I have been studying signal processing and control for a while now, and I use Laplace and Fourier transforms almost everyday. Also another tools such as Nyquist or Bode plots.

However, I had never thought of this until today: what is the physical meaning of a complex number when dealing with frequencies?

This may sound silly, but I was asked this question and I didn't know what to answer. Why do we talk about $j\omega$ and not just $\omega$ in, for example, Fourier transforms and Bode or Nyquist plots? What is the physical sense of the real and imaginary part of a zero or a pole in the Laplace domain?


5 Answers 5


We usually talk of $j\omega$ when we're also interested in the Laplace transform of a signal / system, but want to just talk about the frequency response.

The physical meaning of the imaginary part is that it refers to purely sinusoidal signals and are constant "amplitude". The real part refers to signals for which the "amplitude" decays or grows exponentially.


I think that I started to understand the relationship between zeros/poles and the frequency response. The idea is that you adjust the frequency $w$ of your frequency-domain basis functions $e^{jwn}$ and speed of their decay to match the $z_{zero/pole}$. I mean that that zero/pole can be complex number with amplitude outside the single circle and adjusting the frequency you move your complex number $e^{-jw}$ vector along the single circle in the complex plane but there is no frequency that can make it equal to $z=2$ or $z=j/3$, for instance. So, your basis functions must look like $e^{(k-jw)n}$ to reach any pole/zero in the complex plane. It is interesting because I heard that Fourier basis $e^{-jw}$ can represent any singnal but it seems insufficient and we need Laplace basis $e^{(k-jw)n}$ in filter design.

Now, purely real $z$ means that the "complex exponent" that matches it has no imaginary component. It must decay without oscillations, like $e^{kn}$, in order to respond the zero/pole. Take pole at $z=1$, for instance. You have a system $y_n - y_{n-1} = x_n + x_{n-1} + \ldots$ so that $Y(z) = X(z)/(1-z).$ The pole $z = e^{-jw} =1$ corresponds to frequency $w=0$. Indeed, with $x_n=1$, we have $y_n = y_{n-1} + 1$ which grows unboundedly. Making it oscillating, i.e. setting $w \neq 1$, will break the growth since it will first accumulate, when $x_n=2cos(wn) = e^{jwn} +e^{-jwn}> 0$ and then reduce the accumulation to zero, during the second half of the sine period. This suggests that imaginary poles will give you infinite responses for oscillating functions (components of your inputs signal).

When you have a system $y_n=ay_{n-1}$, you can easily get the pole function by applying a delta impulse at the input. The response observed is the pole. I mean that response is a decaying exponent $y_n = e^{k-jw}n=a^n$. Every clock it is $a = e^{k-jw}=e^ke^{-jw}$ times the previous value. Note, that it (the pole aka feedback coefficient and, hence the response function) is complex it general, which means that your response will oscillate. When you multiply a complex number by another, your number is scaled in length and shifted in phase. The complex part is responsible for the phase shift (the oscillations).

I remember from the system theory that oscillations actually stand for the second order system. Probably, this will answer mine commutation cell question. The idea is that when you have first level controls the increment of the other and the other controls the increment of the first, like the electric inductor and capacitor in harmonic oscillator, $$\begin{cases} \dot{u} & = &i\\ \dot{i} & = &-u\\ \end{cases}$$ enter image description here

is a second order system because can be expanded into $\ddot{u} = \dot{i} = -u$, the famous spring osciallator equation: position negatively controls the accelration. So, two purely real state variables (aka accumulators) do oscillate. I see that the complex plane also consists of two axis, the same two variables. When all energy is concentrated in the first accumulator, you have 1+0j state, when half way back, you have the opposite, state=0+1j, then second accumulator pushed energy backwards, state3=-1+0j, which is ponged to the first into the state4= 0-j and process repeats. These are 4 quarters of traveling along a unit circle in the complex plane and mimicking the harmonic oscillations. So, probably, you will be able to split $1/(1-(a+jb)z)$ into $1/(1-r_0z)\cdot 1/(1-r_1z)$ with real $r_0$ and $r_1$.

Wait, you cannot make that decompose single $z$ into $z^2$ and I recall that complex poles always come in conjugate pairs. That is, if you have pole (a+jb), you also have (a-jb). As i understand, this helps to make output purely real, given real input since feedback (a+jb) means that system evolves as $(a+jb)^n = e^{(k+jw)n}$, the phase rotates in one direction whereas $$(a-jb)^n = e^{(k-jw)n}$$ rotates the phase in the other direction and their sum is $e^{kn}(e^{jw}+e^{-jw})^n$ is purely real. The $x_{n+1}=-x_{n-1}$ system above has solution $X(z)=(x_0+zx_1)/(1+z^2)=(x_0+zx_1)/[(1+jz)(1-jz)]$. Probably you already understand this. I just expanded your question.

The transfer function $1/(1+z^2)$ stands for the sequence $\{1,0,-1,0,1,0,\ldots\}$. There must be a "hidden variable" (yes, it is interesting if complexity of poles is identical to the need for imaginary numbers we need in QM. The position and momenta are complex conjugates, a sort of 90° rotated, of each other and knowing one you can compute the other) hidden variable in order to keep in mind if we are moving to 1 or -1 after 0 state. The complex conjugate is a kind of complimentary, orthogonal accumulator yet real variable, such as inductor current for capacitor voltage, that keeps track for that. I join the question for anybody to clarify why we need two such complements in order to have purely real voltage oscillation and what does single complex oscillation mean.

I see it this way (for the LC-oscillator above)

$$\begin{bmatrix}\text{state} &\text{description}& \text{capacitor [V]} & \text{inductor [I]} \\ 0&\text{all energy is in the capacitor} & 1 + 0j & 0 + j \\ 1&\text{all energy is in the inductor} & 0 + j & 1 + 0 \\ 2&\text{all energy is negaitvely charged cap} & -1 + 0 & 0 + -j \\ 3&\text{all energy is negative current} & 0 + -j & -1 + 0 \\ \end{bmatrix}$$

That is, what you see imaginary voltage is a real current in a parallel frame of reference, i.e. from the inductor's point of view. Because, as I have tolde you, LTI state evolves by multiplying the current state with eigenvalue, we should oscillate between 1 and -1 over the unit circle, which implies j intermediate states. But, what you see as conserved energy in the imaginary space, happens to be just another accumulator. The conjugate accmulaor is just another accumulator. For some reason it happens to be of conjugate kind, as I tried to explain in the commutation cell.

I seem to deviate again. Since harmonic oscillation is a superposition of two evolutions, made by two complex poles $j$ and $-j$, we should have two columns per every conjugate variable. Here is the missing part

$$\begin{bmatrix}\text{state} & \text{capacitor -j [V]} & \text{inductor [I]} \\ 0& 1 & 0 j \\ 1& 0 - j & -1 \\ 2& -1 + 0 & -j \\ 3& 0 + j & +1 \\ \end{bmatrix}$$

The voltage in capacitor is a real value, which is an average of two capacitor columns, $(j^n + (-j)^n)/2 = \cos (n\pi/2)$. The to processes opposite rotations cancel the imaginary components out. In fact, the current flows in one direction but $\ddot(x)=-x$ admits any direction and abstracts it away but the averaging. So, pole alone stands for a concrete process, the flow of current in one direction or the other. And, if you ask what is the complex pole, the answer is it is the factor by which vector [current, voltage] is scaled every clock if we are in the discrete domain (or [di/dt, dv/dt] if we are in the continous domain) where real factor stands for their amplitude, real part $cos w$ of the complex factor $e^{jw}$ stands for voltage evolution and imaginary part $sin w$ stands for current evolution. The current is imaginary because you look from the voltage point of view, $\ddot{v}=-v$. In contrast, the voltage would be imaginary and current real from the current frame of reference, $\ddot{i}=-i$. Hopefully, this is correct and anybody can explain it better.


Laplace transforms can be used to predict a circuit's behavior. The Laplace transform takes a time-domain function $f(t)$, and transforms it into the function $F(s)$ in the $s$-domain. You can view the Laplace transforms $F(s)$ as ratios of polynomials in the $s$-domain. If you find the real and complex roots (poles) of these polynomials, you can get a general idea of what the waveform $f(t)$ will look like.

For example, as shown in this table, if the roots are real, then the waveform is exponential. If they’re imaginary, then it’s a combination of sines and cosines. And if they’re complex, then it’s a damping sinusoid.

All of that comes from the Euler's formula and the definition of Fourier series which is a way to represent a (wave-like) function as the sum of simple sine waves.

  • 1
    $\begingroup$ All the information you give is true. Nevertheless, the questions asked (Why do we use $j\omega$ and not just $\omega$ ? What is the physical sense of the real and imaginary axis in the Laplace domain?) were not answered. $\endgroup$
    – Tendero
    Apr 19, 2016 at 14:57

One answer is very simple: Information.

An AC signal simply can't be quantified with a single number. Sum two 100hz 1V signals and you might get anything between 0 and 2 depending on phase. Complex numbers solve this problem by 'carrying around' two pieces of information all the time.

Poles and zeros are similar - their frequency doesn't tell you everything. Two RC filters create two poles. An LC filter creates two poles. But they're not equal. Complex numbers are able to carry the information that describes the difference.


Theoretical point of you. Please try to take a square root of negative frequencies and it will take you to strange place.

Some 300 years ago it was necessary to introduce a variable called $j$

Nevertheless, Laplace transform, transforms time-domain signal to $s$-domain where $$s=\sigma+j\omega$$

where as Fourier transform, to frequency domain $j\omega$


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