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- 2070 Chaitra · 5+3 marks
Chebyshev order (0.1 dB, 20 dB, 2500 rad/s)
Short answer
The order n must satisfy n ≥ arcosh(√[(10^(0.1αmin)-1)/(10^(0.1αmax)-1)]) / arcosh(ωs/ωp). For αmax=0.1 dB, αmin=20 dB, ωs/ωp=2.5: ε²=0.02329, arcosh(65.1936)=4.8704, arcosh(2.5)=1.5668, so n ≥ 3.109 → n=4.
Derive an expression to calculate the order of Chebyshev low pass filter. Use this formula to estimate the order of Chebyshev low pass filter having following specification: αmax = 0.1 dB, ωp = 1000 rad/s, αmin = 20 dB, ωs = 2500 rad/s
Derivation of the order
The Chebyshev low-pass magnitude response (normalized so that the passband edge is ) is
where is the Chebyshev polynomial of order :
The attenuation in dB is .
Step 1: ripple factor from the passband edge. At , for every , so the attenuation is exactly :
Step 2: stopband condition. At the attenuation must be at least :
Step 3: solve for n. Since , :
The order is the next whole number above this value. (Useful identity: .)
Estimate of the order
Given dB, dB, .
Answer: n = 4 (next integer above 3.109).
A small ripple (0.1 dB) makes small, so a higher order is needed than with 0.5 dB or 1 dB ripple for the same stopband.
✓ Answer checked by Claude Sonnet 5.5
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