Electronics & Communications Test - 2

Concept:

Image Frequency

• It is an undesired input frequency at the receiver end which can also be demodulated by the superheterodyne receivers along with the desired incoming signal. This results in two stations being received at the same time, resulting in interference.
• Image frequency is given by f_i = f_RF + 2f_IF­, where f_RF = frequency of desired incoming signal and

f_IF = Intermediate frequency.

Graphically,

Calculation:

f_i = f_RF + 2f_IF­

f_i = 900 + 2(450) = 1800 kHz

Concept:

The number of flip flops ‘n’ is to be selected such that the number of states N is less than or equal to 2ⁿ, i.e.

N≤ 2ⁿ

With ‘n’ Flip-Flops, the largest count possible is 2ⁿ-1.

Calculation:

2ⁿ-1 = 16,383 (Given)

2ⁿ = 16,384

So, the number of Flip-Flops required will be 14.

Also, the frequency of the output of the last stage (MSB) is:

Concept:

The SOP representation of the circuit is:

F = Σm (minterms)

Minterm: a minterm of n variables is a product of the variables in which each appears exactly once in true or complemented form.

The POS representation of the circuit:

F = ΠM (max terms)

Maxterm: a maxterm of n variables is a sum of the variables in which each appears exactly once in true or complemented form.

Calculation:

Given,

Hence, the function in the form of minterms is expressed as:

f(A, B, C) = Σm (0,1,3,5,7)

Now, we put 0 in each block of the K-map excluding the blocks corresponding to the terms in the above function.

Concept:

f(z) = a₀ + a, (z - a) + a₂ (z - a)² + …+a_-1 (z - a)^-1…

The coefficient of (z – a)-1 in the expansion of f(z) around on isolated singularity is called the residue of f(z) at that point.

Thus, the Laurent series expansion of f(z) around z = a, i.e.

The residue of f(z) at z = a is a_-1.

Application:

Given:

Concept:

Bode plot transfer function is represented in standard time constant form as 

ω_c1, ω_c2, … are corner frequencies.

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Where Z is the number zeros and P is the number of poles

Application:

The initial point of Bode plot is 20 log k

∴ 20 log k = 60

log k = 3

⇒ k = 1000

The initial slope of the magnitude plot is zero. Therefore, there are no poles or zeros at the origin.

The slope change is negative at, ω = 10. This slope is given by

The correct answer is High pass filter

Solution:

At ω = 0

Inductor becomes a short circuit and capacitance becomes an open circuit

∴ here V₀ = 0, we can say a lower frequency signal will not pass

At ω = ∞

Inductor becomes an open circuit and capacitance becomes a short circuit

∴ here V₀ = V_i we can say at a higher frequency signal will pass

∴ this circuit will work as a High pass filter.

Concept:

Typically for an NPN transistor;

V_BEon = 0.7 V

If it is assumed base current (I_B) can be neglected (for high β), then

I_C = I_E 

Calculation:

To determine the states of the diode; perform an open circuit test: