What is the maximum entropy of a source with $M$ equiprobable symbols?

$$H_{max} = \log_2 M \ \text{bits/symbol}$$ achieved when $p_i=\frac{1}{M}$ for all $i$. Any nonuniform distribution gives lower entropy.

Define the self-information (information content) of an event of probability $p$.

$$I = \log_2\frac{1}{p} = -\log_2 p \ \text{bits}$$ Rare events carry more information; a certain event ($p=1$) carries zero information.

Define information rate of a source in terms of entropy and symbol rate.

$$R = r\,H(X) \ \text{bits/second}$$ where $r$ is the symbol rate (symbols/s) and $H(X)$ is entropy (bits/symbol).

Define mutual information $I(X;Y)$ between channel input $X$ and output $Y$.

$$I(X;Y) = H(X) - H(X\mid Y) = H(Y) - H(Y\mid X)$$ It measures the average information about $X$ gained by observing $Y$; it is symmetric and nonnegative.

State the Shannon-Hartley channel capacity theorem for an AWGN channel.

$$C = B\log_2\!\left(1+\frac{S}{N}\right) \ \text{bits/second}$$ where $B$ is bandwidth (Hz) and $S/N$ is the signal-to-noise power ratio. It is the maximum error-free data rate.

Define channel capacity of a discrete channel in terms of mutual information.

$$C = \max_{p(x)} I(X;Y)$$ the maximum mutual information over all input probability distributions, in bits per channel use.

What is the capacity of a binary symmetric channel (BSC) with crossover probability $p$?

$$C = 1 - H(p) = 1 + p\log_2 p + (1-p)\log_2(1-p) \ \text{bits/use}$$ Capacity is 1 at $p=0$ or $p=1$, and 0 at $p=\frac{1}{2}$ (useless channel).

List the three main steps of Pulse Code Modulation (PCM).

1) Sampling (band-limited signal sampled at $f_s \geq 2W$, Nyquist rate). 2) Quantization (mapping amplitudes to $L$ discrete levels). 3) Encoding (each level coded into $n=\log_2 L$ binary bits).

For a uniform quantizer with $L$ levels over range $2V$, give the step size and the maximum SQNR.

Step size $\Delta = \frac{2V}{L}$. Quantization noise power $=\frac{\Delta^2}{12}$. For $n$-bit PCM ($L=2^n$): $$\text{SQNR}_{dB} \approx 6.02n + 1.76\ \text{dB}$$ (sinusoidal input); each extra bit adds about 6 dB.

What is the bit rate of a PCM system?

$$R_b = n \cdot f_s \ \text{bits/s}$$ where $n=\log_2 L$ bits per sample and $f_s$ is the sampling rate. Bandwidth required is at least $\frac{R_b}{2}$.

🇮🇳 GATE E&C Engineering · flashcards

GATE E&C Engineering Communications Flashcards

51 question-and-answer cards covering Communications as it is examined in GATE E&C Engineering. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

51Cards in deck

24Free preview

18Syllabus topics

~212Chars per answer

FreePrice

24 sample cards from the Communications deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

What is the maximum entropy of a source with $M$ equiprobable symbols?
$$H_{max} = \log_2 M \ \text{bits/symbol}$$ achieved when $p_i=\frac{1}{M}$ for all $i$. Any nonuniform distribution gives lower entropy.
Define the self-information (information content) of an event of probability $p$.
$$I = \log_2\frac{1}{p} = -\log_2 p \ \text{bits}$$ Rare events carry more information; a certain event ($p=1$) carries zero information.
Define information rate of a source in terms of entropy and symbol rate.
$$R = r\,H(X) \ \text{bits/second}$$ where $r$ is the symbol rate (symbols/s) and $H(X)$ is entropy (bits/symbol).
Define mutual information $I(X;Y)$ between channel input $X$ and output $Y$.
$$I(X;Y) = H(X) - H(X\mid Y) = H(Y) - H(Y\mid X)$$ It measures the average information about $X$ gained by observing $Y$; it is symmetric and nonnegative.
State the Shannon-Hartley channel capacity theorem for an AWGN channel.
$$C = B\log_2\!\left(1+\frac{S}{N}\right) \ \text{bits/second}$$ where $B$ is bandwidth (Hz) and $S/N$ is the signal-to-noise power ratio. It is the maximum error-free data rate.
Define channel capacity of a discrete channel in terms of mutual information.
$$C = \max_{p(x)} I(X;Y)$$ the maximum mutual information over all input probability distributions, in bits per channel use.
What is the capacity of a binary symmetric channel (BSC) with crossover probability $p$?
$$C = 1 - H(p) = 1 + p\log_2 p + (1-p)\log_2(1-p) \ \text{bits/use}$$ Capacity is 1 at $p=0$ or $p=1$, and 0 at $p=\frac{1}{2}$ (useless channel).
List the three main steps of Pulse Code Modulation (PCM).
1) Sampling (band-limited signal sampled at $f_s \geq 2W$, Nyquist rate). 2) Quantization (mapping amplitudes to $L$ discrete levels). 3) Encoding (each level coded into $n=\log_2 L$ binary bits).
For a uniform quantizer with $L$ levels over range $2V$, give the step size and the maximum SQNR.
Step size $\Delta = \frac{2V}{L}$. Quantization noise power $=\frac{\Delta^2}{12}$. For $n$-bit PCM ($L=2^n$): $$\text{SQNR}_{dB} \approx 6.02n + 1.76\ \text{dB}$$ (sinusoidal input); each extra bit adds about 6 dB.
What is the bit rate of a PCM system?
$$R_b = n \cdot f_s \ \text{bits/s}$$ where $n=\log_2 L$ bits per sample and $f_s$ is the sampling rate. Bandwidth required is at least $\frac{R_b}{2}$.
What is companding in PCM and why is it used?
Companding = compressing then expanding, using nonuniform quantization ($\mu$-law in North America/Japan, A-law in Europe). It gives near-constant SQNR over a wide dynamic range by using finer steps for small signals and coarser steps for large signals.
How does DPCM (Differential PCM) differ from PCM, and what is its advantage?
DPCM quantizes and encodes the difference (prediction error) $e[n]=x[n]-\hat{x}[n]$ between the sample and its predicted value, rather than the sample itself. Since consecutive samples are correlated, the error has smaller variance, needing fewer bits for the same SQNR (bandwidth/bit-rate savings).
What is Delta Modulation, and what are its two characteristic distortions?
Delta modulation is 1-bit DPCM: it transmits only the sign of the difference (a single bit per sample, staircase approximation). Distortions: (1) slope overload — staircase can't keep up with fast signal changes; (2) granular (idle) noise — hunting around slowly varying/constant signals.
What two key advantages do digital modulation schemes offer over analog?
1) Robustness: regeneration of bits removes accumulated noise/distortion (better noise immunity). 2) Flexibility: easy encryption, error-correction coding, multiplexing, and compatibility with digital processing/storage.
Define ASK (Amplitude Shift Keying) and write its binary signal representation.
In binary ASK the carrier amplitude switches with the bit. On-off keying: $$s(t)=\begin{cases}A_c\cos(2\pi f_c t), & \text{bit }1\\ 0, & \text{bit }0\end{cases}$$ Simple but most susceptible to noise/amplitude variations.
Define BPSK and write its two signal waveforms.
Binary PSK shifts carrier phase by $180^\circ$ with the bit: $$s_1(t)=A_c\cos(2\pi f_c t)\ (\text{bit }1), \quad s_2(t)=A_c\cos(2\pi f_c t+\pi)=-A_c\cos(2\pi f_c t)\ (\text{bit }0)$$ The two signals are antipodal, giving optimal noise performance among binary schemes.
Define binary FSK and write its signal representation.
Binary FSK switches carrier frequency with the bit: $$s(t)=A_c\cos(2\pi f_1 t)\ (\text{bit }1), \quad A_c\cos(2\pi f_2 t)\ (\text{bit }0)$$ Constant envelope; can be detected coherently or noncoherently.
Compare the bit error probability ($P_e$) of coherent BPSK, BFSK, and ASK for a given $E_b/N_0$.
BPSK: $P_e=Q\!\left(\sqrt{\tfrac{2E_b}{N_0}}\right)$ (best). Coherent BFSK and coherent ASK: $P_e=Q\!\left(\sqrt{\tfrac{E_b}{N_0}}\right)$. BPSK needs about 3 dB less $E_b/N_0$ than BFSK/ASK for the same $P_e$.
What is QPSK and how does its bandwidth efficiency compare to BPSK?
Quadrature PSK uses 4 phases ($45^\circ,135^\circ,225^\circ,315^\circ$), carrying 2 bits/symbol via I and Q channels. For the same bit rate it needs half the bandwidth of BPSK, while having the same $P_b$; it is twice as bandwidth-efficient.
Define QAM (Quadrature Amplitude Modulation) and give the bits per symbol for M-QAM.
QAM modulates two carriers in quadrature ($\cos$ and $\sin$) with independent amplitudes, varying both amplitude and phase: $$s(t)=A_I\cos(2\pi f_c t)-A_Q\sin(2\pi f_c t)$$ M-QAM carries $\log_2 M$ bits/symbol (e.g., 16-QAM → 4 bits/symbol). Higher density increases data rate but reduces noise margin.
For an ideal (Nyquist) baseband channel of bandwidth $B$, what is the maximum signaling (symbol) rate, and how does the bit rate scale with $M$ levels?
Maximum symbol rate (Nyquist) $=2B$ symbols/s. With $M$-ary symbols: $$R_b = 2B\log_2 M \ \text{bits/s}$$ Increasing $M$ raises bit rate within the same bandwidth at the cost of noise immunity.
Define Inter-Symbol Interference (ISI) and state its main cause.
ISI is the smearing of a transmitted pulse into adjacent symbol intervals, so neighboring symbols interfere at the sampling instant. It is caused by band-limited/dispersive channels (finite bandwidth, multipath) spreading pulses in time, degrading the bit error rate.
State the Nyquist criterion for zero ISI and name the commonly used pulse.
Zero ISI requires the overall pulse $p(t)$ to satisfy $p(nT)=\begin{cases}1,&n=0\\0,&n\neq 0\end{cases}$ i.e. it is zero at all other sampling instants. The raised-cosine pulse (with roll-off factor $\alpha$) is commonly used; its excess bandwidth is $\alpha/2T$ and BW $=\frac{1+\alpha}{2T}$.
State the MAP (Maximum A Posteriori) decision rule and the ML (Maximum Likelihood) rule, and when they coincide.
MAP: choose the symbol $m$ maximizing the posterior $P(m\mid r)\propto p(r\mid m)P(m)$ — minimizes error probability. ML: choose $m$ maximizing the likelihood $p(r\mid m)$, ignoring priors. They coincide when all symbols are equally likely (equal priors $P(m)=\text{const}$).

What this deck covers

The Communications deck follows the GATE E&C Engineering Communications syllabus — 5 chapters and 18 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 10.2 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 212 characters, which is long enough to carry the reasoning and short enough to say out loud.

A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.

Communications flashcards FAQ

How many Communications flashcards are in this GATE E&C Engineering deck?

51 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.

Are these GATE E&C Engineering flashcards free?

Yes. The preview here is free to read with no signup, and the full 51-card deck is free inside the Examius app.

What do the Communications cards cover?

They follow the GATE E&C Engineering Communications syllabus — 5 chapters and 18 topics — so the questions track what is actually examinable.

How should I use these flashcards?

Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.