Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In frequency modulated wave : (1) frequency varies with time (2) amplitude varies with time (3) both frequency and amplitude vary with time (4) both frequency and amplitude are constant

Short Answer

Expert verified
In frequency modulated wave, frequency varies with time.

Step by step solution

01

Understanding Frequency Modulation

Frequency modulation (FM) is a method of encoding information in a carrier wave by varying the frequency of the wave. In FM, the information signal varies the frequency of the carrier wave. This results in a signal where the frequency changes over time depending on the information signal.
02

Examining Amplitude

In FM, the amplitude of the carrier wave remains constant. The information is not encoded by changing the amplitude, but purely by changing the frequency. Thus, the amplitude does not vary with time.
03

Conclusion

Based on the definitions and properties of frequency modulation, the correct statement is that only the frequency varies with time, and the amplitude remains constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

FM Encoding
Frequency Modulation (FM) encoding involves transmitting information by altering the frequency of a carrier wave. Unlike other modulation techniques, such as Amplitude Modulation (AM), FM changes the frequency to encode the data. This means that the encoded signal varies in frequency according to the information being sent.
In FM encoding:
  • The frequency of the carrier wave is varied.
  • The amplitude of the carrier wave remains constant.
Imagine tuning a radio station. Different stations broadcast on different frequencies. When using FM encoding, the information (like music or voice) is transmitted by changing the frequency within a certain range while keeping the amplitude stable.
Carrier Wave
A carrier wave is a high-frequency electromagnetic wave that carries the information signal. In the context of frequency modulation (FM), this carrier wave is modulated to transmit the desired information.
Key aspects of the carrier wave in FM include:
  • It has a consistent amplitude.
  • Its frequency is varied to encode information.
The carrier wave remains at a steady amplitude so that it can travel efficiently through different mediums, like air or cables, without losing its strength. By keeping the amplitude constant, FM can reduce noise and improve the clarity of the transmitted signal. The carrier wave's frequency changes in direct response to the information signal's amplitude, which translates voice or music into variations in frequency.
Amplitude Variation
In frequency modulation (FM), the focus is on varying the frequency, not the amplitude. The amplitude of the carrier wave remains constant, which helps maintain the integrity and clarity of the signal.
Points to remember about amplitude in FM:
  • Amplitude does not change over time during FM encoding.
  • This constancy helps in reducing noise and enhancing signal clarity.
Imagine listening to a song transmitted over FM radio. The volume (amplitude) of the carrier wave stays the same, while the pitch (frequency) varies to create different sounds. This stability in amplitude is what helps FM signals achieve their high-quality audio transmission, even in environments with potential interference. By keeping amplitude constant, FM encoding avoids the common issues associated with amplitude variation, such as signal degradation and noise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The electrode potentials for \(\mathrm{Cu}_{\text {(aq) }}^{2 *}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{\text {(aq) }}^{*} \quad\) and \(\mathrm{Cu}_{\text {(aq) }}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\) are \(+0.15 \mathrm{~V}\) and \(+0.50 \mathrm{~V}\) respectively. The value of \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}\) will be : (1) \(0.500 \mathrm{~V}\) (2) \(0.325 \mathrm{~V}\) (3) \(0.650 \mathrm{~V}\) (4) \(0.150 \mathrm{~V}\)

The value of ' \(\lambda\) ' so that the system of equations $$ \begin{aligned} &x+2 y+3=0 \\ &3 x+4 y+4=0 \\ &7 x-10 y+\lambda=0 \end{aligned} $$ is consistent, is (1) \(-39\) (2) 20 (3) \(-20\) (4) none of these

At a particular temperature, the vapour pressures of two liquids \(A\) and \(B\) are respectively 120 and \(180 \mathrm{~mm}\) of mercury. If 2 moles of \(A\) and 3 moles of \(B\) are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in \(\mathrm{mm}\) of mercury) (1) 156 (2) 145 (3) 150 (4) 108

A plane mirror is placed in \(y-z\) plane facing towards negative \(x\)-axis. The mirror is moving parallel to \(y\)-axis with a speed of 2 \(\mathrm{cm} / \mathrm{s}\). A point object \(\mathrm{p}\) move is moving in front of the mirror with a velocity \((1 \mathrm{~cm} / \mathrm{s})\) \(\hat{i}+(1 \mathrm{~cm} / \mathrm{s}) \hat{j}\). Then which of the following statements is incorrect? (1) The velocity of image is \((-\hat{i}+\hat{j}) \mathrm{cm} / \mathrm{s}\) (2) The velocity of image with respect to mirror is \(-(\hat{i}+\hat{j}) \mathrm{cm} / \mathrm{s}\) (3) The velocity of image with respect to object is \(-(2 \hat{i}) \mathrm{cm} / \mathrm{s}\) (4) The unit vector in the direction of reflected ray is \(\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\)

Light of wavelength \(6000 \AA\) is inciden normally on a slit of width \(24 \times 10^{-5} \mathrm{~cm}\). Finc out the angular position of second minimum from central maximum ? (1) \(15^{\circ}\) (2) \(30^{\circ}\) (3) \(45^{\circ}\) (4) \(60^{\circ}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free