# Why is the frequency complex

When we take the Laplace transform of a function, why is the frequency (s) complex? I can not understand the significance of complex frequency.

• Maybe this will help: ccrma.stanford.edu/~jos/mdft/Projection_Circular_Motion.html If you follow the links up, it is part of a larger textbook available completely online. Jan 25 '21 at 16:18
• not only complex, you'll be surprised when you'll see negative frequency too Feb 25 '21 at 7:42

Why s is complex in the Laplace Transform: Note first that the Fourier Transform is a correlation (multiply and accumulate) of our time domain signal to every possible frequency. You could describe this correlation with sines and cosines (yuck!) or very compactly with exponentials (yes!). The Fourier Series Expansion is an excellent example of this in how it is typically first introduced, we step through every sine and cosine at the possible harmonic frequency locations and perform a correlation to see how much of our input signal in present at each frequency. Later we see the same equations in very compact form by describing our frequencies as single spinning phasors $$e^{j\omega t}$$ instead of carrying two phasors with the sines and cosines, such as: $$\frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$.

See pictures at the bottom of this post to make this very clear, and look again at all the transforms (Fourier, Laplace, Z, etc) to see how in all cases this is a correlation of the form:

$$corr = \int f(t)g^*(t)dt$$

Which measures the linear relationship between two functions (* is complex conjugate). For discrete systems the integral becomes a summation but again we see the correlation relationship.

Why is this important? So in Fourier we compute the correlation to all values of

$$e^{j\omega t}$$

And see the content in our waveform for each frequency given by omega, and we have remapped our time domain waveform into the frequency domain, where certain mathematical challenges can be solved much more simply (or other reason related to observation or manipulation of the signal in frequency).

Some waveforms will not converge in the correlation integral, and therefore do not have a Fourier Transform. However we can "pre-scale" those waveform by multiplying by a decaying exponential which will then allow the solution to converge:

$$e^{st} = e^{(\sigma + j \omega)t} = e^{\sigma t}e^{j \omega t}$$

The result is the Laplace Transform (in discrete systems the z-Transform), where again mapping our time domain waveform to the s-domain allows us to solve mathematical challenges much simpler (specifically Laplace converts differential equations to simple algebra--- yes!!!). It also allows us to solve transient conditions such as step responses and the like. And yes, you can picture the Laplace Transform as a surface showing the correlation of our input function x(t) to all possible spinning phasors (as in Fourier) as well as spinning phasors that grow or decay with time. (Typically this surface is presented as just the poles and zeros showing where the correlation is maximum and zero since that completely describes the rest of the surface not shown.)

Related to time and frequency and complex signals in general: A significant use of complex signals not yet mentioned is in communication systems. The modulation at any carrier for a linear system can be modeled directly at baseband (resulting in much simpler processing as we are no longer concerned with modeling every cycle of a high frequency sinusoidal carrier, but can just model the relatively lower rate of change in amplitude and phase directly of that carrier.)

That said, any waveform where the positive half of the spectrum is not the complex conjugate (not the same magnitude and opposite phase) when viewed at baseband as described above, can be described using complex signals; which covers nearly every modern communications waveform today (simple FM and AM modulations that are double sided are cases where this would not be necessary).

Euler's identity shows this quite well. Any real sinusoid has two complex signals given as a positive frequency and a negative frequency, obeying the rule that the frequencies must be conjugate symmetric. To create just one of these two frequencies (at baseband) we need to have a complex signal (often implemented with "I" and "Q" datapaths one representing the In-phase or real axis and the other representing the Quadrature or imaginary axis). Each complex frequency is a phasor spinning in one direction versus time ($e^{j\omega t}$ ). The pictures below show this relationship in decomposing Euler's and also how a complex frequency is required if the positive half of the spectrum is not equal (complex conjugate) to the lower half:

The time domain signal is shown as a phasor (polar plot with vector representing instantenous time and angle, rotating with time). As those two phasors are equal and spinning in opposite directions (conjugate), a sinusoid that remains only on the Real axis wil result. Thus we see Euler's identity, the concept of positive and negative frequencies, and the reason we use complex frequencies.

Finally, here is a picture showing a case where the signal is modulated. In the picture we are centered at 0 (DC), but we could be equally centered at any carrier frequency with the same phase and amplitude deviation from the carrier versus time as depicted in the IQ (polar plot) diagram on the left. If our spectrum (on the right) does not have a complex conjugate relationship between the upper and lower half, we must use complex signals to describe that signal.

The Laplace transformastion is an extension/generalization of the Fourier transformation. Because the Fourier integral does not converge for all functions we introduce an additional term $e^{-\sigma t}$ under the integral, which approaches zero for increasing time $t$. This ensures convergence of the transform.

Thus, we can combine the expression $e^{-j\omega t}$ from the Fourier integral with this additional exponential function and arrive at $e^{-st}$ with $s=(\sigma+j\omega)$. Hence, the quantity $s$ is a complex number. This "complex frequency" expression describes a sinusoidal signal with increasing (positive $\sigma$) or decreasing (negative $\sigma$) sine wave.

It is interesting to note that the solution in the time domain for a system which is able to show decreasing oscillations (tank circuit) also gives the same complex frequency variable $s$.

• You could just as easily transform the caps to 1/jw as 1/s. Sep 12 '14 at 13:28
• I think, your comment does not apply because this is NOT a transform. Rather, it is simply another symbol for jw. This is one of the major problems students have with "s". Sometimes it is imaginary (as in your example) and some times it is complex. The questioner has asked for the significance of "s".
– LvW
Sep 12 '14 at 14:38
• Can you show a link to where the tank circuit uses "s"? I think we're talking about two different things. Sep 12 '14 at 14:39
• If you are solving the differential equation in the time domain you, of course, will get a damped sinusoidal solution which - in its exponenetial form - shows an exponent "s*t". And this quantity "s" is identical with "our" complex frequency.
– LvW
Sep 12 '14 at 14:44
• I dont think so. The Laplace transform connects the time domain step response with the system function. In both cases the frequency variable is s.
– LvW
Sep 12 '14 at 18:45

I think maybe you need to remember that the background math to a sinewave is complex. Euler brought that idea to the front back in the 1700s: -

OK in EE we use "j" as the complex operator but in math they use "i". Another way of looking at the sine wave is this: -

My preferred explanation about the "complexity" of a sine wave is that a sine wave is constructed from two rotating vectors that rotate in opposite directions (one can be regarded as positive frequency and the other as negative frequency) but the math is all basically the same: -

And thus we end up with a sinewave being $\dfrac{e^{j\theta}-e^{-j\theta}}{j2}$

It's complex numbers right from the start if you really wanted to be accurate.

• But why the complex plane ??? Why not the x-y plane or any other plane ??? Why is the complex plane so special Sep 12 '14 at 13:14
• It has to be the complex plane for the math to work. The complex operator when squared becomes -1 and this is all part of the trig expansion of sin containing alternate negative coefficients. Look up Taylor's series for sin. Sep 12 '14 at 13:47
• Andy - you have nicely explained the meaning and the definition of a complex frequeny, but what is the reason for its introduction/definition and why appears it in the Laplace transform?
– LvW
Sep 12 '14 at 14:41
• @lvw I'm pointing out that things were complex before you do the laplace stuff and why should things be any different after doing laplace. Sep 12 '14 at 15:33
• @Andy, ..."things were complex"? I agree, we CAN use, for example, Eulers identity - but is there any necessity? From the beginning, the quantity "frequency" were not complex.
– LvW
Sep 12 '14 at 15:40

First, let us say that the complex plane is an immediate consequence of the introduction of the embarrassing symbol j (or i in mathematics). But why the introduction? Since every square root of a negative number can be factored as a multiplication of square root of negative one and the (other) real number, that is: $$\\sqrt{-r} = \sqrt{-1\times r} = \sqrt{-1}\times\sqrt{r} = j\sqrt{r},\, r \in \mathbb{R}\$$.

it facilitated mathematicians to introduce j (or i) as a short-hand notation in writing this. More importantly, the complex numbers plane is now more general than the real numbers plane, in that every real number is complex number with no imaginary component, that is:

$$\mathbb{R} \subset \mathbb{C}\,\,\, \equiv \,\,\,\forall{x: x\in \mathbb{R}} \implies x \in \mathbb{C}$$

Now, why is the frequency complex? Simply said, just to facilitate the math. It is much similar to the concept of phasors (sinusoids) used in AC signals analysis, except the complex frequency now can cover all types of inputs and not just sinusoids.

To clarify:

• Consider the input $$\V(t)=V_m\cos(\omega t + \phi)\$$, where by Euler's identity $$\\cos(\omega t + \phi) = \mathrm{Re}\!\left(e^{j(\omega t + \phi)}\right)\$$

• Say now you want a damping input (input that dies after time) of the form $$\V(t) = V_m e^{\sigma t} \cos(\omega t + \phi)\$$, where sigma is negative, of course (otherwise the circuit will explode)

• Rewrite the cosine expression using Euler's identity $$\V_t = \mathrm{Re}\!\left(V_m e^{\sigma t} e^{j(\omega t + \phi)}\right)\$$

• Rearrange to preserve the phasor form $$\V_t = \mathrm{Re}\!\left(V_m e^{j \phi} e^{(\sigma + j\omega)t}\right)\$$

• Make the substitution $$\s = \sigma + j\omega\$$, to be used instead the real frequency (it is more general)

Now you changed nothing in the input except of multiplying it by a real dying exponential. In this way you can generalize the concept of real frequency to complex frequency just as we did above for the complex an real planes to facilitate the math. Magic!

• What does 'w' stand for ??? Is it the amplitude ??? Sep 12 '14 at 16:22
• The angular frequency, the frequency at which the sinusoid oscillates. Remember, w=2*pi*f, where f is the natural frequency. It tells how many wave cycles are there in each second (cycles/second). For example, a 60 Hz frequency system means that there are 60 complete wave cycles every one second. Sep 12 '14 at 16:33

There are two response components in electronic systems, exponential and sinusoidal. The Laplace transform is used to convert a function in time to a function representing the exponential and sinusoidal equivalent. Using Euler's rule the integration comes down to integrating the time function with -est. It's sort of like a trick because the the imaginary part is the exponential component and the real part is the sinusoidal component.

• Horribly important precisation: linear systems! Feb 25 '21 at 7:43
• Well, this is not detailed description. The previous comment provides the details. Apr 4 '21 at 13:26