Signals and Systems

\begin{aligned} y(n)-ay(n-1)&=b_{0}x(n)-b_{1}x(n-2) \\ &\text{By applying ZT,} \\ Y(z)-az^{-1}Y(z)&=b_{0}X(z)-b_{1}z^{-1}X(z)\\ \Rightarrow \, \, H(z)&=\frac{Y(z)}{X(z)}=\frac{b_{0}-b_{1}z^{-1}}{1-az^{-1}} \end{aligned}
By taking right-sided inverse ZT,
h(n)=b_{0}a^{n}u(n)-b_{1}a^{n-1}u(n-1)
By taking left-sided inverse ZT,
h(n)=-b_{0}a^{n}u(-n-1)+b_{1}a^{n-1}u(-n)
Thus system is not necessarily causal.
The impulse response is non-zero at infinitely many instants.

Since pole zero plot of given transfer function

Since, given pole zero is symmetrical about origin hence it is a all pass system.

Since, one pole on the RHS thus unstable.

Since, pole doesnot have complex conjugate poles and zeros present thus system is not real means system is complex.

\begin{aligned} \text{Since, } y(t) &= x(t) + h(t)\\ \text{and }z(t)&=\left | x(t) \right |\times \left | h(t) \right | \end{aligned}
Case-1:


case 2:

\text{then, }y(t)=z(t)


\text{Thus, } z(t)\geq y(t), \text{ for all 't'}

\begin{aligned}x(n)&=\left (\frac{1}{2} \right )^{n} u(n) , \; \; \; \text{ROC of }x(n):\left | z \right | \gt \frac{1}{2} \\ x(n-k)\rightleftharpoons X(z)&=\frac{z^{-k}}{1-\frac{1}{2}z^{-1}} , \; \; \; \text{ROC of }x(n-k): \left | z \right | \gt \frac{1}{2}\\ \text{For } x(n-k) \; \; \; &\text{ROC will be } \left | z \right |\gt \frac{1}{2}\end{aligned}.

It has no poles for any bounded signal that is nonzero in a finite time interval. This is true as we know for finite amplitude finite width signal ROC is entire s plane and ROC never includes any pole.
It implies for such signals there is no poles. Hence the correct answer is option (B).

\begin{aligned} T_0 &=2\pi \Rightarrow \;\omega _0=\frac{2\pi}{T_0}=1 \\ \text{Now, }b_1 &=\frac{2}{T_0}\int_{0}^{T_0}f(t) \sin \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \sin t \; dt\;\;[\because \omega _0=1]\\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \sin t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}2 \sin ^2 t \; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}[1-\cos 2t] \; dt \\ &=\frac{A}{2 \pi} \left [ t-\frac{\sin 2t}{2} \right ]_0^{\pi} \\ &= \frac{A}{2 \pi} [(\pi-0)-(0-0)]=\frac{A}{2}\\ a_1&=\frac{2}{T_0}\int_{0}^{T_0}f(t) \cos \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \cos t \; dt \\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \cos t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi} \sin 2t \; dt=0 \end{aligned}

Steady state value of y(t)
\begin{aligned} &=\lim_{s \to 0}sY(s)\\ &=\lim_{s \to 0}\frac{10s}{s(s^2+s+100\sqrt{2})}\\ &=\frac{10}{100\sqrt{2}}=\frac{1}{10\sqrt{2}} \end{aligned}

a and T represents positive quantities.
u(t) is unit step function.
h(t)=1+e^{-at}u(t), is non-causal
Here '1' is a constant and two sided so the system will be non-causal, because for causal system,
h(t)=0,\;\;\; t \lt 0
h(t) \neq 0,\;\;\; t \gt 0

\begin{aligned} H(s)&=\frac{c_1}{a_2s^2+b_2s+c_2}\\ & as \;\; a_1=b_1=0\\ &=\frac{c_1}{(1+s\tau _1)(1+s\tau _2)}\\ \text{Put }s&=0,\; H(0)=\frac{c_1}{c_2}\\ \text{Put }s&=\infty ,\; H(\infty )=0 \end{aligned}

which represents second order low pass filter.

\begin{aligned} L^{-1}\left ( \frac{s+3}{s^2+2s+1} \right )&=L^{-1}\left ( \frac{s+1+2}{(s+1)^2} \right )\\ &=L^{-1}\left ( \frac{1}{s+1}+\frac{2}{(s+1)^2} \right )\\ &=e^{-t}+2te^{-t} \end{aligned}