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Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates:1.2 converting from Cartesian to polar coordinates:, , , , , 1.3. =, =0, because , .Therefore, =,= =cos. Therefore, =,=, . Therefore, =0,because . =, =1. therefore, =,=. =. Therefore, =,=1.4. The signal xn is shifted by 3 to the right. The shifted signal will be zero for n7. The signal xn is shifted by 4 to the left. The shifted signal will be zero for n0. The signal xn is flipped signal will be zero for n2. The signal xn is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n4. The signal xn is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n0.1.5. x is obtained by flipping x and shifting the flipped signal by 1 to the right. Therefore, x will be zero for t-2. From , we know that x is zero for t-2. Similarly, x is zero for t-1,Therefore, x +x will be zero for t-2. x is obtained by linearly pression x by a factor of 3. Therefore, x will bezero for t1. x is obtained by linearly pression x by a factor of 3. Therefore, x will bezero for t9.1.6 x1 is not periodic because it is zero for t0. x2n=1 for all n. Therefore, it is periodic with a fundamental period of 1. x3n is as shown in the Figure S1.6.-3-141-10-4111-1n5x3nTherefore, it is periodic with a fundamental period of 4.1.7. =Therefore, is zero for 3. Since x1 is an odd signal, is zero for all values of t. Therefore, is zero when 3 and when . Therefore, is zero only when .1.8. 1.9. is a periodic plex exponential. is a plex exponential multiplied by a decaying exponential. Therefore, is not periodic. c is a periodic signal. =. is a plex exponential with a fundamental period of . is a periodic signal. The fundamental period is given by N=m = By choosing m=3. We obtain the fundamental period to be 10. is not periodic. is a plex exponential with =3/5. We cannot find any integer m such that m is also an integer. Therefore, is not periodic.1.10.x=2cos-sinPeriod 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 mon multiple of the periods of the first and second terms. This is equal to .1.11. xn = 1+Period of first term in the RHS =1.Period of second term in the RHS =7 when m=2Period of second term in the RHS =5 Therefore, the overall signal xn is periodic with a period which is the least monMultiple of the periods of the three terms inn xn.This is equal to 35.1.12. The signal xn is as shown in figure S1.12. xn can be obtained by flipping un and thenShifting the flipped signal by 3 to the right. Therefore, xn=u-n+3. This implies that M=-1 and no=-3.0-1-2-3123XnnFigure S 1.121.13y= =Therefore 1.14 The signal x and its derivative g are shown in Figure S1.14.10-1210-1t1-2g2-3-3tFigure S 1.14xTherefore This implies that A=3, t=0, A=-3, and t=1.1.15 The signal xn, which is the input to S, is the same as yn.Therefore ,yn= xn-2+ xn-3 = yn-2+yn-3 =2xn-2 +4xn-3 + =2xn-2+ 5xn-3 + 2xn-4The input-output relationship for S is yn=2xn-2+ 5x n-3 + 2x n-4 The input-output relationship does not change if the order in which Sand S are connected series reversed. . We can easily prove this assuming that S follows S. In this case , the signal xn, which is the input to S is the same as yn.Therefore yn =2xn+ 4xn-1=2yn+4 yn-1 =2+4 =2 xn-2+5xn-3+ 2 xn-4The input-output relationship for S is once again yn=2xn-2+ 5x n-3 + 2x n-41.16 The system is not memory less because yn depends on past values of xn.The output of the system will be yn=0From the result of part , we may conclude that the system output is always zero for inputs of the form , k . Therefore , the system is not invertible .1.17 The system is not causal because the output y at some time may depend on future values of x. For instance , y=x. Consider two arbitrary inputs xand x.x y= xsinx y= xsinLet x be a linear bination of x and x.That is , x=a x+b xWhere a and b are arbitrary scalars .If x is the input to the given system ,then the corresponding output y is y= x sin =a xsin+ xsin=a y+ byTherefore , the system is linear.1.18. Consider two arbitrary inputs xnand xn.xn yn=xn yn=Let xn be a linear bination of xn and xn. That is :xn=axn+b xnwhere a and b are arbitrary scalars. If xn is the input to the given system, then the corresponding output yn is yn= =a+b = ayn+b ynTherefore the system is linear. Consider an arbitrary input xn.Let yn=be the corresponding output .Consider a second input xn obtained by shifting xn in time:xn= xn-nThe output corresponding to this input is yn=Also note that yn- n=.Therefore , yn= yn- nThis implies that the system is time-invariant. If B, then ynB.Therefore ,CB.1.19 Consider two arbitrary inputs x and x. x y= tx x y= txLet x be a linear bination of x and x.That is x=a x+b xwhere a and b are arbitrary scalars. If x is the input to the given system, then the corresponding output y is y= tx = tax+b x= ay+b yTherefore , the system is linear. Consider an arbitrary inputs x.Let y= txbe the corresponding output .Consider a second input x obtained by shifting x in time:x= xThe output corresponding to this input is y= tx= txAlso note that y= x yTherefore the system is not time-invariant. Consider two arbitrary inputs xnand xn. xn yn= xn-2xn yn= xn-2.Let x be a linear bination of xnand xn.That is xn=axn+b xnwhere a and b are arbitrary scalars. If xn is the input to the given system, then the corresponding output yn is yn= xn-2= =axn-2+bxn-2+2ab xn-2 xn-2ayn+b ynTherefore the system is not linear. Consider an arbitrary input xn. Let yn = xn-2be the corresponding output .Consider a second input xn obtained by shifting xn in time:xn= xn- nThe output corresponding to this input is yn= xn-2.= xn-2- nAlso note that yn- n= xn-2- nTherefore ,yn= yn- nThis implies that the system is time-invariant. Consider two arbitrary inputs xnand xn.xn yn= xn+1- xn-1xn yn= xn+1- xn -1Let xn be a linear bination of xn and xn. That is :xn=axn+b xnwhere a and b are arbitrary scalars. If xn is the input to the given system, then the corresponding output yn is yn=xn+1- xn-1=a xn+1+b xn +1-a xn-1-b xn -1 =a+b = ayn+b ynTherefore the system is linear. Consider an arbitrary input xn.Let yn= xn+1- xn-1be the corresponding output .Consider a second input xn obtained by shifting xn in time: xn= xn-nThe output corresponding to this input is yn= xn +1- xn -1= xn+1- n- xn-1- nAlso note that yn-n= xn+1- n- xn-1- nTherefore , yn= yn-nThis implies that the system is time-invariant. Consider two arbitrary inputs x and x.x y= x y=Let x be a linear bination of x and x.That is x=a x+b xwhere a and b are arbitrary scalars. If x is the input to the given system, then the corresponding output y is y=a+b= ay+b yTherefore the system is linear. Consider an arbitrary inputs x.Lety= =be the corresponding output .Consider a second input x obtained by shifting x in time:x= xThe output corresponding to this input is y= =Also note that y= yTherefore the system is not time-invariant.1.20 Givenx= y=x= y=Since the system liner=1/2Therefore=cos=cos we know that=cos2=/2 Using the linearity property, we may once again write= = cosTherefore,=cos2 =cos1.21.The signals are sketched in figure S1.21.t-10-112213x10-112txt4-1321012x0.50.5t3/2-3/2t102xt1012618412Figure S1.211.22 The signals are sketched in figure S1.22a11/2-1/2-1n73210x3- n1.23 The even and odd parts are sketched in Figure S1.2311/2-1/2-1n73210xn-41-12n0x3n+1-111/2-1/2n210x3n2112n0xnun-3=xn-4-1-211/2n20x3- n/2 +nxn/2011n2Figure S1.22-1/2x01/2-1-2021t-201/22tx0t-1/2x01/2-1-2021t1-1/2x01/2-1-2021x0-t/20tt3t/2-3t/20x0Figure S1.237n10-7xon10-1/2-77-1/2nxn1/2n1/2-771n31/2071/2-1nxon103/2-3/2-1/24n1/2xon3/251-5nxenFigure S1.241.24 The even and odd parts are sketched in Figure S1.241.25 periodic period=2/=/2 periodic period=2/=2 x=1+cos/2. periodic period=2/=/2 x=cos/2. periodic period=2/=1/2 x=sinu-sinu/2. Not period.(f) Not period.1.26 periodic, period=7. Not period. periodic, period=8. xn=cos3n/4+cos. periodic, period=8. periodic, period=16.1.27 Linear, stable Not period. Linear Linear, causal, stable Time invariant, linear, causal, stable Linear, stable(g) Time invariant, linear, causal1.28 Linear, stable Time invariant, linear, causal, stableMemoryless, linear, causal Linear, stable Linear, stable Memoryless, linear, causal, stable Linear, stable1.29 Consider two inputs to the system such thatand Now consider a third input n=n+n. The corresponding system outputWill be therefore, we may conclude that the system is additiveLet us now assume that inputs to the system such thatandNow consider a third input x3 n=x2 n+ x1 n. The corresponding system outputWill betherefore, we may conclude that the system is additive Consider two inputs to the system such that and Now consider a third input t=t+t. The corresponding system outputWill betherefore, we may conclude that the system is not additiveNow consider a third input x4 t=a x1 t. The corresponding system outputWill beTherefore, the system is homogeneous. This system is not additive. Consider the fowlingexample .Let n=2n+2+2n+1+2n and n= n+1+ 2n+1+ 3n. The corresponding outputs evaluated at n=0 areNow consider a third input x3 n=x2 n+ x1 n.= 3n+2+4n+1+5nThe corresponding outputs evaluated at n=0 is y30=15/4. Gnarly,y30.This Therefore, the system is homogenous.1.30 Invertible. Inverse system y=xNon invertible. The signals x and x1=x+2give the same outputn and 2n give the same outputd Invertible. Inverse system; y=dx/dt Invertible. Inverse system y=x for n0 and yn=xn for n0 Non invertible. x and x give the same resultInvertible. Inverse system y=x(h) Invertible. Inverse system y=dx/dt(i) Invertible. Inverse system y = x-xn-1(j) Non invertible. If x is any constant, then y=0(k) n and 2n result in yn=0(l) Invertible. Inverse system: y=x(m) Non invertible x1 n=n+n-1and x2 n=n give yn=n(n) Invertible. Inverse system: yn=x2n1.31 Note that x2t=x1 t-x1 t-2. Therefore, using linearity we get y2 =y1 -y1 .this is shown in Figure S1.31Note that x3 =x1 t+x1 t+1. .Therefore, using linearity we get Y3 = y1 +y1 . this is shown in Figure S1.31y2 y3 02-1t-22024tFigure S1.311.32 All statements are true(1) x periodic with period T;y1 periodic, period T/2(2) y1 periodic, period T; bx periodic, period2T(3) x periodic, period T;y2 periodic, period2T(4) y2 periodic, period T; x periodic, period T/2;1.33 True xn=xn+N;y1 =y1 i.e. periodic with N0=n/2if N is even and with period N0=n if N is odd.False. y1 n periodic does no imply xn is periodic i.e. Let xn = gn+hn where Then y1 n = x 2n is periodic but xn is clearly not periodic.True. x n+N =xn; y2 n+N0 =y2 n where N0=2N True. y2 n+N =y2 n; y2 n+N0=y2 n where N0=N/21.34. ConsiderIf xn is odd, xn +x -n =0. Therefore, the given summation evaluates to zero. Let yn =x1nx2n .Theny -n =x1-n x2-n =-x1nx2n =-yn.This implies that yn is odd.Consider Using the result of part , we know that xenxon is an odd signal .Therefore, usingthe result of part we may conclude that Therefore,Consider Again, since xe xo is odd,Therefore,1.35. We want to find the smallest N0 such that m N0 =2k or N0 =kN/m, where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N0, then m/k should be the GCD of m and N. Therefore, N0=N/gcd.1.36If xn is periodic This implies that a rational number .T/T0 =p/q then xn =,The fundamental period is q/gcd and the fundmental frequency is p/gcd periods of x are needed .1.37. From the definition of We have Note from part that This implies that is even .Therefore,the odd part of is zero. Here,and1.38. We know that ThereforeThis implies that The plot are as shown in Figure s3.18.1.39 We have Also, ut11/2/2-/2t0tut12t0tut11/2-t0tut11/2-t0t11-1/2e-t/ut1-0tut11/2-01/2e-t/tFigure s3.18We have Therefore,1.40. If a system is additive ,then also, if a system is homogeneous,then y=x2 is such a systerm . No.For example,consider y with Then x=0for t1,but y=1 for t1.1.41. yn=2xn.Therefore, the system is time invariant. yn=xn.This is not time-invariant because yn- N02xn- N0. yn=xn1+n+1+n-1=2xn.Therefore, the system is time invariant .1.42. Consider two system S1 and S2 connected in series .Assume that if x1 and x2 are the inputs to S1.then y1 and y2 are the outputs.respectively .Also,assume that if y1 and y2 are the input to S2 ,then z1 and z2 are the outputs, respectively . Since S1 is linear ,we may write where a and b are constants. Since S2 is also linear ,we may write We may therefore conclude that Therefore ,the series bination of S1 and S2 is linear.Since S1 is time invariant, we may write and Therefore,Therefore, the series bination of S1 and S2 is time invariant. False, Let y=x+1 and z=y-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z=x which is a linear system. Let us name the output of system 1 as wn and the output of system 2 as zn .Then The overall system is linear and time-invariant.1.43. We have Since S is time-invariant.Now if x is periodic with period T. xt=x. Therefore, we may conclude that y=y.This implies that y is also periodic with T .A similar argument may be made in discrete time .1.44 Assumption : If x=0 for tt0 ,then y=0 for t t0.To prove That : The system is causal.Let us consider an arbitrary signal x1 .Let us consider another signal x2 which is the same as x1for t t0 , x2x1,Since the system is linear,Since for t t0 ,by our assumption =for t t0 .This implies that for t t0 . In other words, t he output is not affected by input values for . Therefore, the system is causal .Assumption: the system is causal . To prove that :If x=0 for t t0 .then y=0 for t t0 .Let us assume that the signal x=0 for t t0 .Then we may express x as ,Where for t t0 . the system is linear .the output to x will be .Now ,since the system is causal . for t t0 .implies that for t t0 .Therefore y=0 for t t0 . Consider y=xx .Now , x=0 for t t0 implies that y=0 for t t0 .Note that the system is nonlinear and non-causal . Consider y=x+1. the system is nonlinear and causal .This does not satisfy the condition of part. Assumption: the system is invertible. To prove that :yn=0 for all n only if xn=0 for all n .Consider .Since the system is linear :. Since the input has not changed in the two above equations ,we require that yn=2yn.This implies that yn=0. Since we have assumed that the system is invertible ,only one input could have led to this particular output .That input must be xn=0 .Assumption: yn=0 for all n if xn=0 for all n . To prove that : The system is invertible .Suppose that and Since the system is linear ,By the original assumption ,we must conclude that .That is ,any particular y1n can be produced that by only one distinct input x1n .Therefore , the system is invertible. yn=x2n.1.45. Consider ,and .Now, consider. The corresponding system output will be Therefore, S is linear .Now ,consider x4=x1.The corresponding system output will be Clearly, y4 y1.Therefore ,the system is not time-invariant.The system is definitely not causal because the output at any time depends on future values of the input signal x. The system will then be linear ,time invariant and non-causal.1.46. The plots are in Figure S1.46.1.47. The overall response of the system of Figure P1.47.=-the response of the system to x1n=Response of a linear system L
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