Quality Factor, Peak Frequency and Bandwidth

The physical meaning of the quality factor $Q$ of a circuit is the ratio between the energy stored in the circuit (in $C$ and $L$) and the energy dissipated (by $R$):

$$Q=2\pi\frac{\mbox{maximum energy stored}}{\mbox{energy dissipated per cycle}}$$

The maximum energy stored in $L$ is:

$$W_L=\int_0^T v(t)\; i(t) dt=\int_0^T i(t) L \frac{di(t)}{dt} dt =L \int_0^{I_p} i \;di=\frac{1}{2}LI_p^2=LI^2_{rms}$$

where $I_p=\sqrt{2}I_{rms}$ is the peak current through $L$. The maximum energy stored in $C$ is:

$$W_C=\int_0^T v(t)\; i(t) dt=\int_0^T v(t) C \frac{dv(t)}{dt} dt =C \int_0^{V_p} v \;dv=\frac{1}{2}CV_p^2=CV^2_{rms}$$

where $V_p=\sqrt{2}V_{rms}$ is the peak voltage across $C$. We show that $W_L=W_C$ at resonant frequency $\omega=\omega_0=1/\sqrt{LC}$:

$$W_L=LI^2_{rms}=L\frac{V^2_{rms}}{(\omega_0L)^2} =L\frac{V^2_{rms}LC}{L^2}=CV^2_{rms}=W_C$$

where $V_{rms}$ is the voltage across $L$ which is the same as that across $C$ when $\omega=\omega_0$. Here the energy $W_L=W_C$ is converted back and forth between megantic energy in $L$ and electrical energy in $C$.

The energy dissipated in $R$ per cycle $T_0=2\pi/\omega_0=1/f_0$ is:

$$W_R=P_R T_0=I^2_{rms} R T_0$$

According to the definition of $Q$ above, we have

$$Q=2\pi\frac{W_L}{W_R}=2\pi\frac{LI^2_{rms}}{I^2_{rms}RT_0 } =2\pi f_0\frac{L}{R}=\frac{\omega_0L}{R}$$

which is the same as the $Q$ defined before.

Relationship between $Q$ and $\zeta$

At the resonance $\omega=\omega_0=1/\sqrt{LC}$, the total impedance of a series RCL circuit becomes:

$$Z_0=Z_{\omega=\omega_0}=R+j(\omega_0 L-\frac{1}{\omega_0 C})=R=\frac{1}{Y_0}$$

If the voltage across $R$ is treated as the output of the circuit, then the frequency response function (FRF) of the system can be expressed as:

$$H_R=\frac{{\bf V}_R}{{\bf V}}=\frac{Z_R}{Z_0} =\frac{R}{R+j\o... ...zeta \omega_0}{(j\omega)^2+j\omega 2\zeta \omega_0+\omega_0^2}$$

Here the denominator is conventionally expressed in the canonical form:

$$(j\omega)^2+j\omega R/L+1/LC=(j\omega)^2+j\omega 2\zeta \omega_0+\omega_0^2$$

where

$$\omega_0\stackrel{\triangle}{=}\frac{1}{\sqrt{LC}},\;\;\;\;\; \zeta\stackrel{\triangle}{=}\frac{R}{2}\sqrt{\frac{C}{L}}$$

On the other hand, we also have the definition of the quality factor:

$$Q\stackrel{\triangle}{=}\frac{\vert Z_L\vert}{R}=\frac{\vert ... ...ga_0L}{R}=\frac{1}{\omega_0CR} =\frac{1}{R}\sqrt{\frac{L}{C}}$$

therefore we get

$$\zeta=\frac{1}{2Q}$$

The quality factor $Q$ can also be used to judget whether a second order system is under, critically or over damped:

$\zeta<1$	$Q>0.5$	under damped
$\zeta=1$	$Q=0.5$	critically damped
$\zeta>1$	$Q<0.5$	over damped

Peak Frequency and Bandwidth

The frequency response function above can be reexpressed as:

$$H_R=\frac{{\bf V}_R}{{\bf V}}=\frac{Z_0}{Z}=\frac{Y}{Y_0} =\... ...omega C)} =[1+j(\frac{\omega L}{R}-\frac{1}{\omega CR})]^{-1}$$

Since $Q=\omega_0L/R=1/\omega_0CR$, we have

$$L/R=Q/\omega_0,\;\;\;\;\;\;\mbo{and}\;\;\;\;\;\;1/RC=Q\omega_0$$

Substituting these into the equation above we get

$$H_R=\frac{Y}{Y_0}=[1+jQ(\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega})]^{-1}$$

At the resonant frequency $\omega=\omega_0$, $H=1$ reaches the maximum. When $\omega$ is either lower or higher than $\omega_0$, $H$ is smaller. The bandwidth is defined as

$$\triangle \omega\stackrel{\triangle}{=}\omega_2-\omega_1$$

where $\omega_2>\omega_1$ are the two cut-off frequencies (or half-power frequency) at which $\vert H\vert=1/\sqrt{2}=0.707$ (power is halved), i.e.,

$$H_R=\frac{1}{1\pm j1},\;\;\;\;\;\mbox{i.e.}\;\;\;\; \frac{\omega}{\omega_0}-\frac{\omega_0}{\omega}=\pm \frac{1}{Q}$$

Therefore the two cut-off frequencies should satisfy

$$\frac{\omega_1}{\omega_0}-\frac{\omega_0}{\omega_1}=-\frac{1}... ...frac{\omega_2}{\omega_0}-\frac{\omega_0}{\omega_2}=\frac{1}{Q}$$

Solving these two equations, we get:

$$\omega_1=\omega_0[\sqrt{1+(\frac{1}{2Q})^2}-\frac{1}{2Q}] =\... ...ac{1}{2Q})^2}+\frac{1}{2Q}] =\omega_0(\sqrt{1+\zeta^2}+\zeta)$$

from which we can find the bandwidth

$$\triangle \omega=\omega_2-\omega_1=\frac{\omega_0}{Q}=2\zeta ... ...;\mbox{or}\;\;\;\;\triangle f=f_2-f_1=\frac{f_0}{Q}=2\zeta f_0$$

i.e., the bandwidth is proportional to $\zeta$ or inversely proportional to $Q$. Also note that the middle point between $\omega_1$ and $\omega_2$ is $\omega'_0=(\omega_1+\omega_2)/2=\omega_0 \sqrt{1+\zeta^2}>\omega_0$, i.e., $\omega_2-\omega_0>\omega_0-\omega_1$.

In practice, $Q=1/2\zeta$ is usually much greater than 1 (typically $Q>10$, i.e., $\zeta<0.05$), we have $\sqrt{1+(1/2Q)^2}=\sqrt{1+\zeta^2} \approx 1$ and

$$\omega_1\approx \omega_0-\frac{\omega_0}{2Q}=\omega_0(1-\zeta... ...\omega_2\approx \omega_0+\frac{\omega_0}{2Q}=\omega_0(1+\zeta)$$

we therefore get these simple relations:

$$\omega_2-\omega_0=\omega_0-\omega_1=\omega_0\zeta,\;\;\;\;\; ... ...le \omega=\omega_2-\omega_1=\frac{\omega_0}{Q}=2\zeta \omega_0$$

If we consider the voltage across each of the three components in the RCL series circuit as the output, then we have the following frequency response functions:

$$H_L=\frac{{\bf V}_L}{\bf V} =\frac{j\omega L}{R+j\omega L+1/j... ...c{(j\omega)^2}{(j\omega)^2+j\omega 2\zeta \omega_0+\omega_0^2}$$

$$H_R=\frac{{\bf V}_R}{\bf V} =\frac{R}{R+j\omega L+1/j\omega C... ...zata \omega_0}{(j\omega)^2+j\omega 2\zeta \omega_0+\omega_0^2}$$

$$H_C=\frac{{\bf V}_C}{\bf V} =\frac{1/j\omega C}{R+j\omega L+1... ...ac{\omega^2_0}{(j\omega)^2+j\omega 2\zeta \omega_0+\omega_0^2}$$

For a parallel RCL circuit with current input, due to the duality between current and voltage, parallel and series configuration, the same derivation of bandwidth can be carried out to obtain the same conclusions.

Summary:

• The resonant frequency of both series and parallel RCL circuits is completely determined by $L$ and $C$: $\omega_0=1/\sqrt{LC}$, independent of the resistance $R$ in the circuit.

• At resonant frequency, the impedance $Z=1/Y$ of a series RCL circuit is real and reaches minimum, and the current through the three components reaches maximum; the admittance $Y=1/Z$ of a parallel RCL circuit is real and reaches minimum and the voltage across the three components reaches maximum.

• In series RCL with voltage input and parallel RCL with current input, the quality factor $Q$ is proportional to the ratio between $L$ and $C$:
  
  $$Q_s=\frac{1}{R}\sqrt{\frac{L}{C}},\;\;\;\;Q_p=R\sqrt{\frac{C}{L}}=\frac{1}{Q_s}$$
  
  

• In series RCL, $Q_s$ is inversely proportional $R$ (the larger $R$, the smaller $Q_s$, the more energy lost and the wider bandwidth), while in parallel RCL, $Q_p$ is proportional to $R$ (the larger $R$, the larger $Q_p$, the less energy lost and the narrower bandwidth).

• At resonant frequency, the impedance of a series RCL circuit reaches minimum, consequently the current ${\bf I}$ reseaches maximum and so does the voltage across the resistor ${\bf V_R}={\bf I}R$. However, the voltage across the inductor ${\bf V_L}={\bf I}Z_L$ reaches maximum at a frequency slightly higher than the resonant frequency as $Z_L=j\omega L$ is proportional to $\omega$, and the voltage across the capacitor ${\bf V_C}={\bf I}Z_C$ reaches maximum at a frequency slightly lower than the resonant frequency as $Z_C=1/j\omega C$ is inversely proportional to $\omega$, as shown in the linear and log-scale plots below.

  

  See this website for more detailed discussions of second-order systems.

Example 1:

A series RCL circuit composed of an inductor $L=80\mu H$ and $R=8\Omega$ and a capacitor $C$ is connected to a voltage source. Find the value of $C$ for this circuit to resonate at $f=400\;kHz$, also find the bandwidth.

$$\omega_0=\sqrt{\frac{1}{LC}},\;\;\;\; C=\frac{1}{\omega_0^2 L}=\frac{1}{(2\pi 400\times 10^3)^2\times 80\times 10^{-6}} =20nF$$

The quality factor is

$$Q=\frac{\omega_0 L}{R}=\frac{2\pi 400\times 10^3\times 80\times 10^{-6}}{8}=25.13$$

The bandwidth is

$$\triangle f=\frac{f_0}{Q}=\frac{400\times 10^3}{25.13}=15.9\;kHz$$

or

$$\triangle \omega=\frac{\omega_0}{Q}=\frac{R}{L}=10^5$$

Example 2:

In reality, all inductors have a non-zero resistance, therefore a parallel resonance circuit should be modeled as shown in the figure.

The admittance is:

$$Y(\omega)=\frac{1}{R+j\omega L}+j\omega C =\frac{R-j\omega L... ...c{1}{R^2+\omega^2L^2}[R-j(\omega L-\omega C(R^2+\omega^2L^2))]$$

As frequency $\omega$ appears in the real part $Re[Y(\omega)]$ as well as in the imagineary part $Im[Y(\omega)]$, the resonant frequency that minimizes $\vert Y(\omega)\vert$ has to be found by

$$\frac{d}{d\omega} \; \vert Y(\omega)\vert =0$$

However, when the quality factor $Q=\omega_0 L/R$ associated with the non-ideal inductor is large enough ($Q > 20$), all previous discussed relations for ideal inductors still hold approximately, and the resonant frequency $\omega_0$ can still be found approximately by the previous approach by letting $Im[Y(\omega)]=0$:

$$\omega_0 L=\omega_0 C(R^2+\omega_0^2L^2),\;\;\;\;\Longrightarrow \;\;\;\;\;\omega_0=\sqrt{\frac{1}{LC}-(\frac{R}{L})^2}$$

For $\omega_0$ to be real, we must have

$$\frac{1}{LC} > (\frac{R}{L})^2, \;\;\;\;\;\;\mbox{i,e,} \;\;\;\;\;\;R<\sqrt{\frac{L}{C}}$$

Typically we have $R \ll \sqrt{L/C}$, and the resonant frequency is

$$\omega_0=\sqrt{\frac{1}{LC}-(\frac{R}{L})^2}\approx \frac{1}{\sqrt{LC}}$$

Note: For the same reason, when considering the transfer function of a series RCL circuit when the output is the voltage across either $C$ or $L$, the peak frequency $\omega_p$ is not exactly the same as the resonant frequency $\omega_0$, which only minimizes the denominator, but the numerator is still a function of $\omega$. Only when the output is the voltage across $R$ (i.e., the numerator is $R$, no longer a function of $\omega$), will the resonant frequency $\omega_0$ be the same as the peak frequency.

Example 3:

Resonant circuuit is widely used in radio and TV receivers to select a desired station from many stations available. The circuit and its model are shown in the figure below. Assume $L=0.3mH$, $R=16\Omega$, and $C$ is variable capacitor, which can be adjusted to match the resonant frequency of the circuit to the frequency of the desired station. Also assume the frequency of the desired station is $640\;kHz$, find the value of $C$. If the induced voltage in the circuit is $e=2 \mu V$ (rms), find the current (rms) in the resonant circuit, and the output voltage (rms) across the capacitor.

Hint: First check if $Q=\omega L/R$ is larger than 20. If so, the resonant frequency can still be found approximately as $\omega_0=1/\sqrt{LC}$ and $jX_L-jX_C=0$.

Solution: At the desired resonant frequency $f=640\;kHz$, the reactance of the inductor is

$$X_L=\omega L=2\pi f L=2\times 3.14\times 640\times 10^3\times 0.3 \times 10^{-3}=1206\Omega$$

and the quality factor $Q$ of this circuit is

$$Q=\frac{\omega L}{R}=\frac{2\times 3.14\times 640\times 10^3\times 0.3\times 10^{-3}}{16}=75 \gg 20$$

The resonant frequency can be expressed approximately as

$$f_0=\frac{1}{2\pi\sqrt{LC}}=640\;kHz$$

Solving this we get

$$C=\frac{1}{(2\pi f_0)^2L}=206\times 10^{-12}F=206\;pF$$

The current in circuit is

$$I_{rms}=e/R=2\times 10^{-6}/16=0.125\times 10^{-6}$$

The output voltage across $C$ is

$$V_C\approx V_L=I_{rms} X_L=0.125\times 10^{-6}\times 1206=151\;\mu V$$