hw 3 solutions

# R 2 r 1 r 2 in terms of poles and zeros this term is

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R 2 / ( R 1 + R 2 ). In terms of poles and zeros, this term is just p 1 /z 1 . What this means for our design is that we will have to add another section to our cascade to cancel out this extra term. Solving for the component values, we get R 1 = 1 /p 1 - 1 /z 1 = 1 / 1 - 1 / 10 = 0 . 9 Ω R 2 = 1 /z 1 = 0 . 1 Ω C 1 = 1 F T hi = p 1 /z 1 = 0 . 1 Using these component values with the circuit above, the synthesized transfer function is 0 . 1( s + 10) / ( s + 1) which is 0 . 1 H 1 ( s ); therefore, we will need an additional circuit with a gain 10 to realize H 1 ( s ). Now let’s look at the second transfer function H 2 ( s ) which has the form H 2 ( s ) = ( s + 100) ( s + 1000) = ( s + z 2 ) ( s + p 2 ) where z 2 = 100 rad / s and where p 2 = 1000 rad / s. Since p 2 > z 2 , we will utilize the first circuit from Table 3.4 shown in Figure 3. Unlike the third circuit, there is no additional term in front. R 3 = 1 /z 2 C 3 = 1 R 4 = 1 / ( p 2 - z 2 ) T ( s ) = ( s + z 2 ) ( s + p 2 ) p 2 > z 2 Figure 3: The lead highpass circuit from Table 3.4 of the text. Solving for the component values, we get R 3 = 1 /z 2 = 1 / 100 = 0 . 01 Ω R 4 = 1 / ( p 2 - z 2 ) = 1 / 900 = 0 . 00111 Ω C 3 = 1 F Using these component values with the circuit above, the synthesized transfer function is ( s +100) / ( s +1000) which is exactly H 2 ( s ). No additional gain circuits are needed for this section. If we cascade the two sections we have designed so far, we will get the overall transfer function given by 0 . 1 ( s + 10) ( s + 1) ( s + 100) ( s + 1000) Dr. Vahe Caliskan 2 of 7 Posted: February 19, 2013

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ECE 412 Introduction to Filter Synthesis Homework #3 Solutions University of Illinois at Chicago Spring 2013 Therefore, we also need to include a circuit with a constant gain of 20 to make the overall product equal to the desired transfer function. This can be done with a noninverting amplifier circuit as shown in Figure 4. The gain of the noninverting amplifier is 1 + R f /R i ; therefore, by choosing the ratio R f /R i = 19 will result in a gain of 20. Since this is the unscaled design, we choose R f = 19 Ω and R i = 1 Ω. Now the overall + + - - R 1 = 0 . 9 Ω R 2 = 0 . 1 Ω C 2 = 1 F R 3 = 0 . 01 Ω R 4 = 0 . 00111 Ω C 3 = 1 F R f = 19 Ω R i = 1 Ω Figure 4: The unscaled circuit that synthesizes H ( s ). transfer function of the three stages is just the product given by parenleftbigg 0 . 1 ( s + 10) ( s + 1) parenrightbigg (1) parenleftbigg ( s + 100) ( s + 1000) parenrightbigg (20) which synthesizes the desired H ( s ). The solution shown in Figure 4 is just fine but we can do a little better. Instead of using the noninverting amplifier at the end we can use it between the two passive sections in place of the buffer. This will use one less opamp than the design of Figure 4 but will use the same noninverting amplifier section as before. The
• Spring '11
• Vahe
• University of Illinois, DR. VAHE CALISKAN

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