Signals and Systems

Given: x(n)=\sum_{k=0}^{N-1}a_ke^{j\frac{2 \pi kn}{N}}
\therefore \; x(n)=a_0+a_1e^{\frac{j2\pi n}{3}}+0+... \;\;\;...(1)
We have,
a_k=a_{k+N}
a_0=a_3
a_1=a_4=1
a_{-3}=a_0=2
Put n = 0 in eq. (1)
x(0)=a_0+a_1=2+1=3
Put n = 1 in eq. (1)
x(1)=a_0+a_1e^{j 2 \pi /3}=2+1\left ( \cos \frac{2\pi}{3} +j \sin \frac{2\pi}{3}\right )=2+(-0.5+j0.866)=1.5+j0.866
These two conditions satisfy by the option (B). Hence, option (B) will be correct.

We know, a linear system follows the law of superposition.
It is a combination of two laws:
(i) Law of additivity:

Both results are same, hence, it follows law of additivity.
(ii) Law of Homogeneity:

Here also both results are same, hence it follows law of Homogeneity.
Therefore, System is linear.
We know, a causal system is independent of future values of input at each & every instant of time them system will be causal.
Given : y(t)=x(e^t)
Put t=0
y(0)=x(e^0)=x(1)
Because its depends on future value.
Therefore, system is non-causal.

Given: h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t
Fourier transform of signal \frac{\sin (10 \pi t)}{\pi t} is given by

Now, impulse response
h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )
Using property, e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)
Therefore, Fourier transform of impulse response

Cut-off frequencies,
\omega_L=30 \pi rad/sec
\omega_H=50 \pi rad/sec
Thus, output of the system =7 \sin 42 \pi t+4 \cos 45 \pi t

Given:
y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)
Taking Laplace transform,
Y(s)+\frac{1}{4}(sY(s))=2X(s)
Now, H(s)=\frac{Y(s)}{X(s)}=\frac{2}{\frac{s}{4}+1}=\frac{8}{s+4}
Taking inverse Laplace, transform,
h(t)=8e^{-4t}u(t)

Fourier transform, F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
\because Peak value of sampling function occurs at \omega=0
Peak value =1

f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t
The following informations are given about f(t) \rightleftharpoons F(\omega).
\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}

As we know,
\begin{aligned} n \cdot a^{n} u(n) & \rightleftharpoons \frac{a z}{(z-a)^{2}} \\ \frac{1}{a} \cdot n \cdot a^{n} u(n) & \rightleftharpoons \frac{z}{(z-a)^{2}} \\ f(n) &=n \cdot a^{n-1} u(n) \rightleftharpoons \frac{z}{(z-a)^{2}}=F(z) \end{aligned}
Time-shifting property,
\begin{aligned} f(n) &\rightleftharpoons F(z) \\ f(n+1) &\rightleftharpoons z \cdot F(z) & \\ x(n)&=(n+1) a^{n} u(n+1) \rightleftharpoons \frac{z^{2}}{(z-a)^{2}}=X(z) & \\ \text{Thus},\qquad \qquad x(n)&=(n+1) a^{n} u(n+1)=(n+1) a^{n} \cdot u(n) \\ &[\because(n+1) u(n+1)=(n+1) u(n) \end{aligned}

\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} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}


Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1


\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.

\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.