What is the main reason high-frequency Carrier signals are used in modulation?
To reduce the physical size of the required antenna.
This question asks why high-frequency carrier signals are used in modulation. The dominant practical reason is the physical size of the antenna required to radiate (and receive) the signal efficiently. For efficient radiation, an antenna's dimension must be comparable to the signal's wavelength — a common rule of thumb is a quarter-wavelength (λ/4) element.
Wavelength and frequency are inversely related:
λ = c / f, where c = 3 × 108 m/s
Consider a baseband audio (speech) signal of about 1000 Hz. Its wavelength would be λ = (3 × 108) / 1000 = 3 × 105 m = 300 km, requiring a λ/4 antenna roughly 75 km long — utterly impractical.
If instead this message is modulated onto a high-frequency carrier, say 100 MHz (FM broadcast band):
λ = (3 × 108) / (100 × 106) = 3 m, so a λ/4 antenna is only about 0.75 m — perfectly practical. Hence the correct answer is to reduce the physical size of the required antenna.
Why the other options are wrong:
A transmission line has a characteristic impedance of 50 ohm and is connected to a load of 100 ohm. The VSWR is _________.
If the effective area of an antenna becomes \(\frac{2A}{3}\) from its initial value of 'A', while keeping its operating frequency same. Then, the antenna gain becomes \(\left(\frac{2x+4}{15}\right)\) times of its initial value. The value of x will be:
Match List I with List II
| LIST I (Type of Aperture Antenna) | LIST II (Beam widtd half power points) |
| A. Uniformly illuminated linear Array | I. \(\frac{58}{D_\lambda}\) |
| B. Uniformly illuminated circular aperture | II. \(\frac{56}{a_{E\lambda}}\) |
| C. Optimum E-plane rectangular horn | III. \(\frac{67}{a_{E\lambda}}\) |
| D. Optimum H-plane rectangular horn | IV. \(\frac{51}{L_\lambda}\) |
Choose the correct answer from the options given below:
Match the following lists in terms of radiation resistances of various antennas :
| List – I | List – II |
| a. Short vertical monopole | i. \(31200\left(\dfrac{\text{Area of Loop}}{\lambda^{2}}\right)^{2}\) |
| b. Small loop antenna | ii. \(80\pi^{2}\left(\dfrac{L}{\lambda}\right)^{2}\) |
| c. Dipole antenna | iii. 73 ohms |
| d. Radiation resistance of half wave dipole | iv. \(400\left(\dfrac{\text{Physical height}}{\lambda}\right)^{2}\) |
Choose the correct answer from the codes given below: