Question

es anh th e c dfe sod aao rt34e k-e: e fi、c ︵ee aod g ti r mdfdpe hrr nd s pt Ide r. csa reu etnn doi ers inas dav si a)rec.z trlar k-te eer 0 dkne nro li ob "bee tdgu sa 9 edeb onfr fr cfr pgeo etg 0 uet tio edl tnd gra .1.1 euds P Hall in re ce capacitor C1 omitted. Therefore, VFB(t) VouT)/32. We assume g -800 What is the actual gain of the amplifier Vour()/V()? Note: Do not use the ideal op-amp model, because g - 800 does not allow the approximation g >oo. [5 points] B. With manufacturer tolerances, the internal gain g can vary between 500 and [5 points] 1000. What is the range of variation for k? C. A loudspeaker is connected to the amplifier. Its resistive (DC-) impedance is 4S2, but the coil and crossover add an inductive impedance of L-200uH To obtain the coil current Is pK(s), Ohm's Law can be applied, but with the combined resistive and inductive impedance: Plot the frequency response (Bode diagram) of the combined transfer function Is pK(s)/V(s), stll without consideration of the capacitor CI. Clearly indicate the asymptotes and the cutoff frequency. How strong is the attenuation of the signal at the higher end of the audio range at 15 kHz, compared to low frequen- cies? Note: Please do not confuse the linear frequency f and the angular frequency points] D. Since a frequency response as seen in C. is clearly unacceptable, the capacitor C1 is used to decrease the feedback signal at higher frequencies. Components are chosen as follows: R1-320kS2, R2-10kS2, C1-4.7nF. The transfer function of the feedback network is therefore 10,000 VouTs R1R2C1s R1 R2 15.5s +330, 000 Determine the combined transfer function Ispk(s)/V(s) when the frequency compensation capacitor C1 is installed and determine the poles and zeros (if [10 points] any) of the new transfer function E. Plot the Bode diagram of the new transfer function. Clearly indicate the asymp- totes and the new cutoff frequency. How strong is the attenuation of the signal 10 points] at the higher end of the audio range at 15 kHz now?

Answer 1

Conside hel an op amp is ideal VE) Ri ket k be -the closed dop in q 800 g. 32306 g00 val

500 K 5) 30469 d0etemine k tahen 9:1000 1DOD Na) K 31007

。Trans{er -function lap VIS) dsp 800 +80o/32) Sx200N6 → 30 구6 x 1 X 10000 3076q.1 dat (S+90000)x2 2(S+ 20000) t 20000 Gurt-o frequenay 20,00 adiser s - single pole. Stope -2ode Idrart 20 log 15384s 20000