Electronics (ECE) - MCQ Practice Questions

H(z) = (1 + z^(-1))/2. Setting numerator to zero: 1 + z^(-1) = 0 gives z = -1.

For a signal, power P = R(0) = autocorrelation at lag 0.

At ω = ωc (cutoff frequency), |H(jωc)| = 1/√(1+1) = 1/√2. This is the -3dB point.

The signal contains frequencies at 500 Hz and 1500 Hz. Maximum frequency is 1500 Hz, so Nyquist frequency required = 3 kHz. But sampling at 4 kHz gives Nyquist frequency = 2 kHz. Since 1500 Hz < 2 kHz, no aliasing occurs. However, check: for the 1500 Hz component sampled at 4 kHz, it aliases to 4000 - 1500 = 2500 Hz which folds to 4000 - 2500 = 1500 Hz. Actually, Nyquist = fs/2 = 2 kHz. The 1500 Hz signal is below 2 kHz, so no aliasing. Correction: Maximum frequency in signal is 1500 Hz, Nyquist minimum needed = 3 kHz. Sampling at 4 kHz gives Nyquist = 2 kHz < 3 kHz, so aliasing WILL occur. Option B is correct.

The autocorrelation at τ=0 gives the total average power: R(0) = 10 + 8cos(0) = 10 + 8 = 18 W. Wait—rechecking: Power = R(0) = 10 + 8 = 18 W. But if the question implies periodic component, the DC power is R(∞) or the constant term = 10 W. Standard definition: total power = R(0) = 18 W. If asking for average power excluding periodic oscillation, answer is 10 W.

Understanding:

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

Formula:

The closed-loop transfer function is:

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

Step 1: Identify parameters from the characteristic equation.

Comparing $s^2+4s+K$ with $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$:

Step 2: Apply critical damping condition $\zeta =1$.

Step 3: Solve for $K$.

Answer:

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

Quick Tip:

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

Understanding:

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

Formula:

The relationship between peak overshoot and damping ratio is:

Step 1: Set up the equation.

Step 2: Take natural logarithm of both sides.

Step 3: Solve for $\zeta$.

Answer:

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

Quick Tip:

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

Understanding:

We need to find the settling time $T_s$ using the $2\%$ criterion.

Formula:

Comparing with the standard form $\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$:

Step 1: Extract system parameters.

Step 2: Compute the product $\zeta {\omega }_n$.

Step 3: Apply the settling time formula.

Answer:

The settling time using the $2\%$ criterion is approximately $1.33\text{s}$.

Quick Tip:

For the $5\%$ criterion, use $T_s\approx 3/(\zeta {\omega }_n)$. The product $\sigma =\zeta {\omega }_n$ is the real part of the closed-loop poles and directly controls the settling speed.

Understanding:

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

Formula:

For a unit ramp, the steady-state error is given by the velocity error constant $K_v$:

where $G_{OL}(s)=K_p\cdot G(s)$.

Step 1: Compute $K_v$.

Step 2: Compute steady-state error.

Answer:

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

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.

Understanding:

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

Formula:

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

Stability requires all first-column elements to be positive.

Step 1: Build the Routh array.

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

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

Answer:

The system is stable for $K>2$.

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.

Understanding:

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

Formula:

The number of asymptotes in the root locus equals:

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.

Answer:

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

Quick Tip:

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

Understanding:

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

Formula:

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

Step 1: Compute $G(0)$.

Step 2: Convert to dB.

Answer:

The initial magnitude at low frequencies is $40\text{dB}$.

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 (${\omega }_c=10\text{rad/s}$ and ${\omega }_c=100\text{rad/s}$) only affect the slope at higher frequencies.

Understanding:

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

Formula:

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

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

Step 2: Apply the FVT.

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 $sY(s)$, 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 $\frac{3}{2}$.

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 $T(s)$.

Understanding:

We must find the hot junction temperature of a thermocouple.

Formula:

The thermocouple EMF relates to the temperature difference by:

so the hot junction temperature is:

Step 1: Compute the temperature difference.

Step 2: Add the cold junction temperature.

Answer:

The hot junction temperature is $525\text{°C}$.

Quick Tip:

Always add the cold junction temperature to the computed $\Delta T$. Forgetting this step is the most common error in thermocouple problems.

Understanding:

We must find the unknown arm $S$ of a balanced Wheatstone bridge.

Formula:

For a balanced Wheatstone bridge, the balance condition is:

Rearranging:

Step 1: Substitute the known values.

Answer:

The unknown resistance $S$ is $300\Omega$.

Quick Tip:

The balance condition $P\cdot S=Q\cdot R$ (cross-multiplication) is the fastest check — no current flows through the galvanometer when this holds.

Understanding:

We must find the resistance of an RTD at $200\text{°C}$.

Formula:

The resistance of an RTD at temperature $T$ is:

Step 1: Substitute values.

Answer:

The resistance of the RTD at $200\text{°C}$ is $177.0\Omega$.

Quick Tip:

The Pt100 RTD (most common type) has exactly $R_0=100\Omega$ and $\alpha =0.00385{\text{°C}}^{-1}$ — memorise these standard values.

Understanding:

We must find the sensitivity of an LVDT given its output voltage and core displacement.

Formula:

Sensitivity of an LVDT is defined as:

Step 1: Substitute the values.

Answer:

The sensitivity of the LVDT is $0.5\text{V/mm}$.

Quick Tip:

LVDT sensitivity is typically expressed in $\text{mV/mm}$ or $\text{V/mm}$. A higher sensitivity means the LVDT can detect smaller displacements more accurately.

Understanding:

We must find the resolution (LSB size) of a 12-bit ADC.

Formula:

The resolution (LSB size) of an ADC is:

Step 1: Compute $2^n$.

Step 2: Compute the LSB size.

Answer:

The resolution of the 12-bit ADC is approximately $1.22\text{mV}$.

Quick Tip:

For an $n$-bit ADC with $V_{ref}=5\text{V}$, the LSB halves with every additional bit. A 10-bit ADC gives $\approx 4.88\text{mV}$; an 11-bit gives $\approx 2.44\text{mV}$; 12-bit gives $\approx 1.22\text{mV}$.

Understanding:

We must find the change in resistance of a strain gauge under applied strain.

Formula:

The gauge factor relates fractional resistance change to strain:

Rearranging:

Step 1: Substitute the values.

Answer:

The change in resistance of the strain gauge is $0.24\Omega$.

Quick Tip:

Metallic strain gauges have $GF\approx 2$, while semiconductor strain gauges have $GF$ in the range of $100$–$150$, making them far more sensitive but also more temperature-dependent.

Understanding:

We must find the output of an instrumentation amplifier given gain, input offset, and sensor voltage.

Formula:

The effective input to the amplifier is the sum of the sensor voltage and the offset. The output is:

Step 1: Compute the effective input.

Step 2: Compute the output.

Answer:

The amplifier output voltage is $-20\text{V}$.

Quick Tip:

Input offset errors are amplified by the full gain of the amplifier. Always null or calibrate the offset before applying high gain in precision instrumentation.