{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW6 Solns

# HW6 Solns - A Figure 59.21(c The total response is the sum...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A Figure 59.21 (c) The total response is the sum of the zero—state and zero-input responses. This is y(t) = 26—‘u(t) — e’2tu(t). 9.21. The pole zero plots for all the subparts are shown in Figure 89.21. (a) The Laplace transform of x(t) is 00 X(s) = / (e‘2t+e'3t)e_”dt o = [—e-Wﬂ/(s + 2m? + [—e“3+3)‘/(s + 3>II3° l 1 2s+5 s+2+s+3=82+5s+6 The region of convergence (ROC) is ’Re{s} > —2. (b) Using an approach similar to that shown in part (a), we have e'4tu(t) (—> 3—, Re{s} > —4. Also, and . 1 e—5te—15tu(t) (5, s + 5 +175, Re{s} > —5. Hour this we obtain _ . 1 _ ‘ . L 5 5t __ 5t 51 —5t — 5t ____ e Sln(5t)U(t) — 517p: 6" ‘6 ‘3 J Mt) ‘ a (5+5)2 +25’ where Re{s} > —5. Therefore, 2 _4t _5t . L .5 +155 + 70 t ——-————— R —5. e u( )+e sm(5t)u(t)(—) s3+1432+903+100’ e{s}> (c) The Laplace transform of :1;(t) is 0 X(s) = / (em + €3t)e_’tdt —oo = he‘s—”Vs ~ 2)]:‘100 + [—e—<s-3>‘/(s — 3)] ‘loo 1 1 25 —— 5 = + : ____._._ 5—2 5—3 32—55+6 The region of convergence (R00) is 'Re{s} < 2. ((1) Using an approach along the lines of part (a), we obtain *2: E 1 —— 'R —2. 8921—1 6 u(t)(—> 5+2, e{s}> ( ) Using an approach along the lines of part (c), we obtain e2‘u(-t) A s i 2, 1245} < 2. (5921—2) Horn these we obtain 2 e—2ltl = e‘2tu(t) + e2‘u(—t) (i; 52 j 4, —2 < ’Re{s} < 2. Using the differentiation in the s—domain property, we obtain d 2.5 252 + 8 t‘zltl 5 —— =——— —2 R . 2. e (———-> (18 32—4 (32—4)2’ < e{e}< (e) Using the differentiation in the s-domain property on eq. (39.21-1), we get (1 l 1 r” t ‘-— = , ’R >~2. e u( ) H d3 [8 +2] (3 + 2)2 e{s} Using the differentiation in the s—domain property on eq. (89.21-2), we get _ _ _ = — 2. te u( t) (—) d3 [5 _2] (3—2)2, Re{s} < Therefore, 1; —4s Itle‘ml = te‘2tu(t) + —teztu(—t) (—2 —2 < Re{s} < 2. (s + 2)2(s — 2)2 ’ 9.22. (a) From Table 9.2, we have :t;(t) = ésin(3t)u(t). (b) From Table 9.2 we know that L s t t +—+ 0. cos(3 )u( ) 52 + 9, ’Re{s} > Using the time scaling property, we obtain 1; s cos(3t)u(—t) +—-% _32 + 9, Re{s} < 0. Therefore, the inverse Laplace transform of X (s) is :r,(t) = —cos(3t)u(——t). (c) From Table 9.2 we know that — 1 Ct cos(3t)u(t) (—1.1) m, R€{S} > 1. Using the time scaling property, we obtain _t c s + 1 e cos(3t)u(—t) 4—) —m, ’Re{s} < —1. Therefore, the inverse Laplace transform of X (s) is z(t) = —e_‘cos(3t)u(—t). ((1) Using partial fraction expansion on X (s), we obtain 2 1 X(s)—3+4-s+3' Horn the given ROC, we know that 2(t) must be a two-sided signal. Therefore, (C(t) = 2e’4tu(t) + e‘3tu(—t). (8) Using partial fraction expansion on X (s), we obtain 2 l X(s)—3+3_s+2' From the given ROC, we know that :1:(t) must be a two-sided signal. Therefore, a:(t) = 2e‘3tu(t) + e‘2tu(—t). (f) We may rewrite X(s) as X“) = ”ST—3%? _ 1+ 3.5 ' (s — 1/2)2 + («i/2)? = 1+3 5—1/2 3/2 (s — 1/2)? + («i/2)? + (s — 1/2)? + («i/2V Using Table 9.2, we obtain \$(t) = 6(t) + 3e-t/2 cos(\/§t/2)u(t) + «ﬁat/2 sin(\/§t/2)u(t). 334 (g) We may rewrite X(s) as 35 X = 1 — . (s) (s + 1)2 From Table 9.2, we know that 1; l tu(t) <——-> 3—2, Re{s} > 0. Using the shifting property, we obtain —t C 1 e tu(t) 6—) (s + D2, ’Re{s} > —1. Using the differentiation property, di:[e-'tu(t)] = e-tuu) — te-‘um «i» 8 (s+1)2’ Re{s} > —1. Therefore, :1:(t) = 6(t) —- 3e“tu(t) -— 3te‘tu(t). 9.23. The four pole-zero plots shown may have the following possible ROCs: oPlot (a): Re{s} < —2 or —2 < Re{s} < 2 or Re{s} > 2. oPlot (b): 'Re{s} < —2 or ’Re{s} > —2. oPlot (c): Re{s} < 2 or ’Re{s} > 2. oPlot (d): Entire s-plane. Also, suppose that the signal z(t) has a Laplace transform X (s) with R00 R. (1) We know from Table 9.1 that e‘3ta;(t) ti) X(s + 3). The ROC R1 of this new Laplace transform is R shifted by 3 to the left. If .7:(t)e~3t is absolutely integrable, then R1 must include the jw axis. oFor plot (a), this is possible only if R was Re{s} > 2. oFor plot (b), this is possible only if R was Re{s} > —2. oFor plot (c), this is possible only if R was Re{s} > 2. oFor plot (d), R is the entire s-plane. (2) We know from Table 9.2 that 1 3+1 e‘tu(t) +5) , ’Re{s} > —1. Also, from Table 9.1 we obtain —t L X(s) :1:(t) * [e u(t)] (~—> 3 +1, If e“u(t) * 22(t) is absolutely integrable, then R2 must include the jw-axis. R2 = Rn [Re{s} > —1]. oFor plot (a), this is possible only if R was —2 < ’Re{s} < 2. oFor plot (b), this is possible only if R was Re{s} > —2. oFor plot (c), this is possible only if R was 'Re{s} < 2. oFor plot (d), R is the entire s—plane. (3) If :c(t) = 0 for t > 1, then the signal is a left—sided signal or a ﬁnite-duration signal. oFor plot (a), this is possible only if R was Re{s} < —-2. oFor plot. (b), this is possible only if R was Re{s} < ~2. 0For plot (c), this is possible only if R was Re{s} < 2. oFor plot (d), R is the entire s—plane. (4) If \$(t) = 0 for t < —1, then the signal is a right—sided signal or a ﬁnite—duration signal oFor plot (a), this is possible only if R was Re{s} > 2. oFor plot (b), this is possible only if R was Re{s} > —2. 0For plot (c), this is possible only if R was Re{s} > 2. 0For plot (d), R is the entire s—plane. (0‘ Cb) "Mb N “Al-““4..— _No (A. w 4. N. N (C) (d) (e) (1}) I430) “'10, wihn with w w —NO 040 £0 9.27. From clues 1 and 2, we know that X (s) is of the form X‘3)=rs:2><T+T>- is —1 + j. Since (C(t) is real, the Furthermore, we are given that one of the poles of X (s) a:1—jandb=1+) poles of X (3) must occur in conjugate reciprocal pairs. Therefore, and A (s) (s+l—])(s+1+j) From clue 5, we know that X (0) = 8. Therefore, we may deduce that A = 16 and . From the pole locations we know that there are two Let R denote the ROC of X (s) —1 or Re{s} > -1. We will now use clue possible choices of R. R may either be Re{s} < 4 to pick one. Note that y(t) = 6mm“) (—5 Y(s) = X(s — 2). The ROC of Y(s) is R shifted by 2 t0 the right. Since it is given that y(t) is not absolutely integrable, the ROC of Y(s) should not include the jw-axis. This is possible only of R is Re{s} > ——1. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

HW6 Solns - A Figure 59.21(c The total response is the sum...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online