Fourier Transform Analysis of Continuous- and Discrete-Time Signals, Exams of Signals and Systems Theory

Download Fourier Transform Analysis of Continuous- and Discrete-Time Signals and more Exams Signals and Systems Theory in PDF only on Docsity! MT2.1 (25 Points) Consider a continuous-time signal x defined as (t+4)x(t) = lOsinc -7- , where the normalized sinc function is defined as sin(m) sine (t) = . m (a) Provide a well-labeled sketch of X(w) , the continuous-time Fourier transform of the signal x. l 7D"-~ ~ I D I _--+----r 70 ..:y., (b) Find the total area under the signal x - the quantity 00Jx(t)dt . -00 2 (c) In the figure below, the signal x modulates a carrier of frequency k rad/s. Sketch the spectra at point a. Finally, design a receiver to recover the signal x from the modulated signal at point a. ax(t) sin(kt) }"'t- -J'Il.~e. -£fA Lt:) ~ x(t) ~I>'\ ~~ = IC it-) ~tb~ - '\f yt~) fll~)} "~II:) 2j Alj~) ~:.n(,,~) . yll) - ,c(f,) Srt~I~!) ~ }<li)(e;";i;j~tr _.L~_-...,...-_.........-t-Jr--~l.J -\c. )~ -jZk.t. It') e. -2.+e -:::~I.t: -If NuJ H{w) +0 1,... 0. LPF "'~ ~ ... ~ of '2. (d) Consider a continuous-time signal y defined as t ItI< 1 yet) = .{o elsewhere Determine Y(w), the continuous-time Fourier transform ofthe signaly. OO • -c­ '( ()P)::. .~rP YLt) e-j'-J tit: J ~ (l t e.-J ~~~ L, - jWi. - j c.Ji: J t ~L $ . + _.-.€.:o.--_ ~. z. -)Co.) to,) (-~_I -it..> ;'-J -j~ . ~ -~+~4-~-~ -j~ -J~ w'Z. ,-<)'2. :::: - ~ u$ L.,) T.:!=- s:", ~ jw jw'L 3 MT2.3 (25 Points) The reconstruction of a continuous-time signal Yc from its samples Yd can be conceptualized as an impulse generator followed by an ideal low-pass filter. In the time domain, this amounts to the sine interpolation step defined as O) '" (). (t - nTs ) RYc Ct = o~ Yd n SillC T~ ,t E . nEZ (a) Show that the set of shifted sinc functions {sine (t - n), n E Z} is orthonormal: <sine(t-k),sine(t-I)> = 5(k -I) You may wish to invoke the orthogonality preserving property of the CTFT, where <D k is the CTFT of the continuous-time signal <A, Vk . ~,,, ~t F \ \ lw\ .:: 1\ ~\(\C.(-t)-;::. to ~"'L~+{W)= ! d ­ 'tL> \l'" 0 I .~""" ~ x(t. ~) ~ --? X(",) ej~--' L~1- cP ... ~$i"c(t-\:.:) r ~ ~K=' r e-~~"" J lw\c.1i l u z:r\Ww:~ 5It"'1\\..,.li & +1 1l _ j"''f:. iwQ. J c:;{t..>(~k/fJl) == ): ~ ... Lw)~:(w) Jw ~ ) e. ~ _'tf 1--T' \P.o. ~ r~ 1d-w := Z tr L'!I { 'U •..., t~-I<.) 1.1- k.=F.l ::> L~ e..) d..., ~ 0 \~ I ~l>;:' ZtT~r\t.-;.1 ~ <4>FI ~Q.>::; cg[~-.(J Jt can be shown that any bandhmited signalf(tJ can be expressed as J(t) = L 00 Cl:nsinc(t - 71.) n=-oo whl?re an = (sinc(t - 11,), f (t)) = /(11.) One generalized interpretation of the sampling theorem is in tem1S of orthonormal bases for bandlimited functions. The Whittaker-Shannon interpolation formula given above is not implementable in real­ time because it is not causal. In practice, an implementable (causal) non-ideal low-pass filter is used. 5 ~ (b) Consider a continuous-time implementable filter with impulse response hand frequency response iI. Here, we seek to minimize the mean-square error in the reconstructed signal: C, = co Jlh(t) - h(t)1 2 dt -co Suppose that we determine the mean-square error in the non-ideal low-pass filter: -co Describe how c/ can be determined from c(J)' 5~1'.CJl. ~ li:\ -~lt") C."'FT 9 k(w) - \-\(w') 'by P"'5~""\ ' s \E t =-E2-rr ~ (c) Instead, consider a discrete-time implementable FIR filter with impulse response J and frequency response fr. Here, we seek to approximate the ideal low-pass filter, 1 lilll ~ f F(ill) = I I ' {o .!!..<ill<Jr2 ­ by minimizing the mean-square error at the following discrete frequencies: 2Jr Jr Jr 2Jr } 5 ... - Jr -- -- 0 - - Jr - { , 3' 3"3'3' . Determine the least-squares solution for the optimallength-4 sequence ] = (J(O) J(l) J(2) ](3)). NOTE: You may leave matrix operations unevaluated in the solution. -h. "",:""_1 :eC2..W""'~ 2:"- \ F{I.»- P[w)\7- ~ "'~~ \\ ~ - D~ \\ .... 4 ~ ­ ~= l\)"iDY' t>~ ~ F{c,));: ± t[o'I] C2..-i<.>~ I'I~O D Fl-~) fl-~) f(O~b. "'" fl\ p(!}) Fe"lt) -::::r. 0 \ t 0 0 f(-~) f(-~) f (0) f(\') F(~) F( tr) - 6 II 'Z.. ~ilf e. ~ ~ \"­ ti~fi1\ a. -\ \ \ \ I .j!.. -j ~ -ti~I e e -\ -)-qt .ti~\ ~ e \ \ .:l -\ A,. \\ " ~ (0) f [\1 ~[tl t(» MT2.4 (25 Points) Consider a discrete-time N-periodic signal x/n) and the length-N signal x(n) obtained by keeping the single period [O,N-l] of x/n). Denote X/k) as the discrete Fourier series (DFS) of xp(n) and X(m) as the discrete-time Fourier transform (DTFT) of x(n). In this problem, use the following DFS equations: N-l 'klo 1 N-l ikE!.n X(k) = Lx(n)e-r Ii x(n) =- LX(k)e N n~O N k=O (a) Express Xp(k) in terms of X(m). Provide at) illustration ofthis relationship. Xlw) ~ 2: )([f\1L-i-' Xlu) 1'\"'0 N~I . 201f X[~1 :. ~ )( Ct" j e.-)'N~ f . ,,::.0 _-+-----1--1-1-1---t-~f,.) For parts (b) - (d), consider the following scheme: DTFT ) X (m) _s_a_mp_le_@_t,_OJ_==::~) X (k) JDFS ) x(n)x(n) (b) Suppose that M = N. Determine an expression for x(n) in terms of x(n). "l.'t(xC,,] =-l "'~ LX(J:.! "­~ 0]'J-;cr-Ic-'" f'J \<';0 N e. := -'-~ [f )([/1 e.-'/;.i t:.J.1 fl') !£-~'" N \1.'=0 ;'::"-00 pO IV-I ;::.l. ~ xCt] L J~ k6·,-.Q.) N ~::...... \1.-:.0 _r\/ ( 0, 00 xT~1:=:'L )(.[... -rN) 7