Answer:
C_h = 0.166 nF
C_L = 0.153 nF
Explanation:
Given:
- Ideal frequency f_o = 500 KHz
- Bandwidth of frequency BW = 40 KHz
- The resistance identical to both low and high pass filter = 2 Kohms
Find:
Design a passive band-pass filter to do this by cascading a low and high pass filter.
Solution:
- First determine the cut-off frequencies f_c for each filter:
f_c,L for High pass filter:
f_c,L = f_o - BW/2 = 500 - 40/2
f_c,L = 480 KHz
f_c,h for Low pass filter:
f_c,h = f_o + BW/2 = 500 + 40/2
f_c,h = 520 KHz
- Now use the design formula for R-C circuit for each filter:
General design formula:
f_c = 1 /2*pi*R*C_i
C,h for High pass filter:
C_h = 1 /2*pi*R*f_c,L
C_h = 1 /2*pi*2000*480,000
C_h = 0.166 nF
C,L for Low pass filter:
C_L = 1 /2*pi*R*f_c,h
C_L = 1 /2*pi*2000*520,000
C_L = 0.153 nF
