Signal Processing and Linear Systems-B.P.Lathi copy

The input voltage is 10 volts starting at t 0 and

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: nsform For an ideal differentiator, the input f (t) and the output yet) are related by df = dt T he Laplace transform of this equation yields For a resistor of R o hms, t he v oltage-current relationship is vet) = R i(t), a nd i ts Laplace t ransform is V (s) = R I(8) [ j (0 - ) = 0 for a causal signal] y es) = s F(s) a nd H (s) y es) = -(s) =s F (6.55) ( c) I deal I ntegrator For an ideal integrator with zero initial state, t hat is y(O-) = 0, yet) = l' and f er) d r 1 Y es) = - ;F(s) Therefore H (s) '" 1 For an LTIC system with transfer function s2 s +5 + 4s+3 (a) Describe the differential equation relating the input f (t) and output yet). ( b) Find the system response yet) to the input f (t) = e - 2t u(t) if the system is initially in zero state. 2 3t 2t Answers: ( a) d y + 4~ + 3 y(t) = '!!.. + 5 f(t) ( b) yet) = ( 2e- t - 3e- + e - ) u (t) "V dt 2 dt E xample 6 .10 shows how electrical networks m ay b e a~alyzed b y writ~ng t he i ntegro-differential equation(s) of t he s ystem a nd t hen solvmg t hese equatiOns. by t he L aplace t ransform. We now show t hat i t is also possible t o a nalyu: electnc~l n etworks d irectly w ithout h aving t o w rite t he i ntegra-differential equ~tiOns. T his p rocedure is considerably simpler because i t p ermits u s t o t reat a n e lectncal network as if i t were a resistive network. For this purpose, we need t o r epresent a network in t he " frequency domain" w here all t he voltages a nd c urrents a re represented by t heir L aplace t ransforms. . .. .. For t he s ake of simplicity, let us first discuss t he case w ith zero mltlal conditIOns. I f vet) a nd i (t) a re t he v oltage across a nd t he c urrent t hrough a n i nductor of L di vet) = L dt T he L aplace t ransform of t his e quation (assuming zero initial c urrent) is V (s) = L sI(s) m L Vj(t) = 0 a nd j=l L ij(t) = 0 j=l (6.57) Now if Vj(t) <==> Vj(s) a nd i j(t) <==> I j(s) t hen k L Vj(s) = 0 a nd j=l dt 6 .4 Analysis of Electrical Networks: T he Transformed Network h enries, t hen k • E xercise E 6.7 H (s) = T hus, in t he "frequency domain," t he v oltage-current relationships of a n i nductor a nd a c apacitor are algebraic; t hese e lements behave like resistors o f "resistance" L s a nd l /Cs, respectively. T he generalized " resistance" o f a n e lement is called its i mpedance a nd is given by t he r atio V (s) I I (s) for t he e lement ( under z ero initial conditions). T he i mpedances o f a r esistor of R o hms, a n i nductor o f L h enries, a nd a c apacitance of C f arads a re R , L s, a nd 1 /Cs, respectively. Also, t he i nterconnection c onstraints ( Kirchhoff's laws) remain valid for voltages a nd c urrents i n t he f requency domain. To d emonstrate t his p oint, l et Vj(t) ( j=l, 2, . .. , k) b e t he v oltages across k e lements in a loop a nd l et i j(t) ( j = 1, 2, . . . , m ) b e t he j c urrents e ntering a node. T hen (6.56) =-:; 399 Similarly, for a capacitor o f C f arads, t he v oltage-current relationship is...
View Full Document

Ask a homework question - tutors are online