【正文】 =3． X(n)=n1 y(o)=1Solution :Find (n)(n)= Consider (0)= =1 then Find h(n) h(n)=and h(n) satisfied the equation h(n+1)+2h(n)=s(n)Hence h(0)= =1 =Where x(n)=n1 By using classical method Let So that From y(n+1)=x(n)2y(n)We got y(1)=x(0)2y(0) =12 =3Hence y(n) =4. y (n+1)+5y(n)=x(0) x(n)=4u(n) y(0)=0Applying the classical method ypn=BSubstitute into equation ,we have B+5B=4 ∴ B=46Then yp(n)=46The plete solution is y(n)=A(5)n+46and y0=A+46=0so that A=46 yn=465n+46by use zero input and state method ∵ y0=0 ∴yzi0=0 Let hn=k15nunFrom hn+1+5hn=δn )When n=1 h0+5h1=δ1 h0=0 h1+5h0=δ0=1 h1=1 h1=K151=1 K1=15hence hn=155nun yzsn=hn*xn =4m=1nhmu(n)|m=0 hm=0 so m=1 =45m=1n5nun =45?55n+11+5u(n) =45?5+5n516un =46465nun95 If the input is x(n)=un and zero state response is yzsn=(n) then find the yzs(n) when input is xn=(n)Solution: let hn=K1δn+ When xn=un Then yzsn=hn*un =K1un+K2m= =K1un++ =K1un+ From yzsn=22? We get K1+2K2=α K2=α K1=αK2= α Hence hn=2δn+(n) When input xn=un We have yzsn=xn*h(n) =2un+2m=(m)?(nm) =2un+2?=0n1u(n) =2un+2?+1u(n) 96 Discrete system as shown in Fig 91. Let initial condition is zero.Write out the differential equation and find impulse response. DD2 q(n+2) q(n+1) q(n) x(n) 1 y(n) 3 2 yn+2+3yn+1+2yn=2x(n+1)+x(n) HZ=2Z+1Z2+3Z+2=2Z+1Z+1（Z+2） HZZ=2Z+1ZZ+1（Z+2）=K1Z+K2Z+1+K3Z+2 K1=2Z+1Z+1(Z+2)|z=0=12 K2=2Z+1Z(Z+2)|z=1=11=1 K3=2Z+1Z(Z+1)|z=2=32=32 97 Given the difference equation : and y0=1 y1=1 xn=un yzin yzsn and y(n). the graph of system imitation. Solution: yzin=A1?1n+A2(z)n and A1+A2=1 A1=1 A1+12A2=1 A2=0 so that yzi n=un hn=K1un+K22nun and hn+23hn+1+2hn=δn+12δ(n) h03h1+2h2=δ12δ2=0 h0=0 h1=3h0+2h1=δ02δ1=1 h1=1 so that here: h0=0 h1=1 accordingly: h0=K1+K2=0 K1=1 h1=K1+2K2=0 K2=1 hn=1+2nun yzsn=un*hn =n+1+12n+112u(n) =n+1+2n+11u(n)DD 1 x(n) 2 y(n) 32 Chapter 10 Problems Determine the solution for given difference equation1. yn+3yn1+2yn2=0 y1=2 y2=1solution:YZ+3Z1[YZ+y1t1]+2Z2[YZ+y1z1+y2z2]=0 or YZ[1+3Z1+2Z2=3Z1y1Z12Z2y1z12Z2y(2)Z2 =64Z12 YZ=84Z11+3Z1+Z2 =84Z1`1+Z11+2Z1=K11+Z1+K21+Z1 K1=84Z11+2Z1|z1=1=8+412=41=4 K2=84Z11+Z1|Z1=1=8+2112=612=12 ∴ yn=yzin=41n122nun or yzin=A11n+A22n y1=A112A2=2 A1=4 y2=A1+14A2=1 A2=12 2。 ，solution: YZ+2Z1(YZ+y1Z3Z3YZ+y1Z+y2Z2+y3Z3=0 YZ+2Z1YZ+2y13Z3YZ3Z2y1—3Zy23y3=0 YZ1+2Z13Z3=3Zy2+3y3 YZ=6Z+31+2Z13Z3=6Z+31p0Z1(1+p1Z1)(1+p1*Z1) P0=1 P1=+==∠176。 P1*==+=∠176。 YZ=K11+p0Z1+K21+p1Z1+K2*1+p1*Z1 K1=6Z+31+pZ11+p*Z1|Z1=1 K2=6Z+31+p0Z11+p*Z1|Z1= K3=K2*3． solution: ∵ y(1)=y(2)=0 ∴ Y(Z) =21+2Z1+2Z2?ZZe YZZ=2Z2ZeZ+p1(Z+p1*)=K1Ze+K2Z+p1+K3Z+p1* K1=2Z2Z2+2Z+2|Z=e=2e2e2+2e+2 K2=2Z2Ze(Z+p1*)|Z=p K3=K2* ∴ y(n)= K1enun+K2p1nun+K2*p1*nu(n) 4． y(n)+(n1)(n2)=10x(n) x(n)=u(n) y(1)=4 y(2)=6solution:YZ+++y1Z+y2Z2=10X(Z)YZ1+=101+++(2) =101Z1++ =101Z1+ Y(Z)= 10 +++ =K11Z1++K31+++K51+ K1=+|Z1=1== K2=+|Z1=10=1093= K3=|Z1=5=106= K5=|Z1=5= =+ The difference equation of the causal system is 1 Find the impulse response of the system Hn=3nun2 ∵ ∴ yzin=A13nu(n) Determine the system function H(Z) and the impulse response h(n),the difference equation are given as following:1. 2. =12[n1n2Zn3]Z=1 =12n1n2u(n) 113 頁 3.(和(4)相同的方程) yn+25yn+1+byn=xn+23x(n) yn5yn1+byn2=xn3x(n2) two equations just same, we only solve the 3. HZ=Z23Z25Z+6=Z23(Z2)(Z+3)=13Z2(12Z1)(1