Question 2 dry(t)- dt2 y(t) + 5y(t) + 6y(t) = x(t) d221) + 3x(t) + 2x(t) = 2x(t) Problem 2.19 Find the output, given the input and initial conditions, for the systems described by the following differential or difference equations: 1 d 1 (a) x(t) = e 'u(t), y(0) = - 2 d2 d dt 4 d 3 (b) x(t) = cos(t)u(t), y(0) 5dry(t)...o = 5 d2 d : dt (c) x[n] = u[n], y[-2] = 8,y[-1] = 0: y[n] - ay[n - 2) = 2x[n] + x[n – 1] (d) x[n] = 2"u[n], [-2] = 26, y[-1) = -1: y[n) – ()vín 11 - (b)vín – 2] = x[n] + ( 5 )a[n – 1]

\begin{array}{l}\text { (a) } x(t)=e^{-t} u(t), y(0)=-\frac{1}{2},\left.\frac{d}{d t} y(t)\right|_{t=0}=\frac{1}{2} \text {. } \\\frac{d^{2}}{d t^{2}} y(t)+5 \frac{d}{d t} y(t)+6 y(t)=x(t) \text {. } \\\end{array}EnterApply Laplace transform on Both sides.We know that,\begin{array}{l}L\left[\frac{d^{2}}{d t^{2}} y(t)\right]=s^{2} y(s)-s y(0)-y^{\prime}(0) \\L\left[\frac{d^{2}}{d t} y(t)\right]=s y(s)-y(0) . \\L[y(t)]=y(s) .\end{array}So,\begin{array}{r}s^{2} y(s)-5 y(0)-y^{\prime}(0)+5(s y(s)-y(0)) \\+6 y(s)=x(s)\end{array}put y(0), \frac{d}{d t} y(0) in above equation.\begin{array}{c}\Rightarrow s^{2} y(s)+\frac{s}{2}-\frac{1}{2}+5\left(s y(s)+\frac{1}{2}\right)+6 y(s)=x(s) \\\Rightarrow y(s)\left[s^{2}+5 s+6\right]+\frac{s}{2}-\frac{1}{2}+\frac{5}{2}=x(s) . \\\Rightarrow y(s)\left[s^{2}+5 s+6\right]=x(s)-\frac{s}{2}-2 \\\text { Given, } x(t)=e^{-t} u(t) \text {, apply L.T } \\x(s)=\frac{1}{s+1} .\end{array}\begin{array}{c}x(s)=\frac{1}{s+1} \\y(s)\left[s^{2}+5 s+6\right]=\frac{1}{s+1}-\frac{s}{2}-2 \\y(s)\left[s^{2}+5 s+6\right]=\frac{2-s^{2}-s-4 s-4}{2(s+1)}\end{array}\begin{array}{l}y(s)=\frac{-s^{2}-5 s-2}{2(s+1)\left(s^{2}+5 s+6\right)} \\y(s)=\frac{-s^{2}-5 s-2}{2(s+1)(s+2)(s+3)}\end{array}Using partial fractions.\begin{array}{c}y(s)=\frac{-s^{2}-5 s-2}{2(s+1)(s+2)(s+3)}=\frac{A}{s+1}+\frac{B}{S+2}+\frac{C}{s+3} . \\A=+\frac{1}{2} \quad B=-\frac{1}{2} \quad C=1 \\y(s)=\frac{1}{2(s+1)}-\frac{1}{2(s+2)}+\frac{1}{s+3} .\end{array}Applying Inverse Laplace transform.\begin{array}{l}y(t)=\frac{1}{2} e^{-t}-\frac{1}{2} e^{-2 t}+e^{-3 t} \\y(t)=\frac{1}{2}\left(e^{-t}-e^{-2 t}\right)+e^{-3 t}\end{array}(b)\begin{array}{l}x(t)=\cos t u(t), y(0)=-\frac{4}{5},\left.\frac{d}{d t} y(t)\right|_{t=0}=\frac{3}{5} . \\\frac{d^{2}}{d t^{2}} y(t)+3 \frac{d}{d t} y(t)+2 y(t)=2 x(t) .\end{array}Applying Laplace transform.\begin{array}{l}s^{2} y(s)-s y(0)-y^{\prime}(0)+3[s y(s)-y(0)] \\+2 y(s)=2 x(s) \\y(s)\left[s^{2}+3 s+2\right]+\frac{4 s}{5}-\frac{3}{5}+\frac{12}{5}=2 x(s) \\x(s)=\frac{s}{s^{2}+1} \\y(s)\left[s^{2}+3 s+2\right]=\frac{2 s}{s^{2}+1}-\frac{4 s}{5}-\frac{9}{5}\end{array}\begin{array}{l}y(s)\left[s^{2}+3 s+2\right]=\frac{10 s-(4 s+9)\left(s^{2}+1\right)}{s\left(s^{2}+1\right)} . \\y(s)=\frac{-4 s^{3}+6 s-9 s^{2}-9}{5\left(s^{2}+1\right)} . \\y(s)=\frac{1^{2} s}{\frac{\text { (d) }}{}\left(s^{2}+1\right)(s+1)(s+2)}-\frac{(4 s+9)}{5(t p+1+1)(s+1)(s+2)}\end{array}(a)y(s)=\frac{1^{2} 0 s}{f^{2}\left(s^{2}+1\right)(s+1)(s+2)}-\frac{(4 s+9)}{5\left(t^{2}+1+\lambda\right)(s+1)(s+2)}Using partial fractions,\begin{array}{c}\frac{2 s}{\left(s^{2}+1\right)(s+1)(s+2)}=\frac{A_{1}}{s+2}+\frac{B_{1}}{s+1}+\frac{C_{1} s+D_{1}}{s^{2}+1} . \\A_{1}=-1.2, B_{1}=0.8, C_{1}=0.4, D_{1}=0.8 . \\\frac{4 s+9}{5(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2} \\A=5, B=-1 \\y(s)=\frac{-1.2}{s+2}+\frac{0.8}{s+1}+\frac{0.4 s+0.8}{s^{2}+1}-\frac{5}{s+1}+\frac{1}{s+2} .\end{array}Apply Inverse Laplace transform.\begin{array}{l} y(t)=-1.2 e^{-2 t}+0.8 e^{-t}+0.4 \cos t+0.8 \sin t \\-5 e^{-t}+e^{-2 t} \\y(t)=-0.2 e^{-2 t}-4.2 e^{-t}+0.4 \cos t+0.8 \sin t\end{array}(c)\begin{array}{l}x[n]=u[n], y[-2]=8, y[-1]=0 \\y[n]-\frac{1}{4} y[n-2]=2[x[n]+x[n-1] \\x(z)=\frac{z}{z-1} .\end{array}Appling z-transform to (1). and we know that\begin{array}{l}z[y[n-2]]=z^{-2} y(z)-z^{-1} y(-1)-y(-2) . \\z[y[n-1]]=z^{-1} y(z)-y[-1] . \\z[y[n]]=y[z] .\end{array}So,\begin{array}{l} y(z)-\frac{1}{4}\left[z^{-2} y(z)-z^{-1} y[-1]-y[-2]\right] \\=2 x(z)+z^{-1} x(z)-g x[-1]\end{array}We know that for unit step signal.\begin{array}{l}u[n]=1, n>0 \\0, n<0 . \\x[-1]=u[-1]=0 .\end{array}put values of y[-1], y[-2].\begin{array}{c}y(z)\left[1-\frac{z^{-2}}{4}\right]=x(z)\left[2+z^{-1}\right]+\frac{z}{4}^{-1}(0)-\frac{8}{4} \\y(z)=\frac{x(z)\left[2+z^{-1}\right]-2}{1-\frac{z^{-2}}{4}} \\y(z)=\frac{\frac{z-1}{z-1}\left[\frac{2 z+1}{z}\right]-2}{\frac{4 z^{2}-1}{4 z^{2}}}\end{array}\begin{array}{l}y(z)=\frac{\frac{2 z+1}{z-1}-2}{\frac{(2 z+1)(2 z-1)}{4 z^{2}}}=\frac{\frac{2 z+1-2 z+2}{z-1}}{\frac{4 z^{2}-1}{4 z^{2}}} \\y(z)=\frac{12 z^{2}}{(z-1)\left(4 z^{2}-1\right)}=\frac{12 z^{2}}{(z-1)(2 z-1)(2 z+1)} \\y(z)=\frac{12 z^{2}}{(z-1)(z z-1)(2 z+1)}=\frac{A}{z-1}+\frac{B}{(2 z-1)}+\frac{c}{(2 z+1)} \\A=4, B=-\infty, C=-\alpha \\y(z)=\frac{4 z}{z-1}-\frac{6 z}{2\left(z-\frac{1}{2}\right)}+\frac{(-2) z}{2\left(z+\frac{1}{2}\right)} \\\end{array}y[n]=4 u[n]-3\left(\frac{1}{2}\right)^{n} u[n]-\left(-\frac{1}{2}\right)^{n} v[n](d)\begin{array}{l}x[n]=2^{n} u[n], y[-2]=26, y[-1]=-1 \\y[n]-\frac{1}{4} y[n-1]-\frac{1}{8} y[n-2]=x[n]+\frac{11}{8} x[n-1]\end{array}Apply z-Transform.\begin{array}{c}y(z)-\frac{1}{4}\left[z^{-1} y(z)-y[-1]\right]-\frac{1}{8}\left[z^{-2} y(z)-z^{-1} y[-1]\right. \\=x(z)+\frac{11}{8}\left[z^{-1} x(z)-x[-2]\right] \\y(-1]] \\y(z)\left[1-\frac{z^{-1}}{4}\right]-\frac{1}{4}-\frac{y(z)}{8} z^{-2}-\frac{z^{-1}}{8}+\frac{26}{8} \\=x(z)\left(1+\frac{11 z^{-1}}{8}\right)-0 . \\y(z)\left[1-\frac{z^{-1}}{4}-\frac{z^{-2}}{8}\right]-\frac{1}{4}-\frac{z^{-1}}{8}+\frac{26}{8}=x(z)\left(1+\frac{11 z^{-1}}{8}\right) \\x(z)=\frac{z}{z-2}\end{array}\begin{array}{l}y(z)=\frac{\frac{z}{z-2}\left(1+\frac{11 z^{-1}}{8}\right)+\frac{z^{-1}}{8}-3}{1-\frac{z^{-1}}{4}-\frac{z^{-2}}{8}} \\y(z)=\frac{\frac{z}{z-2}\left(1+\frac{11 z^{-1}}{8}\right)}{\left(1+\frac{z^{-1}}{4}\right)\left(1-\frac{z^{-1}}{2}\right)}+\frac{\frac{z^{-1}}{8}-3}{\left(1+\frac{z^{-1}}{4}\right)\left(1-\frac{z^{-1}}{2}\right)} \\y(z)=\frac{\left(\frac{z}{z-2}\right)\left(\frac{8 z+11}{8 z}\right)}{\left(\frac{4 z+1}{4 z}\right)\left(\frac{2 z-1}{z z}\right)}+\frac{\frac{1-24 z}{8 z}}{\left(\frac{4 z+1}{4 z}\right)\left(\frac{2 z-1}{z z}\right)} \\Y(z)=\frac{8 z+11}{(z-2)(4 z+1)(2 z-1)}+\frac{(1-24 z) z}{(4 z+1)(2 z-1)} \\y(z)=\frac{6}{5\left(1-2 z^{-1}\right)}-\frac{1}{6\left(1+\frac{z^{-1}}{4}\right)}-\frac{5}{6\left(1-\frac{z^{-1}}{2}\right)} \\-\frac{7}{6\left(1+\frac{z^{-1}}{4}\right)}+\frac{11}{6\left(1-\frac{z^{-1}}{2}\right)} \\\end{array}Apply Inverse z-Transform.\begin{array}{l}y[n]=\left[\frac{6}{5} 2^{n}-\frac{1}{6}\left(-\frac{1}{4}\right)^{n}-\frac{5}{6}\left(\frac{1}{2}\right)^{n}\right. \\\left.-\frac{7}{6}\left(-\frac{1}{4}\right)^{n}+\frac{11}{6}\left(\frac{1}{2}\right)^{n}\right] u[n] \\y[n]=\left[\frac{6}{5} 2^{n}-\frac{4}{3}\left(-\frac{1}{4}\right)^{n}+\left(\frac{1}{2}\right)^{n}\right] u[n] \\\end{array} PLEASE UPVOTE ...