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Signals & Systems (Second Edition) —Learning Instructions (Exercises Answers) Department of Computer Engineering 2005.12

Contents Chapter 1 ······················································· 2 Chapter 2 ······················································· 17 Chapter 3 ······················································· 35 Chapter 4 ······················································· 62 Chapter 5 ······················································· 83 Chapter 6 ······················································· 109 Chapter 7 ······················································· 119 Chapter 8 ······················································· 132 Chapter 9 ······················································· 140 Chapter 10 ·······················································160 1

Chapter 1 Answers 1.1 Converting from polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coordinates: , , , , , 1.3. (a) = , =0, because , .Therefore, = = = , (b) = (c) =cos(t). Therefore, = = = (d) , . Therefore, =0，because < . = = , (e) (f) = = = = , =1. therefore, = = , . Therefore, = = = , . 1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal x[n] is flipped signal will be zero for n<-1 and n>2. (d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4. (e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left. 1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2. (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, (c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be This new signal will be zero for n<-6 and n>0. Therefore, x (1-t) +x(2-t) will be zero for t>-2. zero for t<1. 2 111cos222je111cos()222je2cos()sin()22jjje2cos()sin()22jjje522jjjee42(cos()sin())1442jjje944122jjjee944122jjjee412jje055je22je233jje21322jje412jje2221jje4(1)jje411jje122213jjeE4014tdtePE(2)42()jttxe2()1txE22()dttxdtP211limlim222()TTTTTTdtdtTTtxlim11T2()txE23()dttx2cos()dttP2111(2)1limlim2222cos()TTTTTTCOStdtdtTTt1[][]12nnunx2[]11[]4nunnxE204131[]4nnnxPE2[]nx()28nje22[]nxE22[]nxP211limlim1122121[]NNNNnNnNNNnx3[]nxcos4nE23[]nx2cos()4n2cos()4nP1limcos214nNNnNN1cos()112lim()2122NNnNnN

(d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be (b) x2[n]=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x3[n] is as shown in the Figure S1.6. zero for t<9. 1.6 (a) x1(t) is not periodic because it is zero for t<0. … x3[n … … -3 1 1 ] -4 … … 1.7. (a) 1 4 5 … … … n 0 … … 1 … -1 … … … -1 Therefore, it is periodic with a fundamental period of 4. -1 = Therefore, is zero for >3. (b) Since x1(t) is an odd signal, is zero for all values of t. (c) Therefore, is zero when <3 and when (d) Therefore, is zero only when . 1.8. (a) (b) (c) (d) . 1.9. (a) is a periodic complex exponential. (b) is a complex exponential multiplied by a decaying exponential. Therefore, （c） is not periodic. is a periodic signal. = = . is a complex exponential with a fundamental period of (d) is a periodic signal. The fundamental period is given by N=m( . ) = By choosing m=3. We obtain the fundamental period to be 10. (e) is not periodic. is a complex exponential with =3/5. We cannot find any integer m such that m( ) is also an integer. Therefore, is not periodic. 1.10. x(t)=2cos(10t＋1)-sin(4t-1) Period of first term in the RHS = Period of first term in the RHS = . . Therefore, the overall signal is periodic with a period which the least common multiple of the periods of the first and second terms. This is equal to . 3 1[]vnx1111[][]([][4][][4])22nnununununxx1[]vnx1[]nx2[]vnx11311[][][][3][3]221122vnnnnnununxxx3 []vnxnn1554411()(()())(2)(2)22vtttttututxxxee4()vtxt01{()}22cos(0)tttxe02{()}2cos()cos(32)cos(3)cos(30)4tttttxe3{()}sin(3)sin(3)2tttttxee224{()}sin(100)sin(100)cos(100)2tttttttxeee1()tx101021()jtjttjxee2()tx2()tx3[]nx3[]nx7jnejne3[]nx224[]nx23/510().3m5[]nx5[]nx0w02w5[]nx2105242

1.11. x[n] = 1+ − Period of first term in the RHS =1. Period of second term in the RHS = =7 （when m=2） Period of second term in the RHS = =5 (when m=1) Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35. 1.12. The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipping u[n] and then Shifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that M=-1 and no=-3. 1.13 y(t)= X[n] -3 -2 -1 0 1 2 3 n Figure S 1.12 = = Therefore 1 -1 0 1 2 t 0 x(t) g(t) 1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14. Therefore Figure S 1.14 -2 -1 -3 2 t 1 -3 ) This implies that A =3, t =0, A =-3, and t =1. 1.15 (a) The signal x [n], which is the input to S , is the same as y [n].Therefore , y [n]= x [n-2]+ x [n-3] = y [n-2]+ y [n-3] =2x [n-2] +4x [n-3] + ( 2x [n-3]+ 4x [n-4]) =2x [n-2]+ 5x [n-3] + 2x [n-4] The input-output relationship for S is y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4] 4 74jne25jne7/425/22tdtx)(dtt))2()2((2,022,12,0,ttt224dtEkkktkttg12(3)2(3)(1122221222121211112111111

(b) The input-output relationship does not change if the order in which S and S are connected series reversed. . We can easily prove this assuming that S follows S . In this case , the signal x [n], which is the input to S is the same as y [n]. Therefore y [n] =2x [n]+ 4x [n-1] = 2y [n]+4 y [n-1] =2( x [n-2]+ x [n-3] )+4(x [n-3]+ x [n-4]) =2 x [n-2]+5x [n-3]+ 2 x [n-4] The input-output relationship for S is once again 1.16 (a)The system is not memory less because y[n] depends on past values of x[n]. (b)The output of the system will be y[n]= =0 y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4] (c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form , k ґ. Therefore , the system is not invertible . 1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(- )=x(0). (b) Consider two arbitrary inputs x (t)and x (t). x (t) y (t)= x (sin(t)) x (t) y (t)= x (sin(t)) Let x (t) be a linear combination of x (t) and x (t).That is , x (t)=a x (t)+b x (t) Where a and b are arbitrary scalars .If x (t) is the input to the given system ,then the corresponding output y (t) is y (t)= x ( sin(t)) =a x (sin(t))+ x (sin(t)) =a y (t)+ by (t) Therefore , the system is linear. 1.18.(a) Consider two arbitrary inputs x [n]and x [n]. x [n] y [n] = x [n ] y [n] = Let x [n] be a linear combination of x [n] and x [n]. That is : x [n]= ax [n]+b x [n] where a and b are arbitrary scalars. If x [n] is the input to the given system, then the corresponding output y [n] is y [n]= = =a +b Therefore the system is linear. (b) Consider an arbitrary input x [n].Let = ay [n]+b y [n] 5 12121121112222122212222]2[][nn][kn12111222312312333312121211][001kxnnnnk22][002kxnnnnk312312333][003kxnnnnk])[][(2100kbxkaxnnnnk][001kxnnnnk][002kxnnnnk121

y [n] = be the corresponding output .Consider a second input x [n] obtained by shifting x [n] in time: The output corresponding to this input is x [n]= x [n-n ] y [n]= = = Also note that y [n- n ]= . Therefore , y [n]= y [n- n ] This implies that the system is time-invariant. (c) If

Also note that y [n- n ]= x [n-2- n ] Therefore , y [n]= y [n- n ] This implies that the system is time-invariant. (c) (i) Consider two arbitrary inputs x [n]and x [n]. x [n] y [n] = x [n+1]- x [n-1] x [n ] y [n] = x [n+1 ]- x [n -1] Let x [n] be a linear combination of x [n] and x [n]. That is : x [n]= ax [n]+b x [n] where a and b are arbitrary scalars. If x [n] is the input to the given system, then the corresponding output y [n] is y [n]= x [n+1]- x [n-1] =a(x [n+1]- x [n-1])+b(x [n +1]- x [n -1]) =a x [n+1]+b x [n +1]-a x [n-1]-b x [n -1] Therefore the system is linear. (ii) Consider an arbitrary input x [n].Let y [n]= x [n+1]- x [n-1] = ay [n]+b y [n] be the corresponding output .Consider a second input x [n] obtained by shifting x [n] in time: x [n]= x [n-n ] The output corresponding to this input is Also note that y [n-n ]= x [n+1- n ]- x [n-1- n ] y [n]= x [n +1]- x [n -1]= x [n+1- n ]- x [n-1- n ] Therefore , y [n]= y [n-n ] This implies that the system is time-invariant. (d) (i) Consider two arbitrary inputs x (t) and x (t). x (t) y (t)= x (t) y (t)= Let x (t) be a linear combination of x (t) and x (t).That is x (t)=a x (t)+b x (t) where a and b are arbitrary scalars. If x (t) is the input to the given system, then the corresponding output y (t) is y (t)= = =a +b = ay (t)+b y (t) Therefore the system is linear. (ii) Consider an arbitrary inputs x (t).Let y (t)= = be the corresponding output .Consider a second input x (t) obtained by shifting x (t) in time: The output corresponding to this input is y (t)= = x (t)= x (t-t ) Also note that y (t-t )= Therefore the system is not time-invariant. = y (t) 7 10120210121111222231231233333121211221211112121022210101010102101211d(t) x122(t) x2d312312333d(t) x3(t) xb+(t) ax21dd(t) x1(t) x2d1211d(t) x12)(x-(t) x11t212102(t) x2d2)(x-(t) x22t2)(x-)t-(t x0101tt102)(x-)t-(t x0101tt2

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