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Wiley signals and systems e book TLFe BO 502

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48'7 Appendix A Solutions to the Exercises Solution 8,s tl S ( l ) = -&(LlLj dt dr Substitut,ion: z = (it, - = Q , dl d dr j ci dr [ I case case 0 d dt -=U t < :: yy (( tt jj = = ffI (( tt - r) r) d d rr = = tr 0, 1'1 case I t: yy(t) (t) = J(t - $; - T)tlz = 4t - t Sollltion 8.7 * z:3(t) q ( t )* Q ( ) g/3(t) = Q ( t ) * x:s(t) ;ul(t> = y,z(t) = ?j4(4 = ys(t) 22(1) J.l(t) * Z:i(l) = Q ( t ) * xg(t) yC,(f,) = J g ( t ) * X(j(t) y/7(t) = 5J(t)* z s ( t ) ;YN(t) - Q(t) q(t) yg(t) = q ( t j * Z:s(t) * S o l ~ ~8.8 t~o~ The calculation is perforrned as in Sec 8.4.3, except that ~ ( z is) 1nirrorr:d and shiftecl The xesiilt is, of course, identical to (8.49)

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