Basics of signals and Systems

\begin{aligned} f(t)=1+2 \cos (\pi t)&+3 \sin \left(\frac{2 \pi}{3} t\right)+4 \cos \left(\frac{\pi}{2} t+\frac{\pi}{4}\right) \\ \omega_{1}&=\pi \\ \omega_{2}&=\frac{2 \pi}{3} \\ \omega_{3}&=\frac{\pi}{2} \\ \omega_{0}&=G C D\left(\pi, \frac{2 \pi}{3}, \frac{\pi}{2}\right)=\frac{\pi}{6} \end{aligned}
Fundamental period,
N=\frac{2 \pi}{\omega_{0}}=\frac{2 \pi}{(\pi / 6)}=12

y=a u+b, b \neq 0 is a non-linear system.

Given that, y(t)=\int_{t-T}^{t} x(u) d u
since the given system satisfies both homogeneity and additivity properties, the system is linear.
Check for time invariance:
y\left(t-t_{0}\right)=\int_{t-t_{0}-T}^{t-t_{0}} x(u) d u
When the applied input is x\left(t-t_{0}\right)
\begin{aligned} y_{1}(t) &=\\ \int_{t-T}^{t} x\left(u-t_{0}\right) d u &=\int_{t-t_{0}-T}^{t-t_{0}} x(\tau) d \tau \\ &=y\left(t-t_{0}\right) \end{aligned}
\Rightarrow System is time invariant

Since present output does not depend upon future values of input, the system is causal and also every bounded input produces bounded output, so it is stable.

A signal is said to be periodic if \frac{T}{T_{s}} is a rational number.
Here, T=1.2 T_{s}
\Rightarrow \frac{T}{T_{s}}=\frac{6}{5} \quad Which is a rational number

\begin{array}{l} \text { Energy of } x_{1}[n]=\sum_{n=-\infty}^{\infty}\left|x_{1}[n]\right|^{2} \\ \begin{aligned} =\sum_{n=0}^{\infty} \alpha^{2} &\left(\frac{1}{4}\right)^{n}=\alpha^{2} \cdot \frac{1}{1-\frac{1}{4}}=\alpha^{2} \cdot \frac{4}{3} \\ \text { Energy of } x_{2}[n] &=1.5+1.5=3 \\ \Rightarrow \quad \alpha^{2} \frac{4}{3} &=3 \\ \alpha^{2}=\frac{9}{4} \\ \quad \alpha=1.5 \end{aligned} \end{array}

\begin{array}{l} x(t)=-3 t, \quad-1\lt t \lt 1 \\ x\left(\frac{t-1}{2}\right)=-\frac{3}{2}(t-1) \qquad-1 \lt t \lt 3 \\ \text { and } \quad T=6 \\ \text { Average power }=\frac{1}{6} \int_{-1}^{3}\left(-\frac{3}{2}(t-1)\right)^{2} d t=2 \end{array}

\begin{aligned} x[n] &=\sin \left(\pi^{2} n\right) \\ \omega_{0} &=\pi^{2} \\ \therefore \quad N &=\frac{2 \pi}{\omega_{0}} \cdot m \end{aligned}
where m is the smallest integer that converts \frac{2 \pi}{\omega_{0}} into a integer value.
\therefore \quad N=\frac{2 \pi}{\pi^{2}} \cdot m=\frac{2}{\pi} \cdot m
So, there exists no such integer value of m which could make the N integer, so the system is not periodic.

Step response = Integration of impulse response
\int_{-\infty}^{t} \delta(t-1) d t=u(t-1)
\int_{-\infty}^{t} \delta(t-3) d t=u(t-3)


\begin{aligned} \text{at}\quad t&=2\\ y(t)&=1 \end{aligned}

\begin{array}{l} \omega_{1}=100 \\ \omega_{2}=300 \\ \omega_{3}=500 \end{array}
H.C.F. of \left(\omega_{1}, \omega_{2} \text { and } \omega_{3}\right)= H.C.F. (100,300,500)
\omega=100 \mathrm{rad} / \mathrm{sec}