In most audio editing software(like audacity) waveform can be viewed. Zoomed in these look like actual waveforms. Out of curiosity I downloaded some frequency sweep test files(which are intended to test the linearity of associated devices). In all of these files the amplitude was constant, only the frequency was increasing. But shouldn’t the amplitude be decreasing with increasing frequency inverse squarely for constant intensity(and energy)? Or are these somehow universally equalized? If these are, is there any way of viewing the actual waveform? Or am I missing something?

An ideal loudspeaker generates pressure fluctuations proportional to the voltage and an ideal microphone generates voltages proportional to pressure fluctations. In practice (for non-ideal transducers), there will be frequency-dependent phase differences and a frequency-depenent amplitude as well.

The energy stored in an acoustic wavetrain in air is equivalent to the work needed to compress the air, which is proportional to $(Delta p)^2$.

The local pressure deviation in an acoustic wave traveling in the $z$ direction is proportional to $partial u/partial z$, for a displacement $u$. You can derive that the power must be $propto u_{mathrm{max}}^2 omega^2$. A reason for not specififying a sound wave by its displacement is that this property only applies to a wave in free space. Near a surface, such as the surface of a transducer, the displacement tends to be close to zero.

Your statement:

shouldn't the amplitude be decreasing with increasing frequency inverse squarely for constant intensity (and energy)

also applies to a mass-spring system, where the resonance frequency is $omega_0=sqrt{k/m}$, where $k$ is the spring stiffness and $m$ the mass. In such a system, the stored energy is $U=frac12 omega_0^2 m A^2$ for amplitude $A$, but the only way to increase $omega_0$ at fixed mass is by increasing the stiffness of the spring. In the case of acoustic waves propagating in free space, the stiffness of the air is fixed and there is no resonance.

Answered by Han-Kwang Nienhuys on November 21, 2021

Sweep tones are used from two of the many methods out there, to measure systems (not just sound reproduction systems, but pretty much any kind of equipment). The purpose of those signals is to have predetermined energy at each frequency and use it to excite the system and then measure the output. If you consider a linear system (time invariance would also be welcome but sweeps are also used to measure time variable systems) then the transfer function (or impulse response in the time domain) can be calculated from the measured output and the known input.

The main idea is that, since the system is linear, the input-output relation in the time domain is

$$ y(t) = h(t) * x(t) $$

where $y(t)$ is the *output*, which you will measure, $h(t)$ is the **impulse response** of the system (here we consider the simple case of time invariant system, so this will not change with time), $x(t)$ is the *input* and the $*$ symbol denotes convolution.

Now, if you were to deconvolve the input from the output you would end up with the impulse response. This is most easily done in the frequency domain (or $s$ - Laplace - domain) because convolution is turned into multiplication. Thus, one could easily manipulate algebraically the relation and end up with

$$ F { y(t) } = F { h(t) } F{ x (t) } implies Y(omega) = H(omega) X (omega) implies H(omega) = frac{Y(omega)}{X(omega)}$$

where $F { }$ denotes the Fourier Transform, and $H(omega)$ is termed **transfer function** (which is actually the frequency domain representation of the impulse response).

The operations can be performed with any signal (to be used as $x(t)$), as long as it has enough energy to excite the system to be measured in the frequencies of interest. That being said, most often than not, the signal being used is a sine sweep with logarithmically varying frequency (thus providing a pink-like energy spectrum).

To clarify a bit more, the radiated intensity becomes a characteristic of the system, when the amplitude/energy of the signal is constant. You may argue that the energy of the signal is **not** constant, since we just mentioned a pink-like spectrum (for log-sweeps), but this does not constitute an issue, since we are most interested in the relation between the input and the output. So, if we know that we gave $3 dB$ less energy at $100 Hz$ compared to $50 Hz$, then we can extract all the information we need... The input-output relation of the system.

Hope this helps. For more condensed information on the measurement techniques I would suggest going through *"Chapter 5 - Measuring Transfer Functions and Impulse Repsonses"* of *Signal Processing in Acoustics* by Havelock et al. (editors).

Answered by ZaellixA on November 21, 2021

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