Chapter 9

Solve the homogeneous difference equations. (a) \(y[2+n]-6 y[1+n]+8 y[n]=0\) with \(y \mid 0]=3, y[1]=4\). (b) \(y[2+n]-6 y[1+n]+9 y \mid n]=0\) with \(y[0]=2, y[1]=3\). (c) \(y[2+n]-6 y[1+n]+10 y[n]=0\) with \(y[0]=2, y[1]=4 .\)

Use direct substitution and trigonometric identities to show the following: (a) \(y[n]=x[n]+x[n-1]+x[n-2]\) will "zero-out" \(x[n]=\cos \left(\frac{2 \pi}{3} n\right)\) and \(x[n]=\) \(\sin \left(\frac{2 \pi}{5} n\right)\) (b) \(y[n]=x[n]-x[n-1]+x \mid n-2]\) will "zero-out" \(x[n]=\cos \left(\frac{\pi}{3} n\right)\) and \(x[n]=\) \(\sin \left(\frac{\pi}{3} n\right)\) (c) \(y[n]=x[n]+\sqrt{2} x[n-1]+x[n-2]\) will "zero-out" \(x[n]=\cos \left(\frac{3 \pi}{4} n\right)\) and \(x[n]=\) \(\sin \left(\frac{3 \pi}{4} n\right)\) (d) \(y[n]=x[n]+\sqrt{3} x[n-1]+x[n-2]\) will "zero-out" \(x[n]=\cos \left(\frac{5 \pi}{6} n\right)\) and \(x[n]=\) \(\sin \left(\frac{5 x}{6} n\right)\) (e) \(y[n]=x[n]-\sqrt{3} x[n-1]+x[n-2]\) will "zero-out" \(x[n]=\cos \left(\frac{\pi}{6} n\right)\) and \(x[n]=\) \(\sin \left(\frac{\pi}{6} n\right)\)

(a) Solve \(y[n+2]+y[n]=0\) with \(y[0]=1\) and \(y[1]=0\). (b) Solve \(y[n+2]+y[n]=0\) with \(y[0]=0\) and \(y[1]=1\).

Given the recursion formula \(y[n]=x[n]+x[n-1]+x[n-2]\). (a) Calculate the amplitude response \(A(0.10), A\left(\frac{\pi}{3}\right), A\left(\frac{2 \pi}{3}\right)\), and \(A(2.10)\). (b) Discuss what happens to the filtered signal for the input \(x[n]=\cos (0.10 n)\) \(+\sin (2.10 \mathrm{n})\)

Solve the homogeneous difference equations. (a) Solve \(y[n+2]-\sqrt{2} y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=\sqrt{2}\). (b) Solve \(y[n+2]-\sqrt{2} y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{2}\).

Given the recursion formuls \(y[n]=x[n]-x[n-1]+x[n-2]\). (a) Calculate the amplitude response \(A(0.10), A\left(\frac{\pi}{3}\right), A(1.00)\), and \(A\left(\frac{2 \pi}{3}\right)\). (b) Discuss what happens to the filtered signal for the input \(x[n]=\cos (0.10 n)\) \(+\sin (1.00 n)\)

Solve the nonhomogeneous difference equations. (a) \(y[2+n]-6 y[1+n]+8 y[n]=3^{n}\) with \(y[0]=1, y[1]=3 .\) (b) \(y[2+n]-6 y[1+n]+9 y[n]=2^{n}\) with \(y[0]=2, y[1]=1\). (c) \(y(2+n]-6 y(1+n]+10 y \mid n]=2^{n+1}\) with \(y[0]=1, y[1]=4\).

Solve the nonhomogeneous difference equations. (a) \(y[2+n]-8 y[1+n]+15 y[n]=4^{n}\) with \(y[0]=1, y[1]=4\). (b) \(y[2+n]-8 y[1+n]+16 y[n]=5^{n}\) with \(y[0]=2, y[1]=1\). (c) \(y[2+n]-8 y[1+n]+17 y[n]=2 * 3^{n}\) with \(y[0]=0, y[1]=-1\)

The single-pole low-pass filter is \(y[n]=K x[n]+(1-K) y[n-1]\), where constant \(K\) is between 0 and \(1 .\) (a) Use \(K=\frac{1}{4}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (b) Use \(K=\frac{1}{10}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (c) Use \(K=\frac{1}{16}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\).

(a) Solve \(y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{3} y[n]=0\) with \(y[0]=1\) and \(y[1]=3\). (b) Solve \(y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{8} y[n]=\left(\frac{1}{4}\right)^{n}\) with \(\left.y \mid 0\right]=0\) and \(y[1]=1\).

(a) Solve \(y[n+2]-y[n+1]+\frac{1}{4} y[n]=0\) with \(y[0]=1\) and \(y[1]=1\). (b) Solve \(y[n+2]-y[n+1]+\frac{1}{4} y[n]=\left(\frac{3}{4}\right)^{n}\) with \(y[0]=0\) and \(y[1]=1\)

(a) Solve \(y[n+2]-\frac{5}{5} y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=6\). (b) Solve \(y[n+2]-\frac{5}{5} y[n+1]+y[n]=\left(i^{n}+(-i)^{n}\right)\) with \(y[0]=0\) and \(y[1]=1\)

Solve the difference equation \(y \mid n+1]=a y|n|+b\) with the initial condition \(y[0]=y_{0}\). Use recursion (and mathematical induction) to find the solution. That is, compute \(y|1|=y_{0} a+b, y[2]=y_{0} a^{2}+(1+a) b, y[3]=y_{0} a^{3}+\left(1+a+a^{2}\right) b\), then find the general term.

(a) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=-1\). (b) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{3}\).

(a) Solve \(y[n+2]-y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=1\). (b) Solve \(y[n+2]-y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{3}\).

(a) Construct a filter using the zeros \(e^{\pm i 3 \epsilon / 4}\) and \(e^{-i \pi}=-1\) for "zeroing out" \(\cos \left(\frac{3 x}{4} n\right), \sin \left(\frac{3 x}{4} n\right)\), and \(\cos (\pi n)\). (b) Construct a filter using the poles \(\frac{2}{3} e^{\pm j \pi}\) and \(\frac{1}{3} e^{0 i}=\frac{1}{3}\) for "boosting up" signals near \(\cos \left(\frac{\pi}{3} n\right)\) and \(\sin \left(\frac{\pi}{3} n\right)\) and low- frequency signals. (c) Construct a filter using the zeros and poles in parts (a) and (b).

In the value of an annuity due model \(y[n+1]=(1+r)(y[n]+P)\) use the parameters \(r=\frac{1}{10}, P=1000\) (a) Use the 2 -transform and Tables 9.1-9.3 to find the solution. (b) Use residues to find the solution.

Consider the system of difference equations \(x[n+1]=\frac{\sqrt{2}}{2} x[n]-\frac{\sqrt{2}}{2} y[n]\), and \(y[n+1]=\frac{\sqrt{2}}{2} x[n]+\frac{\sqrt{2}}{2} y[n]\) with the initial conditions \(x[0]=1\), and \(y[0]=0\). (a) Use trigonometric identities to verify that the solution is \(x[n]=\cos \left(\frac{\pi}{4} n\right)\) and \(y[n]=\sin \left(\frac{\pi}{4} n\right)\), (b) Use z-transforms and residues to construct the solution in part (a).