Question 1 |

Which of the following statements is true about the two sided Laplace transform?

It exists for every signal that may or may not have a Fourier transform. | |

It has no poles for any bounded signal that is non-zero only inside a finite time
interval. | |

The number of finite poles and finite zeroes must be equal. | |

If a signal can be expressed as a weighted sum of shifted one sided exponentials,
then its Laplace Transform will have no poles. |

Question 1 Explanation:

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).

It implies for such signals there is no poles. Hence the correct answer is option (B).

Question 2 |

The output response of a system is denoted as y(t), and its Laplace transform is given by

Y(s)=\frac{10}{s(s^2+s+100\sqrt{2})}

The steady state value of y(t) is

Y(s)=\frac{10}{s(s^2+s+100\sqrt{2})}

The steady state value of y(t) is

\frac{1}{10\sqrt{2}} | |

10\sqrt{2} | |

\frac{1}{100\sqrt{2}} | |

100\sqrt{2} |

Question 2 Explanation:

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}

\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}

Question 3 |

A system transfer function is H(s)=\frac{a_1s^2+b_1s+c_1}{a_2s^2+b_2s+c_2}. If a_1=b_1=0, and all other coefficients are positive, the transfer function represents a

low pass filter | |

high pass filter | |

band pass filter | |

notch filter |

Question 3 Explanation:

\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.

which represents second order low pass filter.

Question 4 |

The inverse Laplace transform of

H(s)=\frac{s+3}{s^2+2s+1} \; for \; t\geq 0 is

H(s)=\frac{s+3}{s^2+2s+1} \; for \; t\geq 0 is

3te^{-t}+e^{-t} | |

3e^{-t} | |

2te^{-t}+e^{-t} | |

4te^{-t}+e^{-t} |

Question 4 Explanation:

\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}

Question 5 |

Consider a linear time-invariant system with transfer function

H(s)=\frac{1}{(s+1)}

If the input is cos(t) and the steady state output is A \cos (t+\alpha ), then the value of A is _________.

H(s)=\frac{1}{(s+1)}

If the input is cos(t) and the steady state output is A \cos (t+\alpha ), then the value of A is _________.

0.70 | |

0.26 | |

0.96 | |

1.2 |

Question 5 Explanation:

\begin{aligned} H(s)&=\frac{1}{(s+1)}\\ \text{Put, }s&=j\omega \\ H(j\omega)&=\frac{1}{j\omega+1}\\ |H(j\omega)|&=\frac{1}{\sqrt{\omega ^2 +1}}\\ \because \; \text{ input }x(t)&=\cos (t)\\ \text{Here, } \omega &=1 \text{rad/sec}\\ \text{and }|x(t)|&=1\\ \text{hence, }&\text{steady state output}\\ y(t)&=|H(j\omega)|_{\omega=1} \cos (t+\angle H(j\omega)|_{\omega =1})\\ A&=|H(j\omega)|_{\omega =1}=\frac{1}{\sqrt{2}}=0.707 \end{aligned}

There are 5 questions to complete.