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My router have settings to configure transmission power and as i know energy of a signal is given by $$ E = h\nu $$ where

  • \$E\$ is the energy of the signal,
  • \$h\$ is Planck's constant,
  • \$\nu\$ is the frequency of the signal.

My question is how transmission power can be changed since energy of signal is constant?

Daniele Tampieri
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Ibnjunaid
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    waaaaaait. What exactly do you mean when you write \$E\$? Be very precise, and this will answer itself :) – Marcus Müller Feb 28 '19 at 13:05
  • I'm voting to close this question as off-topic because this is a physics, not an electronics design questions. – Marcus Müller Feb 28 '19 at 13:06
  • @MarcusMüller i updated the equation. please check and answer now – Ibnjunaid Feb 28 '19 at 13:09
  • @MarcusMüller I got this question while reading my High school physics "Communication systems", though they aren't going to ask this question in test but i still wanna know because i like radios and cant find an answer anywhere. Moreover i can't ask my instructor as he probably doesn't know. Rest on you please answer . – Ibnjunaid Feb 28 '19 at 13:23
  • If your instructor doesn't know but is your physics instructor, find a new instructor. – Marcus Müller Feb 28 '19 at 13:30
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    You wrote down a formula for something that is called "Energy". But: The energy of **what** is described by that formula? Again, be precise, and this will answer itself. – Marcus Müller Feb 28 '19 at 13:32
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    Energy carried by a photon of **frequency** v . Does this equation not apply here? – Ibnjunaid Feb 28 '19 at 13:37
  • Often RF circuits use variable-gain amplifiers; review Barry Gilbert's Translinear Principle for one approach. – analogsystemsrf Feb 28 '19 at 13:38
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    Read slowly: Energy carried by **a photon** of frequency v. And your transmitter produces a single photon? – Marcus Müller Feb 28 '19 at 13:43
  • @MarcusMüller I got it, it means that by controlling the transmission power we control number of photons emitted but does changing the transmission power increases the range of if yes then how?? – Ibnjunaid Feb 28 '19 at 13:56
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    @Ibnjunaid now that's a totally different question and you should not move the goal posts. You should also understand that radio-range is determined by transmit power and receiver sensitivity mainly then by antenna configuration and carrier frequency then, atmospheric conditions and interfers. It opens up in a way that your simple new question may not have realized. [This answer](https://electronics.stackexchange.com/questions/83512/long-range-15-km-low-baud-rate-wireless-communication-in-a-mountain-environme) might help. – Andy aka Feb 28 '19 at 14:11
  • @Andy aka. okay – Ibnjunaid Feb 28 '19 at 14:24

2 Answers2

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Energy is an instantaneous unit. Power is a time based unit (joules per unit of time). On the other way, power in electronic/electric circuits can be computed by knowing V and I values (voltage and current respectively). P = VI. So any increase in voltage and/or current will increase the amount of power (joules per second) of the output signal.

Edwin G. Delgado
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As I understand, according to this document

enter image description here

So, the energy of the signal is given by the integral of the squared magnitude of the signal x(t).
The power, however, is the energy computed during a period.

So, you can increase the power of the signal by either increasing the frequency or increasing the magnitude of the signal.

litvinik
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    That isn't really an answer to OPs question! – Marcus Müller Feb 28 '19 at 13:31
  • It's also not correct. Increasing the frequency decreases the denominator in the prefactor, but it also decrease the limits of integration, so the average power remains the same for the same signal amplitude. – The Photon Feb 28 '19 at 17:52