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Ty x cxg p. Signals and Systems P36 (c) Suppose that a second LTI system has the following output y(t) when the input is the unit step x(t) = u(t) y(t) = e'u(t) u(1 t) Determine and sketch the response of this system to the input x(t) shown in. Substitute Y(t) into y’’ p(t)y’ q(t)y = g(t) and determine the coefficients to satisfy the equation There is no solution of the form that we assumed Find a solution of Y(t) Determine the coefficients End N Y Second Order Linear Non Homogenous Differential Equations – Method of Undermined Coefficients –Block Diagram •. C x g E Ёu Ѓt e B A C ^ i V i v ́A e C x g A L y A ̑ i A l ޔh A R e c A z y W A ̔ i ̊ E E ^ c v ܂ B l C E b ̃A e B X g t @ b V u h ̃V E C u R T g ̃X e W E o E 삩 狦 ^ v g ܂ŁA ̓g ^ Ńv f X ܂ B.

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Tkrf(x)k2 will always be positive unless rf(x) = 0, this inequality implies that the objective function value strictly decreases with each iteration of gradient descent until it reaches the optimal value f(x) = f(x) Note that this convergence result only holds when we choose tto be small enough, ie t 1=L This explains. Ό T Y 1960 N ܂ q v l Ƃ̔ I ̍ۂɉ̏ u z o v ̉ 𔭌@ A ^ I ɁA n N ̃T u } l W ۊǂ Ă J Z b g A n 񂪃J I P y މ @ B n N ̉̎ Ƃ Ă̐V Ȉ ʂ _ Ԍ A ̃v C x g S ȏ CD I Ό T Y ̃ h ̗w A n N ̈ D ̗w e } 1 ͐Ό T Y ̃I W i A J o ȁA I ł̉̏ B 1 n N ̃v C x g ƂȂ CD2 g B (C)RS. And f(1) = 5 Since f(x) is a polynomial, f is continuous on the real line We have f( 2) > 0 > f(0) So, by the Intermediate Value Theorem, there exists a number cbetween 2 and 0 such that f(c) = 0 Similarly, there exists a number dbetween 0 and 1 such that f(d) = 0 Note The choices x= 2;0;1 are not the only possibilities 13 Given y= 1 x2.

GDP (Y) is the combination of consumption (C), investment (I), government spending (G), and net exports (exports (X) less imports (M)) The recovery that began in 09 in the United States has been tepid relative to previous recoveries (although better than in most other rich countries) though not because of weak consumption, which in fact has. I K j b N X C c J A G e b N ւ悤 I G e b N I K j b N X C c ́A w q S ĐH ׂ x Ƃ A u S v ɏ i ̑I s Ă ܂ B G e b N ͗L @JAS F H A I K j b N X C c J B L @ i ` r F ؂̃N I e B ͂ A q 낱 Ԗ{ ̔ ڎw i A X S Ă ܂ B. T y s t r e e t ’ss t m a y ’ s f ennel st e n n e l a s y t t r i n i t y w queen stq u e e n r s t king stk i n g s t w e king r st k i n g s t chapel streetc h a p e l m s t r e e t ton stc a t e a t o s t ion app s ta t i o n a p p co r t ion street c o p o rati o n ree t side stree t g a r t s i d e t s t r e e lo wer byrom s t l o w e.

116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l “ K \ B \ Z g J b e k d b”, L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n. ֘A @ \ r \ i Ƃ̋@ \ r \ ߂ ̔̔ X T. KÝ PDF°«x Õh)¦ xiv 04 A¢ àh Ý&4 /Zøüd Z 8!¢ Ýd × ä î 3é\îà#å\9Íh Ý`Î à9°d ñÕh Í 2 G«Àê Ý&Ía;4 ê Za Ý¥X C&!.

D µ è p Í ø x æ ¬ 8 ü ;. For the function z(x,y)=yx^2xy with x(t)=log(t) and y(t)=t^2, we have Example For our introductory example, we can now find dP/dt Implicit Differentiation A special case of this chain rule allows us to find dy/dx for functions F(x,y)=0 that define y implicity as a function of x Suppose x is an independent variable and y=y(x). (R ev ) SE C R E T F E D E R A L B U R E A U O F IN V E S T IG A T IO N b 2 < f y P r e c e d e n c e !.

1977 N 挧 o g A B p H ȑ w( B w/ ) 00 N ɑ ƁB00 N4 A E ɂĕ 킫 s ցB Ј Ƃ ă J ł̏ i Ɩ Ɍg T 01 N C X g ^ Ƃ Ċ J n B10 N1 ɑގ / Ɨ A 킫 s 𒆐S ɖ{ i X ^ g B11 N Ɋ 10 N } B. Signals and Systems P36 (c) Suppose that a second LTI system has the following output y(t) when the input is the unit step x(t) = u(t) y(t) = e'u(t) u(1 t) Determine and sketch the response of this system to the input x(t) shown in. @ @ @ @ @ t Y s Y w O157 u X i b N ̂񂫁v @ @ @ @ @JR Y w @ k 5 d b ԍ F.

= −ˇ=6 (b) f(x;y)=xsin(xy);. X, and e y only one point on the failure surface need to be computed Such point satisfies the following condition Pu =P and = uy y ux x ee ee 0 given load point load point ray b/2 α e y e ux e uy e x y x h/2 The procedure is summarized as follows 4 / 6. References ABeck,FirstOrder Methods in Optimization (17),chapter6 PLCombettesandJChPesquet,Proximal splitting methods in signal processing,inFixedPoint Algorithms for Inverse Problems in Science and Engineering (11) NParikhandSBoyd,Proximal algorithms (13) Theproximalmapping 624.

Tbs $b%f%l%s$,$*fo$1$9$k!v (btv $b%. F „x”= sup y2dom f „xTy f„y”” f isclosedandconvex fromFenchel’sinequality,xTy f„y” f„x”forally andx;therefore f „x” f„x” forallx equivalently,epi f epi f (forany f) if f isclosedandconvex,then f „x”= f„x” forallx equivalently,epi f = epi f (if f isclosedandconvex);proofonnextpage Conjugatefunctions 513. Y = C I G (XM) National income with government interference From the national income, some will be used for consumption, saving, and other’s used for paying taxes With adding of variable taxes on national income, then the formula is Y = C S T To see the balance of national income with the adding of government spending and tax.

This means that x(t) can be written as a weighted integral of functions Cu (Lecture 3) ELE 301 Signals and Systems Fall 1112 9 / 55 Applying the system H to the input x(t), y(t) = H(x(t)) = H Z 1 1 x(˝) ˝(t)d˝ If the system obeys extended linearity we can interchange the order of the system operator and the integration y(t) = Z 1 1 x. Signals and Systems P36 (c) Suppose that a second LTI system has the following output y(t) when the input is the unit step x(t) = u(t) y(t) = e'u(t) u(1 t) Determine and sketch the response of this system to the input x(t) shown in. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor.

12 Let f(x;y) be a di erentiable function, and let u= x yand v= x−yFinda constant such that (fx) 2 (f y) 2 = ((f u) 2 (f v) 2) Solution By the chain rule (fx)2 (f y)2 =(f u f v)2 (f u−f v)2 =2((f u)2 (f v)2)Thus =2 13 Find the directional derivative D ~ufat the given point in the direction indicated by the angle (a) f(x;y)= p 5x−4y;(2;1);. The simplest case, apart from the trivial case of a constant function, is when y is a linear function of x, meaning that the graph of y is a line In this case, y = f(x) = mx b, for real numbers m and b, and the slope m is given by = =, where the symbol Δ is an abbreviation for "change in", and the combinations and refer to corresponding changes, ie. (µ/ý xTª ,— gnomeapplets3364/ 5ustaralbertsgmultiload3533orgMLApanelin Factory Id= InProcess=true Location.

References ABeck,FirstOrder Methods in Optimization (17),chapter6 PLCombettesandJChPesquet,Proximal splitting methods in signal processing,inFixedPoint Algorithms for Inverse Problems in Science and Engineering (11) NParikhandSBoyd,Proximal algorithms (13) Theproximalmapping 624. **** u 񂾂 āv ł͔N Ԓʂ Ă 낢 ȃC x g { Ă ܂ B A ܂肵 O v Ȃ̂ł ̑ ͍L F l U J Â 킯 ɍs A ̎ X ɏ 񂪓` ͈͂Ŏ { Ă ܂ B l Ɍ 肪 C x g ͂ X ł m 点 Ă ܂ B ܂ ` Ȃ 炲 ߂ Ȃ ˁB J I P ̔ \ ɂ́A L s ̗w ɏ 𒆐S N ̔ \ ɎQ Ă ܂ B ЁA C y ɂǂ I. 62 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l “ K \ B \ Z g J b e k d b”, L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n.

From of 1, in this case (tan(x) sec(x))/ (tan(x) sec(x)), the integration could then proceed quite easily ∫ ∫ ∫ =∫ = = u du dx x x x x x dx x x x x xdx x sec tan sec tan sec tan sec tan sec sec sec 2 =ln uC=ln sec xtan xC Now back to the equation µ(t) y′ µ(t)p(t) y = µ(t)g(t) (*) On the right side there is. \ JUDGE & COMMENTS \ i ΂ A ǂ j V 16 N P ɃI v s X g O X z e É ^ V F t Y C u L b ` t ́A x X g u C _ ЁE A g O C X O v ^ c ̃u C _ z e y уu b t F X g B. 116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l “ K \ B \ Z g J b e k d b”, L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n.

The simplest case, apart from the trivial case of a constant function, is when y is a linear function of x, meaning that the graph of y is a line In this case, y = f(x) = mx b, for real numbers m and b, and the slope m is given by = =, where the symbol Δ is an abbreviation for "change in", and the combinations and refer to corresponding changes, ie. May 16, 11 254 CHAPTER 13 CALCULUS OF VECTORVALUED FUNCTIONS (LT CHAPTER 14) Use a computer algebra system to plot the projections onto the xy and xzplanes of the curve r(t) = t cost,tsin t,t in Exercise 17 In Exercises 19 and , let r(t) = sin t,cost,sin t cos2t as shown in Figure 12 y x z FIGURE 12 19. T y c X g ́A { WebPage Ă 鐁 t y c ւ̃ N ł B y W ͂.

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Thus X = x,Y = b−y, and the curvature in question is k = b/a2 3 Let t → (x(t),y(t)) be a closed regular plane curve Let t → (˙x(t),y˙(t)) be the closed regular plane curve formed by the velocity vectors Prove that the integral 1 2π I xd˙ y˙ −yd˙ x˙ x˙2 ˙y2 is an integer Point out geometric interpretations of this integer in. And since x(t) = y(t)F(t), we get x(t) = KeA(t) eA(t) Z t e A(s)b(s)ds is the general solution Example Solve x0= x=(1 t) 2t x(0) = 3 a(t) = 1=(1 t) Thus A(t) = ln(1 t) and F(t) = eA(t) = 1 tSo x(t) = K(1 t) (1 t) Z t 0 2s 1 s ds= K(1 t) 2t 2ln(1 t) The initial data yields, K= 3 so that x(t) = 3 5t 2ln(1 t. From of 1, in this case (tan(x) sec(x))/ (tan(x) sec(x)), the integration could then proceed quite easily ∫ ∫ ∫ =∫ = = u du dx x x x x x dx x x x x xdx x sec tan sec tan sec tan sec tan sec sec sec 2 =ln uC=ln sec xtan xC Now back to the equation µ(t) y′ µ(t)p(t) y = µ(t)g(t) (*) On the right side there is.

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Example 6 X and Y are independent, each with an exponential(λ) distribution Find the density of Z = X Y and of W = Y −X2 Since X and Y are independent, we know that f(x,y) = fX(x)fY (y), giving us f(x,y) = ˆ λe−λxλe−λy if x,y ≥ 0 0 otherwise The first thing we do is draw a picture of the support set the first quadrant (a). ,¿F9Ë€ C' B @;{ A )L Ë€5k _Ù€DE @L7 E @ ÿ A L ×€6m K €/@I· @ ß E '€4 @Y@ 1L ã€7l ^M© )é##@ ã E 3€E § E 2¼è Ü Ð Ä ¸ ª ž ‚ æ ^ 2 Ú ® p D * ö Ü Â ¨ Ž t Z @ & AOÿ@_ß„ ÿ C çZ/ÑFOé€ Le1 ow ƒ @ o@ ›w %€ D_ä@`Þ€ÏyOé@NßJ µBP·€ @J%a> „Ž0 µ@ «€ DùR> @0¡l FQ €m. And since x(t) = y(t)F(t), we get x(t) = KeA(t) eA(t) Z t e A(s)b(s)ds is the general solution Example Solve x0= x=(1 t) 2t x(0) = 3 a(t) = 1=(1 t) Thus A(t) = ln(1 t) and F(t) = eA(t) = 1 tSo x(t) = K(1 t) (1 t) Z t 0 2s 1 s ds= K(1 t) 2t 2ln(1 t) The initial data yields, K= 3 so that x(t) = 3 5t 2ln(1 t.

!%Ý_r í/Id PÓ¨ \ï ãå N h èÕ¥ ÝÝb(Þ` A 9Í(Þ 3h Í 2b©½ ÕÝ º3å#½ (Þ Õ×Íä Ý d 5ËË a;d A ×ÍÝb(Þ ÎØa;ÝxÞ »Anö Unicode3 ÝD ¡ 9Í`Î 3 d ÕÝ ÑΣÍa. Simple and best practice solution for g=cx equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. Hic_et_Hec_òù3_òù3BOOKMOBIßo È&è 0” C¥ M) V _ã i rG {Ò = ŽÆ ˜ ¡F ªÈ ´* ½Ä"Ç $ÐŒ&Ú (ãP*ìÒ,ö(ÿ¯0 ø2 l4 ¯6 %48 Æ 8Q A–> JØ@ T¥B ^ D g=F pìH z4J ƒ L Š N ‘ÝP š^R ¢T £tV ¤¨X ¦tZ ëœ\ û¬^ D` hb œd Å f Í–j Ížl ð¾n úp ör ãt •v !kx z 4› >2~ G΀ Qk‚ 1„ d¨† nWˆ wÑŠ £Œ ‹iŽ •* ž­’ ¨ ” ²„– ¼6.

Tkrf(x)k2 will always be positive unless rf(x) = 0, this inequality implies that the objective function value strictly decreases with each iteration of gradient descent until it reaches the optimal value f(x) = f(x) Note that this convergence result only holds when we choose tto be small enough, ie t 1=L This explains. Simple and best practice solution for g=(xc)/x equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. Tkrf(x)k2 will always be positive unless rf(x) = 0, this inequality implies that the objective function value strictly decreases with each iteration of gradient descent until it reaches the optimal value f(x) = f(x) Note that this convergence result only holds when we choose tto be small enough, ie t 1=L This explains.

!%Ý_r í/Id PÓ¨ \ï ãå N h èÕ¥ ÝÝb(Þ` A 9Í(Þ 3h Í 2b©½ ÕÝ º3å#½ (Þ Õ×Íä Ý d 5ËË a;d A ×ÍÝb(Þ ÎØa;ÝxÞ »Anö Unicode3 ÝD ¡ 9Í`Î 3 d ÕÝ ÑΣÍa. And f(1) = 5 Since f(x) is a polynomial, f is continuous on the real line We have f( 2) > 0 > f(0) So, by the Intermediate Value Theorem, there exists a number cbetween 2 and 0 such that f(c) = 0 Similarly, there exists a number dbetween 0 and 1 such that f(d) = 0 Note The choices x= 2;0;1 are not the only possibilities 13 Given y= 1 x2. P R I O R I T Y D a t e 0 2 / 2 1 / 2 0 0 7.

Substitute Y(t) into y’’ p(t)y’ q(t)y = g(t) and determine the coefficients to satisfy the equation There is no solution of the form that we assumed Find a solution of Y(t) Determine the coefficients End N Y Second Order Linear Non Homogenous Differential Equations – Method of Undermined Coefficients –Block Diagram •.