ECE308_F08_homework_1_solution

# ECE308_F08_homework_1_solution - Problem 1-3(a‘ The...

This preview shows pages 1–7. 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 is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 1-3 (a‘) The signal can be developed in terms of equations as follows: _ 1, mm s 112 mu“) * {0, otherwise _ 1, |r| g 10:2 = 5 _ 0, otherwise This is a rectangular pulse of amplitude 1 between -5 and 5 and 0 otherwise. A sketch will be given at the end of the problem solution. (b) Following a procedure similar to that of (a) one ﬁnds that this is a rectangular pulse of amplitude 1 between -0.05 and 0.05 and 0 otherwise. A sketch will be given at the end of the problem solution. (c )Tliis is a rectangular pulse of amplitude 1 between 0 and l and 0 otherwise. A sketch will be given at the end of the problem solution. ((1) This is a rectangular pulse of amplitude 1 between 0.5 and 4.5 and 0 otherwise. A sketch will be given at the end of the problem solution. (e) The first term of this signal is a rectangular pulse of amplitude 1 between 0 and 2 and 0 otherwise. The second term is a rectangular pulse of amplitude 1 between 0.5 and 1.5 and 0 otherwise. Where both pulses are nonzero, the total amplitude is 2; where only one pulse is nonzero the amplitude is 1. A sketch is provided below. The MATLAB program below uses the special function given in Section 1-6 (page 32) of the text to provide the plots. % Sketches for Problem 1-8 % t = —6:0.0015:6; xa = pls_fn(0. l *t); xb == pls_fn(10*t); xc = pls_fn(t - 0.5); xd = pls_fn((t - 2V5); xe = pls_fn((t - W2) + pls_fn(t - l); subplot(3,2, l),plot(t, xa,'-w'), axis([-6 6 0 l.5]),xlabel(‘t'),ylabel(‘xa(t)') subplot(3,2,2),plot(t, xb,‘-w‘), axis([-.] .1 0 1.5]),xlabel(‘t'),ylabel('xb(t)') subplot(3,2,3).plot(t, xc,'-w'), axis([-1 2 0 1.5]).xlabel('t'),ylabel('xc(t)') subplot(3,2,4),plot(t, xd,'-w'), axis([~l S 0 l.5]),xlabel('l'),ylabel('xd(t)') subplot(3.2,5),plot(t, xe,‘-w‘), axis([—l 3 0 2.5]),xlabel('t'),ylabel(‘xe(t)') —5 0 5 —%.1 —-0.05 O 0.05 0.1 t t 15 1.5 E 1 E 1 9 O 1 2 0 0 2 4 (c ) Single-sided spectra are plotted below. Double-sided amplitude spectra are obtained by halving the lines and taking mirror image; phase spectra are obtained by taking antisymmetric mirror image. 4 _ 2 a \$2 -‘ 0 j —2 O 5 1O 0 ph rad 5 10 Problem 1-9 (3)2193 = 501:, so T0 = lltfu = “25 = 0.04 s. (b) 211‘)“0 = 601:, so To = 1% = 1130 = 0.0333 s (c ) 211:}; =_ 701:, so To = llﬁ, = N35 = 0.0286 s. (d) We have 50a = anfo and 6011: = 2mg" Iwhere m and n are integers and fo is the largest constant that satisﬁes these equations. The largesto’f is 5 Hz with m = S and n = 6. (e) We have 5011: = 21tmﬁ, and 70a = 21tnf0, where m and n are inteogers and f0 is the largest constant that satisﬁes these equations. The largest f0 is 5 Hz with m = 5 and n = 1 Problem 1-10 (3) IA! 24.2426: angletA) = 0.7854 radians; B = 5.0 +J- 3.6601 so Rem) = 5 and MB) = 8 6603 ()A+ =8.0+j11.6603. (c)A»B=-2.0-'5.6603. dA*B=-1090 . - _- Problem 1-14 (a) A sketch is given below: From the figure, it is evident that x0) = u(r) + u(r - 3) - u(r - 5) - u(r - 6). 3 o -1 o 1 2 a 4 5 t (b) The derivative of 1:0) is dx(r)}'dr = 5(0 + 6 (r - 3) - (‘5 (1- 5) - 6 (t - 6) MAW—HQ (a) The integral is zero because the delta function is outside the range of integration. (b) The integral evaluates as follows: 5 fcos(2m)6(: - 2)dt = cos(41't) = l 0 (c ) This integral can be evaluated as 5 feosam) 6(1 — 0.5)d: = (20501:) 0 || | —— (d) The value of this integral is 0: II O In — 2)26(: - 2)dr = (2 — 2F (c) This integral evaluates to [125(I-2)dt = 22 = 4 W (a) Only (1) is pv.‘.'.riodi£:,-‘f1 = 2.5 Hz= 0.5m andf2= 3 Hz = 0.511 where the integers m and n are 5 and 6. respectively. The fundamental frequency is 0.5 Hz and the period is 2 s. (b) Signals (1) and (2) are power signals. Their powers are both 1 W. (c) Only signal (3) is an energy signal; its energy is 1/20 I. Signal (4) is neither energy nor power. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

ECE308_F08_homework_1_solution - Problem 1-3(a‘ The...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online