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I know that this is a very fundamental (and maybe vague) question.

But I'm not sure what period I should use. I read somewhere that a constant signal doesn't have a fundamental period but is periodic. So looking at the formulas for calculating energy and power would that mean that a constant signal is an energy signal?

Thank you for your replies, please let me know if you need more clarification. ( I am new to this topic)

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Suppose x(t) = K is a constant signal.

The energy which will be dissipated on a unit 1 ohm resistor by x(t) can be computed as: $$ \begin{align} E&=\int_{-\infty }^{\infty }\left | x(t) \right |^{2} dt\\ &=K^2\int_{-\infty }^{\infty }dt \\ &= \infty \end{align} $$ The power which will be dissipated on a unit 1 ohm resistor by x(t) can be computed as: $$ \begin{align} P&=\lim _{T->\infty}\frac{1}{2T}\int_{-\infty }^{\infty }\left | x(t) \right |^{2} dt\\ &=K^2\lim _{T->\infty}\frac{1}{2T}\int_{-T }^{T}dt\\ &=K^2\lim _{T->\infty}\frac{2T}{2T}\\ &=K^2 \end{align} $$

Therefore x(t) has finite power but infinite energy. Which implies a constant signal is a power signal.

Refer: Energy and Power Signals

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  • \$\begingroup\$ what about for discrete time? \$\endgroup\$ – bleepblop Apr 10 '18 at 18:31
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    \$\begingroup\$ Apply the discrete time equations. You will get the same result as above. T is irrelavant. \$\endgroup\$ – Mitu Raj Apr 10 '18 at 18:37
  • \$\begingroup\$ The first line in the second equation block should integrate from T to -T, right? \$\endgroup\$ – user253751 Apr 10 '18 at 23:55
  • \$\begingroup\$ @immibis This is the general formula to calculate energy over infinite length of time.If you put -T to T , then limit T tends to infinity will appear in the formula. Both are equivalent. \$\endgroup\$ – Mitu Raj Apr 11 '18 at 2:37
  • \$\begingroup\$ To me, the way it as written seems like it should always evaluate to 0 (any finite value / infinity) or not be defined (infinity / infinity) because the infinity in the limit is not related to the infinity in the integral. \$\endgroup\$ – user253751 Apr 11 '18 at 2:40
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So looking at the formulas for calculating energy and power would that mean that a constant signal is an energy signal?

  • A signal is not a supplier of energy until it starts doing work.
  • A volt is not a joule and neither is an amp but
  • 1 amp x 1 volt is 1 watt or
  • 1 joule per second.
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