信号与系统奥本海姆第4版11章答案

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信号与系统课后答案第四章作业答案_第一次

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信号与系统奥本海姆英文版课后答案chapter

Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coordinates:55j=, 22j e π-=,233jj eπ--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j j eπ-=, 411j jeπ+=- 12e π-= 1.3. (a) E ∞=4014tdt e∞-=⎰, P ∞=0, because E ∞<∞ (b) (2)42()j t t x eπ+=, 2()1t x =.Therefore, E ∞=22()dt t x +∞-∞⎰=dt +∞-∞⎰=∞,P ∞=211limlim222()TTTTT T dt dt TTt x --→∞→∞==⎰⎰lim11T →∞=(c) 2()t x =cos(t). Therefore, E∞=23()dt t x +∞-∞⎰=2cos()dt t +∞-∞⎰=∞,P ∞=2111(2)1lim lim 2222cos()TTTT T T COS t dt dt T Tt --→∞→∞+==⎰⎰(d)1[][]12nn u n x =⎛⎫⎪⎝⎭,2[]11[]4nu n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞=24131[]4nn n x +∞∞-∞===⎛⎫∑∑⎪⎝⎭P ∞=0，because E ∞<∞. (e) 2[]n x =()28n j e ππ-+, 22[]n x =1. therefore, E ∞=22[]n x +∞-∞∑=∞,P ∞=211limlim1122121[]NNN N n Nn NN N n x →∞→∞=-=-==++∑∑.(f) 3[]n x =cos 4nπ⎛⎫ ⎪⎝⎭. Therefore, E ∞=23[]n x +∞-∞∑=2cos()4n π+∞-∞∑=2cos()4n π+∞-∞∑,P ∞=1limcos 214nNN n NN π→∞=-=+⎛⎫∑ ⎪⎝⎭1cos()112lim ()2122NN n Nn N π→∞=-+=+∑ 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. This new signal will be zero for n<-6 and n>0.1.5. (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, Therefore, x (1-t) +x(2-t) will be zero for t>-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of3. Therefore, x(3t) will be zero for t<1. (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be zero for t<9.1.6 (a) x 1(t ) is not periodic because it is zero for t<0. (b) x 2[n ]=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x 3[n1.7. (a)v ε[4])n --Therefore,()1[]vn xεis zero for 1[]n x >3. (b) Since x 1(t ) is an odd signal, ()2[]vn x εis zero for all values of t.(c) (){}11311[][][][3][3]221122v nnn n n u n u n x x x ε-⎡⎤⎢⎥=+-=----⎢⎥⎢⎥⎣⎦⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭Therefore, ()3[]vn x εis zero whenn <3 and when n →∞.(d)()1554411()(()())(2)(2)22vttt t t u t u t x x x ee ε-⎡⎤=+-=---+⎣⎦Therefore,()4()vt x εis zero only whent →∞.1.8. (a) ()01{()}22cos(0)tt t x eπℜ=-=+l (b) ()02{()}cos()cos(32)cos(3)cos(30)4t t t t t x e ππℜ+==+l(c) ()3{()}sin(3)sin(32t t t t t x e e ππ--ℜ=+=+l(d) ()224{()}sin(100)sin(100)cos(1002t t t t t t t x e e e ππ---ℜ=-=+=+l1.9. (a) 1()t xis a periodic complex exponential.(b) 2()t x is a complex exponential multiplied by a decaying exponential. Therefore,2()t x is not periodic.（c ）3[]n x is a periodic signal. 3[]n x =7j n e π=j n e π.3[]n x is a complex exponential with a fundamental period of22ππ=. (d) 4[]n x is a periodic signal. The fundamental period is given by N=m(23/5ππ)=10().3m By choosing m=3. We obtain the fundamental period to be 10.(e) 5[]n x is not periodic. 5[]n x is a complex exponential with 0w =3/5. We cannot find any integer msuch that m(02wπ ) is also an integer. Therefore, 5[]n xis not periodic.1.10. x (t )=2cos(10t ＋1)-sin(4t-1)Period of first term in the RHS =2105ππ=.Period of first term in the RHS =242ππ= .Therefore, the overall signal is periodic with a period which the least commonmultiple of the periods of the first and second terms. This is equal to π . 1.11.x[n] = 1+74j n e π?25j n e πPeriod of first term in the RHS =1. Period of second term in the RHS =⎪⎭⎫ ⎝⎛7/42π=7 （when m=2）Period of second term in the RHS =⎪⎭⎫ ⎝⎛5/22ππ=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 thenShifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that 1.13y (t)= ⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224dt E=2x 1[n-2]+ 5x 1[n-3] + 2x 1[n-4] The input-output relationship for S isy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](b) The input-output relationship does not change if the order in which S 1and S 2 are connected series reversed. . We can easily prove this assuming that S 1 follows S 2. In this case , the signal x 1[n], which is the input to S 1 is the same as y 2[n].Therefore y 1[n] =2x 1[n]+ 4x 1[n-1]= 2y 2[n]+4 y 2[n-1]=2( x 2[n-2]+21 x 2[n-3] )+4(x 2[n-3]+21x 2[n-4]) =2 x 2[n-2]+5x 2[n-3]+ 2 x 2[n-4]The input-output relationship for S is once againy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]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]= ]2[][-n n δδ=0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ][k n -δ, k ∈ ?. 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(-π)=x(0).(b) Consider two arbitrary inputs x 1(t)and x 2(t).x 1(t) →y 1(t)= x 1(sin(t)) x 2(t) → y 2(t)= x 2(sin(t))Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is , x 3(t)=a x 1(t)+b x 2(t)Where a and b are arbitrary scalars .If x 3(t) is the input to the given system ,then the corresponding output y 3(t) is y 3(t)= x 3( sin(t))=a x 1(sin(t))+ x 2(sin(t))=a y 1(t)+ by 2(t)Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] → y 1[n] =][001k x n n n n k ∑+-=x 2[n ] → y 2[n] =][02k x n n n n k ∑+-=Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 1[n]+b x 2[n]where a and b are arbitrary scalars. If x 3[n] is the input to the given system, then the corresponding outputy 3[n] is y 3[n]=][03k x n n n n k ∑+-==])[][(2100k bx k ax n n n n k +∑+-==a ][001k x n n n n k ∑+-=+b ][02k x n n n n k ∑+-== ay 1[n]+b y 2[n]Therefore the system is linear.(b) Consider an arbitrary input x 1[n].Lety 1[n] =][01k x n n n n k ∑+-=be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n-n 1]The output corresponding to this input isy 2[n]=][02k x n n n n k ∑+-== ]n [1100-∑+-=k x n n n n k = ][01011k x n n n n n n k ∑+---=Also note that y 1[n- n 1]=][01011k x n n n n n n k ∑+---=.Therefore , y 2[n]= y 1[n- n 1] This implies that the system is time-invariant.(c) If ][n x <B, then y[n]≤(2 n 0+1)B. Therefore ,C ≤(2 n 0+1)B.1.19 (a) (i) Consider two arbitrary inputs x 1(t) and x 2(t). x 1(t) → y 1(t)= t 2x 1(t-1)x 2(t) → y 2(t)= t 2x 2(t-1)Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= t 2x 3 (t-1)= t 2(ax 1(t-1)+b x 2(t-1))= ay 1(t)+b y 2(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x 1(t).Let y 1(t)= t 2x 1(t-1)be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input is y2(t)= t2x2(t-1)= t2x1(t- 1- t)Also note that y1(t-t)= (t-t)2x1(t- 1- t)≠y2(t)Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x1[n]and x2[n]. x1[n] →y1[n] = x12[n-2]x2[n ] →y2[n] = x22[n-2].Let x3(t) be a linear combination of x1[n]and x2[n].That is x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then the corresponding outputy3[n] is y3[n] = x32[n-2]=(a x1[n-2] +b x2[n-2])2=a2x12[n-2]+b2x22[n-2]+2ab x1[n-2] x2[n-2] ≠ay1[n]+b y2[n]Therefore the system is not linear.(ii) Consider an arbitrary input x1[n]. Let y1[n] = x12[n-2]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time:x 2[n]= x1[n- n]The output corresponding to this input isy 2[n] = x22[n-2].= x12[n-2- n]Also note that y1[n- n]= x12[n-2- n]Therefore , y2[n]= y1[n- n]This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x1[n]and x2[n].x 1[n] →y1[n] = x1[n+1]- x1[n-1]x2[n ]→y2[n] = x2[n+1 ]- x2[n -1]Let x3[n] be a linear combination of x1[n] and x2[n]. That is :x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then thecorresponding output y3[n] is y3[n]= x3[n+1]- x3[n-1]=a x1[n+1]+b x2[n +1]-a x1[n-1]-b x2[n -1]=a(x1[n+1]- x1[n-1])+b(x2[n +1]- x2[n -1])= ay1[n]+b y2[n]Therefore the system is linear.(ii) Consider an arbitrary input x1[n].Let y1[n]= x1[n+1]- x1[n-1]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time: x2[n]=x 1[n-n]The output corresponding to this input isy 2[n]= x2[n +1]- x2[n -1]= x1[n+1- n]- x1[n-1- n]Also note that y1[n-n]= x1[n+1- n]- x1[n-1- n]Therefore , y2[n]= y1[n-n]This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x1(t) and x2(t).x 1(t) →y1(t)= dO{}(t)x1x 2(t) →y2(t)= {}(t)x2dOLet x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= d O {}(t) x 3={}(t) x b +(t) ax 21d O=a d O {}(t) x 1+b {}(t) x 2d O = ay 1(t)+b y 2(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x 1(t).Lety 1(t)= d O {}(t) x 1=2)(x -(t) x 11t -be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input isy 2(t)= {}(t) x 2d O =2)(x -(t) x 22t -=2)(x -)t -(t x 0101t t --Also note that y 1(t-t 0)= 2)(x -)t -(t x 0101t t --≠ y 2(t)Therefore the system is not time-invariant. 1.20 (a) Givenx )(t =jt e 2 y(t)=tj e 3x )(t =jte2- y(t)=tj e3-Since the system liner+=tj et x 21(2/1)(jt e 2-) )(1t y =1/2(tj e3+tj e3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jte 2+jejt e 2-)/2Using the linearity property, we may once again write x 1(t)=21( j e -jt e 2+j e jt e 2-))(1t y =(je-jt e 3+je jt e 3-)= cos(3t-1)Therefore,x 1(t)=cos(2(t-1/2)))(1t y =cos(3t-1)1.21.The signals are sketched in figure S1.21.(f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) x[n]=(1/2)[cos(3πn/4+cos(πn/4)). periodic, period=8. (e) periodic, period=16. 1.27 (a) Linear, stable(b) Not period. (c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable (f) Linear, stable(g) Time invariant, linear, causal 1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable(f) Memoryless, linear, causal, stable (g) Linear, stable1.29 (a) Consider two inputs to the system such that[][][]{}111.S e x n y n x n −−→=ℜand [][][]{}221.Se x n y n x n −−→=ℜNow consider a third inputx 3[n]= x2[n]+x 1[n]. The corresponding system outputWill be [][]{}[][]{}[]{}[]{}[][]33121212e e e e y n x n x n x n x n x n y n y n ==+=+=+ℜℜℜℜtherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that andNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n]. The corresponding system output Will betherefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that()()()()211111Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦and ()()()()222211S dx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦Now consider a third inputx 3[t]= x2[t]+x 1[t]. The corresponding system outputWill betherefore, we may conclude that the system is not additiveNow consider a third input x 4 [t]= a x 1 [t]. The corresponding system output Will beTherefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let δ[n]=2δ[n+2]+ 2δ[n+1]+2δ[n] andx2[n]=δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluated at n=0 areNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n].= 3δ[n+2]+4δ[n+1]+5δ[n]The corresponding outputs evaluated at n=0 is y 3[0]=15/4. Gnarly, y 3[0]≠]0[][21y y n +.ThisTherefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x 1(t)=x(t)+2πgive the same output (c) δ[n] and 2δ[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n ≥0 and y[n]=x[n] for n<0 (f) Non invertible. x (n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0 (k) δ[n] and 2δ[n] result in y[n]=0 (l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x 1 [n]= δ[n]+ δ[n-1]and x 2 [n]= δ[n] give y[n]= δ[n] (n) Invertible. Inverse system: y[n]=x[2n]1.31 (a) Note that x 2[t]= x 1 [t]- x 1 [t-2]. Therefore, using linearity we get y 2 (t)= y 1 (t)- y 1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is2220201.34. (a) ConsiderIf x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x 1[n]x 2[n] .Theny [-n] =x 1[-n] x 2[-n] =-x 1[n]x 2[n] =-y[n]. This implies that y[n] is odd.(c)ConsiderUsing the result of part (b), we know that x e [n]x o [n] is an odd signal .Therefore, using {}1[][0][][]n n x n x x n x n ∞∞=-∞==++-∑∑22[][]e o n n n n x x ∞∞=-∞=-∞=+∑∑222[][][]e on n n n n n x x x ∞∞∞=-∞=-∞=-∞==+∑∑∑the result of part (a) we may conclude thatTherefore,(d)ConsiderAgain, since x e (t) x o (t) is odd,Therefore,1.35. We want to find the smallest N 0 such that m(2π /N) N 0 =2πk or N 0 =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 N 0, then m/k should be the GCD of m and N. Therefore, N 0=N/gcd(m,N).1.36．(a)If x[n] is periodic 0(),0..2/j n N T o ewhere T ωωπ+= This implies that 022o T kNT k T T Nππ=⇒==a rational number . (b)T/T 0 =p/q then x[n] =2(/)j n p q e π,The fundamental period is q/gcd(p,q) and the fundmental frequencyis(c) p/gcd(p,q) periods of x(t) are needed .1.37.(a) From the definition of ().xy t φWe have (b) Note from part(a) that()().xx xx t t φφ=-This implies that ()xy t φis even .Therefore,the odd part of ().xx t φis zero.(c) Here, ()().xy xx t t T φφ=-and ()().yy xx t t φφ= 1.38.(a) We know that /22(2)().t t δδ=V V Therefore This implies that(b)The plot are as shown in Figure s3.18.1.39 We have Also,2[][]0eon n n x x ∞=-∞=∑222[][][].e on n n n n n xx x ∞∞∞=-∞=-∞=-∞==+∑∑∑2220()()()2()().eoet dt t dt t dt t t dt x x x x x ∞∞∞∞-∞-∞-∞-∞=++⎰⎰⎰⎰0()()0.et t dt x x ∞-∞=⎰222()()().e ot dt t dt t dt xx x ∞∞∞-∞-∞-∞=+⎰⎰⎰0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===/21lim (2)lim ().2t t δδ→∞→∞=V V V V(b) y(t)=x 2(t) is such a systerm . (c) No.For example,consider y(t) ()()ty t x d ττ-∞=⎰with ()()(1).x t u t u t =--Then x(t)=0for t>1,but y(t)=1 for t>1.1.41. (a) y[n]=2x[n].Therefore, the system is time invariant.(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N 0]≠(2n-1)2x [n- N 0]. (c) y[n]=x[n]{1+(-1)n +1+(-1)n-1}=2x[n].Therefore, the system is time invariant .1.42.(a) Consider two system S 1 and S 2 connected in series .Assume that if x 1(t) and x 2(t) arethe inputs to S 1..then y 1(t) and y 2(t) are the outputs.respectively .Also,assume thatif y 1(t) and y 2(t) are the input to S 2 ,then z 1(t) and z 2(t) are the outputs, respectively . Since S 1 is linear ,we may writewhere a and b are constants. Since S 2 is also linear ,we may write We may therefore conclude thatTherefore ,the series combination of S 1 and S 2 is linear. Since S 1 is time invariant, we may write andTherefore,Therefore, the series combination of S 1 and S 2 is time invariant.(b) False, Let y(t)=x(t)+1 and z(t)=y(t)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then The overall system is linear and time-invariant. 1.43. (a) We haveSince S is time-invariant.Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b)1.44 (a) Assumption : If x(t)=0 for t<t 0 ,then y(t)=0 for t< t 0.To prove That : The system is causal.Let us consider an arbitrary signal x 1(t) .Let us consider another signal x 2(t) which is the same as x 1(t)fort< t 0. But for t> t 0 , x 2(t) ≠x 1(t),Since the system is linear, Since ()()120x t x t -=for t< t 0 ,by our assumption =()()120y t y t -=for t< t 0 .This implies that()()12y t y t =for t< t 0 . In other words, t he output is not affected by input values for 0t t ≥. Therefore, thesystem is causal .Assumption: the system is causal . To prove that :If x(t)=0 for t< t 0 .then y(t)=0 for t< t 0 .Let us assume that the signal x(t)=0 for t< t 0 .Then we may express x(t) as ()()12()x t x t x t =-, Where()()12x t x t = for t< t 0 . the system is linear .the output to x(t) will be()()12()y t y t y t =-.Now ,since the system is causal . ()()12y t y t = for t< t 0 .implies that ()()12y t y t = for t< t 0 .Therefore y(t)=0 for t< t 0 .(b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t 0 implies that y(t)=0 for t< t 0 .Note that the system is nonlinear and non-causal .(c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a). (d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider[]0[]x n y n =→. Since the system is linear :2[]02[]x n y n =→.Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 .Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that andSince the system is linear ,By the original assumption ,we must conclude that 12[][]x n x n =.That is ,any particular y 1[n] can be produced that by only one distinct input x 1[n] .Therefore , the system isinvertible.(e) y[n]=x 2[n].1.45. (a) Consider ,and()222()()s hx x t y t t φ→=.Now, consider ()()()312x t ax t bx t =+. The corresponding system output will be Therefore, S is linear .Now ,consider x 4(t)=x 1(t-T).The corresponding system output will beClearly, y 4(t)≠ y 1(t-T).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(t).(b) The system will then be linear ,time invariant and non-causal.1.46. The plots are in Figure S1.46.1.47.(a) The overall response of the system of Figure P1.47.(a)=(the response of the system to x[n]+x 1[n])-the response of the system to x 1[n]=(Response of a linear system L to x[n]+x 1[n]+ zero input response of S)- (Response of a linear system L to x 1[n]+zero input response of S)=( (Response of a linear system L to x[n]).。

奥本海姆版信号与系统课后答案

第七章7.6 解：见 8.17.8 解： (a) )]()([)21()(50πωδπωδπωk k j j X n k +--=∑= 信号截止频率 πω5=m采样频率 m s T ωπππω2102.022====对于正弦信号，会发生混叠 (b) ππω5==T c所以输出信号 )sin()21()(40t k t y k k π∑== 所以j e e t g tjk t jk k k 2)21()(40ππ-=-=∑ ∑-==44k t jk k e a π其中，⎪⎪⎩⎪⎪⎨⎧≤≤-=≤≤-=+-+14)21(0041)21(11k j k k j a k k k 7.10 解：(a) 错信号时域为矩形波，频域为sinc 函数，无论怎么样都会混叠 (b) 符合采样定理，对(c) 符合采样定理，对7.15 解：要求 76N 2,76273ππππω>=⨯>即s 237max =<∴N N 取 7.16 解：易见ππn n 2sin2满足性质1, 3对性质2，考虑时域乘积得频域卷积，易见2))2/sin((4][n n n x ππ=7.19 解：设x[n]经零值插入后得输出为z[n] (a) 531πω≤时， ⎪⎩⎪⎨⎧><=1101)(ωωωωωj e X ⎪⎪⎩⎪⎪⎨⎧>≤<=30531)(11ωωπωωωj e Z所以 ⎪⎪⎩⎪⎪⎨⎧><=3031)(11ωωωωωj e W因此可得，n n n w πω/)3(sin ][1=又由 ]5[][n w n y =可得 )5/()35(sin][1n n n y πω= (b) 531πω>时 ⎪⎪⎩⎪⎪⎨⎧>><=53031)(11πωωωωωj e Z)/()5(sin ][n nn w ππ=∴][51)5/()(sin ][n n n n y δππ== 7.21 解：采样频率m s Tωππω2200002>== 即πω10000<m 时，可以恢复 (a) 可以(b) 不可以(c) 不能确定(d) 可以 (e) 不可以 (f) 可以 (g) 可以7.22 解：)(*)()(21t x t x t y = 则有πωωωω10000)()()(21>==j X j X j Y πω1000=∴m 因而 πωω20002=>m s故 s T s 3102-=<ωπ 7.23 解：见 8.27.24 解：见 8.37.29 解：见 8.107.31 解：见 8.157.35 解：见 8.247.38 解：见 8.97.41 解：见 8.197.45 解：见 8.17。

奥本海姆信号与系统习题解答Chapter 4

Chapter 44.1 Solution: (a). ⎰∞∞--=dt e t x jw X jwt )()(⎰∞∞-----=dt e t u e jwt t )1()1(2 ⎰∞∞-----=dt e t u e jwt t )1()1(2⎰⎰∞--∞---==12)2(1)1(2dt e e dt e e t jw jwt tjwe jw e e jw e e jw jw t jw +=---=--=---∞--222)2(21)2(24.2 Solution: (a). ⎰∞∞--=dt e t x jw X jwt )()(⎰∞∞++=-j w t -1)]e -(t 1)(t [dt δδ⎰⎰∞∞∞∞-++=-j w t-j w t-1)e -(t 1)e (tdt dt δδw e e jw jw cos 2=+=-+(b). ⎰∞∞--=dt e t x jw X jwt )()(dt e t u t u dtdjwt -∞∞--+--=⎰)}2()2({ dt e t t jwt -∞∞--+---=⎰)}2()2({δδ dt e t dt e t jwt jwt -∞∞--∞∞--+---=⎰⎰)2()2(δδw j e ew j wj 2s i n 222-=+-=-+4.3 Solution: (a). )42sin(ππ+t =je et j t j 2)42()42(ππππ+-+-=je e ee tj jtj j22424ππππ---t j e π2 −→←FT)2(2ππδ-wt j e π2- −→←FT)2(2ππδ+w∴ )42sin(ππ+t −→←FT)}2()2({44πδπδπππ+---w ew e jjj4.4 Solution: (a). dw e jw X t x jwt)(21)(⎰∞∞-=π dw e w w w jwt )}4()4()(2{21ππδππδπδπ++-+=⎰∞∞- })4()4()(2{21dw e w dw e w dw e w jwt jwtjwt ππδππδπδπ++-+=⎰⎰⎰∞∞-∞∞-∞∞-)4cos(1)}4cos(22{21}2{2144t t e e t j t j ππππππππ+=+=++=- 4.5 Solution:dw e e jw X dw e jw X t x jwt jw X j jwt )()(21)(21)(<∞∞-∞∞-⎰⎰==ππdw e ew u w u jwt w j )23()}3()3({221ππ+-∞∞---+=⎰dw ee dw ee wt j j jwtw j )23(33)23(331--+--⎰⎰==ππππ)23(1)23(1)23(3)23(333)23(--⋅-=-⋅-=-----t j e e t j e t j t j wt j ππ)23()23(3sin 2)23()23(3sin 21---=--⋅-=t t j t j t j ππ If0)(=t x ,then ⎪⎪⎩⎪⎪⎨⎧≠-±±==-0)23(,...2,1,0,.........)23(3t k k t πThat is 0., (2)33≠+=k k t π 4.6 Solution:Accorrding to the properties of the Fourier transform, we ’ll get: (a). )(t x −→←FT)(jw X∴)1(t x - −→←FT jwe jw X --)()1(t x -- −→←FT jwe jw X +-)(then)1()1()(1t x t x t x --+-=−→←FTw jw X e jw X e jw X jw X jw jw cos )(2)()()(1-=-+-=--(b). )(t x −→←FT)(jw X)(b at x + −→←FTwa bj e aw j X a )(1∴)63()(2-=t x t x −→←FTw j e wj X jw X 22)3(31)(-= (c).)(t x −→←FT)(jw X )1(-t x −→←FTjw e jw X -)()(22t x dtd −→←FT)()(2jw X jw∴)1()(223-=t x dtd t x −→←FTjw e jw X w jw X --=)()(234.7. (a). neither,neither Solution: )2()()(1--=w u w u jw X∴ )()(*11jw X jw X -≠,that is )()(*t x t x ≠so )(t x is not real.)()(*11jw X jw X -≠-,that is )()(*t x t x -≠so )(t x is not imaginary.And )()(11jw X jw X -≠,that is )()(t x t x -≠so )(t x is not even.)()(11jw X jw X --≠,that is )()(t x t x --≠so )(t x is not odd.4.10. (a). s olution:suppose t t t f πsin )(= −→←FT⎪⎩⎪⎨⎧><=1,.......01,........1)(w w jw Fthen 21)]([)(t f t f =−→←FT)()(21jw F jw F *π=)(1jw F )()(21)(1jw F jw F jw F '*='π)]1()1([)(21--+*=w w jw F δδπ)]}1([)]1([{21--+=w j F w j F π⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤-<≤-=others w w ..........,.........020...,.........2102...,.........21ππ∴)(1jw F ⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤--<≤-+=others w w w w ..........,.........020..., (22)02...,.........22ππ∴ )()(1t tf t x = −→←FT)()(1jw F dw d j jw X =⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤-<≤-=others w j w j..........,.........020...,.........202...,.........2ππ (b). Solution:dw jw X dt t t t A 222)(21sin ⎰⎰∞∞-∞∞-=⎪⎭⎫⎝⎛=ππ32222214421221πππππ===⎰-dw j 4.11 A=1/3, B=3 Solution: )()()(t h t x t y *= −→←FT)()()(jw H jw X jw Y =and )3()3()(t h t x t g *=)3(t x −→←FT)3(31w j X )3(t h −→←FT)3(31w j H then )3(31)3(31)(w j H w j X jw G ⋅=)3(91wj Y =−→←FT)3(31)(t y t g =∴3,31==B A4.13. Solution:(a). According to the property of FT, )(t tx −→←FT)(jw X dwdj te- −→←FT212w+∴tte- −→←FT()()222221412212w wj w w j w dw d j+-=+⋅-=⎪⎭⎫ ⎝⎛+(b). According to the duality of FT, If)(t x −→←FT)(jw X then )(jt X −→←FT)(2w f -πandtte- −→←FT()2214w wj +-tjte- −→←FT()2214w w+∴()2214t t+ −→←FT wwwej ew j ----=-ππ2)(24.14 Solution: From (1), we know )(t x is real and 0)(≥t x ;From (2), we know : )(2t u Ae t- −→←FT)()1(jw X jw +And we also know )(2t u Ae t - −→←FT2+jw ASo )()1(jw X jw +=2+jw AThat is21)2)(1()(+-++=++=jw Ajw A jw jw A jw X−→←FT )()()(2t u Ae t u Ae t x t t ---=From (3), we know: π2)(2=⎰∞∞-dw jw XBut dt e e A dt t x dw jw X t t 22222)(2)(2)(--∞∞∞-∞∞--==⎰⎰⎰ππdt e e e At t t )2(243202---∞+-=⎰π∞----+---=04322)4322(2tt t e e e A π22122)413221(2A A ππ=+-= So 2122A π=π2, that is 122=A, 12±=AWhile 0)(≥t x ,12=AThen)()()(2t u Ae t u Ae t x t t ---=)()(122t u e e t t ---=4.15 Solution:From (3), we know:tet - −→←FT)}(Re{jw X)}(Re{jw X 2)()(jw X jw X *+=−→←FT2)()(t x t x -+*From (1)、(2), we know: )(t x is real and )()()(t u t x t x =∴2)()(t x t x -+*=2)()(t x t x -+=te t -∴ )()]()([t u t x t x -+*=)()]()([t u t x t x -+===)()()(t x t u t x )(2t u e t t-then=)(t x )(2t u te t -4.16. Solution:(a).)(sin )4()4()4sin()(t g t t k t k kt x k ⎪⎭⎫ ⎝⎛=-=∑∞-∞=ππδππ∴ )4()(πδπk t t g k -=∑∞-∞=(b).)4(sin )(sin )(πδπππk t t t t g t t t x k -⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=∑∞-∞=⎪⎭⎫ ⎝⎛=t t t f πsin )(−→←FT⎩⎨⎧≤=others w jw F ...,.........01....,.........1)()4()(πδπk t t g k -=∑∞-∞=−→←FT)8(8)4/2(4/2)(k w k w jw G k k -=-=∑∑∞-∞=∞-∞=δπππδπππ∴ ))8((4)()(21)(k w j F jw G jw F jw X k -=*=∑∞-∞=π∴ In one period, ⎪⎩⎪⎨⎧≤<≤=41.,.........01.,.........4)(w w jw X 4.21. (g).Solution: ⎰∞∞--=dt e t x jw X jwt )()(⎰⎰⎰------++-=211112dt e dt tedt ejwt jwtjwt=21111112}{1jw ejw e te jwjwe jwt jwt jwt jwt -+-------------222)(jw e e jw e e e e e e jwjw jw w j jw jw w j jw ----++++-=----222)(jw e e jw e e jwjw w j w j ---+=--jw wjww j w w j jw w 2cos 2sin 2sin 22cos 222-=----=4.22 Solution:(a). According to the duality of FT: )(jt X −→←FT)(2w x -π, we will gett Wt πsin −→←FT⎪⎩⎪⎨⎧>≤=Ww W w jw X ,...0,...1)(∴when )2()]2(3sin[2)(ππ--=w w jw X , ⎪⎩⎪⎨⎧>≤=3.......,.........03.,.........)(2t t e t x tj π(b). 22)34cos()(4343)34()34(wj jwj jw j w j e e ee e ew jw X --+-+-=-=+=πππππ∴2)4()4()(33--+=-t e t e t x jjδδππ(d). )]2()2([3)]1()1([2)(πδπδδδ++-++--=w w w w jw X∴ t t j e e e e t x t j t j jt jt πππππππ2cos 3sin 2)(23)(1)(22+=++-=-- 4.24. (a). Solution:for )(t x is real, we have:(1). 0)}(Re{=jw X02)()()}(Re{=+=*jw X jw X jw X−→←FT02)()(2)()()}({=-+=-+=*t x t x t x t x t x Ev ∴To satisfy 0)}(Re{=jw X , )()(t x t x --=, )(t x is real and odd.∴ figure (a), (d) satisfy the condition.(2). 0)}(Im{=jw X02)()()}(Im{=-=*jw X jw X jw X−→←FT02)()(2)()()}({=--=--=*t x t x t x t x t x Ev ∴To satisfy 0)}(Im{=jw X , )()(t x t x -=, )(t x is real and even.∴ figure (e), (f) satisfy the condition.(3). There exists a real αsuch that )(jw X e w j α is real)(jw X e w j α is real∴ 0)}(Im{=jw X e w j α, and )(jw X e w j α −→←FT)(α+t x ∴ To satisfy 0)}(Im{=jw X e w j α, )(α+t x is real and even. ∴ figure(a) and (b) )0(=α, (e) and (f) )0(≠αsatisfy the condition.(4). 0)(=⎰∞∞-dw jw X)0(x =0)(=⎰∞∞-dw jw X∴ If 0)(=t x when 0=t , then it can satisfy 0)(=⎰∞∞-dw jw X∴ figure(a) , (c), (d) and (f) satisfy the condition.Note: (b) is not satisfies the condition.)1(2-t δ −→←FT2 (5). 0)(=⎰∞∞-dw jw wX)(t x dtd −→←FT)(jw jwX∴ To satisfy0)(=⎰∞∞-dw jw wX , it only needs0)(0==t t x dt d)1(2-t δ −→←FT2, 02=⎰∞∞-wdw∴ figure (b), (c), (e), (f) satisfy this condition.(6). )(jw X is periodicIf )(jw X is periodic, )(t x is discrete in time domain.So only (b) satisfies the condition.4.25. Solution:(a). Suppose )(1t x =)1(+t x , then we ’ll find )(1t x is real and even. So )(1jw X is real and even, which means 0)(1=<jw X .)(1jw X =jw e jw X )(∴ )(jw X =jw e jw X -)(1∴ ⎩⎨⎧≤-≥-=<0)(,......0)(,.........)(11jw X w jw X w jw X π (b). dt t x X j X )()0()0(⎰∞∞-== is the integral area of )(t xthen, 718122124)0(=-=**-*=X (c). πππ422)0(2)(=*==⎰∞∞-x dw jw X (e).dt t x dw jw X 22)(2)(⎰⎰∞∞-∞∞-=π⎪⎭⎫ ⎝⎛++-+=⎰⎰⎰⎰-dt dt t dt t dt 2322212102012)2(22ππ376=4.27. Solution: (1).)(t u −→←FTjww 1)(2+πδ∴ )3()2(2)1()(-+---=t u t u t u t x−→←FT)2(1)(2)(32wj w j jw e e e jw w jw X ---+-⎪⎪⎭⎫ ⎝⎛+=πδ)2(132w j w j jwe e e jw---+-=(2).dt e t x T dt et x T a t T jk Tt Tjk T k ππ22)(1)(~1--⎰⎰==)2(1)(12Tjk X T dte t x T t T jk ππ==-∞∞-⎰4.30. Solution: (a).⎩⎨⎧≤=others w jw G ., (02).,.........1)( ∴ tt t t t t g ππcos sin 22sin )(==t t x t g cos )()(=∴ ttt x πsin 2)(=4.32 Solution: (a).)()()(11t h t x t y *=)1())1(4sin()(--=t t t h π −→←FT ⎩⎨⎧≤=-others w e jw H jw .........,.........04....,.........)()26cos()(1π+=t t x −→←FT)}6()6({)(121-++=w w ejw X jδδππ∴ )()()(11jw H jw X jw Y =0=(b).)()()(22t h t x t y *=)1())1(4sin()(--=t t t h π −→←FT⎩⎨⎧≤=-others w e jw H jw .........,.........04....,.........)()3sin()21()(02kt t x k k ∑∞==−→←FT)}3()3({)21()(02k w k w j jw X k k --+=∑∞=δδπ∴ ⎪⎪⎩⎪⎪⎨⎧=--+=-==--others k w w e j k w w e j jw H jw X jw Y jw jw ........,.........01)},...3()3({210....)},.......()({)()()(22δδπδδπ⎪⎪⎩⎪⎪⎨⎧=--+==-o t h e r s k w e j w e j k j j ........................................................,.........01..),........3(21)3(210.......................................................,.........033δπδπ∴ only when 1k =, 2()Y j ω has output impulse .∴ )1(3sin 212214141)()1(3)1(333332-=--=-=-----t j e e e e j e e j t y t j t j t j t j 4.34 Solution: (a).)()()(jw H jw X jw Y =65)(4564)(22+++=+-+=jw jw jw jw w jw jw H∴ }4){(}65)){((2+=++jw jw X jw jw jw Y∴)(4)()(6)(5)(22t x t x dt dt y t y dt d t y dt d +=++ (b).65)(4564)(22+++=+-+=jw jw jw jw w jw jw H3122)3)(2(4+-++=+++=jw jw jw jw jw∴ )()(2)(32t u e t u e t h t t ---=(c). )()()(44t u te t u e t x t t ---=∴ 22)4(3)4(141)(++=+-+=jw jw jw jw jw X)3)(2(4564)(2+++=+-+=jw jw jw jw w jw jw H∴ )3)(2(4)4(3)()()(2+++++==jw jw jw jw jw jw H jw X jw Y)2(2/1)4(2/1)2)(4(1+++-=++=jw jw jw jw∴ ))()((21)(42t u e t u e t y t t---=4.35 Solution: (a).0........,.........)(>+-=a jwa jwa jw Hthe magnitude is: 1)(2222=++=+-=wa w a jwa jwa jw Hthe phase is: ⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=<a w arctg a w arctg a w arctg jw H 2)( the impulse response is:jwa jwjw a a jw a jw a jw H +-+=+-=)()()(2))(()()(t t u ae t u e dtd t u ae t h at atat δ-=-=--- (b). )3cos(cos )3/cos()(t t t t x ++= +-++=)3/1()3/1({)(w w jw X δδπ)}3()3()1()1(-+++-+++w w w w δδδδandarctgw j e jwjwjw H 211)(-=+-=so )()()(jw H jw X jw y =+-++=---)3/1()3/1({)3/1(2)3/1(2w e w e a r c t g j a r c t gj δδπ)1()1()1(2)1(2-+++---w e w e arctg j arctg j δδ)}3()3()3(2)3(2-+++---w e w e arctg j arctg j δδ+-++=-)3/1()3/1({3/3/w e w e j j δδπππ)1()1(2/2/-+++-w e w e j j δδππ)}3()3(3/23/2-+++-w e w e j j δδππThen )3/23cos()2/cos()3/3/cos()(πππ-+-+-=t t t t yWe can find that y(t) is the sum of a phase shift for each x(t)’s component. 4.36 Solution: (a). )(][)(3t u e e t x t t --+=∴ )3)(1()2(23111)(+++=+++=jw jw jw jw jw jw X )(]22[)(4t u e e t y t t ---=∴ )4)(1(64212)(++=+-++=jw jw jw jw jw Y∴ )2)(4()3(3)()()(+++==jw jw jw jw X jw Y jw H(b). According to (a)’s result,22/342/3)2)(4()3(3)(+++=+++=jw jw jw jw jw jw H∴ )()(23)(24t u e e t h t t--+=(c).86)(93)2)(4()3(3)(2+++=+++=jw jw jw jw jw jw jw H∴ )}93){(}86)){((2+=++jw jw X jw jw jw Y∴ )(9)(3)(8)(6)(22t x t x dt dt y t y dt d t y dtd +=++4.37 Solution:(a). Supposing )(t x −→←FT)(jw X ,and from the differentiation property of FT,we canget22()d x t dt −→←FT 2()()j X j ωω22()(1)2()(1)d x t t t t dt δδδ=+-+-∴ 2/2/222()()2()(2sin(/2))j j j j j X j e e ee j ωωωωωωω--=-+=-=Finally, 22sin(/2)()[]X j ωωω=(b).(c). It is obvious that another signal g(t) sketched as following figure can reduce:~()()(4)k x t g t t k δ∞=-∞=*-∑(d).(4)()22k k t k k ππδδω∞∞=-∞=-∞-↔-∑∑From the convolution property of FT,we can get: ~~()()(4)()()()222k k kx t g t t k X j G jk πππδωδω∞∞=-∞=-∞=*-↔=-∑∑~~()()(4)()()()222k k kx t x t t k X j X jk πππδωδω∞∞=-∞=-∞=*-↔=-∑∑ So, we can know that,although ()G j ω is different from ()X j ω,()()22kkG j X jππ= forall integers k .4.43 Solution:We can draw a system as figure (a) ,where ,sin ttπ −→←FT ()H j ω2tshown in figure(c).Extra problems:We have known )(t x −→←FT)(jw X , prooftt x dt d π1)(* −→←FT)(jw X w ⋅ Proof: )(t x −→←FT)(jw X∴)(t x dtd −→←FT)(jw jwX)(t sign −→←FTjw2According to the duality of FT, we will get t π1 −→←FT)()(w jsign w jsign -=- ∴ tt x dt d π1)(* −→←FT)]([)(w jsign jw jwX -⋅)()()(w sign w jw X w wsign ⋅==Then, we havett x dt d π1)(* −→←FT)(jw X w ⋅)(t s )(t y )(t x t)(a )(b。

奥本海姆信号与系统习题参考答案

.第三章作业解答3.1解：420ππω==T ， j a a 4*33-==- 则：t j t j t j t j k tjk ke a e a e a e a ea t x 00000333311)(ωωωωω----∞-∞=+++==∑-)243cos(84cos 443sin 84cos 4)](21[8)(2144422434344434344πππππππππππππ++=-=--⨯++⨯=-++=------t t tt e e je e jejeeet j t j t j t j t jt jt j t j3.3解：)35sin(4)32cos(2)(t t t x ππ++= 则3)32cos(1=→T t π 56)35s i n (2=→T t π故：6],[21==T T lcm T 320ππω==T )(214)(21235353232t j t j t j t j e e je e ππππ---⨯+++=则：20=a 2122==-a a 25j a -= 25j a =- 3.9x[n]波形如下图所示：0 1 4 5 n…- 4 -3则：N=4，220ππω==N ]84[41]}1[8][4{41][41][122302300πππωδδjk n jk n n jk n n jk N n k e e n n e n x e n x N a --=-=->=<+=-+===∑∑∑即：2112133210j a a j a a +=-=-==3.15解：6π=T ，1220==Tπω )(ωj H 如下图所示：则：⎩⎨⎧>≤=9||08||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k ktjk k k ea ea jk H t y 00880)()(ωωω∑∑-=∞-∞===而：)()(t y t x =，即：t jk k k tjk k k e a t y ea t x 0088)()(ωω∑∑-=∞-∞====故：当9||≥k 时，0=k a3.22解：（a ）2=T ，ππω==T20 ]|[12121)(11111110dt e te jk dt te dt e t x T a tjk t jk t jk T t jk k ⎰⎰⎰---------===πππωπkjk t jk t jk k j k j k k je k j e jk te k j )1(k ]02[21]|1|[211111-=⎪⎪⎩⎪⎪⎨⎧-=--=---=-----πππππππππ为奇数为偶数021110==⎰-dt t a(注意：与性质验证，由于x(t)是实奇函数，则a k 为纯虚的奇函数，满足： *k k k a a a -=-=- 且：00=a ） (d) 2=T ，ππω==T20 ])1(21[21]21[21)]1(2)([21)(1200k jk t jk T tjk k e dt e t t dt e t x T a --=-=--==---⎰⎰--ππωδδ21)]1(2)([21200-=--=⎰--dt t t a δδ3.28（b ）解：)(21)(21)2cos()32sin(][223232nj n j n jnje e eejn n n x ππππππ--++== )(416/76/6/6/7n j n j n j n j e e e e j ππππ----+=12/2.712/2.12/2.12/2..7(41ππππn j jn jn n j e e e e j----+=⎪⎪⎪⎩⎪⎪⎪⎨⎧++=-++==othersrN rN k j rN rN k j a k 05,11417,141 则：⎪⎩⎪⎨⎧++++==othersrN rN rN rN k a k 05,11,7,141||⎪⎪⎪⎩⎪⎪⎪⎨⎧++=++=-=∠othersrN rN k rN rN k a k 05,1127,12ππ 3.34解：(b)∑∞-∞=--=n nn t t x )()1()(δ其波形如下图所示：其周期T=2，基波频率为：ππω==T20 ⎩⎨⎧=--=-=--==---⎰⎰--是偶数是奇数k 01])1(1[21]1[21)]1()([21)(1200k e dt e t t dt e t x T a k jk t jk T tjk k ππωδδ而：⎪⎩⎪⎨⎧<>==--00)(44||4t et e et h t tt则：240401684141)()(s s s dte e dt e e dt e t h s H st t st t st -=++-=+==--∞-∞--∞∞-⎰⎰⎰故：2)(168)(ππjk jk H -=故：⎪⎩⎪⎨⎧-==∑∞-∞=为偶数为奇数（k k e jk ea jk H t y tjk tjk k k 0)168)()(200πωπω3.357π=T ，1420==Tπω 解：)(ωj H 如下图所示：则：⎩⎨⎧<>=17||017||1)(0k k jk H ωtjk k kea t x 0)(ω∑∞-∞==tjk k k tjk k k ea ea jk H t y 0018||0)()(ωωω∑∑∞=∞-∞===而：)()(t y t x =，即：tjk k ktjk k kea t y ea t x 0018||)()(ωω∑∑∞=∞-∞====故：当18||<k 时，0=k a3.44解：（1）*k k a a =- （2）6=T ，320ππω==T （3）⎩⎨⎧===其他，不为02||1||0k k a k（4）k jk k k a e b t x a t x π--=→--→)3()(k jk k a ea π--= 则：当为偶数k a k 0=结合（3）则：⎩⎨⎧==其他不为01||0k a k（5）帕斯瓦尔关系式：21||21||||12121=⇒=+-a a a （6）211=a 211=-a 则t e e ea e a t x t j t j t j tj 3cos )(21)(333131πππππ=+=+=--- 故：03,1===C B A π。

奥本海姆信号与系统第一章部分习题答案

(e)
x[n], n 1

y[n] 0,
n0
x[n 1], n 1

(e)
Байду номын сангаас[n], n 1

y[n] 0,
n0
x[n 1], n 1

(g )
y[n] x[4n 1]
+++
1.31 在本题中将要说明线性时不变性质的重要结果之一，即一旦知道了一个线性
∴ 1 = 3，1 = 0，2 = −3，2 =1（或-1）
1.19判定下列输入-输出关系的系统是否具有线性性质、时不变性质，或两者俱有。
线
性： 3 = 1 + 2
时不变性： 2 = 1 ( − 0 )
(a) = 2 ( − 1)
∴ 是线性的
∴ 不是时不变的
基波周期0 ：使[] = + 成立的最小正整数。

离散时间复指数信号的周期： 0 ， 0 = 是有理数，则是周期的，
2
2 0

且和无公因子时，基波周期为，角频率为 =
常数通常不讨论它的周期性，但可以认为周期为1。
1
4
2
=
= ，1 = 7
2 7 × 2 7
1
3
1
[] = [cos + cos( )]
2
4
4
N1
2
* m 8, m 3
3 / 4
N1
∴ 是周期的，基波周期为 =8
2
* m 8, m 1
/4
+ + + 1.27 这一章介绍了系统的几个一般性质，这就是一个系统可能是或不是：

信号与系统奥本海姆答案

系统（第二版）-学习说明（练习答案）系计算机工程系2005.12目录17第35章第62章第83章第109章第119章第132章第140章160章答案1.1从极坐标转换：1.2从笛卡尔极坐标转换：limlim dtdtdt=cos（t）。

因此，信号翻转限制信号对，所以因此，我们知道（2）线性压缩，因为线性压缩。

因此，基态周期奇信号，所有值为零时为零只有当周期复指数时。

10 10复数指数乘以衰减指数。

因此，周期信号。

复指数基本周期信号。

fundamentalperiod我们得到fundamentalperiod complexexponential=3/5。

找不到任何整数整数。

因此，定期1.10。

x（t）=2cos（10t＋1）-sin（4t-1）周期第一项第一项，整个信号周期至少有多个第二项。

-3-1-1-2-3-3-3第一项第二项第二项整个信号周期，至少在35.1.12中有多个共同的三项。

图1.12。

翻转信号对，所以，no=-3.1.13其导数图1.14。

因此[n-3]=2x[n-2]+4x[n-3]+4x[n-4]）=2x[n-2]+5x输入输出关系y[n]=2x[n-2]+5x[n-3]2x[n-4]输入输出关系的连接序列是反向的。

我们可以很容易地证明[n-3]）+4（x输入-输出关系在y[n]=2x[n-2]+5x[n-3]2x[n-4]1.16无记忆性，因为过去值我们可能总是得出系统输出，因为有时可能取决于考虑两个任意输入（sin（t））的未来值，让线性组合任意标量给系统相应的输出线性。

1.18.（a）考虑两个任意输入线性组合任意标量。

给定系统，相应的输出随机输入相应的输出。

考虑第二个输入输出对应的Alsonote+1）B。

因此+1）B.1.19考虑两个任意输入（t-1）让线性组合任意标量。

给定系统，相应的输出为线性。

（ii）考虑相应输出的任意输入。

考虑第二输入输出相应的输出。

考虑两个任意输入[n-2]。

信号与系统第四章(Lec11)

理想低通滤波器在物理上是不可实现的，只能想方设法去逼近佩利－维纳准则是必要，而非充分条件

周期正弦信号激励下的系统响应
非正弦周期信号激励系统的响应

H(jw)如图所示,其相位特性为零.输入信号如图b,求输出信号y(t)=?
非正弦周期信号的激励
非正弦周期信号的激励
傅里叶级数法
傅里叶级数法
例题

一个周期矩形脉冲信号作用于组成的串连谐振电路中,求电路中的电流.(设 E=1V,T=6.28微秒)
理想低通滤波器的转移函数理想低通滤波器的冲激响应理想低通滤波器的阶跃响应佩利-维纳准则

理想低通滤波器
理想低通滤波器的冲激响应
1.h(t)的求取(p171) 2.说明： a.波形产生了失真。
b.h(t)的主峰发生在t=t0处。 c.系统违背了因果律。
理想低通滤波器的冲激响应
系统的物理可实现性—佩利.维纳准则
单位阶跃响应ru(t)的求取：
单位阶跃响应ru(t)的求取
正弦积分的说明
正弦积分的说明
单位阶跃响应ru(t)
Gibbs现象
b.预冲和过冲。
a. t → −∞的振荡,为非因果; t → ∞的振荡为Gibbs波纹。
矩形脉冲信号通过RC电路的Z.S.R
1 . RC低通滤波器对阶跃信号的响应
b.求联系响应与激励的系统的转移函数
佩利-维纳准则只是物理可实现系统的必要条件, 而不是充分条件。
系统的物理可实现性—佩利.维纳准则
3.推论: a.幅度函数H(jw)在某些离散频率处,可以是零, 但在一有限频带内不能为零。
都是不能实现的.
系统的物理可实现性—佩利.维纳准则

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��
= 0 = �� 2 − 2�� − 5 => �� = 1 ± √6
jω−1 5jω−jω3 +4ω2 −2 (2−ω2 )2 +9ω2
= (jω+1)(jω+2)
5ω − ω3 = ��(√5 + ��)(√5 − ��) = 0 ω = 0, 则
= 1+��(��)��(��)
��(��)
11.5
Q(z) = 1+��(��)��(��) =
1 �� (��)
1 1 1−2��−1 1 1+ 1 −1 (1−�� −1 ) 1−2��
=
�� −1
1 1 (1− �� −1 )(1+ �� −1 ) 3 2
= (
5
6
1
1 3
1− �� −1
−
1 1+ �� −1
1 2
)
➔q[n] = (( ) − (− ) ) ��[��] 5 3 2 (b) Q(z) =
2 1 −1 − �� 3 6 2 1 −1 ��−1 1+( − �� )( 1 −1 ) 3 6 1−2��
−1 K
(与实轴交点)
{ −1 1 ω = ∓√5, 则 K = 3 4ω2 − 2 = 2(2ω2 − 1) = 2(√2�� + 1)(√2�� − 1) = 0 (与虚轴交点)
= −2
1
{ 1 1 −K < −2
−K > 3
1
1
则:2>K>-3
11.9
K(s + 1)(s + 3) + (s + 2)(s + 4) = (K + 1)�� 2 + (4�� + 6)�� + (3�� + 8) = 0 �� 2 �� 0 �� + 1 3�� + 8 4�� + 6 3�� + 8
3 3 3
1
G(ejω )H(ejω )|ω=π,3π,5π = −1
3 3 3
此处必须− K ≥ 1 或 − K ≤ −1系统才稳定即：1 > K > −1
1
1
11.19 11.20
∠H(jωx )G(jωx ) = ∠ −ω
jωx +1
x 2 +jω x +1
= −π ➔ωx = ？
增益裕度为：G�� = −20 log10 |��(�� )��(�� )| dB 现在考虑稳定性：对于H(��)G(s) = s2+s+1 =
6
1 ��
1 ��
= 6 (−3 +
1
7 1+ �� −1
1 3
)
q[n] = − ��[��] + (− ) ��[��] 2 6 3
1
7
1 ��
11.24
(a) G(s)H(s) =
1 1+��
(b) G(s)H(s) =
1 (��−1)(��+3)
��(��)
(a) Q(s) = ��(��) = (b) Q(s) = ��(��) = (c) Q(s) = ��(��) =
��(��) ��(��)
= (��+2)2➔ q(t) = te−2�� (��) = (��+2)2 + (��+2)2➔ q(t) = (1 − t)e−2�� (��)
11.22
��(��) ��(��)
= 1+��(��)��(��)
��(��)
1 (��+1)(��+3) 1 1+ (��+1)(��+3) 1 (��+3) 1 1+ (��+1)(��+3) 1 2 1 −��/3 1+ �� 2
= 2��−1−��
1
要稳定必有：-1-b>0 即 b<-1
11.4
S(s) = ��(��) = ��−2 +��−1 +1 = G(s) = �� −1
Y(s) ��−1
1 ��+1 1 1+ ��(��) 1+��
(��+1)2 ��3
=
��2 (��+3)(��+1) ?
= 0, ��0 = −1, ��1,2 = 0, ��3 = −3
−��3
G(jω)H(jω) =
−��2 +2��+1
=
−��2 −2��+�� 3
=2∗
1
�� − −
1 �� 4 2
要稳定必须|− 4 − 2| < 1 即：− < �� <
2 5 3 2
11.6
Q(z) = 1−��(��)��(��) =
��(��) 1−�� −�� 1+�� −1
(c) G(s)H(s) = ��2 +��+1
��
1
= −2�� − 1 = 0➔s = −
1 2
(d) G(s)H(s) =
��+1 ��2
(e) G(s)H(s) =
��
�� 1
1
11.23
Q(z) =
��(��) ��(��)
=
��(��) 1+��(��)��(��)
(a) Q(z) =
��−1 1 1−2��−1 ��−1 2 1 1+ 1 −1 ∙( − �� −1 ) 3 6 1−2��
4 1 4 1
− + �� 4 0
1
|K − | < 1 =>
−3 4
< �� <