Part a Step 1 Find the general transfer function by tracking the rate of change

# Part a step 1 find the general transfer function by

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Part (a) Step 1 Find the general transfer function by tracking the rate-of-change (ROC): The plot starts with , indicating that there is no pole or zero at the origin. The first change in ROC occurs at . At , we have . This implies a single zero: ( ) At , we have . This implies a quadruple pole, as each pole contributes , such that ( ) . Thus, we have a term: ( ) There are no further poles or zeros. The final transfer function is: ( ) ( ) ( ) ( . / . / . / . / . / ) ( . / ) Step 2 Match the absolute level of gain. Since there are no zeros or poles at the origin, it is simple to match the gain level. We can pick any point in the leftmost region where and see that the absolute level of the graph is . In the above ( ) , if : ( )( ) ( ) Part (b) Step 3 Group poles and zeros, divide gain, and form the circuit This transfer function has one zero and four repeated poles. We can use one general pole-zero-gain (PZG) block plus three pole-gain (PG) blocks to realize this transfer function. We also have a very large additional gain; it makes sense to distribute it among the various stages in the absence of other criteria. In a real-world implementation, we would have to be more careful we must keep the signal level at the output of each stage within the supply limits of the op-amp used, and above a minimum level (or else the signal is too small to work with, and becomes lost in normal noise).

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18/92 Step 4 Pick reasonable component values. In general, we start picking with capacitors and inductors where possible. However, we have also decided to divide the gain between stages, so each stage needs to implement gain: For stage one, we have a standard P-Z-G stage: ( ) ( ) ( * ( * ( ) If we were to pick and first, this would fix our choices for and , meaning we cannot achieve the necessary gain. This can be confirmed: Picking capacitors first, we have: ( ) ( ) So, we instead pick our resistor values first. Since the ratio is not large, we attempt to pick values near the middle of the total range of to . Also, since the gain required is greater than 1 and less than 100, a good rule of thumb is to set and ,
19/92 Using these values, it is a simple matter to find and from the crossover frequencies: ( ) ( ) For the remaining blocks, we have the three identical transfer functions: ( ) ( ) ( * ( ) ( ) We assume that , , and , as there is no reason not to have the circuits be identical. Let’s try picking the capacitor value first. Since : ( ) Finally, since : ( ) Thus: All selected component values are within the reasonable ranges for real components.

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20/92 Question 12. To solve the above circuit, we need only two equations. We label the unknown node : (1) (2) Solving Equation (2) for and substituting into Equation (1): ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * , ( ) - , -
21/92 ( ) ( ) . / . / ( * Substituting in the given component values yields: ( ) ( )( ) This can be factored into ( ) form: ( ) . / . / . / . / ( ) This yields a zero at the origin, plus two real poles at and , plus a constant gain of: For the zero at the origin, it crosses at , with a constant ROC for all , thus at (where the given graph range starts), it would normally contribute .

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