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.