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Example-1

Example-1
  1. Design a low-pass active filter at a cut-off frequency of 1kHz with a pass band gain of 2. Using the frequency scaling technique, convert this filter to a low-pass filter of cut-off frequency 1.6 kHz.
  2. Plot the frequency response of this low-pass active filter.
Solution
  1. Here, fc = 1 kHz, AF = 2; Let, C = 0.01 μF.
Description: 4680.png
 
∵ AF = 2 = Description: 4701.png ⇒ Rf = R1 = 10 kΩ
 
So, the complete circuit is shown in Fig. (a).
 
Description: 1568.png
(a) Circuit of Example 1
 
To change the cut-off frequency from 1 kHz to 1.6 kHz, we multiply the 15.9 kΩ resistor by
Description: 4707.png
 
∵ New resistor, R = 15.9 × 0.625 = 9.94 kΩ
  1. To plot the frequency–response, the data are obtained from the equation,
Description: 4716.png
   
Frequency (Hz) Gain Gain (in dB)
10 2 6.02
100 1.99 5.98
200 1.96 5.85
700 1.64 4.29
1,000 1.41 3.01
3,000 0.63 –3.98
7,000 0.28 –10.97
10,000 0.20 –14.02
30,000 0.07 –23.53
100,000 0.02 –33.98
 
Description: 4722.png
(b) Filter characteristics of Example 1
 




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