Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Electronics and Communication Engineering (ECE) MCQ

Answer: c
Explanation: In TE waves, the electric field strength will be transverse to the wave direction. Thus the TE waves are also called H waves.

Answer: b
Explanation: TE₁₀ is the dominant mode in the rectangular waveguide. This is because it gives the minimum cut off frequency required for transmission.

Answer: d
Explanation: The number of modes is given by m = V²/2, where V is the v number. On substituting for V = 10, we get m = 100/2 = 50.

Answer: b
Explanation: For a waveguide, the dimension a should be greater than b. Here a = 3 and b = 5.5, thus such waveguide does not exist.

Answer: a
Explanation: Dominant modes are the modes having least cut off frequency. This implies they have highest cut off wavelength.

Answer: a
Explanation: The intrinsic impedance of a TE wave is given by η_TE = η/cos θ, where cos θ is given by √(1- (fc/f)²). On substituting for fc = 6 GHz, f = 7.5 GHz and η = 377, we get the intrinsic impedance as 628.33 units.

Answer: d
Explanation: The cut off frequency in dominant mode will be fc = mc/2a. On substituting for c = 3 x 10⁸ and a = 0.03, we get the cut off frequency as 5 GHz.

Answer: d
Explanation: The propagation constant is given by γ = α + jβ, where α and β are the attenuation and phase constants respectively. For a lossless line, the attenuation constant is zero. Thus γ = jβ. It is clear that γ is complex and equal to β.

Answer: c
Explanation: The attenuation of a wave is given by α = R/2Z₀. On substituting for R = 100 and Z₀ = 50, we get α = 100/(2 x 50) = 1 unit.

Answer: a
Explanation: The cut off frequency of a TE wave in any other medium is mc/2a√εr. On substituting for a = 0.035 and εr = 2.2, we get the cut off frequency as 2.88 GHz.

Answer: a
Explanation: The phase constant is given by β = ω√LC. Thus the relation is β is directly proportional to ω.

Answer: d
Explanation: The attenuation constant, resistance and conductance are zero for a lossless line. Only the phase constant is non zero.

Answer: c
Explanation: In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves

Answer: d
Explanation: The modes TM₁₀, TM₀₁ and TM₂₀ does not exist in any waveguide. The TM₁₁mode is the dominant mode in the waveguide.

Answer: b
Explanation: The number of modes in a waveguide is given by m = V²/2. On substituting for V = 20, we get m = 400/2 = 200 modes.

Answer: a
Explanation: The number of modes in a waveguide is given by m = V²/2. On substituting for m = 120, we get V = √(2 x 120) = 15.5.

Answer: c
Explanation: The intrinsic impedance of the transverse magnetic wave is given by η_TM = η √(1-(fc/f)²). Here the term √(1-(fc/f)²) is always lesser than unity. Thus the intrinsic impedance of the TM wave is lesser than 377 ohms.

Answer: d
Explanation: The modes TM_mo and TM_on does not exist. These modes are said to be evanescent mode.

Answer: c
Explanation: Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution.

Answer: b
Explanation: Modes in the format of TM_mo and TM_on does not exist. The given mode is in the form of TM_mo, which is does not exist. It is an evanescent mode.

Answer: a
Explanation: The Fresnel- Fraunhofer boundary is related by the wavelength as R = 2L²/λ. On substituting for L = 12 and λ = 3, we get R = 2 x 12²/3 = 96 units.

Answer: b
Explanation: When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves.

Answer: c
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength.

Answer: a
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. On substituting for Vmin = 2cm, we get λg = 2Vmin = 2 x 0.02 = 4cm.