Selectivity of Frequency

If we plot the magnitude of the current $I = E/Z_T$ versus frequency for a fixed applied voltage E, we obtain the curve shown in [Fig. 1], which rises from zero to a maximum value of $E/R$ (where $Z_T$ is a minimum) and then drops toward zero (as $Z_T$ increases) at a slower rate than it rose to its peak value. The curve is actually the inverse of the impedance-versus-frequency curve. Since the $Z_T$ curve is not absolutely symmetrical about the resonant frequency, the curve of the current versus frequency has the same property. There is a definite range of frequencies at which the current is near its maximum value and the impedance is at a minimum. Those frequencies corresponding to $0.707$ of the maximum current are called the band frequencies, cutoff frequencies, or half-power frequencies. They are indicated by $f_1$ and $f_2$ in [Fig. 1]. The above condition is derived using the fact that and The smaller the bandwidth, the higher the selectivity. The shape of the curve, as shown in [Fig. 2], depends on each element of the series RLC circuit. If the resistance is made smaller with a fixed inductance and capacitance, the bandwidth decreases and the selectivity increases. Similarly, if the ratio L/C increases with fixed resistance, the bandwidth again decreases with an increase in selectivity.

What is the difference between selectivity and bandwidth?

In terms of $Q_s$, if $R$ is larger for the same $X_L$, then $Q_s$ is less, as determined by the equation $Q_s = w_s L/R$. These conditions are shown in Fig. 3, indicating that the cutoff frequencies are then equidistant from the resonant frequency. For any $Q_s$, the preceding is not true. The cutoff frequencies $f_1$ and $f_2$ can be found for the general case (any $Q_s$) by first employing the fact that a drop in current to $0.707$ of its resonant value corresponds to an increase in impedance equal to $1/0.707 = \sqrt{2}$ times the resonant value, which is R.
Substituting $\sqrt{2} R$ into the equation for the magnitude of $Z_T$, we find that becomes or, squaring both sides, that and Taking the square root of both sides gives us or Substituting $wL$ for $X_L$ and $1/wC$ for $X_C$ and bringing both quantities to the left of the equal sign, we have or which can be written Solving the quadratic, we have and Let us first consider the case where $X_L > X_C$, which relates to $f_2$ or $w_2$. Also Eq.(2) relates $R = X_L-X_C$ to be a +ive value ($R=X$). Hence Eq.(3) becomes, with The negative sign in front of the second factor was dropped because ${1 \over 2} \sqrt{{R^2 \over L^2}+{4 \over LC}}$ is always greater than $R/(2L)$. If it were not dropped, there would be a negative solution for the radian frequency w. If we repeat the same procedure for $X_C > X_L$, which relates to $w_1$ or $f_1$ such that Also $R = X_L-X_C$ in Eq. (2) results in a -ive value ($R=-X$), the solution $f_1$ becomes The bandwidth (BW) is and Substituting $R/L = w_s/Q_s$ from $Q_s = w_sL/R$ and $1/2 \pi = f_s /w_s$ from $w_s = 2 \pi f_s$ gives us or which is a very convenient form since it relates the bandwidth to the $Q_s$ of the circuit. As mentioned earlier, Equation (6) verifies that the larger the $Q_s$, the smaller the bandwidth, and vice versa. Written in a slightly different form, Equation (6) becomes The ratio (${f_2 - f_1 \over f_s}$) is sometimes called the fractional bandwidth, providing an indication of the width of the bandwidth compared to the resonant frequency.
It can also be shown through mathematical manipulations of the pertinent equations that the resonant frequency is related to the geometric mean of the band frequencies; that is, Geometric mean of any numbers can be found by using formula