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

This preview shows page 2 - 4 out of 7 pages.

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
Image of page 2

Subscribe to view the full document.

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
Image of page 3
Image of page 4
  • Spring '11
  • Vahe
  • University of Illinois, DR. VAHE CALISKAN

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern