Introduction of C.T. and D.T. Signals

\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)

\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'}

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

x(t)=r(t-1)-r(t-2)-r(t-3)+r(t-4)

\begin{aligned} Ex(t)&=\int_{-\infty }^{\infty }|x(t)|^2 dt \\ &=\int_{1}^{2}|(t-1)^2| dt +\int_{2}^{3}1^2 dt\\&+\int_{3}^{4}(-(t-4))^2dt\\ &= \frac{1}{3}+1+\frac{1}{3}=\frac{5}{3} \end{aligned}

Mean square value = power of f(t)
Mean square value
\begin{aligned} &=\frac{1}{T_0}\int_{T_0} |f(t)|^2 dt\\ &= \frac{1}{4}[4^2 \times 1+2^2 \times 2]\\ &= \frac{16+8}{4}=6 \end{aligned}

Given relationship,
y(n)=[1+(-1)^n]x(n)
Time invariance test:

Since, y(n-1)\neq y'(n)
So, the system is time invariant.
Invertibility test:

Thus, we are getting many to one mapping between input and output. So, the system is non-invertible.

To find the value of \int_{-\infty }^{\infty }e^{-t}\delta (2t-2)dt
Since, \delta (2t-2)=\frac{1}{2}\delta (1t-1) above integral can be written as
\int_{-\infty }^{\infty }e^{-t}\frac{1}{2}\delta (t-1)dt=\frac{1}{2}e^{-1}=\frac{1}{2e}

The given function can be realized as follow:

Combining the function (i0, (ii), (iii) and (iv), we get the given function

\begin{aligned} \text{Therefore, } f(t)&=u(t)-u(t-T)+\frac{(t-T)}{T}\\ &-u(t-T)-\frac{(t-2T)}{T}u(t-2T)u(t)\\&-u(t-T)+ \frac{1}{T}r(t-T)-\frac{1}{T}r(t-2T)\\ &OR \\ &u(t)-u(t-T)+\frac{1}{T}(t-T)u(t-T)\\&-\frac{1}{T}(t-2T)u(t-2T) \end{aligned}

y[n]=x[3-4n]=x[-4n+3]
So to obtain y[n], we first advance x[n] by 3 unit.
i.e. z_1[n]=x[n+3]
Now we will take every fourth sample of z_1[n]
i.e. z_2[n]=z_1[4n]=x[4n+3]
Now reverse (time reverse) z_2[n] will give
y[n]=z_2[-n]=x[-4n+3]