2792 Question 14 Part a For the circuit given we have negative feedback through

2792 question 14 part a for the circuit given we have

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27/92 Question 14. Part (a) For the circuit given, we have negative feedback through the R3 // C3 network. We do not have positive feedback, as R1 is terminated to ground. As such, we assume that the output is set such that , where is a single unknown node voltage. We can thus write two equations: (1) (2) Solving Equation (1) for : ( * ( ) ( ) , - ( * Simplify Equation (2): ( * ( * ( * ( * ( ) ( ) ( * ( * ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Substitute solved Equation (1) into Equation (2): ( * ( ) ( * ( ) ( ) ( ) * + , - , - ,( )( )- , - ,( )( )- , - , - ( ) Substituting in and :
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28/92 ( ) Part (b) ( * [ ( * ] ( * ( * ( ) ( ) [ ( * ] ( * ( *
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29/92 Question 15. Part (a) Given our goal transfer function: ( ) . / . / . / Start by writing a node equation at the inverting input of the op-amp, solve for ( ) : ( * [ ] [ ] ( ) ( )( ) . / . / . / You could also solve this quickly by taking ( ) , where | and . Part (b) To write the magnitude of the transfer function in long form, we break ( ) into the multiplied contributions of each pole, zero, and constant gain: ( ) . / . / . / [ ] , - [ ] [ ] Substituting in , we obtain the following.
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30/92 ( ) [ ] , - [ ] [ ] Taking the magnitude: | ( )| | | | | | | | | (| ( )|) (| |* (| |) (| |) (| |) (| ( )|) ( * ( ) (| |* (| |* Part (c) Substitute in the given values to find pole and zero corner frequencies. Note that with rounding, all corner frequencies are separated in decade increments we will align our graph on multiple of ten of the lowest frequency, rather than as usual. If you use a standard graph, it will be more work but the graph should look the same. ( ) ( ) ( )( ) Gain at : ( ( )( ) * ROC at : -140 -120 -100 -80 -60 -40 -20 0 20 40 60 80 100 120 140 62.83184 628.3184 6283.184 62831.84 628318.4 6283184 Frequency (rad/sec) Magnitude (dB) 20Log(1/R2C1) 20Log(w) -20Log(jw+6283) -20Log(jw+628318) |H(jw)|
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31/92 Part (d) This circuit is a band-pass filter. It has a first-order numerator, second-order denominator. The lower and upper cut-off frequencies clearly visible, creating a pass-band where gain is close to , creating no attenuation. Center frequency of around , given clearly defined band-pass region. Part (e) ( ) ( * ( * The above can be obtained directly from the transfer function by noting the pole locations. They can also be visually obtained from our plotted estimate of the transfer function. Part (f) Note: The follow form is correct but not covered in the newer notes. It would be sufficient to use the upper and lower cut-off frequencies for an approximation. From the general form of the transfer function for a band-pass filter: ( ) ( *( * ( * ( * ( *
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32/92 Question 16. Note: The following is taken from an old version of the course (when filters spanned more of the course content). We do not cover notch filters in as much detail. Some other elements of this solution may also be unfamiliar, so do not worry too much. Try to understand as much as you can, given the current course contents.
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39/92 Question 17. The sub-circuit which comprises the complex impedance can be modeled separately first. To find its equivalent impedance, we can examine just the relevant sub-circuit: One unknown node, which we can relate to : , - , - ( * No current flows into the op-amp non-inverting input, so is completely given by the current through resistor : . / ( * ( * ( * The equivalent input impedance is thus: . / 𝑉 ?? 𝑉 ? 𝐼 ??
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40/92 We can now write our Kirchoff’s equations for the rest of the circuit: (1) (2) (3) Before we start, notice that there is no net current through , due to our feedback assumption that for the op-amp. As such, all current through flows directly through , and all current from flows directly through . This can also be seen more directly by examining the equations;
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