From the bottom op amp From the top right op amp For the top left op amp we

From the bottom op amp from the top right op amp for

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From the bottom op-amp: From the top-right op-amp: For the top-left op-amp, we have the weighted summer from the notes: Where: || Solving the above three equations: ( ) ( ) ( ) 𝑉 ? 𝑉 ?
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87/92 Question 46. Find the transfer function of the following circuit: For the above circuit, we will need two equations, and . Equation 1 (Inverting Input) Equation 2 ( ) Solving: ( ) 𝑉 ?
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88/92 Question 47. Find the transfer function of the following circuit. To solve this circuit, we will need three equations. We assume that . Equation 1 (Inverting Input) Equation 2 ( ) ( ) Equation 3 ( ) Solving: ( ) ( ) 𝑉 ?? 𝑉 ? 𝑉 ?
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89/92 Question 48. Find the transfer function of the following circuit. To find the transfer function, we will need three equations, assuming : Equation 1 (Non-Inverting Input) Equation 2 ( ) Equation 3 (Inverting Input) Solving: ( ) ( ) 𝑉 ?? 𝑉 ??? 𝑉 ? 𝑉 ?
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90/92 Question 49. Find the equivalent impedance looking into the input of the following circuit. To find an equivalent impedance, the simplest method is to find an expression for both input voltage, , and input current, , and take the ratio to find . For the above circuit, we can write the following equations, assuming that for the op-amp: Equation 1 (Non-Inverting Input) Equation 2 (Inverting Input) Eliminating and solving for ratio of : Thus, this circuit behaves much like an inductor with negative impedance.
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91/92 Question 50. Find the equivalent impedance of the following circuit. Find the transfer function, ( ) . To begin, we need to find , which in this case is ratio of the voltage difference across the circuit (input to output) to the input current: We can write two equations for this circuit as follows, assuming that : Equation 1 (Non-Inverting Input) ( ) ( ) Equation 2 (Node of , , and ) Substituting in the result from the previous equation: ( ) Solving: ( ) ( ) || This is to be expected if the op-amp has . To find ( ) , we can use the first equation from the previous part of the question: ( ) This circuit is called a capacitance multiplier. It can be used to amplify the smoothing effect of the capacitor when used as a filter. 𝑖
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92/92 Question 51. Find the transfer function of the following circuit. Ignore and . Assume that at and at , and is set at 50%. To begin, we are given that follows a step function; it is zero before , and after . The function in the frequency domain is: ( ) * ( )+ Since no current flows into Pin 3: ( ) ( * Virtual ground for op-amp : Equation 1 (Pin 2) We also have that and :
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93/92 Equation 2 (Pin 6) We have : Solving Equation 1 and Equation 2: ( ) ( ) Substituting values: ( ) ( ) The above output function is a bit unusual compared to our normal functions in that it contains a term which is not dependent on ; we cannot solve for . This is caused by the effect of a step input on from when power is applied to . If we were to consider the effect only of this transient, we would have: ( ) ( ) ( ) ( ) , - ̅ Examining the effect of this input, we see that the output decays to a constant in steady-state.
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