Test: ESE Electrical - 6 - Electrical Engineering (EE) MCQ

After killing all source,

If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent

Let V_o be the voltage across current source

V_o(20 + j10) - (20 + j40) V_x = j600

 ​

   .......(i)
 ........(ii)

As the line voltage V_RY = V∠0⁰ is taken as a reference phasor. Then the source voltage V_BR is V∠-240⁰.

The term power is defined as the product of square of current and the impedance. So the power in the B phase = 40² x 0 = 0W.

The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.

The electric field intensity is the product of the velocity and the magnetic flux density ie, E = v x B = 12 x 8.75 = 105 units.

Given: h(t) = u(t)
x(t) = e^–at u(t)

∴ Y(s) = X(s) H(s)
=   

y(t) = 1/a (1 – e^{– at}) u(t)

First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z) = z2/(z2-1.5z + 0.5)
The poles of X(z) are p1 = 1 and p2 = 0.5. Consequently, the expansion will be
(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)
⇒ X(z)= 2z/(z-1)-z/(z-0.5).

To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,
X(z)=(z(z+1))/(z^-2-z+0.5)
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j).

CASE 1: Let us assume that x(t) is real, then x(t) = x*(t). This implies that for x(t) to be real, ak = a*_-k. Since this is not true. So, x(t) is not real.
CASE 2: Let us assume that x(t) is even, then x(t) = x(-t) and a_k = a_–k. Since this is true for this case, hence x(t) is even.
CASE 3: Let us assume

Therefore,

Since bk is not even, So, g(t) is not even.

We know that,

Given form is,

So, we can write it as,

Comparing both equations, we get,

Taking Inverse Fourier Transform,

Comparing with given signal, g(t) = Ay (Bt)
A = 1/3 and B = 3

Final value theorem,

Calculation:
We have:

and x(t) = 2 + 5sin(100πt)
sampling frequency, f_s = 400 Hz
put t = nT_s, the output of the sampling process is,
x(nT_s) = 2 + sin(100πnT_s)
x(nT_s) = 2 + 5 sin

where,
ω₀ = π/4

The z-transform of x(n) is,


From equation (i)

using final value theorem,


y(∞) = 0

From the question, it is given that
f₁(n)=x(2n)
f₂(n)=x(2n+1) ,n=0,1,2…N/2-1

X(k) = F₁(k)+WNk_N^k F₂(k)

We know that the rectangular window of length M-1 is defined as
w(n)= 1, n=0,1,2…M-1
=0, else where.

• The direct form realization follows immediately from the non-recursive difference equation given by 
• We observe that this structure requires M-1 memory locations for storing the M-1 previous inputs.

The CLTF is

Using mason’s gain formula transfer functionThe given signal flow graph can be represented by the transfer function:

T(s) = (abcd + efg) / (1 - cd - fg - cdfg)

We can rearrange this equation as follows:

T(s) = (abcd + efg) / (1 - (cd + fg) - cdfg)

We need to find the transfer function using Mason's Gain formula. Mason's Gain formula is given by:

T(s) = Σ(Pk * Δk) / Δ

where T(s) is the transfer function, Pk is the gain of the kth forward path, Δk is the determinant of the kth forward path, and Δ is the overall determinant.

Step 1: Identify the forward paths and their gains
- Forward path 1: abcd, gain = abcd
- Forward path 2: efg, gain = efg

Step 2: Identify the loops and their gains
- Loop 1: cd, gain = cd
- Loop 2: fg, gain = fg

Step 3: Calculate the overall determinant (Δ)
Δ = 1 - (sum of loop gains) + (sum of product of gains of non-touching loops) - ...
Δ = 1 - (cd + fg) + (0) - ...
Δ = 1 - cd - fg

Step 4: Calculate the determinants for each forward path (Δk)
Δ1 = 1 - fg (since loop 1 touches forward path 1)
Δ2 = 1 - cd (since loop 2 touches forward path 2)

Step 5: Calculate the transfer function using Mason's Gain formula
T(s) = (abcd * Δ1 + efg * Δ2) / Δ
T(s) = (abcd * (1 - fg) + efg * (1 - cd)) / (1 - cd - fg)

Now, let's compare the given options with the calculated transfer function:

Option A: abcd + efg / (1 - cd - fg - cdfg)
Option B: abcd(1 - fg) + efg(1 - cd) / (1 - cd - fg)
Option C: abef + bcd / (1 - cd - fg - cdfg)
Option D: adcdefg / (1 - cd - fg - cdfg)

The correct answer is Option B, as it matches the calculated transfer function. from signal flow graph can be calculated which relates the forward path gain to the various paths and loops.

The characteristic equation of the given closed loop system is

Comparing above equation with

Thus, setting time for 2% tolerance band is

or, 

Given characteristic equation is,
s² + 2Ks² + (K + 2)s + 4 = 0
Routh’s array is:

For stability, 2K > 0 or K > 0

Now, K² + 2K - 2 = 0 or, K = 0.73, -2.73
For K > 0.73, K² + 2 K - 2 > 0
(since K should be > 0)
Hence, system will be stable if K > 0.73.

Here, P = 4,Z = 1, P - Z = 3
No. of branches of RL terminating at zero = 1.
No. of branches of RL terminatina at infinity = 3

Since out of the given option, real axis between s = -2 and s = -4 is only a part of root locus, therefore breakaway point will lie between -2 and -4.

Given, G(s) = 1/(1+Ts)
It is a type-0 and order-1 system, therefore its polar-plot will be as shown below.

Hence, the polar plot will be semicircular in shape.