GATE EE 2008

Number of branches =b =6
No. of nodes = n= 4
No. of chords =b-(n-1) =6-(4-1)=3

Thevenin's Impedance:
Z_0=2.38-j0.667\Omega
as real part is not zero, so Z_0 has resistor
Img[Z_0]=-j0.667
CASE-I:
Z_0 has capacitor (as Img[Z_0] is negative)
CASE-II:
Z_0 has both capacitor and inductor, but inductive reactance \lt capacitive reactance.
At, \omega=5 rad/sec
For minimal realization case-I is considered. Therefore, Z_0 will have a resistor and a capacitor.

x is uniformly distributes in [0,1]
Therefore, probability density function
\begin{aligned} f(x)&=\frac{1}{b-a} =\frac{1}{1-0}=1\\ \therefore \; f(x) &=1\;\;\;0 \lt x \lt 1 \\ &=0\;\;\;\text{elsewhere} \\ \text{Now, } E(x^3)&=\int_{0}^{1}x^3f(x)\; dx \\ &= \int_{0}^{1}x^3 \times 1 \times dx\\ &= \left [ \frac{x^4}{4} \right ]_0^1=\frac{1}{4} \end{aligned}

If characteristic equation is
\lambda ^3+\lambda ^2+2\lambda +1=0
Then by cayley- hamilton theorem,
P^3+P^2+2P+I=0
I=-P^3-P^2-2P
Multiplying by P^{-1} on both sides,
P^{-1}=-P^2-P-2I=-(P^2+P+2I)

If rank of (5 x 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as will as 4 linearly independent columns.

Taking (L.T.) on both sides
Y(s)(s+1)=1
\therefore \;\; Y(s)=\frac{1}{s+1}
Taking inverse laplace transform
y(t)=e^{-t}u(t)

V_P=\frac{10}{10+10}\times 10sin\omega t
\;\; =5sin\omega t
Since maximum voltage across at the point P may be 5 V, hence voltage across the diode always will be less than or equal to zero. So it will be reversed always.

\therefore V_0=\frac{10}{10+10}\times 10sin\omega t=5sin\omega t

Single slope integrating type ADC utilize digital counter techniques to measure time required for a voltage ramp to rise from zero to the input voltage.
If conversion time for input voltage 5V=T_A
So, conversion time for input voltage 2.5V=T_A/2
Conversion time in successive type ADC does not depend on input voltage. So, conversion time for input voltage 2.5V is also T_B.