信号与系统7.2

137Chapter 7 Answers7.1 From the Nyquist sampling theorem , we know that only if X (j w)=0 for |w| > w s /2 will be signal be recoverable from its samples. Therefore, X(jw)>5000л.7.2 From the Nyquist theorem ,we know that the sampling frequency in this case must be at least w s =2000п.In other words ,the sampling period should be at most T=2п/ (w s )=1＊10-3.Clearly ,only (a) and (e) satisfy this condition.7.3 (a) We can easily show that X(j w)=0 for |w| >4000п.Therefore, the Nyquist rate for this signal is w N =2(4000п)=8000п.(b)From the Tables 4.1 and 4.2 we know that X(j w) is a rectangular pulse for which X(j w)=0 for |w| > 4000п.Therefore, the Nyquist rate for this signal is w N =2(4000п)=800п.(c) From the Tables 4.1 and 4.2 we know that X(j w) is the convolution of two rectangular pulses each of which is zero for |w| > 4000п.Therefore ,X(j w)=0 for |w| >8000пand the Nyquist rate for this signal is w N =2(8000п)=16000п.7.4 If the signal x(t) has a Nyquist rate of w o ,then its Fourier transform X (j w)=0 for |w| > w o /2. (a) From chapter 4,y(t) = x (t) + x (t-1) −→←FTY (jw) = X (jw) + e -jwt X (jw).Clearly, we can only guarantee that Y (jw) =0 for |w| > w o /2. Therefore, the Nyquist rate for y(t) is also w o . (b) From chapter 4,y(t) = dtt dx )( −→←FTY (jw)= jw X(jw).Clearly, we can only guarantee that Y (jw) =0 for |w| > w o /2. Therefore, the Nyquist rate for y(t) is also w o . (c) From chapter 4,y(t) =x 2(t) −→←FTY (jw)= (1/2п)[X(jw)*X(jw)]Clearly, we can only guarantee that Y (jw) =0 for |w| > w o . Therefore, the Nyquist rate for y(t) is also 2w o . (d) From chapter 4,y(t)=x(t)cos (w o t) −→←FTY (jw)= (1/2)X(j(w- w o )) +(1/2)X(j(w+ w o )).Clearly, we can guarantee that Y (jw) =0 for |w| > w o + w o /2. Therefore, the Nyquist rate for y(t) is 3w o. 7.5 Using Table 4.2,p(t) −→←FT Tπ2∑∞-∞=-K T K )/2(πωδFrom Table 4.1 p(t-1) −→←FT Tπ2 e -jw T jk k eTk ππωδ2)2(-∞-∞=∑-. Since y(t)=x(t)p(t-1),we haveY (jw)= (1/2п)[X(jw)*FT{P(t-1)}]=(1/T)T jk K e Tk j X ππω2))2((-∞-∞=∑-Therefore, Y(j ω) consists of replicates of X(j ω) shifted by k2π/T and added to earth other (see Figure⎩⎨⎧≤=otherwiseT j H c ,0||,)(ωωωWhere (2/0ω)<c ω<(2π/T) - (2/0ω).7.6 Consider the signal w(t)=x 1(t)x 2(t).The Fourier transform W(j ω) of w(t) is given by W(j ω)=π21[])(*)(21ωωj X j X .Since 0)(1=ωj X for |ω|≥1ωand X 2(j ω)=0 for |ω|≥2ω, we may conclude that W(j ω)=0 for |ω|≥1ω+2ω.Consequently ，the Nyquist rate for w(t) iss ω=2(1ω+2ω).Therefore ,the maximum sampling138period which would still allow w(t) to be recovered is T=2π/(s ω)=π/(1ω+2ω). 7.7 We note thatx 1(t) =h 1(t)*{∑∞-∞=-n nT t nT x )()(δ}Form Figure 7.7 in the book ,we know that the output of the zero-order hold may be written as x 0(t)=h 0(t)* {∑∞-∞=-n nT t nT x )()(δ}where h 0(t) is as shown in Figure S7.7 By taking the Fourier transform of the two above equations, we have X 1(j ω)=H 1( j ω)X p ( j ω)X 0(j ω)=H 0( j ω) X p ( j ω)We now need to determine a frequency response H d ( j ω) for a filter which produces x 1(t) at its output when x 0(t) is its input. Therefore, we needX 0(j ω) H d ( j ω)= X 1(j ω)The triangular function h 1(t) may be obtained by convolving two rectangular pulses as shown in Figure S7.7Therefore,h 1(t)={(1/T ) h 0(t+T/2)}*{( 1/T ) h 0(t+T/2)} Taking the Fourier transform of both sides of the above equation, H 1( j ω)=T1e T j ω H 0( j ω) H 0( j ω) ThereforeX 1(j ω)= H 1( j ω) X p ( j ω)=T 1e T j ω H 0( j ω) H 0( j ω) X p ( j ω) =T1e Tj ω H 0( j ω) X 0(j ω)ThereforeH d ( j ω)=T1eTj ω H 0( j ω)=e2/jwT TT ωω)2/sin(2 7.8 (a) Yes, aliasing does occur in this case .This may be easily shown by considering the sinusoidal term of x(t) for k=5. This term is a signal of the form y(t)=(1/2)5sin(5πt).If x(t) is sampled as T=0.2, then we will always be sampling y(t) at exactly its zero-crossings (This is similar to the idea presented in Figure 7.17 of your textbook). Therefore ,the signal y(t) appears to be identical to the signal (1/2)5sin(0πt) for frequency 5π is a liased into a sinusoid of frequency 0 in the sampled signal.(b) The lowpass filter performs band limited interpolation on the signal ∧x(t).But since aliasing has alreadyresulted in the loss of the sinusoid (1/2)5sin(5πt),the output will be of the formx γ(t)=k k )21(40∑= sin(k πt)The Fourier series representation of this signal is of the form139x γ(t)=∑-=44k k a e )/(t k j π-Where a k =-j(1/2)1+kj(1/2)1+-k7.9 The Fourier transform X(jWe know from the results on impulse-train sampling thatG(jw)=∑∞∞--ωωk j X T ((1s )),Where T=2π/s ω=1/75.therefore,G(jw) is as shown in Figure S7.9 .Clearly, G(jw)=(1/T)X(j ω)=75 X(j ω) for |ω|≤50π.7.10 (a) We know that x(t) is not a band-limited signal. Therefore, it cannot undergo impulse-train sampling without aliasing.(b) Form the given X(j ω) it is clear that the signal x(t) which is bandlimited. That is, X(j ω)=0 for |ω|>0ω.Therefore, it must be possible to perform impulse-train sampling on this signal without experiencing aliasing. The minimum sampling rate required would bes ω=20ω,This implies that thesampling period can at most be T=2π/s ω=π/0ω(c) When x(t) undergoes impulse train sampling with T=2π/0ω,we would obtain the signal g(t) with Fourier transformG(jw)= T1∑∞-∞=-k T k j X ))/2((πωFigure S7.10It is clear from the figure that no aliasing occurs, and that X(jw) can be recovered by using a filter with frequency response T 0≤ωω≤0 H(jw)= 0 otherwiseTherefore, the given statement is true. 7.11 We know from Section 7.4 thatX d (ωj e )= T1∑∞-∞=-k cT k j X ))/2((πω(a) Since X d (ωj e) is just formed by shifting and summing replicas of X(jw),we may argue that ifX d (ωj e ) is real , then X(jw) must also be real(b) X d (ωj e) consists of replicas of X(jw) which are scaled by 1/T,Therefore,if X d (ωj e) has amaximum of 1, then X(jw) must also be real.(c) The region πωπ≤≤||4/3in the discrete-time domain corresponds to the regionT T /||)4/(3πωπ≤≤ in the discrete-time domain. Therefore ,if X d (ωj e )=0 forπωπ≤≤||4/3,then X(jw)=0 for πωπ2000||1500≤≤,But since we already have X(jw)=0 for140πω2000||≥,we have X(jw)=0 for πω1500||≥(d) In this case, sinceπ in discrete-time frequency domain corresponds to 2000π in the continuous-time frequency domain, this condition translates to X(jw)=(j(ω-2000π))7.12 Form Section 7.4 ,we know that the discrete and continuous-time frequencies Ω and ω are related by Ω=ω.Therefore, in this case for Ω=43π,we find the corresponding value of ω toω=43πT1=3000π/4=7500π7.13 For this problem ,we use an approach similar to the one used in Example 7.2 .we assume thatx c (t)=tT t ππ)/sin(The overall output isy c (t)= x c (t-2T)= )2()]2)(/sin[(T t T t T --ππForm x c (t). We obtain the corresponding discrete-time signal x d [n] to be x d [n]= x c (nT)= T1][n δalso, we obtain from y c (t),the corresponding discrete-time signal y d [n] to be y d [n]= y c (nT) =)2()]2(sin[(--n T n ππWe note that the right-hand side of the above equation is always zero when n ≠2.When n=2 ,wemay evaluate the value of ratio using L ,Hospital ,s rule to be 1/T ,Thereforey d [n]= T1]2[-n δWe conclude that the impulse response of the filter is h d [n]= ]2[-n δ7.14 For this problem ,we use an approach similar to the one used in Example 7.2.We assume that x c (t)= tT t ππ)]/sin[(The overall output isy c (t)=)2(T t x dt d c -=)2/()]2/()/[()/(T t T t T COS T ---πππ-2))2/(()]2/)(/sin[(T t T t T --πππForm x c (t) , we obtain the corresponding discrete-time signal x d [n] to be x d [n]= x c (nT)= T1][n δAlso, we obtain from yc(t),the corresponding discrete-time signal y d [n] to beY d [n]=y c (nT)=)2/1()]2/1(cos[)/(--n T n T πππ- )2/1()]2/1(sin[--n T n ππThe first term in rig πht-hang side of the above equation is always zero because cos[π(n-1/2)]=0, therefore, y d [n]= )2/1()]2/1(sin[--n T n ππWe conclude that the impulse response of the filter is h d [n]= )2/1()]2/1(sin[--n T n ππ7.15. in this problem we are interested in the lowest rate which x[n] may be sampled without the possibility of aliasing, we use the approach used in Example 7.4 to solve this problem. To find the lowest rate at which x[n] may be sampled while avoiding the possibility of aliasing, we must find an N such that (22≥Nπ)73πN ≤7/37.16 Although the signal x 1[n]=2sin(πn/2)/( πn) satisfies the first tow conditions, it does not satisfy the thirdcondition . This is because the Flurries transform X 1(e j ω) of this signal is rectangular pulse which is zero for π/2<|ω|<π/2 We also note that the signal x[n]=4[sin(πn/2)/(πn)]2 satisfies the first tow conditions. Fromour numerous encounters with this signal, we know that its Fourier transform X(e j ω) is given by the periodic141convolution of X 1(e j ω) with itself. Therefore, X(e j ω) will be a triangular function in the range 0≤|ω|≤π. This obviously satisfies the third condition as well. T therefore, the desired signal is x[n]=4[sin(πn/2)/(πn)]2.7.17 In this problem .we wish to determine the effect of decimating the impulse response of the given filter by a factor of 2. As explained in Section 7.5.2 ,the process of decimation may be broken up into two steps. In the first step we perform impulse train sampling on h[n] to obtain H p [n]∑∞-∞=k h[2k]δ[n-2k]The decimated sequence is then obtained using h 1[n]=h[2n]=h p [2n]Using eq (7.37), we obtain the Fourier transform H p (e j ω) of h p [n] to beH 1(e j ω)=H p (e jω/2)In other words , H 1(e j ω) is H p (e j ω/2) expanded by a factor of 2. This is as shown in the figure above. Therefore, h 1[n]=h[2n] is the impulse response of an ideal lowpass filter with a passband gain of unity and a cutoff frequency of π/27.18 From Figure 7.37,it is clear interpolation by a factor of 2 results in the frequency response getting compressed by a factor of 2. Interpolation also results in a magnitude sealing by a factor of 2. Therefore, in this problem, the interpolated impulse response will correspond to an ideal lowpass filter with cutoff frequency π/ and a passband gain of 2.7.19 The Fourier transform of x[n] is given by1 |ω|≤ω1X(e j ω)= 0 otherwiseThis is as shown in Figure 7.19.(a) when ω1 ≤3π/5, the Fourier transform X 1(e j ω) of the output of the zero-insertion system is shown inFigure 7.19. The output w(e j ω) of the lowpass filter is as shown in Figure 7.19. The Fourier transform of theoutput of the decimation system Y(e j ω) is an expanded or stretched out version of W(e j ω). This is as shown in Figure 7.19.therefore, y[n]=51nn πω)3/5sin(1(b) When ω1＞3π/5, the Fourier’s transform X 1(e j ω) of the output of the zero-insertion system is as shownin Figure 7.19 The output W(e j ω142The Fourier transform of the output of the decimation system Y(e j ω) bis an expanded or stretched outversion of W(e j ω) .This is as shown in Figure S7.19. Therefore,y[n]=][51n δ7.20 Suppose that X(e j ω) is as shown in Figure S7.20, then the Fourier transform X A (e j ω) of the output of theoutput of S A , the Fourier transform X 1(e j ω) of the output of the lowpass filter , and the Fourier transform X B (e j ω) of the output of S B are all shown in the figures below. Clearly this system accomplishes the filtering task .Figure S7.20(b) Suppose that X(e j ω) is as shown in Figure S7.20 ,then the Fourier transform X B (e j ω) of the output ofS B ,the Fourier transform X 1(e j ω)of the output of the first lowpass filter ,the Fourier transfore X A (e j ω) of theoutput of S A ,the Fourier transform X 2(e j ω) of the output of the first lowpass filter are all shown in the figure below .Clearly this system does not accomplish the filtering task. 7.21(a) The Nyquist rate for the given signal is 2×5000π=10000π. Therefore in order to be able to recover x(t)from x p (t) ,the sampling period must at most be T max =2π/10000π=2×10-4 sec .Since the sampling period used is T=10-4＜T max ,x(t) can be recovered from x p (t).(b) The Nyquist rate for the given signal is 2×15000π=30000π. Therefore in order to be able to recover x(t)from x p (t) ,the sampling period must at most be T max =2π/30000π=0.66×10-4 sec .Since the sampling period used is T=10-4＞T max , x(t) can not be recovered from x p (t).(c) Here,I m {X(j ω)} is not specified. Therefore, the Nyquist rate for the signal x(t) is indeterminate. Thisimplies that one cannot guarantee that x(t) would be recoverable from x p (t).(d) Since x(t) is real,we may conclude that X(j ω)=0 for |ω|＞5000. Therefore the answer to this part isidentical to that of part (a)(e) Since x(t) is real, X(j ω)=0 for |ω|＞15000π. Therefore the answer to this part is identical to that of part(b)(f) If X(j ω)=0 for |ω|＞ω1,then X(j ω)*X(j ω)=0 for |ω|＞2ω1,Therefore in this part X(j ω)=0 for |ω|＞7500. The Nyquist rate for this signal is 2×7500π=15000π. Therefore in order to be able to recover x(t) from x p (t) ,the sampling period must at most be T max =2π/15000π=1.33×10-4 sec .Since the sampling period used is T=10-4＜T max , x(t) can be recovered from x p (t). (g)If |X(j ω)|=0 for ω＞5000π,then X(j ω)=0 for |ω|＞5000π. Therefore the answer to this part is identical to that of part (a).7.22 Using the properties of the Fourier transform, we obtain Y(j ω)=X 1(j ω)X 2(j ω).Therefore, Y(j ω)=0 for |ω|＞1000π.This implies that the Nyquist rate for y(t) is2×1000π=2000π.Therefore, the sampling period T can at most be 2π/(2000π)=10-3sec. Therefore we have to use T<10-3sec in order to be able to recover y(t) from y p (t). 7.23(a) We may express p(t) asP(t)=p 1(t)-p 1(t-△);Where p 1(t)=∑∞-∞=∆-k k t )2(δnow,143P 1(j ω)=∆π∑∞-∞=∆-k )/(πωδTherefore,P(j ω)= P 1(j ω)-e -j ω∆P 1(j )ωIs as shown in figure S7.23. Now,X p (j ω)=)](*)([21ωωπjP j XTherefore, X p (j ω) is as sketched below for △<π/(2ωM ),The corresponding Y(j ω) is also sketched in figure S7.23.(b) The system which can be used to recover x(t) from x p (t) is as shown in FigureS7.23. (c) The system which can be used to recover x(t) from x(t) is as shown in FigureS7.23.(d) We see from the figures sketched in part (a) that aliasing is avoided when ωM ≤π/△.therefore, △max =π/ωM.7.24 we may impress s(t) as s(t)=s(t)-1,where s(t) is as shown in Figure S7.24 we may easily show thats (j )ω= ∑∞-∞=-∆k T k kT k )/2()/2sin(4πωδπFrom this, we obtainS(j =-=)(2)()ωπδωωj S∑∞-∞=-∆k T k k T k )/2()/2sin(4πωδπ-2)(ωπδ Clearly, S(j ω) consists of impulses spaced every 2π/T.(a) If △=T/3, thenS(j =)ω∑∞-∞=-k T k kk )/2()3/2sin(4πωδπ-2)(ωπδNow, since w(t)=s(t)x(t),πω21)(=j W ∑∞-∞=--k X T k j X kk )(2))/2(()3/2sin(4ωππωπTherefore, W(j ω)consists of replicas of X(j ω) which are spaced 2π/T apart. Tn order to avoid aliasing,ωW should be less that π/T. Therefore, T max =2π/ωW. (b) If △=T/3, then(a)(b)()jw Figure S7.24x144S(j =)ω∑∞-∞=-k T k k k )/2()4/2sin(4πωδπ-2)(ωπδ we note that S(j ω)=0 for k=0,±2, ±4,…..This is as sketched in Figure S7.24.Therefore, the replicas of X(j ω)in W(j ω) are now spaced 4π/T apart. Tn order to avoid aliasing,ωW should be less that2π/T. Therefore, T max =2π/ωW. 7.25 Here, x T (kT) can be written asX T (kT)= ∑∞-∞=--k nT x n k n k )()()](sin[ππNote that when n ≠k,0)()](sin[=--n k n k ππAnd when n=k,1)()](sin[=--n k n k ππ Therefore,x τ(kT)=x(kT)7.26. We note thatp(j ω)=Tπ2δ(ω－k2π/T)Also, since x p (t)=x(t)p(t).X p (j ω)=12π{ x(j ω) * P(j ω)}=1Tx(j(ω－k2π/T))Figure S7.26Note that as T increase, Tπ2－ω2 approaches zero. Also, we note that there is aliasingWhen2ω1-ω2<Tπ2－ω2<ω2If 2ω1-ω2≥0(as given) then it is easy to see that aliasing does not occur when 0≤Tπ2－ω2≤2ω1－ω2For maximum T, we must choose the minimum allowable value for Tπ2－ω2 (which is zero).This implies that T max =2π/ω2. We plot x p (j ω) for this case in Figure S7.26. Therefore, A=T, ωb =2π/T, and ωa =ωb －ω11457.27.(a) Let x 1(j ω) denote the Fourier transform of the signal x 1(t) obtained by multiplyingx(t) with e -j ω0t Let x 2(j ω) be the Fourier transform of the signal x 2(t) obtained at the output of the lowpass filter. Then, x 1(j ω), x 2(j ω),and x p (j ω),are as shown in Figure S7.27(b) The Nyquist rate for the signal x 2(t) is 2×(ω2－ω1)/2=ω2－ω1.Therefore, thep 7.28. (a) The fundamental frequency of x(t) is 20π rad/sec.From Chapter 4 we know that the Fourier transform of x(t) is given byX(j ω)=2πk ∞=-∞∑a k δ(ω－20πk).This is as sketched below. The Fourier transform x c (j ω) of the signal x c (t) is also Sketched in Figure S7.28. Note thatP(j ω)=2510π⨯3(2/(510))k k δωπ∞-=-∞-⨯∑Andx p (j ω)=12π[ x c (j ω)* p(j ω)]Therefore, x p (j ω) is as shown in the Figure S7.28.Note that the impulses from adjacentReplicas of x c (j ω) add up at 200π.Now the Fourier transform x(e j Ω) of the sequence x[n] is given byx(e j Ω)= x p (j ω)|ω=ΩT. This is as shown in the Figure S7.28.Since the impulses in x(e j ω) are located at multiples of a 0.1π,the signal x[n] is146(b) The Fourier series coefficients of X[n] aT π2(12)k , k=0,±1,±2,….,±9 a k =4T π(12)10 , k=10 7.29. x p (j ω)=1T((2/))k x j k T ωπ∞=-∞-∑x(jwe ), Y(jwe ), Y p (j ω),and Y c (j ω) are as shown in Figure S7.29. 7.30. (a) Since x c (t)=δ(t),we have()c dy t dt+y c (t)= δ(t) Taking the Fourier transform we obtainj ωY(j ω)+ Y(j ω)=1 Therefore , Y c (j ω)=11j ω+, and y c (t) =e -t u(t). (b) Since y c (t) =e -tu(t) , y[n]= y c (nT)= e -nT u[n].Therefore, j ωH(e j ω)=()()j W e Y e ω=11/(1)T j e e ω---=1－e －T e －j ωTherefore,h[n]= δ[n]－e －T δ[n －1]7.31. In this problem for the sake of clarity we will use the variable Ωto denote discretefrequency. Taking the Fourier transform of both sides of the given difference equation we obtainH(j e Ω)=()()j j Y e X e ΩΩ=1112j e -Ω-Given that the sampling rate is greater than the Nyquist rate, we have147x(j eΩ)=1Tx c (j Ω/T), for －π≤Ω≤π Therefore,Y(j eΩ)=1(/)12c j x j T T e -ΩΩ-For －π≤Ω≤π.From this we getY(j ω)= Y(jw eT)= =1()12c j Tx j T e ωω--For －π/T ≤ω≤π/T. in this range, Y(j ω)= Y c (j ω).Therefore,H c (j ω)=()()c c Y j X j ωω=1/112j TT e ω--7.32. Let p[n]=[14]k n k δ∞=-∞--∑.Then from Chapter 5,p(jwe )= e -j ω24π(2/4)k k δωπ∞=-∞-∑=2π2/4(2/4)j k k k eπδωπ∞--=-∞∑Therefore, G(jw e )=()1()()2j j p e x e d πθωθπθπ--⎰=32/4(2/4)01()4j k j k k e x e πωπ--=∑jwjwFigure S7.32Clearly, in order to isolate just x(jwe ) we need to use an ideal lowpass filter with Cutoff frequency π/4 and passband gain of 4. Therefore, in the range |ω|<π, 4, |ω|<π/4H(e j ω)= 0, π/4≤|ω|≤π7.33. Let y[n]=x[n][3]k n k δ∞=-∞-∑.ThenY(e j ω)=3(2/3)1()3j k k x eωπ-=∑Note that sin(πn/3)/(πn/3) is the impulse response of an ideal lowpass filter with cutoff frequency π/3 and passband gain of 3.Therefore,we now require that y[n] when passed through this filter should yieldx[n].Therefore, the replicas of x(e j ω) contained in Y(e j ω) should not overlap with one another. This ispossible only if x(e j ω) =0 for π/3≤|ω|≤π.7.34. In order to make x(e j ω) occupy the entire region from －πto π,the signal x[n]148must be downsampled by a factor of 14/3.Since it is not possible to directly downsample by a noninteger factor, we first upsample the signal by a factor of 3. Therefore, after the upsampling we will need toreduce the sampling rate by 14/3× 3=14. Therefore, the overall system for performing the sampling rate conversion isy[n][]2nx ,n=0,±3,±6,… y[n]=p[14n] ω[n]= 0, otherwise Figure S7.34)(e xp)(ωj d e x 7.36. (a) Let us decnote the sampled signaled signal by x p (t). We have∑∞-∞=-=n pnT t nT x t x )()()(δSince the Nyquist rate for the signal x(t) is T /2π,we can reconstruct the signal from x p (t). From Section 7.2,we know that)(*)()(t h t x t x p = whereTt T t t h /)/sin()(ππ=Thereforedtt dh t x dtt dx p )(*)()(=Denoting dtt dh )( by g(t),we have∑∞-∞=-==n pnT t g nT x t g t x dtt dx )()()(*)()(Therefore,2)/sin()/cos()()(tT t T tT t dtt dh t g πππ-==(b) No.7.37. We may write p(t) asp(t)=p 1(t)+p 1(t-∆),where∑∞-∞=-=k W k t t p )/2()(1πδTherefore,)()1()(1ωωωj p e j p j ∆-+= where∑∞-∞=-=k kW w j p )()(1ωδω149Let us denote the product p(t)f(t) by g(t).Then,)()()()()()()(11t f t p t f t p t f t p t g ∆-+== This may be written as)()()(11∆-+=t bp t ap t g Therefore,)(()(1)ωωωj p be a j G j ∆-+= with )(1ωj p is specified in eq.(s7.37-1). Therefore [])()(kw be a w j G k w jk -+=∑∞-∞=∆-ωδωWe now have)()()()(1t f t p t x t y = Therefore,[])(*)(21)(1ωωπωj x j G j Y =This give us[]))((2)(1kW j x be a Wj Y wjk -+=∑∆-ωπωIn the range 0<ω<W, we may specify Y 1(j ω) as[]))(()()()(2)(1W j x be a j x b a w j Y w jk -+++=∆-ωωπωsince )()()(112ωωωj H j Y j Y =, in the range 0<ω<W we may specify Y 2(j ω) as []))(()()()(2)(2W j x be a j x b a jW j Y W j -+++=∆-ωωπωSince ),()()(3t p t x t y =in the range 0<ω<W we may specify Y 3(j ω) as []))(()1()(22)(3W j x e j x W j Y W j -++=∆-ωωπωGive that 0<W △<π,we require that )()()(32ωωωj kx j Y j Y =+ for 0<ω<W. That is[][])())(()1(2)()(2ωωπωπj kx W j x e W j x jb ja a Ww j =-++++∆-This implies that01=+++∆-∆-W j Wj jbe ja e Solving this we obtainA=1, b= -1, When W △=π/2. More generally, we also geta=sin(W △)+)tan())cos(1(∆∆+W W and )sin()cos(1∆∆+-=W W bexcept when 2/π=∆W Finally, we also get [])2/(12jb ja Wk ++=π。

第7章采样第8章通信系统第9章拉普拉斯变换第10章Z变换第11章线性反馈系统第7章采样7.2连续时间信号x（t）从一个截止频率为的理想低通滤波器的输出得到，如果对x（t）完成冲激串采样，那么下列采样周期中的哪一些可能保证x（t）在利用一个合适的低通滤波器后能从它的样本中得到恢复？7.3在采样定理中，采样频率必须要超过的那个频率称为奈奎斯特率。

试确定下列各信号的奈奎斯特率：7.4设x（t）是一个奈奎斯特率为ω0的信号，试确定下列各信号的奈奎斯特率：7.5设x（t）是一个奈奎斯特率为ω0的信号，同时设其中。

7.6在如图7-1所示系统中，有两个时间函数x1（t）和x2（t）相乘，其乘积W （t）由一冲激串采样，x1（t）带限于ω17.7信号x（t）用采样周期T经过一个零阶保持的处理产生一个信号x0（t），设x1（t）是在x（t）的样本上经过一阶保持处理的结果，即7.8有一实值且为奇函数的周期信号x（t），它的傅里叶级数表示为7.9考虑信号x（t）为7.10判断下面每一种说法是否正确。

7.11设是一连续时间信号，它的傅里叶变换具有如下特点：7.12有一离散时间信号其傅里叶变换具有如下性质：7.13参照如图7-7所示的滤波方法，假定所用的采样周期为T，输入xc（t）为带限，而有7.14假定在上题中有重做习题7.13。

7.15对进行脉冲串采样，得到若7.16关于及其傅里叶变换7.17考虑理想离散时间带阻滤波器，其单位脉冲响应为频率响应在条件下为7．18假设截止频率为π/2的一个理想离散时间低通滤波器的单位脉冲响应是用于内插的，以得到一个2倍的增采样序列，求对应于这个增采样单位脉冲响应的频率响应。

7.19考虑如图7-11所示的系统，输入为x[n]，输出为y[n]。

零值插入系统在每一序列x[n]值之间插入两个零值点，抽取系统定义为其中W[n]是抽取系统的输入序列。

若输入x[n]为试确定下列ω1值时的输出y[n]：7.20有两个离散时间系统S1和S2用于实现一个截止频率为π/4的理想低通滤波器。

第二部分课后习题第7章采样基本题7.1已知实值信号x（t），当采样频率时，x（t）能用它的样本值唯一确定。

问在什么ω值下保证为零？解：对于因其为实函数，故是偶函数。

由题意及采样定理知的最大角频率即当时，7.2连续时间信号x（t）从一个截止频率为的理想低通滤波器的输出得到，如果对x（t）完成冲激串采样，那么下列采样周期中的哪一些可能保证x（t）在利用一个合适的低通滤波器后能从它的样本中得到恢复？解：因为x（t）是某个截止频率的理想低通滤波器的输出信号，所以x（t）的最大频率就为＝1000π，由采样定理知，若对其进行冲激采样且欲由其采样m点恢复出x（t），需采样频率即采样时间问隔从而有（a）和（c）两种采样时间间隔均能保证x（t）由其采样点恢复，而（b）不能。

7.3在采样定理中，采样频率必须要超过的那个频率称为奈奎斯特率。

试确定下列各信号的奈奎斯特率：解：（a）x（t）的频谱函数为由此可见故奈奎斯特频率为（b）x（t）的频谱函数为由此可见故奈奎斯特频率为（c）x（t）的频谱函数为由此可见，当故奈奎斯特频率为7.4设x（t）是一个奈奎斯特率为ω0的信号，试确定下列各信号的奈奎斯特率：解：（a）因为的傅里叶变换为可见x（t）的最大频率也是的最大频率，故的奈奎斯特频率为0 。

（b）因为的傅里叶变换为可见x （t）的最大频率也是的最大频率．故的奈奎斯特频率仍为。

（c）因为的傅里叶变换蔓可见的最大频率是x（t）的2倍。

从而知x 2（t）的奈奎斯特频率为2（d）因为的傅里叶变换为，x（t）的最大频率为，故的最大频率为，从而可推知其奈奎斯特频率为7.5设x（t）是一个奈奎斯特率为ω0的信号，同时设其中。

当某一滤波器以Y（t）为输入，x（t）为输出时，试给出该滤波器频率响应的模和相位特性上的限制。

解：p（t）是一冲激串，间隔对x（t）用p（t-1）进行冲激采样。

先分别求出P（t）和P（t-1）的频谱函数：注意0ω是x（t）的奈奎斯特频率，这意味着x（t）的最大频率为02ω，当以p（t-1）对x（t）进行采样时，频谱无混叠发生。

《信号与系统教案》课件第一章：信号与系统导论1.1 信号的概念与分类讲解信号的定义和特性介绍常见信号的分类，如连续信号、离散信号、模拟信号和数字信号等1.2 系统的概念与分类讲解系统的定义和特性介绍常见系统的分类，如线性系统、非线性系统、时不变系统等1.3 信号与系统的研究方法讲解信号与系统的研究方法，如数学分析、仿真实验等第二章：连续信号与系统2.1 连续信号的基本性质讲解连续信号的定义和特性，如连续性、周期性、对称性等2.2 连续信号的运算介绍连续信号的基本运算，如加法、乘法、积分等2.3 连续系统的基本性质讲解连续系统的基本性质，如线性、时不变性等第三章：离散信号与系统3.1 离散信号的基本性质讲解离散信号的定义和特性，如离散性、周期性、对称性等3.2 离散信号的运算介绍离散信号的基本运算，如加法、乘法、求和等3.3 离散系统的基本性质讲解离散系统的基本性质，如线性、时不变性等第四章：模拟信号处理4.1 模拟信号处理的基本方法讲解模拟信号处理的基本方法，如滤波、采样、量化等4.2 模拟滤波器的设计与分析介绍模拟滤波器的设计方法，如巴特沃斯滤波器、切比雪夫滤波器等讲解滤波器的频率响应、阶数等特性分析4.3 模拟信号处理的应用讲解模拟信号处理在实际应用中的案例，如音频处理、通信系统等第五章：数字信号处理5.1 数字信号处理的基本方法讲解数字信号处理的基本方法，如离散余弦变换、快速傅里叶变换等5.2 数字滤波器的设计与分析介绍数字滤波器的设计方法，如IIR滤波器、FIR滤波器等讲解滤波器的频率响应、阶数等特性分析5.3 数字信号处理的应用讲解数字信号处理在实际应用中的案例，如图像处理、语音识别等第六章：信号与系统的时域分析6.1 线性时不变系统的时域特性讲解线性时不变系统的时域特性，如叠加原理和时移特性6.2 常用时域分析方法介绍常用时域分析方法，如单位脉冲响应、零输入响应和零状态响应6.3 时域分析在实际应用中的案例讲解时域分析在实际应用中的案例，如信号的滤波、去噪等第七章：信号与系统的频域分析7.1 傅里叶级数与傅里叶变换讲解傅里叶级数的概念和性质介绍傅里叶变换的定义和性质，包括连续傅里叶变换和离散傅里叶变换7.2 频域分析方法介绍频域分析方法，如频谱分析、滤波器设计等7.3 频域分析在实际应用中的案例讲解频域分析在实际应用中的案例，如通信系统、音频处理等第八章：信号与系统的复频域分析8.1 拉普拉斯变换和Z变换讲解拉普拉斯变换的概念和性质介绍Z变换的定义和性质8.2 复频域分析方法介绍复频域分析方法，如系统函数分析、滤波器设计等8.3 复频域分析在实际应用中的案例讲解复频域分析在实际应用中的案例，如数字通信系统、信号的调制与解调等第九章：信号与系统的状态空间分析9.1 状态空间模型的概念和性质讲解状态空间模型的定义和性质，如状态向量、状态方程和输出方程等9.2 状态空间分析方法介绍状态空间分析方法，如状态预测、状态估计等9.3 状态空间分析在实际应用中的案例讲解状态空间分析在实际应用中的案例，如控制系统的设计和分析等第十章：信号与系统的应用案例分析10.1 通信系统中的应用讲解信号与系统在通信系统中的应用，如信号的调制与解调、信道编码与解码等10.2 音频处理中的应用讲解信号与系统在音频处理中的应用，如音频信号的滤波、均衡等10.3 图像处理中的应用讲解信号与系统在图像处理中的应用，如图像的滤波、边缘检测等重点解析信号与系统的基本概念及其分类信号与系统的研究方法连续信号与系统的性质和运算离散信号与系统的性质和运算模拟信号处理的基本方法和应用数字信号处理的基本方法和应用信号与系统的时域分析方法及其应用信号与系统的频域分析方法及其应用信号与系统的复频域分析方法及其应用信号与系统的状态空间分析方法及其应用信号与系统在不同领域中的应用案例分析难点解析信号与系统理论的数学基础和抽象概念的理解不同信号与系统分析方法的相互转换和应用信号与系统在实际工程应用中的复杂性和挑战高频信号处理和数字信号处理的算法优化和实现状态空间分析方法的数学推导和系统设计的实践应用。