Conversão de Coordenadas Polares para Cartesianas e Análise de Sinais

Converting from polar to Cartesian coordinates ej heta cos heta jsin heta ej heta cos heta jsin heta cos heta jsin heta E sqrtl2 m2 l j heta an1fracml ej1i 1 i The signal rn is flipped The flipped signal will be zero for n 4 and n 2 The signal xn is flipped and the flipped signal is shifted by 2 to the right This new signal will be zero for n 2 and n 4 The signal yn is flipped and the flipped signal is shifted by 2 to the left This new signal will be zero for n 6 and n 0 Therefore rn will be zero for n 0 Consider the arbitrary input xn and yn The inputoutput relationship does not change if the order in which S1 and S2 are cascaded in series is reversed Similarity if yn is the input to S2 it means that xn is the same as yn Therefore the inputoutput relationship for S2 is also equal to yn Therefore the system is linear gt ut st 0 xt yt therefore We want to find the smallest N0 such that fracddm N0 2k for k N0 where k is an integer If N has to be an integer then frac2km and m must be an integer This implies that m is a divisor of both N and N0 Also if we want the smallest possible N0 then m should be the GCD of N and N0 Therefore N0 N ext gcdm N Also we have limn o infty un0 limn o 0 un0 0 The signals xt and yt are plotted in Figure S28 yt begincases 1 2 t 1 3 1 t 0 0 extotherwise endcases For n 0 sn frac231frac32 for n 0 sn 0 214 a We first determine ht is absolutely integrable as follows 220 a We have yt 0t xτht τdτ 0t e05τut τdτ 0 t 1 226 a We have yn xn hn k xkhn k n 3 We now require that for n 2 B left frac32 right 2 left frac32 rightn1 left frac32 rightn1 Now hypothesizing that yt A ekt B ekt for t 0 Assuming initial rest y0 A B 1 Now we also have Now note that y1 0 for t L it must be true that for a causal system yt 0 for t 0 However the results of part a and b show that this is not true Therefore the system is not causal We have yt xt ut u0 At 0 to t xτAt τdτ 0 to t xτAt τdτ Also xt ut yt c In this case xt ut eαt eγt Therefore This may be written as yt a0 eβt eγt 1 Note that ddt eαt n αeαt n Given that we know that Aα 2Aα 1 must yield yt yt 2α 1 at the output From the given information we may conclude that Aα eα 1 and Aα 1 Hence we get yt from n0 to anAt n Thus yt At bnAt for n 0 if bn 0 and the system is causal Therefore the output will be either ut bnAt for all n where bn0 as n Using the results of Problem 283 we know that the homogeneous solution of the differential equation d2ydt2 adydt by ht will have terms of the form Ker1t Ker2t where r1 and r2 are roots of the equation Using the results of Problem 253 we know that the homogeneous solution of the differential equation yt ayt byt ht will have terms of the form Ker1t Ker2t where r1 and r2 are roots of the equation Therefore yn sn yhn A0D 100D Using the initial condition y0 100000 we have A 100000 100D Chapter 3 Answers 31 Using the Fourier series synthesis eq 338 yn 12Jk gn kT 2cos14 cos2πnT8 sin14 cos3πnT2 and 4 cosπn2 32 Using the Fourier series synthesis eq 338 zn 1 gn kT 122cos2πnfT sin4πnfT 33 The given signal is yn 12Jk gn k 2fk cos2πfnT8 34 Since a0 2 T 2sinπ2T Therefore now T 35 Both zn and zn k are periodic with fundamental period T 2 Since xn is linear transformation of zn and zn kT it is also periodic with fundamental period 36 Comparing zn with the Fourier series synthesis eq 338 we obtain the Fourier series coefficients of zn to be 37 Given that zn 12 38 Since xn is periodic with period 4 the signal yn can be written as 310 Since the Fourier series coefficients repeat every N we have a0 a6 a2 and ak a7 311 Since the Fourier series coefficients repeat every N 10 we have a0 a2 a4 a5 312 Using Parsevals relation we have 313 Let us evaluate Fourier series coefficients of gn Clearly since xn is real and odd it is purely imaginary and odd 314 The signal xn is periodic with period N 4 Its Fourier series coefficients are 315 From the results of Section 38 we have 316 The given signal xn is periodic with period N 2 and its Fourier series coefficients in the range 0 k 1 are 317 Since complex exponentials are eigen functions of LTI systems the input xn ejωt has to produce an output of the form yn 318 By using an argument similar to the one used in part a of the previous problem we conclude that xn is definitely not LTI 319 Voltage across inductor 320 Current through the capacitor 321 Using the Fourier series synthesis xn 2gn 2 2sinkπn8 322 ai T 6 a0 a2 323 b First let us consider a signal yn with FS coefficients 324 Since one period is xn 1 input must be periodic with one period 325 The output obtained by passing yn through the filter with frequency response Therefore 324 a We have c Using the multiplication property we know that 347 Considering xt to be periodic with period 1 the nonzero FS coefficients of xt are a0 a1 12 If we now consider xt to be periodic with period 3 then the nonzero FS coefficients of xt are a0 a3 12 a Using the results of part a the FS coefficients of xn xn 1 are given by ak frac1N sumn0N1 xn ejfrac2piNkn a The PS coefficients are given by ak frac1N sumn0N1 xn ejfrac2piNkn The following is an eigen function with an eigen value of 1 phit sumninftyinfty Mn ejomegan t The following is an eigen function with an eigen value of 12 phit sumninftyinfty Mn ejomegan t The following is an eigen function with an eigen value of 2 phit sumninftyinfty Mn ejomegan t The dc component of the input is 0 The dc component of the output is 2pi 363 The average energy per period is frac1T int0T yt2 dt sum0N1 cdots The fundamental period of the input is T 2 pi The signals are as shown in Figure S362 The output is yt htxt Therefore the system is linear Now consider xt ett0 Then yt int0t ht au x au d au int0t cdots d au The Fourier transform Xjomega is given by Xjomega int xtejomega t dt Therefore Xjomega as shown in Figure S41 This implies that int xtdt is equal to sin2omegat Using the time reversal property Sec 435 we have xt et1 ejt1 and thus yt cdots as shown in Figure S42 Consider a signal yt whose magnitude of the Fourier transform is Yjomega A and whose phase of the Fourier transform is angle Yjomega extSign left yt right Since Yjomega left Yjomega right we conclude that the signal yt is real See Table 41 Property 43 The signal xt is plotted in Figure S49 We see that this signal is very similar to the one considered in the previous problem In fact we may express the signal xt in terms of the rectangular pulse yt as follows We know that gt is periodic with a period of 8 Using the multiplication property we know that Xjomega 12piGjomega if we denote T as extFTgt Therefore xt xt 2ut 2ut 4 Taking the inverse Fourier transform we obtain yt t ut 2 sin2t ut Therefore Yjω XjωHjω 12 jω13jω The Fourier transform Xjω is Xjω 12 jω 13jω Using the multiplication property we have Hu HaHn Ha n Ha n Hn Therefore the frequency response of the inverse is Since the system is not invertible because Hjω is not defined for all ω we have b Let xn un 2 un 2 Using the Fourier transform analysis equation 59 the Fourier transform Xejω of this signal is Xejω n xn ejωn 2 sin2ω 53 We note from Section 52 that a periodic signal xn with Fourier series representation xn k ak ej2πTkn has a Fourier transform Xejω k ak δω 2πTk Therefore in the range π ω π we obtain Xejω 2 sinω 2 ω 2 b The given signal is zm sin5πm3 cos7πm3 sinm3 2πm3 j0 ejπm3 Therefore Xejω 13 jω 13 jω ω π j zm is periodic with period 6 The Fourier series coefficients of zm are given by an 16n zmej 2πnm6 Therefore from the results of Section 52 Xejω 16sum f The given Fourier transform may be written as Xejω Σ 16 ejπω2 n Comparing each of the two terms in the righthand side of the above equation with the Fourier transform analysis eq 59 we obtain zm 21 16 Σ ejωm iii In this case Xejω 1 2ejω Therefore Yejω 1 Taking the inverse Fourier transform we obtain yn δn 212un iv We have Yejω 1 2ejω 1 ejω Taking the inverse Fourier transform we obtain yn 32n 112un 1 312un 534 a Since the two systems are cascaded the frequency response of the overall system is Hejω H1ejωH2ejω Therefore the Fourier transforms of the input and output of the overall system are related by Xejω 11 ejω b We may rewrite the overall frequency response as Hejω 43 1 j331 12ejω Taking the inverse Fourier transform we get hn 3212un 1 j32ejω This is an sketch in Figure 535 a The frequency responses are related by the following expression Hejω2 hn2Xejω2 b Here Hejω 1 ejω1 ejω and yn 13un c The frequency response of the given system is Gejω2 491 2312ejω Therefore gn 13un 1 537 Given that yn xn an Then yn ak xn k 5401 Using the convolution property of the Fourier transform Yejω XejωAejω Now let Aejω ejω Then Hejω 2Xejω Substituting in the righthand sides of equations 5401 and 5402 and equating them bk hn 11 2n 1 1 1 232sinω2 Therefore we have shown that n 1 is valid for n if it is valid for n 3 n 4 and so on 545 We know from Problems 543 that the inverse Fourier transform of Xejw is the sequence xn 12πXejωejωndω where n 0 1 2 The evenindexed samples of xn are x2n n 0 1 2 The oddindexed samples of xn are x2n 1 n 0 1 2 This sampling preserves the impulse response of an LTI system Therefore the given statement is true Taking the Fourier transforms of both equations and calculating Hejw we obtain Hejw 14 32 12ejw 12ejw Crossmultiplying and taking the inverse Fourier transform we obtain hn 13un 16un 2 12un 1 Therefore the frequency response of the second filter is obtained by shifting The frequency response of the first filter is a passband filter and the first filter acts as a passband between tolerances with the same area as the second filter The first system has a frequency response H1jomega while the second system has a frequency response H2jomega Therefore the frequency response will be no oscillatory dynamics The frequency response will be Hjomega fracJomega2 omega 1 Thus jomega responses are derived using the previously obtained expressions for H1jomega and H2jomega From Table 42 we have ht sinleft fractT right Therefore At sinleft fractT right 6241 and ht ht ut gt ht b We may write the frequency response of iv as frac110 frac11 jomega10 Therefore Hjomega frac1010 1 and Hjomega are as shown in Figure S628 The Bode plot for this frequency response is shown in Figure S632 b One possible choice for the compensated frequency response is Hjomega frac50jomega 1jomega 1jomega 1 Therefore the overall frequency response is Hjomega frac1jomega 1jomega 1 We may write hn as H1ejomega sum h1n ejomega n sum h2n ejomega n sum hn ejomega n Therefore Hejomega corresponds to a bandstop filter Taking the inverse Fourier transform we obtain hn frac54 cosfracn pi2 sinn Therefore Hejω 12 jω 1 Hejω Therefore the step response is un 1 110n un Therefore Hejω is the frequency response of an ideal lowpass filter with cutoff frequency ωc 659 a We have Een Hejω Hejω unejωnnn Therefore en hn Ahn b Noting that Een is the Fourier transform of en we may use Parsevals theorem to obtain c We have 664 a We have Gejω Hejω Hejω unejωnnn Therefore b We have c Starting from Table 51 we know that when a signal is real and even then its Fourier transform is also real and even Therefore duality now says that if the Fourier transform of a signal is real and even then the equal signal is real and even 79 The Fourier transform Xω of xt is as shown in Figure S79 Gejω 1π Xk ejωk where k 0 1 2 Therefore Xω 17TX0 T17X07k show in Figure S79 Clearly Gejω 1TXω 7Xk for ω 50 Hejω Wejω Hejω Therefore Yejω XejωHejω A system that can be used to recover xt from xn is shown in Figure S727 Let yn xn summ inftyinfty hn m Then Yejomega frac12pi sumk inftyinfty Xejomega 2kpi Note that extsincx fracsinpi xpi x is the impulse response of an ideal lowpass filter with cutoff frequency frac12 and passband gain 1 Therefore we require that yn when passed through this filter should yield yn This is possible only if Xejomega should not overlap with one another This is possible only if Xejomega occupies the entire region from frac12 to frac12 and is downsampled by a factor of 12 Since it is not possible to directly downsample by a noninteger we need to perform the sampling twice ie use the upsampling process to reduce the sample rate by 12 Therefore the overall system for performing the sampling rate conversion is shown in Figure 374 The signals xn and yn are sketched in Figure 375 b X1ejomega and X2ejomega are sketched in Figure 375 a Let us denote the sampled signal by yn We have yn sumk inftyinfty yn k Since the Nyquist rate for the signal yn is frac2T we can reconstruct the signal yn From Section 72 we know that where gt extsinct and fracddt gt gt fracddt t Therefore denoting h by gt we have fracddt gt gt gn T n Therefore gt g0 tfracddt g0 b Now we may write pt p0t p1t Delta where p0t sumdeltat 2kT Therefore Pf 1 ejsTp0t where Pf W sumk kDelta Let us denote the product pt pf by gt Then gt p0tp1t p0t Delta tt This may be written as Gs psgf with Fps is specified in S737 Therefore Gjomega W suma b jomega Xjomega We now have Yjomega zjomegapjomega Therefore this gives us Yf W sumxjomegajomegaXjomega jcdots jDelta In the range 0 omega W we may specify yjomega as Yjomega Ypleft i cdots jright Since Yjomega cdotsyjomega in the range 0 omega W we may specify Yjomega as Yjomega yjJjomega W Given that 0 omega W we require that Yjomega KXy for 0 omega W This implies that Solving this when W frac12 More generally we get G extsincWAleft1 frac1 cosWAtanWright except when W frac12 Finally we also get K frac12j a j