Question 1 |

Two discrete-time linear time-invariant systems with impulse responses h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ] and h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ] are connected in cascade, where \delta \left [ n\right ] is the Kronecker delta. The impulse response of the cascaded system is

\delta \left [ n-2\right ]+\delta \left [ n+1 \right ] | |

\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ] | |

\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ] | |

\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ] |

Question 1 Explanation:

\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)

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)

Question 2 |

Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces
output y(t), so that x(t)*h(t)=y(t). Further, if |x(t)|*|h(t)|=z(t), which of the following
statements is true?

For all t \in (-\infty ,\infty ), z(t)\leq y(t) | |

For some but not all t \in (-\infty ,\infty ), z(t)\leq y(t) | |

For all t \in (-\infty ,\infty ), z(t)\geq y(t) | |

For some but not all t \in (-\infty ,\infty ), z(t)\geq y(t) |

Question 2 Explanation:

\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'}

Case-1:

case 2:

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

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

Question 3 |

The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?

e^{+at}u(t) | |

e^{-a(t+T)}u(t) | |

1+e^{-at}u(t) | |

e^{-a(t-T)}u(t) |

Question 3 Explanation:

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

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

Question 4 |

The signal energy of the continuous-time signal

x(t) =[(t -1) u(t -1)] -[(t -2)u(t -2)] -[(t -3)u(t -3)] +[(t -4)u(t -4)] is

x(t) =[(t -1) u(t -1)] -[(t -2)u(t -2)] -[(t -3)u(t -3)] +[(t -4)u(t -4)] is

11/3 | |

7/3 | |

1/3 | |

5/3 |

Question 4 Explanation:

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}

\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}

Question 5 |

The mean square value of the given periodic waveform f(t) is_________

12 | |

24 | |

6 | |

3 |

Question 5 Explanation:

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}

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}

Question 6 |

Consider the system with following input-output relation y[n]=(1+(-1)^{n})x[n]

where, x[n] is the input and y[n] is the output. The system is

where, x[n] is the input and y[n] is the output. The system is

invertible and time invariant | |

invertible and time varying | |

non-invertible and time invariant | |

non-invertible and time varying |

Question 6 Explanation:

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.

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.

Question 7 |

The value of \int_{-\infty }^{+\infty }e^{-t}\delta (2t-2)dt, \; where \; \delta (t) is the Dirac delta function, is

\frac{1}{2e} | |

\frac{2}{e} | |

\frac{1}{e^{2}} | |

\frac{1}{2e^{2}} |

Question 7 Explanation:

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}

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}

Question 8 |

The function shown in the figure can be represented as

u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t-2T) | |

u(t)+\frac{t}{T}u(t-T)-\frac{t}{T}u(t-2T) | |

u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t) | |

u(t)-u(t-T)+\frac{(t-T)}{T}u(t-T) -\frac{(t-2T)}{T}u(t-2T) |

Question 8 Explanation:

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}

Question 9 |

A zero mean random signal is uniformly distributed between limits -a and +a
and its mean square value is equal to its variance. Then the r.m.s value of the
signal is

\frac{a}{\sqrt{3}} | |

\frac{a}{\sqrt{2}} | |

a\sqrt{2} | |

a\sqrt{3} |

Question 10 |

Given a sequence x[n], to generate the sequence y[n]=x[3-4n], which one of the
following procedures would be correct ?

First delay x[n] by 3 samples to generate z_{1}[n], then pick every 4^{th} sample of z_{1}[n] to generate z_{2}[n], and than finally time reverse z_{2}[n] to obtain y[n]. | |

First advance x[n] by 3 samples to generate z_{1}[n], then pick every 4^{th} sample of z_{1}[n] to generate z_{2}[n], and then finally time reverse z_{2}[n] to obtain y[n] | |

First pick every fourth sample of x[n] to generate v_{1}[n], time-reverse v_{1}[n] to obtain v_{2}[n], and finally advance v_{2}[n] by 3 samples to obtain y[n] | |

First pick every fourth sample of x[n] to generate v_{1}[n], time-reverse v_{1}[n] to obtain v_{2}[n], and finally delay v_{2}[n] by 3 samples to obtain y[n] |

Question 10 Explanation:

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]

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]

There are 10 questions to complete.