Design a basic band pass filter comprising of a series combination of T-section high pass filter and a T-section low pass filter to pass the frequencies between 10kHz and 100kHz. take K=100 for HPF and K=400 for LPF
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Please see a simple design for a series combination. Actually there should be some isolation or matching of impedance (load ) between the two filters.
We could also do a design of a band pass filter (constant K) with T-network of L and C.
For the simple design as shown:
HPF:
K = √(Z1 Z2) = √[L₂ / C₂] = 100
=> L₂ = 10,000 * C₂
f₀ = cut of frequency = 10 kHz
ω₀ = 2 π * 10 kHz = 1 / [ 2√(L₂ C₂) ]
=> L₂ C₂ = 1 / (16 π² f₀²) = 63.325 * 10⁻¹²
Solving for C₂ and L₂, we get C₂ = 0.0795 μF and L₂ = 795 μH
LPF: design:
K = 400 = √(L/C)
=> L₁ = 160,000 * C₁
f₀ = cut of frequency =100 kHz
ω₀ = 2 π * 100 kHz = 2 /√(L₁ C₁)
=> L₁ C₁ = 10.13 * 10⁻¹²
Solving for C₁ we get 7.95 pF
L₁ = 1.273 mH
=============
Alternately a design for Band pass filter using T-sections is available in http://www.brainly.in/question/744122 .
Please see that too.
We could also do a design of a band pass filter (constant K) with T-network of L and C.
For the simple design as shown:
HPF:
K = √(Z1 Z2) = √[L₂ / C₂] = 100
=> L₂ = 10,000 * C₂
f₀ = cut of frequency = 10 kHz
ω₀ = 2 π * 10 kHz = 1 / [ 2√(L₂ C₂) ]
=> L₂ C₂ = 1 / (16 π² f₀²) = 63.325 * 10⁻¹²
Solving for C₂ and L₂, we get C₂ = 0.0795 μF and L₂ = 795 μH
LPF: design:
K = 400 = √(L/C)
=> L₁ = 160,000 * C₁
f₀ = cut of frequency =100 kHz
ω₀ = 2 π * 100 kHz = 2 /√(L₁ C₁)
=> L₁ C₁ = 10.13 * 10⁻¹²
Solving for C₁ we get 7.95 pF
L₁ = 1.273 mH
=============
Alternately a design for Band pass filter using T-sections is available in http://www.brainly.in/question/744122 .
Please see that too.
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