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ECEN 215 homework 8 key, Study notes of Electrical Engineering

Texas A&M University (A&M)Electrical Engineering
Prof. John Tyler

tamu ECEN215 hw 8 solutions John Tyler

Typology: Study notes

2024/2025

Uploaded on 04/09/2025

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ECEN 215 Spring 2025 Solutions to HW 8 Page 1 of 3
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ECEN 215 Spring 2025 Solutions to HW 8 P13.8 = The inverting amplifier configuration is shown in Figure 13.4 in the text. The voltage gain is given by 4 =—R&/R, the input impedance is equal to Ri, and the output impedance is zero. P13.9 = This is an inverting amplifier having a voltage gain given by A, --&/R —-3. Thus, we have v,(t) — -3 x [2cos(2000z¢)] Sketches of vin(#) and v7) are Wadd) So t (ms) 2 13) aN . #(ms) . LW P13.11 Because of the summing-point constraint, the voltages across the two resistors of value Rare equal. Thus, the currents in the resistors of value R are equal as indicated: 4 ieR Then applying KVL, we have v,, =2R(2/)+A/ and v, =-152/. Solving, we find 4 =“ =-3. Page lof 3