81 d1 12 1 w000055 hfreqsndw

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .005:5; >> h=freqs(n,d,w); >> plot(w,abs(h)) >> grid >> xlabel('Frequency r/s') >> ylabel('magnitude response') (c) Normalized frequency: 1.115Hz @ normalized gain (1.875*0.707 1.325625) (i) 600 2 3381.86 1.115 (ii) 1 14.788 10 · 3381.86 20 10 (iii) ·, 14.79 Ω , ·, 14.79 Ω , , , , , · · 20 20 Page 3 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo (d) Optional *****Netlist for hspice simulation***** .option post node list vin vin 0 1 ac 1 R1 vin vf 14790 R2 vf vp 14790 C1 vf vout 20n C2 vp 0 20n RA vn 0 10000 RB vn vout 10000 E1 vp vn vout 0 100000000K gain) .ac DEC 1024 1 100MEG .print ac v(vout) .end **for an ideal opamp(Exx vin+ vin- vout+ vout- ( freq 526 Hz) Page 4 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 34. (a) h(t) = 2u(t − 1) + u(t − 2) − 2u(t − 3) − u(t − 5) 3 2.5 2 h(t) 1.5 1 0.5 0 -6 -4 -2 0 t 2 4 6 (i) f1(t) = δ(t)-1.5δ(t − 1) When a function is convoluted with an impulse function, Kδ(t−T) ∗ f(t) = Kf(t−T) that function will be scaled by the strength of the impulse function and shifted by the amount the impuls...
View Full Document

Ask a homework question - tutors are online