Match the following lists : Choose the correct answer from the codes given below:List - I List - II a) 
i) Frequency modulated detector b) 
ii) Pre-emphasis c) 
iii) Amplitude modulator d) 
iv) Amplitude detector
a-iv, b-i, c-iii, d-ii
Each circuit is identified by the one element that gives its function away.
a → iv. Amplitude detector. A series diode followed by a parallel R–C is the classic envelope detector. The diode rectifies the modulated carrier and the capacitor holds the peak between cycles, so the output follows the envelope — i.e. the message. The time constant must satisfy
\(\dfrac{1}{f_c}\ll RC\ll\dfrac{1}{f_m}\)
fast enough to follow the modulation but slow enough to smooth the carrier; too large an RC produces diagonal clipping.
b → i. Frequency modulated detector. The giveaway is the tuned transformer feeding the diode. An FM detector must first convert frequency variations into amplitude variations, and a tuned circuit operated on the slope of its response does exactly that; the diode and RC then recover the resulting envelope. This slope/ratio-detector arrangement is the standard FM demodulator.
c → iii. Amplitude modulator. Here the diode is driven from a resistive divider and its output feeds an L–C tuned tank. The non-linear diode mixes the carrier and modulating signals, generating sum and difference frequencies, and the tank selects the wanted band around the carrier — producing AM rather than detecting it. Non-linearity plus a tuned output is always a modulator.
d → ii. Pre-emphasis. The audio enters a transistor whose collector load is an inductor in series with a resistor. Since \(X_L=\omega L\) rises with frequency, so does the stage gain: the circuit deliberately boosts the treble before modulation. At the receiver a matching de-emphasis network restores the balance and, in doing so, attenuates the high-frequency noise that FM detection produces — improving the output signal-to-noise ratio. The standard time constant is 75 μs.
Recognition rule: diode + RC → detector; tuned transformer ahead of the diode → FM detector; non-linear device with a tuned output → modulator; a series L in the load → a deliberate high-frequency boost, i.e. pre-emphasis.
Hence, the correct matching is a-iv, b-i, c-iii, d-ii.
Match the following lists :
| List – I | List – II |
| a. PCM | i. Slope overload distortion |
| b. DM | ii. Constant carrier frequency |
| c. AM | iii. Encoding |
| d. TDM | iv. Commutator |
The correct match is :
Match the following :
| List – I | List – II |
| a. DSB-SC modulation | i. envelope detection |
| b. SSB-modulation | ii. Foster Seeley |
| c. AM-demodulation | iii. Weaver's method |
| d. Phase-shift detection | iv. Balanced modulator |
Codes :
Match the following :
| List – I | List – II |
| a. Power efficient transmission | i. SSB-SC |
| b. Most bandwidth efficient transmission of voice signal | ii. VSB |
| c. Simplest receiver | iii. FM |
| d. Bandwidth efficient transmission of signals with significant d.c. component | iv. AM |
Codes :
Match List - I with List - II.
| List - I | List - II |
| (A) White noise | (I) Reduces antenna height |
| (B) RMS noise voltage | (II) \(P_{C}\left(\dfrac{m^{2}}{2}\right)\) |
| (C) Modulation | (III) Inversely proportional to \(\sqrt{T}\) |
| (D) PDSBSC | (IV) Johnson Noise |
Choose the correct answer from the options given below :
Match the following lists :
| List -I | List - II |
| a. Power efficient transmission of signals | i. AM |
| b. Bandwidth efficient transmission of signals with significant dc signals | ii. VSB |
| c. Most bandwidth efficient transmission of voice signals | iii. SSB-SC |
| d. Simplest Receiver Circuit | iv. FM |
Correct Codes are :
Arrange in order of the increasing Bandwidth value for the below mentioned systems :
(a) DSB-SC (b) VSB (c) FM (d) SSB
Codes :
During the heterodyne process in the receiver, the modulation of the signal __________
Which of the following modulations is used in India for radio transmission?
Which of the following statements are correct?
A. DSB‐SC modulation is well suited for point to point communication involving one transmitter and one receiver.
B. VSB modulation is a linear modulation scheme.
C. SSB is a non‐linear modulation scheme.
D. FM is a linear modulation scheme.
Choose the correct answer from the options given below: