537 questions and answers
Draw the block diagram of a super heterodyne receiver. Explain the function of each block. Explain the significance of the name super hetero dyne. How RF sections suppress the image channel? Why is up – conversion used in super heterodyne receive? Define image Rejection Ratio.
Super heterodyne receivers was proposed as an alternative to TRF receiver which suffers from selectivity problem .The basic super heterodyne receiver is most widely used receivers at present. Super heterodyne principal is also used in television and radar receiver. In the super heterodyne receivers, the received RF signal voltage is combined with the local oscillator voltage and is converted into a signal of lower fixed frequency.
"FM and PM are different but inseparable." – Justify the statement.
The expression for FM and PM are almost identical. FM and PM cannot be distinguished at a constant modulating frequency.
What are Narrowband FM and Wideband FM?
A narrow band FM is the FM wave with a small bandwidth. The modulation index β of narrow band FM is small as compared to one radian. Hence, the spectrum of narrowband FM consists of the carrier and upper side band and lower side band.
Explain with proper expression
The modulation index in AM wave is defined as the ratio between amplitude of the modulating signal and amplitude of the carrier signal. The mathematical representation is given below
The maximum deviation allowed in an FM broadcast system is 75 kHz. If the modulating signal is a single tone sinusoid of 10 kHz, find the bandwidth of the FM signal. What will be the change in the bandwidth, if the modulating frequency is double? Determine the bandwidth when modulating signal amplitude is also double.
Given that ∆f=75KHz and fm=10KHz According to the Carson’s thumb rule the bandwidth BW=2(∆f+fm )=2(75+10)=170KHz
1. a. State sampling theorem. What is Nyquist rate of sampling?
b. What is pulse amplitude modulation
c. What is meant by flat top sampling? Why is it more preferred than natural sampling?
d. Explain the demodulation of PAM signal using a hold circuit
e. What is the bandwidth required for transmission of PAM signal?
a. The sampling theorem state that A band limited signal of finite energy, which has no frequency components higher than fm Hz, is completely described by its sampled values less than or equal to 1/(2fm ) second apart.
At t=0, the instantaneous value of a 50 Hz, sinusoidal current is 5 Amp and increases in magnitude further. Its R.M.S value is 10 Amp.
The R.M.S value of sinusoidal current is IR.M.S=10 Amp, thus the peak value of current Imax=√2 ×IR.M.S=10√2 Amp.
For the circuit shown bellow, determine the current i1,i2,i3 using nodal analysis
Considering two nodes V1 and V2, we have (V1-24)/2+V1/6+V2/8=3 Or 16V1+3V2=360 …(1) The difference between two nodes is
For the circuit shown below, find the potential difference between a and d:
Applying KVL in 1st mesh, we have 4I1-3=0 Or I1=3/4 Amp. Vab=2I1=2×3/4=3/2 V
State and explain Thevenins theorem. Find the Thevenins equivalent of the circuit of Fig.18(a) as shown at terminal XY.
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Fig. 18(a)
Thevenin’s theorem states that any two terminal bilateral linear d.c circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.
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