This can also be seen more directly by examining the equations the following

This can also be seen more directly by examining the

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. This can also be seen more directly by examining the equations; the following workings assume no simplification, however, the equation can be constructed through simple substitution (and greatly reduces the work needed): Non-simplified solution; solving Equation (1) for : ( * ( * ( * ( * Substitute into Equation (3) and solve for : 𝑉 ?? 𝑉 ? 𝑉 ? 𝑉 ? 𝑉 ??? ? 𝑒𝑞
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41/92 [( * ( * ] ( * ( * [ ] ( * [ ] ( * [ ] ( * ( * ( * Substitute both of the above into Equation (2): ( ) ( * ( ) ( * ( ) ( * ( * ( ) ( * ( ) ( * ( * ( ) ( * ( * ( *( ) ( * ( * ( * ( * ( ) ( * ( * ( * ( * ( ) ( * ( * ( * ( ) ( *( * ( ( ) ) ( ( ) ) ( )
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42/92 Question 18. The given transfer function must first be converted to ( ) form: ( ) ( ) ( ) ( )( ) . / . / . / . / . / . / . / . / . / . / . / Drawing the Bode Magnitude Plot: ROC is initially , as there is no pole/zero at the origin ROC becomes at the first pole, with ROC becomes at the double zero, with ROC becomes zero again after the final pole at No shift is required, as Drawing the Bode Phase Plot: Initial value is zero degrees phase shift, with . At ( ) , the first pole takes effect, and . At ( ) , the double zero becomes active, and . At ( ) , the first pole’s effect ends, and . At ( ) , the second pole takes effect, and . At ( ) , the double zero ends, and . At ( ) , the second pole ends, and .
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43/92
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44/92 Question 19. To begin, we must first write a transfer function to implement. From the graph, the initial ROC is , indicating that there is a zero at the origin. One can also notice that at , the magnitude, | | . There are no poles or zeros active before , and thus there is no additional gain in this system (i.e., ). At , the ROC becomes zero, indicating the presence of a pole. The ROC remains constant until , at which point , indicating another pole. The ROC does not change any more, indicating no further poles or zeros. ( ) ( )( ) . / . / . / . / With the above, if we want | | at , we need . However: Thus, we need : ( ) ( )( ) We can implement the zero at the origin and the pole at using one op-amp. We can implement the remaining pole (which is many orders of magnitude in difference) using a second op- amp. For the first pole and zero at the origin, we use a block with: ( ) Where: ( ) ( *( ) We can readily select values for and , given that ( ) ( ) : Let (maximum) ( )( )
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45/92 This is at the outer range of what is allowable; however, at low frequencies, inductors would not provide a good alternative, and so this is the best we can do. We also need 1000000 total additional gain, however, with , the highest gain we can set using this stage is: Thus, we take as well. Unless we can readily implement the required gain using the second stage, we may need to add a third stage which is pure gain. If this is the case, the second and third stages could share the required gain, resulting in a gain per stage of: The second stage implements just a pole plus gain. Thus: || ( ) ( * ( * ( ) Again, we can start with the pole cut-off frequency: Let for ( )( ) We also need 1000000 additional gain from this stage: ( ) All selected values are within the ranges given, so our design is complete. Notice that in the end we did not need to add an additional stage to achieve the desired gain. However, if you used more stages or components than this, marks would not be deducted.
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46/92 Question 20.
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47/92
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48/92 Question 21.
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