Bandwidth and Cutoff Frequencies MCQ Quiz - Objective Question with Answer for Bandwidth and Cutoff Frequencies - Download Free PDF

Concept:

The quality factor Q is defined as the ratio of the resonant frequency to the bandwidth.

$Q=\frac{f_r}{BW}=\frac{1}{R}\sqrt{\frac{L}{C}}$

fr = Resonant frequency

BW = Bandwidth of the resonant circuit

At resonance, the voltage across the capacitor is Q times the supply voltage.

Vc = QVs

Analysis:

Given half-power frequencies F₁ and F₂ as 64 kHz and 81 kHz

Resonant frequency $F_r=\sqrt{F_1{F}_2}$

$=\sqrt{64\times 81}kHz$

= 72 kHz

Bandwidth = F₂ - F₁

= 81 - 64 kHz

= 17 kHz

Concept:

The quality factor Q is defined as the ratio of the resonant frequency to the bandwidth.

$Q=\frac{f_r}{BW}=\frac{1}{R}\sqrt{\frac{L}{C}}$

fr = Resonant frequency

BW = Bandwidth of the resonant circuit

At resonance, the voltage across the capacitor is Q times the supply voltage.

Vc = QVs

Analysis:

Given half-power frequencies F₁ and F₂ as 64 kHz and 81 kHz

Resonant frequency $F_r=\sqrt{F_1{F}_2}$

$=\sqrt{64\times 81}kHz$

= 72 kHz

Bandwidth = F₂ - F₁

= 81 - 64 kHz

= 17 kHz

Concept:

The graph between impedance Z and the frequency of parallel RLC circuit:

Here,

f₁ is the lower cutoff frequency

f₂ is the upper cutoff frequency

f_r is the resonant frequency

BW is the bandwidth

Formula:

BW = f₂ – f₁

$f_1=f_r-\left(\frac{BW}{2}\right)$

$f_2=f_r+\left(\frac{BW}{2}\right)$

Calculation:

Given

Lower cutoff frequency f₁ = 3.89KHz

Upper cutoff frequency f₂ = 4.1KHz

Bandwidth BW = f₂ – f₁

= 4.1 – 3.89

BW = 0.21KHz

$f_1=f_r-\left(\frac{BW}{2}\right)$

$3.89=f_r-\left(\frac{0.21}{2}\right)$

3.89 = f_r – 0.105

f_r = 3.995KHz _~ 4KHz

At cutoff frequencies

Impedance Z = 0.707 Z_r

$\frac{Z_r}{Z}=1.414$

We know that

V = I Xc

Where

V is the voltage across the capacitor

I is current through the capacitor

Xc is the capacitive reactance

V is directly proportional to Xc                                                             

$\frac{V_r}{V}=\frac{Z_r}{Z}$

Where

V_r is the voltage at resonance

V is the voltage at cutoff frequencies

Z_r is impedance at resonance

Z is impedance at the cutoff frequency

$\frac{V_r}{1.4}=1.414$

V_r = 2V

So, the voltage at 4 kHz is 2V

Points to be remembered:

• Frequency below resonant frequency acts as an inductive circuit in parallel RLC circuit
• Frequency above resonant frequency acts as a capacitive circuit in parallel RLC circuit

Equivalent admittance $\mathrm{Y}\left(\mathrm{s}\right)=\frac{1}{\mathrm{R}+\mathrm{s}\mathrm{L}}+\mathrm{s}\mathrm{C}$

$\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{\left(\mathrm{R}-\mathrm{j}\omega \mathrm{L}\right)}{{\mathrm{R}}^2+{\left(\omega \mathrm{L}\right)}^2}+\mathrm{j}\omega \mathrm{C}$

For resonance, Imaginary part = 0

$\begin{array}{l}\therefore \frac{\omega L}{R^2+{\left(\omega L\right)}^2}=\omega C\\ \therefore {\mathrm{f}}_0=\frac{1}{2\pi }\sqrt{\frac{1}{\mathrm{L}\mathrm{C}}-\frac{{\mathrm{R}}^2}{{\mathrm{L}}^2}}\\ {\left(2\pi \times 15\times {10}^3\right)}^2+\frac{{\left({10}^3\right)}^2}{{\left(0.2\right)}^2}=\frac{1}{0.2\mathrm{C}}\end{array}$

$\mathrm{C}=561.88\mathrm{P}\mathrm{F}$

$\begin{array}{rl}& y=\frac{1}{R_L+j{\omega }_L}+\frac{1}{R_C+\frac{1}{j{\omega }_C}}\\ & =\frac{\left(R_L-j{\omega }_L\right)}{R_L^2+{\left(\omega L\right)}^2}+\frac{\left(R_C+\frac{j}{{\omega }_C}\right)}{{\left(R_C^2+\frac{1}{\omega C}\right)}^2}\end{array}$

At resonance → Imaginary part = 0

$\begin{array}{rl}& \frac{{\omega }_{o}L}{R_L^2+{\left({\omega }_{o}L\right)}^2}=\frac{\frac{1}{{\omega }_{o}C}}{R_C^2+\frac{1}{{{(}_{{\omega }_{o}C})}^2}}\\ & {\omega }_{o}L\cdot R_C^2+\frac{{\omega }_{o}L}{{\left({\omega }_{o}C\right)}^2}=\frac{R_L^2}{{\omega }_{o}C}+\frac{{\left({\omega }_{o}L\right)}^2}{{\omega }_{o}C}\\ & R_C^2\cdot {\omega }_{o}L+\frac{L}{{\omega }_o{C}^2}=\frac{R_L^2}{{\omega }_{o}C}+\frac{L^2\cdot {\omega }_o}{C}\\ & {\omega }_o\left[L\cdot R_L^2-\frac{L^2}{C}\right]=\frac{1}{{\omega }_{o}C}\left[R_L^2-\frac{L}{C}\right]\\ & {\omega }_o=\frac{1}{\sqrt{LC}}\sqrt{\frac{\left(R_L^2-\frac{L}{C}\right)}{\left(R_C^2-\frac{L}{C}\right)}}\\ & {\omega }_o=\frac{1}{1}\sqrt{\frac{15}{3}}=\sqrt{5}rad/sec\end{array}$