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cricket20 [7]
3 years ago
5

An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a pass

ive bandpass filter to do this. Do this by combining a high pass and a low pass filter. For our purposes, create a pass band width of exactly 40kHz and is centered at the ideal frequency. R = 2kOhms. Select C[lowpass], enter value in terms of nF =
Engineering
1 answer:
Finger [1]3 years ago
7 0

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          

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the ratio of the etched to the original crack tip radius is 30.24

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Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

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2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

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