Relationship between time constant and frequency

Question

I have seen the relationship that the RC time constant (\$\tau\$) is equal to the inverse of the -3dB angular frequency (\$\omega\$). The time constant is in units of seconds, while the angular frequency is in units of radians per second.

My question is why the time constant (in seconds) isn't equal to the inverse of frequency (in 1/seconds) rather than the inverse of angular frequency (in radians per second)?

Answer 1

The short story is this: that \$rad\$ is a ghost.

Here's why: in the SI angles can be considered either fundamental quantities - with their own unit (the radian) - or dimensionless quantities as ratios of lengths (and, for solid angles, as ratios of areas). But both cases must respect dimensional homogeneity in the series expansion of functions that employ angles.

Dimensions and units

In the former case (angles as fundamental quantity with their own unit), since the argument of any analytic function - such as exponential and trigonometric functions - HAS TO BE dimensionless (because otherwise one would end up mixing apples with oranges), a conversion factor HAS TO BE introduced to leave only the magnitude of the angle. In the same way chemists have pressures inside logarithms by dividing by a reference pressure, one should be using a unit angle (that some, like Gibbins (1) call \$ \beta_0 \$) that will divide angle quantities wherever they appear. This will make the [\$\alpha\$] dimension disappear and from the argument of any analytic function, avoiding the orange-apple paradox.

In the latter case (angles as ratios, hence dimensionless), the paradox is avoided from the start since angles are pure numbers to begin with. Some author (for example Szirtes (2)) prefer to view dimensionless quantities as quantities with dimension \$[1]\$, so that the any power of such quantity will still have dimension \$[1]^n=[1]\$ (as opposed to dimension \$[\alpha]^n\$ , thus saving the day and restoring homogeneity.

In practice, people will not use the conversion factor \$\beta_0\$ but will still consider angles dimensionless while using the unit \$rad\$ for the following reason illustrated by the Bureau International des Poids et Mesures (BIPM):

In practice, with certain quantities, preference is given to the use of certain special unit names, or combinations of unit names, to facilitate the distinction between different quantities having the same dimension. When using this freedom, one may recall the process by which the quantity is defined. [...] The SI unit of frequency is given as the hertz, implying the unit cycles per second; the SI unit of angular velocity is given as the radian per second; and the SI unit of activity is designated the becquerel, implying the unit counts per second. Although it would be formally correct to write all three of these units as the reciprocal second, the use of the different names emphasises the different nature of the quantities concerned. Using the unit radian per second for angular velocity, and hertz for frequency, also emphasizes that the numerical value of the angular velocity in radian per second is 2 pi times the numerical value of the corresponding frequency in hertz.

Excerpt taken from SI Brochure: The International System of Units (SI) [8th edition, 2006; updated in 2014]. Emphasis mine.

But all trig, exp and log functions need to have dimensionless arguments. Period.

So, if you use dimensionless angles, \$\omega\$ will have dimension \$[T]^-1\$ and unit \$s^{-1}\$; while if you use angles of dimension \$[\alpha]\$ and unit \$rad\$, by dividing by the conversion factor \$\beta_0\$, you will still end up with the same result. The 'ghost' \$rad\$ you put 'as a reminder' in the \$\omega\$ case is there to remember you that there is a \$2 \pi\$ that derives from the fact that trig functions are periodic of period \$2 \pi\$.

The \$2\pi\$ factor

And why in one case you have 1 and in the other \$2 \pi\$?

Because, in one case the solution of your problem is a real exponential that has no periodicity, while in the other you are basically doing frequency analysis which rely on imaginary exponentials which are in essence trigonometric functions of period \$2 \pi\$. The \$2 \pi\$ has nothing to do with \$\tau\$, it is instead introduced as the angular frequency of the generic (co)sine wave used as a stimulus in frequency analysis. (I'll be back with a few drawings).

I would add that for the impulse (and step) response, we are using a real-valued exponential - so there's really no 'period' here since the function is not periodic. When you dump in that i or j in the exponential stimulus to your RC network, you enter another Realm where, Euler taught us, a complex (actually imaginary) exponential turns into trigonometric functions. And as we all know, trig functions are periodic of period \$2 \pi\$, so we need a rescaling to fit them to our desired periods. That's were the \$2 \pi\$ come from.

Bandwidth and time constant (rise time)

As a side note, the numerical relationship between rise time and bandwidth has its roots in the addition of a \$2 \pi\$ factor. In a single pole RC network the step response rise time is linked to the time constant \$\tau\$ by:

\$ t_{r} = ln(90/10) \tau \approx 2.2 \tau\$

but \$\tau\$ is linked to the cutoff angular frequency of the frequency response by \$ \tau = 1/\omega \$ and \$ \omega = 2\pi f \$, so:

\$ t_{r} \approx 2.2 \tau = \frac{2.2}{\omega} = \frac{2.2}{2 \pi f} = \frac{0.35}{f} \$

Quite a familiar result.

References

(0) Bureau International des Poids et Mesures (BIPM)

(1) J. C. Gibbins, "Dimensional Analysis", Springer (2011)

(2) T. Szirtes, "Applied Dimensional Analysis and Modeling" 2nd edition, Elesevier (2007)

Answer 2

Radians per second is the natural unit of frequency in all kinds of situations. The reason is that whatever values determine the frequency end up as the parameters of a cosine function:

$$v(t) = A \cos (stuff \cdot t)$$

The cosine parameter is an angle, and angles are measured in degrees or radians. Radians are more useful from a geometric standpoint, so that's what we use. Ultimately, it comes down to the fact that we describe oscillation in terms of the geometry of a circle.

Now I'm wondering whether oscillation could be described in another way. Trigonometric functions are solutions to second-order differential equations, so this seems to be baked into mathematics at a very low level.

Answer 3

Consider a series \$\small RC\$ circuit where the input voltage is across the \$\small RC\$ series combination and the output voltage is taken across \$\small C\$.

The transfer function of this arrangement is: \$\small \dfrac{1}{1+RCs}\$, and the impulse response is: \$\small\dfrac{1}{RC}e^{-t/RC}\$, and thus \$\small RC\$ is identified as the time constant, \$\tau\$. Hence the transfer function may be written: \$\small \dfrac{1}{1+\tau s}\$, and the impulse response is \$\small \dfrac{1}{\tau}e^{-t/\tau}\$

To find the frequency response of this circuit, we use the substitution \$s\rightarrow j\omega\$, where \$\omega\$ is the angular frequency in \$rad\: s^{-1}\$. This gives the frequency response: \$\small \dfrac{1}{1+j\omega \tau}\$, from which the gain is \$\small \dfrac{1}{\sqrt{1+\omega ^2\tau ^2}}\$.

The -3dB frequency, \$\omega = \omega _c\$, is defined when the gain is \$\small \dfrac{1}{\sqrt{2}}\$, and therefore we must have \$\omega_c \tau=1\$ or \$\omega_c=\dfrac{1}{\tau}\$

Answer 4

Rise time or max slew rate and Bandwidth are mathematically related. But visually you can verify for yourself that the fastest rise time of a pulse is equivalent to sine wave where Tr=1/(3f)

This is not a pure (fundamental only) sinewave.

This one is even a worse drawing of a sine wave often seen in publications from rookie illustrators.

Take note the slope is closer to a triangle wave with 30% rounded at top and NEVER like above.

Answer 5

This is more of a math question than a electrical engineering question.

I think about this in terms of the complex exponent:

\$ e^{i \alpha} \$

\$\alpha\$ is in radians: \$2\pi\$ gets you back to where you started. If we instead look about this for a real negative exponent (\$\tilde{\alpha} = i\alpha\$) and plug in, we get the normal expression:

\$ e^{-\tilde{\alpha}}\$

But, we haven't done anything to change the units of \$\alpha\$, which are still in "radians" as much as we can use that as the name of a unit.

When we look at your problem, we get:

\$ e^{-\tfrac{t}{\tau}} \$

Since we want time to be in "seconds", we need \$1/\tau\$ to be in "radians/second" not "1/seconds".

We can do this another way, as you mention in the comments:

\$ e^{i 2\pi \tfrac{t}{\tau}}\$

This makes the units of \$\tau=\$"seconds", and now you are in a system where \$1/\tau = f\$. By convention this isn't the case, but there isn't anything fundamentally wrong with it. You won't have the same definition of \$\tau\$, your definition is now \$RC/2\pi\$.

This is a very hand-wavy explanation, but hopefully it is sufficient.