Combined with the constant gain of the absolute value of the magnitude graph at

Combined with the constant gain of the absolute value

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Combined with the constant gain of , the absolute value of the magnitude graph at this point is ( ) . The graph has a ROC, as no other poles or zeros are yet active. Tracking the ROC across the graph: The ROC continues until the first pole becomes active, at a critical frequency of . At this point, it begins to contribute , for a total ROC of zero. The zero ROC continues until the second pole becomes active at , resulting in a ROC for the remainder of the graph. Likewise for the phase graph, due to the zero at the origin, the phase shift starts at with a zero ROC. Tracking the ROC again: The zero ROC continues until . At this point, the phase ROC becomes for two decades (i.e., the pole phase contribution ends at . The second pole also becomes active at , overlapping with the first, and resulting in a ROC. The ROC continues until . At this point, the ROC becomes just .
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22/92 This ROC continues until , at which point the ROC becomes and stays zero. The total phase shift ends at , as the transfer function has one zero and two poles.
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23/92 Question 13. To begin, let’s construct the transfer function by tracking the ROC. The graph starts with a zero ROC, indicating that there are no poles or zeros at the origin. The first change in ROC happens at , where the ROC becomes . This indicates that is the corner frequency of a zero. The ROC of continues until , at which point the ROC becomes zero again. This indicates the presence of a pole with a corner frequency of . The general transfer function is thus: ( ) ( ) ( ) Transforming this into ( ) form: ( ) . / . / * + ( ) Since there are no poles or zeros at the origin, any point to the left of the lowest corner frequency ( ) should have a magnitude of | | . However, this region of the graph is at , indicating that: ( ) ( ) ( * This transfer function might be able to be realized with a single P-Z-G block using some careful component selection:
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24/92 ( ) ( ) ( * ( * ( ) Let’s pick the capacitors first – we know that the zero corner frequency is , which by our rule of thumb should yield a capacitor value around . Thus: ( ) Now, typically we would pick for the pole’s corner frequency of . However, we also desire , so it makes sense to select one order of magnitude lower than normal this will have the effect of increasing the value of we need (remember, the relation is ). Thus: ( ) This value is at the upper range of our resistor values (which is why we would normally pick based on our rule-of-thumb, yielding ). However, this does work perfectly for our gain: Note that it is relatively difficult to select a set of values which works. Since we were given no restrictions on the number of parts, and equally valid solution might be to use two op-amps and implement the poles/zeros and gain separately. Using inductors would not be a useful solution, as we are dealing with low corner frequencies, rather than high frequencies. ,
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25/92 , Alternate Solution 1: Find the transfer function ( ) . This is a first order filter (the slope is +20dB/decade): | | | | Realize the transfer function by RC-op-amp circuit: Assign proper value to the components: | | | | | | Therefore, must satisfy the following conditions: + _ R 2 R 1 V in V out C 2 C 1
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26/92 Alternate Solution 2: Realize the transfer function by RC circuit: Assign proper value to the components: ( * | | | | ( ) ( ) ( ) ( ) | | : R 1 V in C 1 V out C 2 R 2
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