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581, we've a similar consequence might be bought by way of employing L'Hospitalls rule to Eq. (3. 58). three. four. exhibit that if x ( t ) is a right-sided sign and X(s)converges for a few price of s, then the R O C of X ( s ) is of the shape equals the utmost genuine a part of any of the poles of X ( s ) . the place amax give some thought to a right-sided sign x(t) in order that and X(s) converges for Re(s) = a,. Then hence, X(s) converges for Re(s) = a , and the ROC of X(s) is of the shape Re($) > a ( , . because the ROC of X(s) can't comprise any poles of X(s), we finish that it's of the shape Re( s > ~ m a x the place a,,,,, equals the utmost genuine a part of any of the poles of X(s). three. five. locate the Laplace rework X ( s ) and cartoon the pole-zero plot with the ROC for the next indications x( t ): ( a ) x ( t )=e-"u(t) +eP3'u(t) ( b ) x ( t ) =e-"u(t) + e2'u(-t) (c) x ( t ) = e 2 ' u ( t )+ e - three ' u ( - t ) (a) From desk 3-1 LAPLACE rework AND CONTINUOUS-TIME LTI structures [CHAP. three (b) Fig. 3-11 We see that the ROCs in Eqs. (3. fifty nine) and (3. 60) overlap, and therefore, five From Eq. (3. sixty one) we see that X(s) has one 0 at s = - and poles at s s = - three and that the ROC is Re(s) > - 2, as sketched in Fig. 3-1 l(a). (b) From desk 3-1 = - 2 and = - three and We see that the ROCs in Eqs. (3. sixty two) and (3. sixty three) overlap, and therefore, (c) From Eq. (3. sixty four) we see that X(s) has no zeros and poles at s = 2 and s that the ROC is - three < Re(s) < 2, as sketched in Fig. 3-1 l(b). From desk 3-1 - 1 Re(s) < - three (3. sixty six) s +3 We see that the ROCs in Eqs. (3. sixty five) and (3. sixty six) don't overlap and that there's no universal ROC; therefore, x(t) has no rework X(s). e - three r ~ -(t ) - - CHAP. 31 three. 6. LAPLACE remodel AND CONTINUOUS-TIME LTI platforms allow locate X(s)and comic strip the zero-pole plot and the ROC for a > zero and a < zero. The sign x ( t ) is sketched in Figs. 3-12(a) and ( b ) for either a > zero and a < zero. considering the fact that x ( t ) is a two-sided sign, we will convey it as x ( t ) =e-"u(t) +ea'u(-r) word that x ( t ) is constant at t =0 earu( - t ) (3. sixty seven) and x(O-) =x(O) = x ( O + ) = 1 . From desk 3-1 1 H - s-a (c) Fig. 3-12 Re(s) < a (3. sixty nine) LAPLACE rework AND CONTINUOUS-TIME LTI structures [CHAP. three If a > zero, we see that the ROCs in Eqs. (3. 68)and (3. sixty nine) overlap, and hence, 1 1 - 2a x ( s )= - -= - a < Re(s) < a s+a s-a sZ-aZ From Eq. (3. 70)we see that X ( s ) has no zeros and poles at s = a and s = - a and that the ROC is - a < Re(s) < a , as sketched in Fig. 3-12(c). If a < zero, we see that the ROCs in Eqs. (3. sixty eight) and (3. sixty nine) don't overlap and that there's no universal ROC; therefore, X ( I ) has no rework X ( s ) . homes OF THE LAPLACE rework three. 7. confirm the time-shifting estate (3. 161, that's, x(t - t o )H e-"oX(S) R1=R by means of definition ( three . three ) by means of the switch of variables T =t - I , we receive with an analogous ROC as for X ( s ) itself. for that reason, the place R and R' are the ROCs earlier than and after the time-shift operation. three. eight. determine the time-scaling estate (3. 181, that's, via definition ( three . three ) = a t with a > zero, now we have I w ( x a )= x(r)e-('/")'dr = - X a -, a by way of the switch of variables 7 ( ) - RP=aR notice that as a result of the scaling s / a within the rework, the ROC of X ( s / a ) is aR.

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