# Pass Band Filters   Whatsapp  A number of methods are used to establish the pass-band characteristic of Fig. 1. One method employs both a low-pass and a high-pass filter in cascade, as shown in Fig. $2$. Fig. 1: Low-Pass filter. Fig. 2: Pass-band filter.
The components are chosen to establish a cutoff frequency for the high-pass filter that is lower than the critical frequency of the low-pass filter, as shown in Fig. 3. A frequency $f_{1}$ may pass through the low pass filter but have little effect on $V_{o}$ due to the reject characteristics of the high-pass filter. Fig. 3: Pass-band characteristics.
A frequency $f_{2}$ may pass through the high-pass filter unmolested but be prohibited from reaching the high-pass filter by the low-pass characteristics. A frequency $f_{o}$ near the center of the passband will pass through both filters with very little degeneration.
The network of Example $1$ will generate the characteristics of Fig. 3. However, for a circuit such as the one shown in Fig. 4, there is a loading between stages at each frequency that will affect the level of $\mathbf{V}_{o}$. Fig. 4: Pass-band filter.
Through proper design, the level of $\mathbf{V}_{o}$ may be very near the level of $\mathbf{V}_{i}$ in the pass-band, but it will never equal it exactly. In addition, as the critical frequencies of each filter get closer and closer together to increase the quality factor of the response curve, the peak values within the pass-band will continue to drop. For cases where $V_{o_{max }} \neq V_{i_{max }}$ the bandwidth is defined at $0.707$ of the resulting $V_{o_{max }}$.
Example 1: For the pass-band filter of Fig. 4:
a. Determine the critical frequencies for the low- and high-pass filters.
b. Using only the critical frequencies, sketch the response characteristics.
c. Determine the actual value of $V_{o}$ at the high-pass critical frequency calculated in part (a), and compare it to the level that will define the upper frequency for the pass-band. Solution:
a. High-pass filter:
$$f_{c}=\frac{1}{2 \pi R_{1} C_{1}}=\frac{1}{2 \pi(1 \mathrm{k} \Omega)(1.5 \mathrm{nF})}=\mathbf{1 0 6 . 1} \mathbf{k H z}$$
Low-pass filter:
$$f_{c}=\frac{1}{2 \pi R_{2} C_{2}}=\frac{1}{2 \pi(40 \mathrm{k} \Omega)(4 \mathrm{pF})}=\mathbf{9 9 4 . 7 2} \mathbf{~ k H z}$$
b. In the mid-region of the pass-band at about $500 \mathrm{kHz}$, an analysis of the network will reveal that $V_{o} \cong 0.9 V_{i}$ as shown in Fig. 5. The bandwidth is therefore defined at a level of $0.707\left(0.9 V_{i} \right)=0.636 V_{i}$ as also shown in Fig. $5$. c. At $f=994.72 \mathrm{kHz}$
$$X_{C_{1}}=\frac{1}{2 \pi f C_{1}} \cong 107 \Omega$$ Fig. 5: Pass-band characteristics for the filter of Fig. 4.
and
$$X_{C_{2}}=\frac{1}{2 \pi f C_{2}}=R_{2}=40 \mathrm{k} \Omega$$
resulting in the network of Fig. 6. Fig. 6: Network of Fig. 4 at f = 994.72 kHz.
The parallel combination $R_{1} |\left(R_{2}-j X_{C_{2}} \right)$ is essentially $0.976 \mathrm{k} \Omega \angle 0^{\circ}$ because the $R_{2}-X_{C_{2}}$ combination is so large compared to the parallel resistor $R_{1}$. Then
$$\mathbf{V}^{\prime}=\frac{0.976 \mathrm{k} \Omega \angle 0^{\circ}\left(\mathbf{V}_{i}\right)}{0.976 \mathrm{k} \Omega-j 0.107 \mathrm{k} \Omega} \cong 0.994 \mathbf{V}_{i} \angle 6.26^{\circ}$$
with
$$\mathbf{V}_{o}=\frac{\left(40 \mathrm{k} \Omega \angle-90^{\circ}\right)\left(0.994 \mathbf{V}_{i} \angle 6.26^{\circ} \right)}{40 \mathrm{k} \Omega-j 40 \mathrm{k} \Omega}$$
$$\mathbf{V}_{o} \cong 0.703 \mathbf{V}_{i} \angle-39^{\circ}$$
so that
$$V_{o} \cong \mathbf{0 . 7 0 3} V_{i} \quad \text{ at } f=994.72 \mathrm{kHz}$$
Since the bandwidth is defined at $0.636 V_{i}$ the upper cutoff frequency will be higher than $994.72 \mathrm{kHz}$ as shown in Fig. $5$.
The pass-band response can also be obtained using the series and parallel resonant circuits discussed in Chapter 18:Resonance. In each case, however, $V_{o}$ will not be equal to $V_{i}$ in the pass-band, but a frequency range in which $V_{o}$ will be equal to or greater than $0.707 V_{max }$ can be defined. For the series resonant circuit of Fig. 7, $X_{L}=X_{C}$ at resonance, and
$$V_{o_{max }}=\frac{R}{R+R_{l}} V_{i} \quad f=f_{s}$$   Fig. 7: Series resonant pass-band filter.   Fig. 8: Parallel resonant pass-band filter.
and
$$f_{s}=\frac{1}{2 \pi \sqrt{L C}}$$
with
$$Q_{s}=\frac{X_{L}}{R+R_{l}}$$
and
$$B W=\frac{f_{s}}{Q_{s}}$$
For the parallel resonant circuit of Fig. 8, $Z_{T_{p}}$ is a maximum value at resonance, and
$$V_{o_{max }}=\frac{Z_{T_{p}} V_{i}}{Z_{T_{p}}+R}$$
with
$$Z_{T_{p}}=Q_{l}^{2} R_{l}$$
and
$$f_{p}=\frac{1}{2 \pi \sqrt{L C}}$$
﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿For the parallel resonant circuit
$$Q_{p}=\frac{X_{L}}{R_{l}}$$
and
$$B W=\frac{f_{p}}{Q_{p}}$$
As a first approximation that is acceptable for most practical applications, it can be assumed that the resonant frequency bisects the bandwidth.

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250