H})$$
Verification Summary Table
Parameter
Theoretical Expression
Resonant Frequency (
ω0)
1
Bandwidth (β)
1
π
πΆ
Target / Calculated Value
100ππππ/π (≈ 15. 915ππ»π§)
πΏπΆ
5ππππ/π (≈ 795. 77π»π§)
Quality Factor (π)
πΆ
πΏ
π
=
ω0
20
β
Resistor (π
)
Chosen
1πβ¦
Capacitor (πΆ)
1
βπ
200ππΉ
Inductor (πΏ)
1
0. 5ππ»
2
ω0πΆ
Q5:
Nmos transf ch
Op ch
Pmos op
Pmos tf ch