Question 1 |

Which of the following is true?

A finite signal is always time bounded | |

A time bounded signal always possesses finite energy | |

A bounded signal is always zero outside the interval [-t_0,t_0
] for some t_0
| |

A time bounded signal is always finite |

Question 2 |

x(t) is a real valued function of a real variable with period T. Its trigonometric
Fourier Series expansion contains no terms of frequency \omega = 2\pi(2k)/T; k=1,2... Also, no sine terms are present. Then x(t) satisfies the equation

x(t)=-x(t-T) | |

x(t)=x(T-t)=-x(-t) | |

x(t)=x(T-t)=-x(t-T/2) | |

x(t)=x(t-T)=x(t-T/2) |

Question 2 Explanation:

Since trigonometric fourier series has no sine terms and has only cosine terms therefore this will be an even signal i.e. it will satisfy.

x(t)=x(-t)

or, we can write,

x(t-T)=x(-t+T)

but signal is periodic with period T.

therefore x(t-T)=x(t)

therefore, x(t)=x(T-t)\;\;...(i)

Now, since signal contains only odd harmonics i.e. no terms of frequency

\omega =\frac{2 \pi \times 2k}{T},\;\;\;k=1,2,3,4

i.e. no even harmonics.

This means signal contains half wave symmetry

this implies that,

x(t)=-x\left ( t-\frac{T}{2} \right )\;\;...(ii)

From eq. (i) and (ii),

x(t)=x(T-t)=-x\left ( t-\frac{T}{2} \right )

x(t)=x(-t)

or, we can write,

x(t-T)=x(-t+T)

but signal is periodic with period T.

therefore x(t-T)=x(t)

therefore, x(t)=x(T-t)\;\;...(i)

Now, since signal contains only odd harmonics i.e. no terms of frequency

\omega =\frac{2 \pi \times 2k}{T},\;\;\;k=1,2,3,4

i.e. no even harmonics.

This means signal contains half wave symmetry

this implies that,

x(t)=-x\left ( t-\frac{T}{2} \right )\;\;...(ii)

From eq. (i) and (ii),

x(t)=x(T-t)=-x\left ( t-\frac{T}{2} \right )

Question 3 |

In the figure the current source is 1\angle 0 A, R=1\Omega, the impedances are Z_{C} =-j \Omega and Z_{L}= 2j \Omega. The Thevenin equivalent looking into the circuit across X-Y is

\sqrt{2}\angle 0V , (1+2j)\Omega | |

2\angle 45^{\circ}V , (1-2j)\Omega | |

2\angle 45^{\circ}V , (1+j)\Omega | |

\sqrt{2}\angle 45^{\circ}V , (1+j)\Omega |

Question 3 Explanation:

To calculate Thevenin's impedance, current-souce is open-circuited

Z_{th}=R+Z_L+Z_c

\;\;=1+2j-j

\;\;=1+j \Omega

Open-circuit voltage at terminals X-Y

=I \times Z_{th}

\;\;=1\angle 0 \times (1+j)

\;\;=\sqrt{2}\angle 15^{\circ} volts

Z_{th}=R+Z_L+Z_c

\;\;=1+2j-j

\;\;=1+j \Omega

Open-circuit voltage at terminals X-Y

=I \times Z_{th}

\;\;=1\angle 0 \times (1+j)

\;\;=\sqrt{2}\angle 15^{\circ} volts

Question 4 |

The three limbed non ideal core shown in the figure has three windings with
nominal inductances L each when measured individually with a single phase AC
source. The inductance of the windings as connected will be

very low | |

L/3 | |

3L | |

very high |

Question 5 |

Which of the following statement holds for the divergence of electric and magnetic flux densities ?

Both are zero | |

These are zero for static densities but non zero for time varying densities. | |

It is zero for the electric flux density | |

It is zero for the magnetic flux density |

Question 5 Explanation:

The divergence of magnetic field is always zzero because magnetic flux makes always a closed path.

So, \bigtriangledown \cdot B=0 (Maxweel's equation)

while divergence of electric field,

\bigtriangledown \cdot \vec{E}=\frac{\rho _v}{\in}

So, \bigtriangledown \cdot B=0 (Maxweel's equation)

while divergence of electric field,

\bigtriangledown \cdot \vec{E}=\frac{\rho _v}{\in}

There are 5 questions to complete.