# Significance of signal energy in audio?

Probably a dumb question but I can't find the answer anywhere. It seems to me that the energy carried by a sound wave should be proportional to frequency squared. For example, the waves $$x(t)=\sin(t)$$, $$x(t)=\sin(100\cdot t)$$, and $$x(t)=\frac{1}{\sqrt{2}}$$ carry different amounts of energy - the last one carries no energy at all and sounds like silence. But in DSP they are all considered to have the same signal energy (integral of $$x(t)$$ squared). What's the significance of that?

It seems to me that the energy carried by a sound wave should be proportional to frequency squared.

Why? What makes you think that this is the case ?

But in DSP they are all considered to have the same signal energy (integral of x(t) squared)

The concept of associating power or energy with a a digital signal is somewhat tricky. Digital signals are just a bunch of numbers so assigning actual physical properties to them doesn't really work unless you carefully define the context.

Mostly the concept is used assuming that the digital signal is an accurate representation of an analog signal that has actual physical properties. If this relationship is well defined and constant you can compare the sum-of-squares of two different digital signals and draw some conclusions on the energy difference of the actual analog situations. However, that really requires "all things being equal" and there are actually a lot of things that need to be equal for this to hold.

STEP 1: basic physics of power

Power, intensity or energy are typically defined or calculated as the product of two field quantities (not one). In electricity power is defined as the product of of voltage and current. Simple example: if your signals were defined as "voltage over a one Ohm resistor", than you can indeed calculate the power by summing the squares and it would be the same power for all of your three examples since the current is proportional to the voltage.

If it were the voltage over an ideal inductor or capacitor then the answer would be wrong. Average power over one of these would be zero in all cases. For an ideal inductor voltage and current are 90 degrees out of phase and the average power is therefore zero.

STEP 2: physics of sound

The two field quantities that make up a sound wave are sound pressure and particle velocity. First you need to define which one your digital signal actually represents. Most of the time it's pressure but many second order microphones (cardioid, dipoles) will create a signal that's proportional to the particle velocity. If you are far enough away from the sound source than pressure and velocity are in-phase and proportional to each other and the proportionality factor is the free field impedance of air (density times speed of sound). In this case you can actually calculate the intensity at the microphone position by simply squaring and scaling. If you are close to a sound source that doesn't work anymore and you actually need to measure both signals.

As others have said: per it's physical definition sound can't exist at 0 Hz. Sound pressure is a variation of pressure around a steady-state average and if there is no variation there is no sound.

Step 3: human perception

Sound energy and perceived loudness are only loosely related. Human perception is very dependent on frequency and its quite non-linear. Building good loudness models is quite difficult. Simple example: 1 Watt of sound at 1kHz radiated into a typical residential room would extremely loud and quite painful. At 50 kHz it would be utterly inaudible, even though it's the exact same amount of physical sound power.

• I always thought that during playback, the signal represented neither pressure nor velocity, but the position of some physically moving thing (loudspeaker piston, phonograph needle...) And moving the thing and the surrounding air requires energy proportional to square of velocity. That's why energy of mechanical waves is proportional to frequency squared times amplitude squared. There are many sources for that, e.g. hyperphysics.phy-astr.gsu.edu/hbase/Waves/imgwav/powstr3.gif – Vladimir Slepnev Apr 11 at 22:12
• You can (and sometimes do) represent the displacement (amount of movement) as a digital signal, but the most common convention is pressure. I.e. a wave file from a CD is intended to be proportional to sound pressure. For some sound sources like an electro-dynamic loudspeaker the ratio radiated energy to displacement is indeed proportional to the square of of the frequency (at least in the piston band). However (above resonance) input voltage is roughly proportional to force on the cone which $F=m \cdot a$ makes it proportional to acceleration, not displacement. – Hilmar Apr 12 at 7:05
• Thanks, that's very informative. There's still something I don't understand though. If acceleration is sin(f*t), then velocity is -cos(f*t)/f, so radiated energy will be inversely proportional to frequency squared. Is that right? – Vladimir Slepnev Apr 12 at 8:20
• And physically too, I don't see why air pressure is cone acceleration. It seems like it should be velocity - the faster the cone moves, the more it compresses air, even if acceleration is zero. – Vladimir Slepnev Apr 12 at 8:30
• Never mind, I think I've figured out why pressure is acceleration. Thanks! Accepting the answer. – Vladimir Slepnev Apr 12 at 13:14

your complete system, if you wanna consider your ear another microphone with it's own preamp is:

$$\qquad \qquad x(t) \rightarrow$$ loudspeaker $$\rightarrow$$ air $$\rightarrow$$ microphone

both the loudspeaker and microphone are transducers that convert between an electrical signal and an acoustic wave. as components, both of those devices have no gain (that is $$-\infty$$ dB) at DC. that is they're transfer function at DC is zero:

$$H(j\omega) \Bigg|_{\omega=0} = 0$$

$$x(t) = \tfrac{1}{\sqrt{2}}$$ is a purely DC signal. if you applied it to a resistor, energy would be transferred and dissipated in the resistor. but applied to a loudspeaker, all that does is move the piston away from its equilibrium position without wiggling it and generating an acoustic wavefront, which is what loudspeakers, as transducers, are supposed to do.