Electronics (ECE) - MCQ Practice Questions

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.

Understanding:

We must compute the FM modulation index.

Formula:

Maximum frequency deviation:

Modulation index:

Step 1: Compute maximum frequency deviation.

Step 2: Compute the modulation index.

Answer:

The FM modulation index is $25$.

Quick Tip:

A common error is to confuse $A_m$ (amplitude, in volts) with $f_m$ (frequency, in Hz). Always extract both from the message signal $m(t)=A_m\cos (2\pi f_{m}t)$ before substituting.

Understanding:

We must find the maximum trackable slope in a delta modulation system to avoid slope overload distortion.

Formula:

In delta modulation, the staircase approximation rises or falls by $\delta$ every sampling interval $T_s=1/f_s$. The maximum slope the system can follow is:

Step 1: Compute the maximum slope.

Answer:

The maximum slope the DM system can track is $6400\text{V/s}$.

Quick Tip:

Slope overload occurs when the message signal changes faster than $\delta f_s$. To avoid it, either increase $\delta$ or increase $f_s$. Adaptive delta modulation (ADM) solves this by varying $\delta$ dynamically.

Understanding:

We must evaluate the argument of the $Q$-function for BPSK bit error probability.

Formula:

For BPSK:

Step 1: Compute the argument.

Step 2: Express the answer.

Since $\sqrt{16}=4$, both $Q(\sqrt{16})$ and $Q(4)$ represent the same numerical value. Option D writes the argument before simplification as $Q(\sqrt{16})$, and option A writes it after simplification as $Q(4)$. The question asks which expression directly follows from substituting $E_b/N_0=8$ into the formula:

This matches the form $Q(\sqrt{16})$.

Answer:

The bit error probability is $Q(\sqrt{16})=Q(4)$, expressed directly from the formula substitution.

Quick Tip:

For BPSK with $E_b/N_0=8(\approx 9\text{dB})$, $P_e=Q(4)\approx 3.2\times 10^{-5}$. BPSK is the most power-efficient binary modulation scheme — remember it requires 3 dB less power than BFSK for the same $P_e$.