Question

1. Use Eq. 1 to derive an expression for the expected output waveform from an ideal differentiator circuit having input waveform Vin=lsin[(21)1000t] V. Let RF1.5 k12 and C=10 nF. 2. Use Eq. 3 to find the peak-peak output amplitude of the ideal differentiator of question 1 for a 2 Vpp sine wave input at 1 kHz and 2 kHz. Put the results in the Calculated Output column of Table 1 in Appendix A. 3. Use the indefinite integral version of Eq. 6 given in Eq. 11 to derive an expression for the expected output waveform from an ideal integrator circuit having input waveform Vin=l sin((21)1000t] V. Let Ri=10 kN and C=10 nF. Assume ve(0) = 0. v.(t)= at vin (t)dt +v. (0 (11) RC 4. Use Eq. 8 to find the peak-peak output amplitude of the ideal integrator of question 3 for a 2 Vp- sine wave input at 1 kHz and 2 kHz. Put the results in the Calculated Output column of Table 3 in Appendix A.

Answer 1

Rf 47 0 RC= 1.5x10²xloxl09 ON. I. > Solveis ideal differentiator> cirwit → Vin Vout using vertual ground concept: (v=v=0) so Vin to Vout branch current will Remain same le-c & Ve & Vin-o= dvin o-Vout ica dt Rt so Vout: (- Rf C dvin Re: 1.5kr & Vin: 1. sim [(2) 1000t] v c = lont so = 1.5xlos second dve dt Vin also 2