Notes 04 largely plagiarized by %khc

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1 Noes 04 largely plagiarized by %khc Convoluion Recap Some ricks: x() () =x() x() (, 0 )=x(, 0 ) R ț x() u() = x( )d x() () =ẋ() This hen ells us ha an inegraor has impulse response h() =u(), and ha a differeniaor has impulse response h() = (). Convoluion is associaive and commuaive. Convoluion also disribues over addiion. Exercise Prove i. Convoluion wih h() is boh linear and ime invarian. [x () x 2 ()] h() = [x ()] h() [x 2 ()] h() = [x () h()] [x 2 () h()] = y () y 2 () x(, 0 ) h() = [x() (, 0 )] h() = [x() h()] (, 0 ) = (, 0 ) = y(, 0 ) This should no surprise you. If is he oupu of some LTI sysem wih inpu x(), hen =x() h(). 2 More Convoluion Fun We now dig ino he old ee20 archives and drag ou some old problems o illusrae convoluion. [Skeches of he oupu for each of he pars are available. Please see Figure.] P A. An LTI sysem has impulse response h() =e,=2 u(), inpu x() =x () n=, (, n), and oupu. If x () =u( 0:0), u(, 3:0), deermine he oupu. Firs deermine x(). Noice ha x() is jus an impulse rain muliplied by a pulse from =,0:0 o = 3:0. This jus picks ou he four impulses a = 0, =, = 2, and = 3. Then use superposiion. Convoluion wih shifed impulses gives four shifed copies of h(). So =e,=2 u() e,(,)=2 u(, ) e,(,2)=2 u(, 2) e,(,3)=2 u(, 3). B. x() =e,2 u(). h() =2(, ) ( ). Same ricks as above. =2e,2(,) u(, ) e,2() u( ). P C. x() = n=, (, 0:0n). h() =Π(200). x() is a rain of impulses, each separaed from he nex by 00. h() is a pulse of heigh from =, 400 o = 400. Convolving h() wih an impulse cenered a = 0 gives h(, 0 ). Since here are infiniely many shifed impulses, here are infiniely many copies of h(), whose ceners are separaed from ha of heir neares neighbors by 00.So =P n=, Π[200(, n 00 )].

2 EE20: Signals and Sysems; v x () impulse rain x() (a) waveforms for A e 4 /00 (b) oupu for B. (c) oupu for C. /200 /e 2 (d) oupu for D. /2π /2 /2 7/2 9/2 2 (e) oupu for E. (f) oupu for F. Figure : Skeches of he example problems. Knowing wha he oupu looks like is definiely as good as having a mahemaical formula. D. x() =e, u(). h() =r(, )Π(, 3 2 ). Le s choose x() o flip and shif. There are hree places where he inegral is differen. For <, here is no overlap, so =0. For <<2, For >2, = = Z e,(,) (, )d Z = e, e (, )d = e, (e, 2e )j =, 2 e, Z 2 e,(,) (, )d Z 2 = e, e (, )d = e, (e, 2e )j 2 = e,(,) E. x() =Π(). h() =Π(,2 4 ). You can do flip and shif, make he pulses ino sums of uni seps and convolve, or jus use your inuiion.

3 EE20: Signals and Sysems; v = 8 >< >: 0 for <, 2 2 for, 2 << 2 for 2 << , for 7 2 << 9 2 F. x() =u(, ). h() =cos(2)u(). Convolve u() wih h(). This gives = sin(2)u(). Now, since he uni sep is acually delayed by, 2 delay by. So = sin[2(, )]u(, ). 2 3 Sysem Inerconnecions There are hree major ways of puing sysems ogeher. Check ou Figure 2 and see if he equivalens make sense o you [wo of hem already should, since you saw hem on ps2, problem 6]. The reason why we haven fully alked abou he feedback configuraion is ha he analysis becomes much easier when we hi he Fourier and Laplace ransforms. Wai.5 monhs. g() h() g() h() (a) wo sysems in series; impulse responses are g() and h() (b) he equivalen of he series connecion in (a); noe he poor convoluion sign g() h() g() h() (c) wo sysems in parallel (d) he equivalen of he parallel connecion in (c) g() h() (e) wo sysems in feedback connecion (g) unforunaely, we re no prepared o alk abou he equivalen of he feedback connecion ye in a simple fashion, bu we will Figure 2: The hree major ways we have of composing sysems. 4 Eigenfuncions Consider an LTI sysem wih inpu x(), impulse response h(), and oupu. Wha funcion can we pu ino he sysem so ha we will ge he same funcion ou, scaled by a consan? Such funcions are called eigenfuncions and heir associaed consans are called eigenvalues. In symbols, for a sysem performing operaion H on is inpu, we have H[f ()] = f (), wheref() is he eigenfuncion and is is eigenvalue. Le s ry x() =e j!.then = x() h() Eigen is German for self, i hink. Well, on good days, ha is.

4 EE20: Signals and Sysems; v Z = x( )h(, )d, Z = x(, )h( )d, Z = e j!(,) h( )d,z = e j! e,j! h( )d, = e j! H(!) where H(!) is defined as R, e,j!) h( )d. Noe ha H(!) could be complex. Tha means ha we can also wrie: =jh(!)je j(! 6 H(!)) Wha if we ried a cosine? If we assume ha he sysem performs operaion H on is inpu: = H[x()] = H[cos(!)] = H[ 2 (ej! e,j! )] = 2 H[ej! ] 2 H[e,j! ] 6 6 = 2 jh(!)jej!j H(!) 2 jh(,!)je,j!j H(,!) 6= cos(!) Oh well. Because here is no guaranee ha jh(!)j is even and 6 (H(!)) is odd, we have o delee cosine from he lis of candidaes. Similar reasoning allows us o delee sine. Exercise Prove ha sin(!) is no an eigenfuncion. Anyway, we sill have an imporan resul. If you gave me a sysem wih impulse response h() and inpu x() and old me o find, i could always convolve and give you an answer. Bu since e j! is an eigenfuncion for he convoluion operaor, if i can. represen he inpu x() as he sum of complex exponenials 2. deermine H(!) for he impulse response h() hen i can give you he oupu as he sum of complex exponenials, scaled by H a he appropriae values of!. Tha is, if hen = = X x() = X n e jn!0 n=, X X n H(n! 0 )e jn!0 n=, X n=, X n jh(n! 0 )je j(n!0 6 H(n! 0)) Noice ha i didn have o do any convoluion. Wha a feaure.

5 EE20: Signals and Sysems; v Sinusoidal Seady-Sae Response of Real-World LTI Sysems If h() is real, hen we can definiely say some useful hings abou cosine and sine being inpus ino LTI sysems. If h() is real, hen h() =h (). So Z H (!) = [ h()e,j! d], Z = h ()e j! d, Z = h()e,j(,!) d, Rewriing H (!) and H(,!) in polar form gives: If we equae he wo polar forms, = H(,!) H (!) = [jh(!)je j 6 H(!) ] = jh(!)je,j 6 H(!) H(,!) = jh(,!)je j 6 H(,!) jh(,!)j = jh(!)j 6 H(,!) =,6 H(!) In oher words, for a real h() he magniude of he frequency response is even and he phase of he frequency response is odd. Le s now go back o rying a cosine as an inpu ino an LTI sysem. = H[cos(!)] = 2 jh(!)jej!j 6 H(!) 2 jh(,!)je,j!j 6 H(,!) = 2 jh(!)jej!j H(!) 2 jh(!)je,j!,j H(!) = jh(!)j cos[! H(!)] So if our impulse response is real, hen a cosine as inpu comes back ou sill looking like a cosine, bu is ampliude is scaled by he magniude of he frequency response, and is phase is shifed by he phase of he frequency response. Bu all real-world sysems are real, so his should work on any sysem i care o ake down o he lab and hrow cosines ino. In fac, his gives me a good way o figure ou wha H(!) is; i can generae all sors of cosines wih ampliude a various frequencies and record he oupu cosines ampliude and phase shif o consruc H(!). This procedure is acually used in he real world. Exercise Prove ha if h() is real, hen sine as inpu gives a sine as oupu. 6 A Look Ahead Fourier series you have already seen in differenial equaions. We re going o use i as a sepping sone ino he Fourier ransform. Afer developing he FT, we ll find ou ha we have a shorcu o convoluion. This will be exceedingly cool.