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soex2-3723-06f

soex2-3723-06f - Problem 1 Derive the transfer function V(S...

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Unformatted text preview: Problem 1: Derive the transfer function V(%(S) for the given RC ladder circuit given below where e is the input source and Vis the output response (note R, at R2 at i R” and C1: C2 ¢-—- :2 CH)” l 111%. 47/: 177%.. *2 v(:— as: :i/Ly—Lm ) I \$32!? [ raids“ ) l at) [‘Crleif .7. V141!“ {2}} Vl':(7i_ C P- V ,0 g; : V2 i ) R1 ’0‘» (qr ’ ’5 V1 2 1C, 1K2) VI )H-l—f 3) ﬁrm 1004.21): 034/1: 3 ’9UU~C|€ 5):019YL:——1—— (33 34% > 73— % v3 3 33 I71 1'02K35 m VII 77/)", W11 Vﬁﬂ- : U (at R 5)=Ev:?,=ﬂ':._l__n- 2 T‘CM’E") «4 its“) 4' “4 ’Vn—i kW ‘VZVH W W. EU) 45) 11-10) DEW 17,0} EU) *— ’ .ei__._J___, Vang/'5 t—Cg'tw (’cym l-CJﬁj Problem 2: Obtain an analogous electrical circuits (using force—current analogy) for the mechanical system shown below. WWW HWW-ﬁé , Pro/alum wa/mt/q /Yll>f<'l Tblki "l" QLX\KXL);O W¢§G ﬂap}; "(elm 7(2):. 0 "Vs-Wat“ ’92“; ’UAVI’Wﬁ/mlojv’ we Amt/a H ' “L” "L L "’ : Gﬂﬂzw 1m + L2 96. 92> a cl%+ Jig/(#2 “Pi—acc’éf’L‘zﬁo Ndl’a wast" yze/ N3 MVZ. ' re alt: fJ/Ze‘Jc—Jggzidti-i—zg (<7, z) o Page 3 Problem 3: X0 (5) 19(5) response xu(t) when the input 13(1) is a pulse signal given by Derive the transfer function of the mechanical system shown below. Then obtain he X,, 0 < r < I] x, (I) = . Assume that xn(0—) = 0. 0, elsewhere . 950%ij «lac/WW4 m y/slzm: k: (IQ—)9) 3% (990”3') 1 EM ed): ha 195?. >26 +1?! X6 g:le +1’Qy' by + ('23 =£2xlt Ndh/ (MAT X(o~)£0/ 2 (0-7,:0 (fa/43, 475m? WW (lawman): Wig/143+ we”) wig EMT“): 92 521(5) Etymwwhg TO); Luz/Aqu deg—eat mam: leg,- 0) T425 ==> EU) _ kud925+éﬂ EU) Cwmw-Hla, [42, L2 320W 1725* k2, Page 4 Problem 4: Consider the electrical circuits shown below. Assume that the input is sinusoidal, e, (t) = E, 005 wt, what is the steady state current i0) "? Please derive the formula fopr steady state response when the system is subject to an input of E, cos (at. grmvr Stui’e l/Mﬂ-Wﬁb _ jut xljkjﬁ +o~2 WINE paws, Eli—5a (St 2/97“ = H97“) 573??” E A¥9HLS)S{:;; Waving”) 1 awn») Page 5 Problem 5: Consider the mechanical vibratory system shown below. Assume that displacement x is measured from the equilibrium position in the absence of the sinusoidal excitation force. The initial conditions are x(0) = Oand M0) = 0, and the input force p(t) = PSiI’ltO! is given at t = 0. The numerical values are given as m = 2 kg, [3 = 24 N-s/rn, k = 200 N/mi P = 5 N and e) = 6 rad/s. Obtain the complete solution xU) . input force p(.‘) = Psin cor adaﬁ'pm (1+ WWI—tine Tkxa/oajp’Pm/itfi‘ CXC°)~’0/ >60)ij iavtm Wmva =9 (WW/rt) Zw—“Pw Wm’zr L374, [\$200, 17:5, was 5 fé‘e PZES); ﬁ—LS;*_Z_ ...
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