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Electronics (ECE) - MCQ Practice Questions

Electronics and Communication questions reward anyone who is comfortable moving between the time domain and the frequency domain. This set covers network theory, analog and digital circuits, signals and systems, control systems, communication systems, and electromagnetics. Numerical solutions keep the units visible at every step, since a dropped factor is the most common reason a correct method still produces a wrong option.

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Q.1Medium

A unity feedback control system has an open-loop transfer function $G (s) = \frac{K}{s ( s + 4 )}$ . For the system to be critically damped, the value of $K$ is:

Answer: B

Understanding:

We need to find $K$ such that the closed-loop system is critically damped (damping ratio $ζ = 1$ ).

•Open-loop transfer function:

G (s) = \frac{K}{s ( s + 4 )}

•Unity feedback system

Formula:

The closed-loop transfer function is:

T (s) = \frac{G ( s )}{1 + G ( s )} = \frac{K}{s ^{2} + 4 s + K}

For a standard second-order system $\frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ , critical damping requires:

ζ = 1

Step 1: Identify parameters from the characteristic equation.

Comparing $s^{2} + 4 s + K$ with $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ :

ω_{n}^{2} 2 ζ ω_{n} = K = 4

Step 2: Apply critical damping condition $ζ = 1$ .

2 (1) ω_{n} ω_{n} = 4 = 2

Step 3: Solve for $K$ .

K = ω_{n}^{2} = 2^{2} = 4

Answer:

The value of $K$ for critical damping is 4.

K = 4

Quick Tip:

For a second-order system, the condition $ζ = 1$ gives repeated real poles. Any $K > 4$ makes it underdamped and $K < 4$ makes it overdamped for this system.

Q.2Medium

For a second-order system, the unit step response has a peak overshoot of $16.3%$ . The damping ratio $ζ$ is approximately:

Answer: A

Understanding:

We need to find the damping ratio $ζ$ from the percentage peak overshoot (% PO).

•% PO

= 16.3%

Formula:

The relationship between peak overshoot and damping ratio is:

% P O = e^{- π ζ / 1 - ζ^{2}} \times 100

Step 1: Set up the equation.

0.163 = e^{- π ζ / 1 - ζ^{2}}

Step 2: Take natural logarithm of both sides.

ln (0.163) - 1.814 = \frac{- π ζ}{1 - ζ ^{2}} = \frac{- π ζ}{1 - ζ ^{2}}

Step 3: Solve for $ζ$ .

1.814 1.81 4^{2} 3.29 (1 - ζ^{2}) 3.29 3.29 ζ^{2} ζ = \frac{π ζ}{1 - ζ ^{2}} = \frac{π ^{2} ζ ^{2}}{1 - ζ ^{2}} = 9.87 ζ^{2} = (9.87 + 3.29) ζ^{2} = 13.16 ζ^{2} = 0.25 = 0.5

Answer:

The damping ratio corresponding to $16.3%$ peak overshoot is $0.5$ .

ζ = 0.5

Quick Tip:

The standard result $% P O \approx 16.3%$ corresponds to $ζ = 0.5$ and is a frequently tested value. Memorise: $ζ = 0.707$ gives $\approx 4.3%$ overshoot (Butterworth), and $ζ = 1$ gives $0%$ overshoot.

Q.3Medium

The closed-loop transfer function of a system is $T (s) = \frac{25}{s ^{2} + 6 s + 25}$ . The settling time (using the $2%$ criterion) is:

Answer: A

Understanding:

We need to find the settling time $T_{s}$ using the $2%$ criterion.

•Closed-loop transfer function:

T (s) = \frac{25}{s ^{2} + 6 s + 25}

Formula:

Comparing with the standard form $\frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ :

T_{s} \approx \frac{4}{ζ ω _{n}} (2% criterion)

Step 1: Extract system parameters.

ω_{n}^{2} 2 ζ ω_{n} = 25 ⟹ ω_{n} = 5 rad/s = 6 ⟹ ζ = \frac{6}{2 \times 5} = 0.6

Step 2: Compute the product $ζ ω_{n}$ .

ζ ω_{n} = 0.6 \times 5 = 3

Step 3: Apply the settling time formula.

T_{s} = \frac{4}{ζ ω _{n}} = \frac{4}{3} \approx 1.33 s

Answer:

The settling time using the $2%$ criterion is approximately $1.33 s$ .

T_{s} \approx 1.33 s

Quick Tip:

For the $5%$ criterion, use $T_{s} \approx 3/ (ζ ω_{n})$ . The product $σ = ζ ω_{n}$ is the real part of the closed-loop poles and directly controls the settling speed.

Q.4Medium

A proportional controller with gain $K_{p}$ is used in a unity feedback system with plant $G (s) = \frac{1}{s ( s + 2 )}$ . The steady-state error to a unit ramp input is:

Answer: A

Understanding:

We need to find the steady-state error $e_{ss}$ for a unit ramp input $r (t) = t u (t)$ with a Type-1 plant and proportional controller.

•Plant:

G (s) = \frac{1}{s ( s + 2 )}

•Controller:

K_{p}

•Input: unit ramp

Formula:

For a unit ramp, the steady-state error is given by the velocity error constant $K_{v}$ :

e_{ss} = \frac{1}{K _{v}}, K_{v} = s \to 0 lim s \cdot G_{O L} (s)

where $G_{O L} (s) = K_{p} \cdot G (s)$ .

Step 1: Compute $K_{v}$ .

K_{v} = s \to 0 lim s \cdot \frac{K _{p}}{s ( s + 2 )} = s \to 0 lim \frac{K _{p}}{s + 2} = \frac{K _{p}}{2}

Step 2: Compute steady-state error.

e_{ss} = \frac{1}{K _{v}} = \frac{1}{K _{p} /2} = \frac{2}{K _{p}}

Answer:

The steady-state error to a unit ramp input is $\frac{2}{K _{p}}$ .

e_{ss} = \frac{2}{K _{p}}

Quick Tip:

A Type-1 system has zero steady-state error to a step input but a finite error to a ramp. A Type-2 system would give zero error to a ramp. Adding an integrator (I-controller) increases system type by 1.

Q.5Medium

The characteristic equation of a system is $s^{3} + 4 s^{2} + K s + 8 = 0$ . Using the Routh–Hurwitz criterion, the system is stable for:

Answer: A

Understanding:

We need to determine the range of $K$ for which all roots of the characteristic equation lie in the left-half $s$ -plane.

•Characteristic equation:

s^{3} + 4 s^{2} + K s + 8 = 0

Formula:

The Routh array for $a s^{3} + b s^{2} + cs + d$ is:

s^{3} s^{2} s^{1} s^{0} a b \frac{b c - a d}{b} d c d 0

Stability requires all first-column elements to be positive.

Step 1: Build the Routh array.

s^{3} s^{2} s^{1} s^{0} 14 \frac{4 K - 8}{4} 8 K 80

Step 2: Simplify the $s^{1}$ row entry.

\frac{4 K - 8}{4} = K - 2

Step 3: Apply stability conditions (all first-column elements $> 0$ ).

s^{2} : 4 > 0 ✓ s^{1} : K - 2 > 0 ⟹ K > 2 s^{0} : 8 > 0 ✓

Answer:

The system is stable for $K > 2$ .

K > 2

Quick Tip:

A sign change in the first column of the Routh array indicates unstable roots. The number of sign changes equals the number of roots in the right-half $s$ -plane.

Q.6Medium

The open-loop transfer function is $G (s) = \frac{K ( s + 3 )}{s ( s + 1 ) ( s + 2 )}$ . Using the root locus technique, the number of asymptotes is:

Answer: A

Understanding:

We need to find the number of asymptotes in the root locus of the given open-loop transfer function.

•

G (s) = \frac{K ( s + 3 )}{s ( s + 1 ) ( s + 2 )}

•Number of poles:

P

, Number of zeros:

Z

Formula:

The number of asymptotes in the root locus equals:

N_{A} = P - Z

where $P$ = number of finite open-loop poles and $Z$ = number of finite open-loop zeros.

Step 1: Count finite poles.

Poles are at $s = 0, - 1, - 2$ , so $P = 3$ .

Step 2: Count finite zeros.

Zero is at $s = - 3$ , so $Z = 1$ .

Step 3: Compute number of asymptotes.

N_{A} = P - Z = 3 - 1 = 2

Answer:

The number of asymptotes in the root locus is $2$ .

N_{A} = 2

Quick Tip:

The asymptote angles are $θ_{k} = \frac{( 2 k + 1 ) \times 180°}{P - Z}$ for $k = 0, 1, \dots, (P - Z - 1)$ . For $P - Z = 2$ , the angles are $90°$ and $270°$ (i.e., $\pm 90°$ ).

Q.7Medium

In the Bode plot of a transfer function $G (s) = \frac{100}{( 1 + 0.1 s ) ( 1 + 0.01 s )}$ , the initial magnitude (at very low frequencies) in dB is:

Answer: A

Understanding:

We need to find the DC gain (magnitude at $ω \to 0$ ) of $G (s) = \frac{100}{( 1 + 0.1 s ) ( 1 + 0.01 s )}$ in dB.

Formula:

The DC gain is found by setting $s = j ω$ and letting $ω \to 0$ , then converting to dB:

∣ G ∣_{d B} = 20 lo g_{10} ∣ G (0) ∣

Step 1: Compute $G (0)$ .

G (0) = \frac{100}{( 1 + 0 ) ( 1 + 0 )} = 100

Step 2: Convert to dB.

∣ G (0) ∣_{d B} = 20 lo g_{10} (100) = 20 \times 2 = 40 dB

Answer:

The initial magnitude at low frequencies is $40 dB$ .

40 dB

Quick Tip:

On a Bode magnitude plot, the low-frequency asymptote is determined solely by the DC gain $K$ . The corner frequencies of the lag terms ( $ω_{c} = 10 rad/s$ and $ω_{c} = 100 rad/s$ ) only affect the slope at higher frequencies.

Q.8Medium

A system has the transfer function $T (s) = \frac{s + 3}{( s + 1 ) ( s + 2 )}$ . The final value of the step response $y (\infty)$ is:

Answer: A

Understanding:

We need to find $y (\infty)$ when the input is a unit step $R (s) = 1/ s$ .

•Transfer function:

T (s) = \frac{s + 3}{( s + 1 ) ( s + 2 )}

Formula:

By the Final Value Theorem (FVT), provided all poles of $s Y (s)$ are in the left-half plane:

y (\infty) = s \to 0 lim s \cdot Y (s) = s \to 0 lim s \cdot T (s) \cdot R (s)

Step 1: Form $s Y (s)$ for a unit step input.

s Y (s) = s \cdot T (s) \cdot \frac{1}{s} = T (s)

Step 2: Apply the FVT.

y (\infty) = s \to 0 lim T (s) = \frac{0 + 3}{( 0 + 1 ) ( 0 + 2 )} = \frac{3}{1 \times 2} = \frac{3}{2}

Step 3: Verify FVT is applicable.

Poles of $Y (s) = T (s) / s$ are at $s = 0, - 1, - 2$ . The pole at $s = 0$ is cancelled by the $s$ in $s Y (s)$ , and the remaining poles $s = - 1, - 2$ are in the left-half plane, so FVT applies.

Answer:

The final value of the step response is $\frac{3}{2}$ .

y (\infty) = \frac{3}{2}

Quick Tip:

The final value of the step response equals the DC gain $T (0)$ of the transfer function, provided the system is stable. This is a quick check: just substitute $s = 0$ into $T (s)$ .

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