homework4 - F(w) - lO/(jw+ 1)2. a) E - [fm|f(t)2dt]/R -...

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Unformatted text preview: F(w) - lO/(jw+ 1)2. a) E - [fm|f(t)2dt]/R - 2fmt2e‘2‘dt - 0.5 J. _., o 2 100 1 2 - b) 2[*J___ dw] .. i + Etan 1(1) - 81.8%. 0 c) 2 [—Z—J———IQQ—— dw} =- % [tan'1(5) - tan-1(3)] - 1—: - 10.6%. 3 65 O a 1 a)P-zifi£8f(w)dw-z%f[ H + 600-2)] dw - — + 1 - 0.818 w. o w2+l 2 E 1.1 b) fiofg [130; + 6(w~2)] dw - %[can-1(1.1) - tan'1(0.9)] = 0.0319 w. 5.1 i 1 +6 -2 dw-lc’12.1-c‘11.9+1=0331w c) bl» 1J9 (w "I: an ( ) an ( ) ] , . d) Using the results of Ex. 4.2.1, the two impulse functions can be associated with a signal 42/11’ cos 2t. Because this is a power spectral density, we cannot associate the term (t1.)2+l)'1 directly with an exponential signal. From Eq. (4.21), we see that it can be associated with a constant (1) power spectral density through 3. RC low-pass filter with RC=1. Because the phase is not included, these results are not unique, and other solutions are possible. ) 11020:) - [(10mV)/10]2 - 1 x 10-6 V2. Also, we can write: n°2(t,_%f_1mflz_dw__n__ o (02 + 1/(RC)2 “RC Setting these two results equal, we get: 71 - ARC x 10'6 W/Hz. b) v°(t) - ——1fl§l—— ficosnonc - tan'1(307rRC)]; J(30«)2 + 1/(RC)2 1 (RC 2 2 . 10 v2 t - ___Z__L___ _ 10 mv ; solvm , RC .. _ _ 1,051 _ ° ( ) (3O1r)2+1/(RC)2 ( ) 8 3” sec P4.3.3 a) Same answer as a) of Problem 4.3.2; no changes. O 1 1 1 Rc)2 1 b) V3“) ' FL?” m d” ' arm ' (1° “‘V’z- 0 Solving, we get: RC - 5 x 103 sec. / ra.4.1 T/Z a) Rim - .,}_, f/zFl‘e"”°‘F1e‘”°“"’dt - |F1|2e‘”°'; '1 s50») - 5mm} - 21r|F1126(w- we). 1'l2 — t' ' Z t + 2 b) 112(1) .. %—J;2[F1te duo + sze J 00 MFleJuom 1) + erj 0°(t+r)] dt Rim - IF1|2e3“°' + Ileze‘2“°'; stag) - 2«|F1|26(w—wo) + ZKIFZIZMw- 2%). 0 c) The cross-terms integrate to zero and we get: R£(r) - X anlzedmof; “3-0 I) Sim) - 211’ 2 [Fn|26(w-nwu). [See Eq. (4.20)] Etyme correlation fns. below are zero outside of intervals given. a For -l<1<0: rx(r)-jf+1dt - 1+1; For 0<r<1z rx(r) —f1dt - l-r. Combining these results, owe have: rx(r) - A(r). T b) For -1<¢<o: ry(7) — {mcu- 7) dt - %(2 - r)(r+1)2; 1 For 0<r<1: rym- ft(t-r)dt-16-(r3-3r+2). T +1 c)For-1<r<0: rxy(r)-{ (c-r)dc-12-(1-12); 1 For O<r<1: rum - {(t-r)dt-%(r-1)2. Graphs of these results are shown. ...
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homework4 - F(w) - lO/(jw+ 1)2. a) E - [fm|f(t)2dt]/R -...

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