- - t D -!> TO. z. $1 V\c..t. - - To. e- 0 ~ "'2'('\" ~~1( ---->-... .;. s-c ().)e.. t..o.y\ w ('\ ~-e._. To :::. \ ~ ~ = T() c~> f.::.
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1 j_, \i~ \_t>~ Tc ~ 1-a... i-e_ -ztl ~\G\.Y\S? ' e.j. Ti> ::::. l.sec:. 9 "f"'lft 0. t L 'A -zrr \f''4el..4(' c... Tc Tb=- ;1o 11.. c_ => J () 't '2Tlv-~~ec. - - To $,'b \j'-o t-o,}rt.\m - 'Z. Tr - - CQ. \.l ~l.s L..-l) - '2.\1 z. $1 V\c..t. To 'ro~/;ec e- 0 ~ "'2'('\" ~~1( ---->-... t D -!> TO ~ ~ ~. ~To l f\.r u """be'(- C)f c.~ c\e~/jec... = - 70 To :::. \ ~ ~ = T() c~> f.::. lo t+s.;. s-c ().)e.. t..o.y\ w ('\ ~-e._ )0 i< L \:) -::- A~ ( w 0 -b) W\i V\_ Wot.:. 0 Mrl '2..lT f I) t-.::: e -= f ~tq_ 1Je"'c.'1 J~ rteat-5!$ z,4 Pa I (p
2 % Table of frequency and Period clc,clear all format rat % Fractions frequency = [ ] ; period= 1./[frequency]; % c = rdivide(a,b) and c = a./b perform right-array division %by dividing each element of a by the corresponding element of b. %If inputs a and b are not the same s i ze, %one of them must be a scalar value. % % One Way to do it - display and align A=[frequency;period] '; % Make 2X6 Array disp( ' frequency-hz period-seconds' ) disp (A) % Better way Make a table varnames={ 'Frequency', 'Period' } T=table(frequency',period', 'VariableNames',{ 'Frequency', 'Period' }) frequency-hz period-seconds 1 1 / 10 1 / / / / varnames = lx2 cell array { 'Frequency' } {'Period'} T 6x2 table Frequency Period le+os le-05 Published with MATLAB R20 I Ba
3 18 CHAPTER 2 SINUSOIDS 2-4 I.. ' rcn~ '~. - I t - I t - I 0 I t (a) (b) (c) / Figure 2-8 Illustration of time-shifting: (a) the triangular signal s(t); (b) shifted to the right by 2 s, x1 (t) = s(t - 2); (c) shifted to the left by 1 s, x2(t) = s(t + 1). j EXERCISE 2.2 Derive the equations for the shifted signal x2(t) = s(t + 1). We will have many occasions to consider time-shifted signals. Whenever a signal can be expressed in the form x 1 (t) = s(t - t 1 ), we say that x 1 (t) is a time-shifted version of s(t). If t 1 is a positive number, then the shift is to the right, and we say that the signal s(t) has been delayed in time. When t 1 is a negative number, then the shift is to the left, and we say that the signal s(t) was advanced in time. In summary, time shifting is essentially a redefinition of the time origin of the signal. In general, any function of the forms (t - t 1 ) has its time origin moved to the location t = t 1 For a sinusoid, the time shift is defined ~ith respect to a zero-phase cosine that has a positive peak at t = 0. Since a sinusoid has many positive peaks, we must pick one to define the time shift, so we pick the positive peak of the sinusoid that is closest tot = 0. In the plot of Fig. 2-6, the time where this positive peak occurs is t 1 = s.'since the peak in this case occurs at a positive time (to the right oft = 0), we say that the time shift is a delay of the zero-phase cosine signal. Now we can relate the time delay to phase. Let x 0 (t) = A cos(wot) denote a cosine signal with zero phase. A delay of x 0 (t) can be converted to a phase <p by making the following comparison: vjf;l\ C-~ \ xo(t - t i ) =A cos (wo(t - t1)) =A cos(wot + <p). c; l ~ f'd.:r> => cos(w 0 t - wot1) = cos(wot + <p) e o..'f'i J-. f 'fi '-S~ince the second equation must hold for all t, we conclude that <p = -w 0 t 1 Notice that -\\'JV'-. \o ~ '&' the ph~j.s _!!egative when the time sb!!t t 1 is positive (a del'!y ~ We can also express the Q ~~ ~ v\ 1:. phase in terms of the period (To = 1 /Jo) wliere we get the more intuitive formula S ~ ~ -'::>c'0 v S"'~ ( t ) ~ 1 -r(l '1 '\11'-Q._. -S t f\ <p = -w 0 t 1 = - 2rr _..!... (2.7a) ~ ~ '\"'~ 11 1~1 which states that the phase is 2rr times the fraction of a period given by the ratio of the "W time shift to the period. If we need to get the time shift from the phase, we solve for t 1 DEMO Sine Drill <p <p <pto t, = - - = --- = -- Wo 2rrfo 2rr (2.7b) Since the pos1t1ve peak nearest to t = 0 must always lie within the interval [ -~ To, ~ T 0 ], the phase calculated in (2.7a) will always satisfy - rr < <p ::: rr. However,
4 J:>-e /t:a-t..tt.~ q ~ ~n. ;e.'-'v\cls ( Sc } ~ c..-\ 't 7-. '--/l 'f'g<.d / ~ Yl.-' -= I"' t U>~ ::.. Pe..r-10~ -=. '1 C ycj ~ ).f.6.m.tf.._ <=. qc ~ UQ\'\Cl1 S <D _J -::= +--dr.lom-1 t-~ ~-c. ~'fv"+ 1.e. +t:,ml -z._tr T CZ-.~~MPl--\: z.rr <'tt.~ j ~ f ~ / uo\ ; ;,~, ~'t 1n +-C.. S\q\-t fuf' A CA-a.-( 2-rr+ - {!) ~A ~&::t-t-t,j] l~-v\l. f-~j ~ 0 T= t.s.ee- ~ Sl-.14- f;-,t,t,jo..{ ~ ~ 77 lc T ~ ~ I:. ~He_ e.-1,..eq: i 27T:. <1jLJ J,, 11/2.-- ~ r:::. )4>ec ~A~ Grr(~ -Vf)l 1il Yz_ I ' '
5 +=- 6)141\ \) o Sc~ ' ~ ~ (._,,... 2-rr ~~ \. C1 2, 1 p ~ t {) \o ~l 271 Cfltot -b 1 1t rr) - 0 i 2.. ~ (),,~7_ c z ( l 7 Y/;(_ s ) =- oyts tf.vt ~
6 ;oms 0 ' G.C'S ~LV1 l 0 z-3 SINUSOIDAL SIGNALS Yr-t... Y-r'2-15 ~GJ 'll-- $& I' so we define the phase to be cp = cp' - TC /2 in (2.2). For simplicity and to prevent confusion, we often avoid using the sine function. t. ( c) w 0 is called the radian frequency. Since the argument of the cosine function must be in radians, which is dimensionless, the quantity wot must likewise be dimensionless. Thus, w 0 must have units of rad/sift has units of seconds. Similarly, Jo= w 0 /2n is called the cyclic frequency, and Jo must have units of s- 1, or hertz. ssions netric ntities.tu ting tmple,.ty 1 book on for s 96l, Sinusoids EXAMPLE 2-1 Plotting Sinusoids Figure 2-6 shows a plot of the signal x(t) = 20cos(2n(40)t - 0.4n) (2.3) In terms of our definitions, the signal parameters are A = 20, wo = 2n ( 40) rad/s, Jo = 40 Hz, and cp = -0.4n rad. The signal size depends on the amplitude parameter A; its maximum and minimum values are +20 and -20, respectively. In Fig. 2-6 the maxima occur at and the minima at t =..., -0.02, 0.005, 0.03,......, , , ,... The time interval between successive maxima in Fig. 2-6 is s, which is equal to 1 / J 0. To understand why the signal has these properties, we will need to do more analysis. naking lowing (2.2) 1endent es how +1 and gument unction 1e, e.g.,. use the Relation of Frequency to Period IO x(t) 0-10 The sinusoid plotted in Fig. 2-6 is a periodic signal. The period of the sinusoid, denoted by T 0, is the time duration of one cycle of the sinusoid. In general, the frequency of the sinusoid determines its period, and the relationship can be found by applying the definition of periodicity x(t + T 0 ) = x(t) as follows: Vd-.,n \[ e,ef' '1,t}.,LJ> _,_~... ~~...,,.G "--... ~~.L.-~-'--'-'-'-~~ ~10 i'\? Ti,me t (s) 5..-vi> Ju~V\'f Jo 2.Tf' F Q, ~ ::-XU <<Jl_ ~l~!),,\ Figure 2-6 Sinusoidal signal with parameters A= 20, wo = 2rr(40), fo = 40 Hz, and <p = -0.4rr rad. Pvuo~'c... t:- :::-_.; nt.s to /l1~~t Tb -~ '/(i:> )t:q_ f<u\..e_ ~ Fo""
7 % Plot of cosine and shifted cosine % 20*cosine(2*pi* *pi) vs 20*cosine(2*pi*40) % TO= 1/40 = 25ms. Consider a time axis from to seconds taxis= -. 04:. 001:.04; xl=cos (2*pi*40*taxis) ; x2=cos(2*pi*40*taxis -0.4*pi); subplot(2,1,1); plot(taxis,xl) subplot(2,1,2); plot(taxis,x2) grid on Published with MATLAB R20 I Ba
8 LA l3eg~ ~(N ~ F=-1 ~ L> r<..c,...t od L5-0.5 X:O Y: ,,_.'-----'-~~-'-~-"-"--'--~----'~~-'--"-"-~-'--~~..,......J O.D '------"'..L...l..~~---L-~~...l...""'-L ~~----'-~~'...l..._~~.l..._~----J Time in Seconds - DSPf PG 15
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