Electronics (ECE) - MCQ Practice Questions

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.

Understanding:

We must find the applied pressure from the measured capacitance of a capacitive sensor.

Formula:

The change in capacitance is related to pressure by:

so:

Step 1: Compute the change in capacitance.

Step 2: Compute the applied pressure.

Answer:

The applied pressure is $30\text{kPa}$.

Quick Tip:

Capacitive sensors are preferred over piezoresistive ones for low-frequency or static pressure measurements due to their low power consumption and high sensitivity.

Understanding:

We must find the unknown voltage $V_x$ measured by a dual-slope ADC.

Formula:

In the dual-slope integration method, the unknown voltage is related to the reference by:

Step 1: Substitute the known values.

Answer:

The unknown input voltage is $6.5\text{V}$.

Quick Tip:

The key advantage of dual-slope integration is that the result depends only on the ratio $T_2/T_1$, making it immune to variations in the integrator's RC time constant.

Understanding:

We must find the percentage of the final value reached by a first-order instrument at $t=4\text{s}$.

Formula:

The step response of a first-order system is:

The percentage of final value is:

Step 1: Compute $t/\tau$.

Step 2: Evaluate the exponential term.

Step 3: Compute the percentage.

Answer:

After $4\text{s}$ (i.e., $2\tau$), the output reaches approximately $86.5\%$ of the final value.

Quick Tip:

Memorize these standard milestones for first-order systems: at $1\tau$ → $63.2\%$, at $2\tau$ → $86.5\%$, at $3\tau$ → $95.0\%$, at $5\tau$ → $99.3\%$.

Understanding:

We must find the total power of an AM signal.

Formula:

The total power of a standard AM (DSB-FC) signal is:

Step 1: Substitute the given values.

Answer:

The total power of the AM signal is $450\text{W}$.

Quick Tip:

The sideband power alone is $P_c{\mu }^2/2$. For $\mu =1$ (100% modulation), total power is $1.5P_c$ — a commonly tested value.

Understanding:

We must find the image frequency for a superheterodyne receiver with high-side LO injection.

Formula:

With high-side injection, the local oscillator frequency is:

The image frequency is located at:

Step 1: Compute $f_{LO}$.

Step 2: Compute the image frequency.

Answer:

The image frequency is $921.4\text{MHz}$.

Quick Tip:

The image frequency is always $2f_{IF}$ away from the desired carrier — on the same side as the LO. Image rejection is the primary reason for the RF bandpass filter before mixing.

Understanding:

We must find the bandwidth of an FM signal using Carson's rule.

Formula:

Carson's rule gives the approximate FM bandwidth as:

Step 1: Identify the modulation index.

This is wideband FM.

Step 2: Apply Carson's rule.

Answer:

The approximate bandwidth of the FM signal is $180\text{kHz}$.

Quick Tip:

For commercial FM broadcasting, $\Delta f=75\text{kHz}$ and $W=15\text{kHz}$ are the standard values, giving exactly $180\text{kHz}$ — a number worth memorising for GATE and ESE.

Understanding:

We must compare the output SNR of a standard AM (DSB-FC) receiver to that of a DSB-SC receiver for equal total transmitted power at $\mu =1$.

Formula:

For DSB-SC, the figure of merit (output SNR / channel SNR) is $1$.

For standard AM with modulation index $\mu$, the fraction of total power in the sidebands is:

The figure of merit of AM relative to DSB-SC is $\eta$.

Step 1: Compute sideband power fraction at $\mu =1$.

Step 2: Interpret the result.

Only $1/3$ of the total transmitted power is in the information-bearing sidebands; the rest ($2/3$) is wasted in the carrier. Hence the output SNR of AM is $1/3$ that of DSB-SC for the same total power.

Answer:

The figure-of-merit of standard AM relative to DSB-SC at $\mu =1$ is $\frac{1}{3}$.

Quick Tip:

At 50% modulation ($\mu =0.5$), $\eta =0.25/1.25=1/5$, making AM even less efficient — demonstrating why DSB-SC and SSB are preferred for power-limited links.

Understanding:

We must find the minimum bandwidth for a binary PCM channel.

Formula:

The bit rate of the PCM signal is:

For binary (2-level) Nyquist signalling, the minimum bandwidth is:

Step 1: Compute the bit rate.

Step 2: Compute the minimum bandwidth.

Answer:

The minimum channel bandwidth required is $32\text{kHz}$.

Quick Tip:

Nyquist's theorem states that a channel of bandwidth $B$ can support at most $2B$ symbols per second. For binary signalling this gives $B_{\min }=R_b/2$. For $M$-ary signalling, $B_{\min }=R_b/(2{\log }_{2}M)$.

Understanding:

We must find the Shannon channel capacity.

Formula:

Shannon–Hartley theorem:

Step 1: Evaluate the logarithm.

Step 2: Compute the capacity.

Answer:

The channel capacity is $16\text{kbps}$.

Quick Tip:

Memorize that ${\log }_2(2^n)=n$. Examiners often choose SNR values that are powers of $2$ minus $1$ (e.g., $3,7,15,31$) to make the log term a clean integer.

Understanding:

We must find the hold-in (lock) range of a first-order PLL.

Formula:

For a first-order PLL (no loop filter, $F(s)=1$), the open-loop DC gain is:

The hold-in range in Hz is:

Step 1: Compute the loop gain $K_L$.

Step 2: Convert to frequency (Hz).

So the hold-in range is $\pm 5\text{kHz}$.

Answer:

The hold-in range of the first-order PLL is $\pm 5\text{kHz}$.

Quick Tip:

The hold-in range equals the open-loop DC gain $K_L$ expressed in Hz. For higher-order PLLs with large loop filter DC gain, the hold-in range can be much larger than the pull-in range.