JNTU
JNTU B.Tech III ANALOG COMMUNICATIONS Previous Question Paper (February - 2022 - R13)
Answer any five questions. All questions carry equal marks.
Main
Explain the generation of DSB-SC signal with the help of balanced modulator using diodes.
Answer
A DSB-SC (Double Sideband Suppressed Carrier) signal is generated by multiplying the message signal m(t) with the carrier signal c(t) = A cos(ωct). A balanced modulator using diodes is a common circuit implementation that achieves this by canceling the carrier component while retaining the sidebands.
The circuit consists of two identical transformers and four diodes arranged in a bridge configuration. The message signal is applied in series with the center tap of the input transformer, while the high-frequency carrier signal is applied across the primary winding. The diodes act as switches that are turned on and off by the carrier signal.
The operation can be described as follows:
- During the positive half-cycle of the carrier, two diodes conduct, allowing the message signal to pass through to the output transformer.
- During the negative half-cycle of the carrier, the other two diodes conduct, effectively reversing the polarity of the message signal at the output.
- This switching action is equivalent to multiplying the message signal by a square wave carrier, which contains the fundamental frequency and its odd harmonics.
- The output transformer acts as a bandpass filter or simply provides the differential output, which effectively cancels the carrier frequency component and leaves the product of the message and the carrier.
The resulting output voltage is given by:
Because the diodes are matched and the transformer is center-tapped, the carrier signal is suppressed at the output due to the balanced nature of the circuit. The output contains only the upper and lower sidebands, satisfying the condition for DSB-SC modulation. The efficiency of the suppression depends on the precision of the diode matching and the symmetry of the transformer windings.
Derive an expression for AM wave and its total power relation.
Answer
An Amplitude Modulated (AM) wave is generated by varying the amplitude of a high-frequency carrier signal in proportion to the instantaneous value of a low-frequency message signal. Let the message signal be m(t) = Am cos(2πfmt) and the carrier signal be c(t) = Ac cos(2πfct).
The expression for the AM wave s(t) is given by:
where μ = Am / Ac is the modulation index. Expanding this expression using trigonometric identities:
This shows the AM wave consists of the carrier component, an upper sideband, and a lower sideband.
To derive the total power, we calculate the mean square value of s(t). Since the components are orthogonal, the total power Pt is the sum of the powers of the individual components:
Given the carrier power Pc = Ac2 / 2, the powers of the sidebands are:
Summing these, the total power is:
This expression demonstrates that the total power in an AM wave increases with the modulation index μ, with the maximum power occurring at μ = 1.
Explain how a SSBSC signal is represented in time and frequency domain.
Answer
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An AM transmitter of 1KW power is fully modulated. Calculate the power transmitted, if it is transmitted as SSB.
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Draw the block diagram for the generation and demodulation of a VSB signal and explain the principle of operation.
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Explain the principle of indirect method of generating a wide-band FM signal with a neat block diagram.
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A single tone modulating signal cos(10π103t) frequency modulates a carrier of 10MHz and produces a frequency deviation of 75 KHz. Find the modulation index and phase deviation produced in the FM wave. If another modulating signal produces a modulation index of 100 while maintaining the same deviation, find the frequency and amplitude of the modulating signal, assuming Kf=10kHz/V.
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Discuss about noise effect in PM and obtain expression for figure of merit.
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Explain different types of noise involved in communication system.
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When a super heterodyne receiver is tuned to 555 KHz, its local oscillator provides the mixer with an input at 1010 KHz what is the image frequency? The antenna at receiver is connected to mixer via a tuned circuit whose loaded Q is 40. What will be rejection ratio for the calculated image frequency?
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With the aid of the block diagram explain TRF receiver.
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