then one can notice that final op amp U4 forms a summing amplifier We will omit

Then one can notice that final op amp u4 forms a

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, then one can notice that final op-amp, U4, forms a summing amplifier. We will omit the intermediate transfer functions for space:
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12/92 ( * ( * ( * ( * ( * ( * This particular circuit is an extremely flexible filter circuit. Using (relatively) few components, it can realize a wide variety of combinations of low-pass, high-pass, and band-pass filtering simultaneously. At a minimum, it typically uses far fewer components than constructing each filter individually, and it maintains good flexibility in pole location for each filter despite the small number of components.
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13/92 Question 10. Consider the following circuit: Find the general transfer function of the circuit. Use the following set of component values to plot the transfer function for the circuit: Solution . / . / ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 𝑉 ?𝐿 𝑉 ? 𝑉 𝑉
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14/92 [ ( ) ] * ( ) ( ) ( ) ( ) + * ( ) + * ( ) ( ) + ( ) ( ( ) )( ( )) ( )( ) ( ) ( ) . / . / ( * . / . / . / . / . / 0 1 ( * . / . / Notice that through this transfer function, the circuit can realize almost any second-order quadratic zero and pole combination. Most of the coefficients of ‘s’ contain at least one unique component, allowing each to be adjusted independently of the others. Using the given component values: ( ) ( )( ) ( )( ) . / . / . / . / ( )( ) ( )( ) ( )( ) We have two poles, two zeros, with crossover frequencies: ( * ( * ( ) ( ) ( )( ) To create the Bode Magnitude plot, we (1) order the poles and zeros from smallest crossover frequency to largest, (2) track the rate-of-change (ROC) in order to draw the curve, and then (3) use the gain, , to shift the curve to the correct absolute location. * +
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15/92 Magnitude Plot Since there are no poles or zeros at the origin, the starting ROC is . This continues until the first cross-over frequency at . This corresponds to a zero, which begins to contribute for any frequency , resulting in a total ROC of . This ROC persists until , at which point two poles begin to contribute each, resulting in a total . This ROC persists until the final crossover frequency at , after which the second zero begins to contribute , for a total ROC of zero. In the above graph, the red line is the actual magnitude plot, and the blue line is the asymptotic plot we produced using the procedure above. Notice that the graphs are most different near the corner frequencies. At , the magnitude of the difference will be per pole/zero. Thus, at and , we have a difference, and at , we have a difference (due to the double pole with ). ?? ?????? ?? ??????
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16/92 Phase Plot To produce the phase plot, we need to make a new set of critical points. Recall that a pole or zero will produce a total of of phase shift over a range starting one decade before , and ending one decade after . Thus, for every pole and zero we have two critical points: * + Alternatively, we can track phase change by ranges: from to ( ) from to ( ) from to ( ) from to ( ) Since these ranges overlap, we must be aware of this when drawing the phase plot: Notice that the sketched phase plot appears significantly less accurate than the magnitude plot; this is reflected when attempting to ‘correct’ the phase plot – the process is significantly more involved.
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17/92 Question 11.
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