Principles of Communication Engineering

Answer

Correct Option: (ii) $2f_m$

Explanation

Amplitude Modulation: $s(t)=A_c[1+\mu m(t)]\cos\omega_c t$ with baseband $m(t)$ bandwidth $f_m$.

Spectrum: carrier at $f_c$ plus upper sideband $f_c+f_m$ and lower $f_c-f_m$.

Total transmission bandwidth:
$$B_T = (f_c+f_m)-(f_c-f_m)=2f_m$$

If $f_m=5$ kHz, $B=10$ kHz.

Other options: $f_m$ would be SSB, $f_c\pm f_m$ are sideband frequencies, not bandwidth.

Final Answer

AM bandwidth $=2f_m$ → Option (ii).

Answer

Correct Option: (ii) Frequency

Explanation

Frequency Modulation: Instantaneous frequency $f_i(t)=f_c+k_f m(t)$ varied proportional to message $m(t)$. Amplitude $A_c$ constant, phase $\theta(t)=2\pi\int f_i dt$. PM varies phase, FM varies frequency (integral of PM).

Final Answer

Frequency varied → Option (ii).

Answer

Correct Option: (iii) Both power and bandwidth

Explanation

DSB-SC transmits both sidebands ($2f_m$) with suppressed carrier → saves power vs AM but still $2f_m$ bandwidth.

SSB suppresses one sideband + carrier → saves both $50\%$ bandwidth ($f_m$) and carrier power.

Final Answer

Both power and bandwidth → Option (iii).

Answer

Correct Option: (iii) TV video transmission

Explanation

VSB compromises: one sideband fully + vestige of other. Bandwidth $f_m+f_v$ ($f_v\approx0.75$ MHz). SSB filter too sharp for video $4.2$ MHz, DSB wastes $8.4$ MHz. VSB saves ~50% vs DSB with practical filter.

TV video uses VSB (AM-VSB). AM radio uses AM, FM radio uses FM, satellite uses FM.

Final Answer

TV video → Option (iii).

Answer

Correct Option: (iii) Requires less bandwidth

Explanation

SSB vs DSB-SC (both suppressed carrier): DSB $2f_m$, SSB $f_m$ → 50% saving. Power: SSB saves half sideband power but main saving is bandwidth. Both save carrier power vs AM. Better noise immunity not primary; SSB still needs coherent detection.

Final Answer

Less bandwidth → Option (iii).

Answer

Correct Option: (i) $15$

Explanation

FM index $\beta=\Delta f/f_m=75\text{kHz}/5\text{kHz}=15$.

Final Answer

$15$ → Option (i).

Answer

Frequency-domain representation shows signal spectral content via Fourier.

Transform

$X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt$.

• Periodic: Fourier Series → line spectrum at $nf_0$.
• Aperiodic: Continuous $X(f)$.
• AM: $s(t)=A_c[1+\mu m(t)]\cos\omega_c t$ → impulses at $\pm f_c$ plus sidebands $M(f\pm f_c)$.
• FM: Bessel sidebands $J_n(\beta)$.

Importance

Reveals bandwidth, filter design, modulation analysis.

Final: Time → frequency via Fourier series/transform.

Answer

SSB vs VSB

SSB: Suppress carrier + one sideband. BW = $f_m$. Generation via sharp filter or phase shift (Hilbert). Coherent detection. Saves 50% BW and power. Need sharp filter, carrier reinserted.

VSB: One sideband fully + vestige of other. BW = $f_m+f_v$ (TV: $f_v\approx0.25$ MHz). VSB filter gradual. Used for TV video (SSB filter too sharp for 4.2 MHz video).

	SSB	VSB
BW	$f_m$	$f_m+f_v$
Filter	Very sharp	Gradual
Complexity	High	Moderate
Use	Telephony	TV

Final: SSB max saving, VSB practical for wideband.

Answer

Angle Modulation: $s(t)=A_c\cos(\omega_c t+\phi(t))$. FM: $\phi(t)=k_f\int m$, $s_{FM}=A_c\cos(\omega_c t+k_f\int m)$, $\beta=\Delta f/f_m$. PM: $\phi=k_p m$, $s_{PM}=A_c\cos(\omega_c t+k_p m)$. Bessel sidebands, Carson $B=2(\Delta f+f_m)$.

Answer

Given $P_t=10$kW $\mu=0.8$: $P_t=P_c(1+\mu^2/2)$ → $P_c=10/1.32=7.58$kW, $P_{SB}=2.42$kW total, each 1.21kW. Efficiency $\mu^2/(2+\mu^2)=24.2\%$.

Answer

FM Generation: Direct VCO (varactor) or Armstrong (NBFM then multiply). Detection: Slope discriminator (FM→AM+envelope), PLL tracking, zero-crossing. Block limiter→discriminator→LPF→$m(t)$.

Answer

Given $c=10\cos2\pi10^6t$, $m=5\cos2\pi10^3t$, $A_m=5$, $A_c=10$, $\mu=A_m/A_c=0.5$ (50%). Envelope $A_c[1+0.5\cos\omega_m t]$ max15 min5. $P_c=50$W, $P_{SB}=6.25$W total.

Answer

AM vs FM: AM BW $2f_m$ narrow, power 33% max carrier wasteful, noise poor envelope. FM BW $2(\Delta f+f_m)$ wide, constant envelope efficient, capture effect, high noise immunity ~10dB better, complex VCO/discriminator. FM for VHF high fidelity, AM MW simple.

Answer

Superheterodyne: Converts RF to fixed IF $f_{IF}=|f_{LO}-f_{RF}|$ via mixer+LO. IF amp 455kHz AM, 10.7MHz FM, sharp IF filter gives selectivity, sensitivity. RF amp→Mixer←LO→IF amp→Detector→AF. Image $f_{im}=f_{RF}+2f_{IF}$ needs preselection.