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If r2 = x2 + y2 + z2 then the value of is
The potential (scalar) distribution is V = 10y4 + 20x3. If ∈0 be the permittivity of free space, then the charge density 'p' at the point (0, 2) is
The skin depth of copper at a frequency of 3 GHz is 1 micron (10-6 meter). At 12 GHz for a non-magnetic conductor whose conductivity is 1/9 times that of copper, the skindepth would be
Shortest length of a 42 Ω air line required toproduce a reactance of j73Ω at 1 MHz when line is short circuited is
A transmission line of characteristic impedance 50Ω is terminated by a resistor 100Ω.
Q. The impedance at the voltage maximum position is
A transmission line of characteristic impedance 50Ω is terminated by a resistor 100Ω.
Q. The impedance at the voltage minimum position is
An EM wave travels in free space with the electric field component, E. = 100 ei(a866 + a5z)ax V/m and ω = 3 x 108 rad/sec. The time average power in the wave is
Standard equation is,
A 50Ω characteristic impedance line is connected to a load which shows a reflection coefficient of 0.268. If Vin = 15 V, then the net power delivered to the load will be
A TEM wave impinges obliquely on a dielectric-dielectric boundary (,
). The angle of incidence for total reflection is
Match List-I (Nature of Polarization) with
List-II (Relationship Between X and Y components) for a propagating wave having cross-section in the XY plane and propagating along Z-direction and select the correct answer:
List-I
A. Linear
B. Left circular
C. Right circular
D. Elliptical
List-II
X and Y components are in same phase
X and Y components have arbitrary phase difference
Consider the following statements:
A 1 kHz sinusoidal signal is ideally sampled at 1400 samples/sec and the sampled signal is passed through an ideal low-pass filter with cut-off frequency 800 Hz. The output signal has the frequency kHz
Given that,
fm = 1 kHz
fs = 1.4 kHz
So, the output signal coli[..]iris two frequencies and they are
⇒(1.4 + 1) kHz and (1.4 - 1)I\Hz
⇒(2.4) kHz and (0.4) kHz
But, LPF has fc = 0.8 kHz
So, only 0.4 kHz will appear in output
In a SSB modulation, the power spectral density of m(t) is given as
where a and W are constants. White Gaussian noise of zero mean and PSD of N0/2 is added to SSB modulator wave at the receiver input. The (SNR) at the output of the receiver is,
(Where Ac = Carrier signal amplitude)
Average signal power = Pav
= aW
By using SSB modulation in the transmitter and coherent detection in the receiver, the (SNR)output is given by
(SNR)output
(where Ac carrier signal amplitude)
An angle modulated signal with carrier frequency ωc = 2π x 105 is described by the equation QAm (t) = 10 cos (ωc t + 5 sin 3000 t + 10 sin 2000 πt)
The maximum frequency deviation is
The instantaneous frequency ω1 is given by
ω1 = ωc+ 15000 cos 3000 t + 20000π cos 2000πt
Maximum frequency deviation,
Δω = 15000 20000 π
Hence,
= 12388 Hz
= 12.38 kHz
A frequency modulated wave is 100% modulated by 10 kHz modulating signal m(t). Then the bandwidth required is
In a FM wave, we are concerned only with frequency.100% modulation corresponds to the maximum frequency deviation which is equal to 75 kHz.
∴ B an dwidth(B.W)
= 2(Δf + fm)
= 2(75 + 10) kl-lz
= 170 kHz
A sinusoid x(t) = A cos2πfot is sampled at 5 times the Nyquist rate for 2.5 sec. A total of 250 samples are acquired value of fs and fm are given by
Sampling frequency,
fs = 5 x (N quilt rate)
= 5 x f0= 10f0
No of samples = = T x fs
⇒ 250 = 2,5 x fs
The angle-modulated signal with carrier frequency ωc = 27c x 106 is discribed by the equation
v(t) = 10 cos(ωct + 10sin200t + 20si n400πt) The deviation ratio is
A PCM system uses a uniform quantizer followed bya 7-bit binary encoder. The bit rate of the system is equal to 50 x 106 bps.
Q. The maximum message bandwidth for which the system oeprates satisfactorily, is
Maximum message bandwidth =
The approximate output signal - to (quantization) noise ratio when a full - load sinusoidal modulating wave of frequency 1 MHz is applied to the input, is
Output signal to (quantization) noise ratio
where, 0 2n (where n = number of bits)
For an AM wave, the maximum voltage was found to be 10 V and the minimum voltage was found to be 5 V. The modulation index of the wave would be
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