Exams › GATE › Technical › MAIN PAPER - EC
14 questions with worked solutions.
Q1. Which of the following is NOT associated with a p-n junction?
Answer: (D) Channel Length Modulation
Channel Length Modulation is a phenomenon related to field-effect transistors (FETs) and does not pertain to the characteristics of a p-n junction, which primarily involves capacitance effects.
Q2. Which of the following is true?
Answer: (A) A silicon wafer heavily doped with boron is a p+ substrate
A silicon wafer that is heavily doped with boron has an excess of holes, which makes it a p+ substrate, indicating a high concentration of p-type carriers.
Answer: (D) ∇×E = 0, ∇·B = 0
The correct option states that the curl of the electric field is zero, indicating that the electric field is static and conservative, while the divergence of the magnetic field is also zero, consistent with the absence of magnetic monopoles in a source-free medium.
Answer: Ix = Ibias
In this circuit configuration, since M2 is in saturation and the output is unloaded, the current flowing through M1 (Ix) is equal to the bias current (Ibias) provided to the circuit, as there are no additional current paths or loads affecting the current flow.
Answer: (D) Q goes to 0 at the CLK transition and goes to 1 when D goes to 1.
The correct option is (D) because when the clock transitions from 1 to 0, the output Q captures the current state of D, which is 0. After this transition, when D changes to 1, Q will then update to reflect this new value.
Answer: (A) 6.25 GHz
For TE11, fc=(c/2)*sqrt((1/a)^2+(1/b)^2) with a=0.04 m, b=0.03 m. This gives (1.5e8)*sqrt(625+1111.1)=1.5e8*41.67=6.25 GHz. The stored 3.75 GHz is wrong; the answer is 6.25 GHz.
Answer: (D) Inductive
When a transmission line is short-circuited at one end, the input impedance at the other end can be determined by considering the line's length and frequency. At 3GHz, the wavelength is short enough that the line behaves like an inductor, resulting in an inductive input impedance.
Answer: (C) 0.5
The reflection coefficient for a wave incident on a dielectric interface can be calculated using the formula R = |(η1 - η2) / (η1 + η2)|, where η1 is the impedance of free space and η2 is the impedance of the dielectric. Given the dielectric constant εr = 9, the impedance of the dielectric is lower than that of free space, resulting in a reflection coefficient of 0.5.
Q9. At 20 GHz, the gain of a parabolic dish antenna of 1 meter diameter and 70% efficiency is
Answer: (D) 45 dB
At 20 GHz, lambda=0.015 m, so G=0.7*(pi*1/0.015)^2 about 30700, which is 10*log10(30700) about 44.9 dB ~ 45 dB. The stored 35 dB is wrong.
Answer: p³ + 3p²(1-p)
The correct option accounts for the scenarios where the transmitted sequence of three 1s is received with at least two bits still being 1. This includes the case where all three bits are received correctly (probability (1-p)³) and the cases where exactly two bits are received correctly and one is flipped (which can occur in three different ways, hence the factor of 3), leading to the total probability of error being p³ + 3p²(1-p).
Answer: 7W
The minimum bandwidth required for TDM is determined by the sum of the bandwidths of all individual messages. In this case, the total bandwidth is W + W + 2W + 3W, which equals 7W.
Answer: 12.5
The modulation index in frequency modulation is determined by the ratio of the peak frequency deviation to the modulation frequency. In this case, the peak frequency deviation is given by the sum of the amplitudes of the modulating signals (5 + 7.5 = 12.5), and since the carrier frequency is 10⁵ Hz, the modulation index is calculated as 12.5.
Q13. The signal cos ω_ct + 0.5 cos ω_mt sin ω_ct is
Answer: AM only
The given signal contains a term that modulates the amplitude of the carrier wave (cos ω_ct) with another cosine function, which is characteristic of Amplitude Modulation (AM). The presence of the sine function does not contribute to frequency modulation (FM), confirming that the signal is AM only.
Answer: 32 kHz
The minimum bandwidth required for distortion-free transmission of a signal is determined by the Nyquist theorem, which states that the bandwidth must be at least twice the highest frequency of the signal. Since the speech signal is band limited to 4 kHz, the minimum bandwidth needed is 2 times 4 kHz, resulting in 8 kHz. However, when considering the bipolar pulse transmission, the effective bandwidth doubles, leading to a required bandwidth of 32 kHz.