so my question is the following : "Why are all real frequencies located on the imaginary axis of the s-plane. I'd like to understand intuitively why this is true?"

My guess would be that any "real" frequency has a constant amplitude in the time-domain, which is only true for frequencies with real part equal to zero.

Is this right? Or how should I know intuitively that the statement about real frequencies lying on the imiganiry axis is true?

  • $\begingroup$ Also note that if the real frequency represents that of a strictly real signal, that location on the imaginary axis also has a complex conjugate mirror image. $\endgroup$
    – hotpaw2
    Aug 30, 2016 at 16:50

3 Answers 3


Take the complex (Laplace transform) variable $s$, and look at the basic exponential signal


With $s=\alpha+j\omega$ we have

$$x(t)=e^{(\alpha+j\omega)t}=e^{\alpha t}e^{j\omega t}=e^{\alpha t}(\cos(\omega t)+j\sin(\omega t))\tag{2}$$

From $(2)$ you see that the real part $\alpha$ takes care of the exponential amplitude, whereas the imaginary part $\omega$ determines the frequency of the signal. If $\alpha=0$ (i.e., $s=j\omega$ is on the imaginary axis) you have an undamped sinusoidal signal with frequency $\omega$.

  • $\begingroup$ Indeed, with $\alpha=0$ we have two orthogonal sinusoidals, or a complex exponential. $\endgroup$
    – msm
    Aug 31, 2016 at 11:15

There is no "real" frequency since there is no imaginary frequency either. The reason we refer to the imaginary axis as the "frequency"-axis is that it represents $\omega$ and by definition, $\omega=2\pi f$. Therefore, if you want a pure "frequency", then it is right on the $\omega$ axis where $$s=0+j\omega=j2\pi f$$ In any other place on the $s$-plane you have $$s=\alpha+j\omega=\alpha+2j\pi f$$ So it won't be a pure "frequency".

  • $\begingroup$ In your first equation, $2\pi\omega$ should be just $\omega$. $\endgroup$
    – Matt L.
    Aug 31, 2016 at 11:18

i cannot completely decode the meaning of your question but i will try to answer a different question that, i believe, might be at the root of your question. it is:

Why do we, when determining the frequency response of an "analog" linear, time-invariant system (LTI), substitute "$s\leftarrow j\omega$" into the transfer function $H(s)$?

note that $j\omega$ lies on the imaginary axis.

the reason why we do this is because

$$ x(t) \triangleq e^{st} $$

is an eigenfunction to an LTI system. an "analog" LTI system (or a better term is "continuous-time" LTI system) has output $y(t)$ that is the impulse response $h(t)$ convolved with the input $x(t)$:

$$ y(t) = \int\limits_{-\infty}^{+\infty} h(u) x(t-u) \ du $$

that is true and totally general for any input $x(t)$ as long as the LTI system remains linear and time-invariant. note that the impulse response $h(t)$ totally describes the LTI system.

so if $e^{st}$ (for some constant $s$) is substituted for the input $x(t)$, then

$$\begin{align} y(t) &= \int\limits_{-\infty}^{+\infty} h(u) x(t-u) \ du \\ &= \int\limits_{-\infty}^{+\infty} h(u) e^{s(t-u)} \ du \\ &= \int\limits_{-\infty}^{+\infty} h(u) e^{st} e^{-su} \ du \\ &= \left( \int\limits_{-\infty}^{+\infty} h(u) e^{-su} \ du \right) e^{st} \\ \\ &= H(s) \ e^{st} \\ \\ &= H(s) \ x(t) \\ \end{align}$$

so you can see that when an exponential $x(t)=e^{st}$ goes into the LTI, then what comes out of the LTI is $H(s)$ times $x(t)$, where $H(s)$ is a constant with regard to $t$, and that is precisely what we mean by the exponential function, $e^{st}$, being an eigenfunction to an LTI system.

now $s$ is a finite and complex value. normally, when we consider frequency response of an LTI system, we want to drive it with a sinusoid and see what comes out. Euler tells us that

$$ e^{j \theta} = \cos(\theta) + j\sin(\theta) $$


$$\begin{align} x(t) &= A \cos(\omega t + \theta) \\ &= \frac{A}{2} \left(e^{j(\omega t + \theta)} + e^{-j(\omega t + \theta)} \right) \\ &= \frac{A}{2} e^{j(\omega t + \theta)} + \frac{A}{2} e^{-j(\omega t + \theta)} \\ &= \frac{A}{2} e^{j\theta} e^{j\omega t} + \frac{A}{2} e^{-j\theta} e^{-j\omega t} \\ \end{align} \ .$$

(here $\omega$ is real and is in radians per unit time.)

so now we have exponential input going in (actually two exponentials) where $s=j\omega$ for one and $s=-j\omega$ for the other, and with scaling factors. because it's LTI, we can consider each separately and add the results along with their scaling factors.

if the impulse response $h(t)$ is entirely real, then it turns out that the transfer function $H(s)$ has this property on the imaginary axis:

$$ H(-j\omega) = H^*(j\omega) $$

where $H^*$ means the complex conjugate.

$$\begin{align} y(t) &= \frac{A}{2} e^{j\theta} H(j\omega) e^{j\omega t} + \frac{A}{2} e^{-j\theta} H(-j\omega) e^{-j\omega t} \\ &= \frac{A}{2} e^{j\theta} |H(j\omega)| e^{j\phi} e^{j\omega t} + \frac{A}{2} e^{-j\theta} |H(j\omega)| e^{-j\phi} e^{-j\omega t} \\ &= |H(j\omega)| \frac{A}{2} \left( e^{j\theta} e^{j\phi} e^{j\omega t} + e^{-j\phi} e^{-j\theta} e^{-j\omega t} \right) \\ &= |H(j\omega)| A \cos(\omega t + \theta + \phi) \\ \end{align}$$

where the complex $H(j\omega)$ is expressed in magnitude/phase form

$$ H(j\omega) \triangleq |H(j\omega)| e^{j\phi} $$


$$ H(-j\omega) = H^*(j\omega) = |H(j\omega)| e^{-j\phi} $$

so concluding, if you have an LTI linear system with real impulse response $h(t)$ and associated transfer function $H(s)$ defined as

$$ H(s) \triangleq \int\limits_{-\infty}^{+\infty} h(t) e^{-st} \ du $$

and if you drive that LTI system with sinusoidal input $x(t)$

$$ x(t) = A \cos(\omega t + \theta) $$

having amplitude $A$, angular frequency $\omega$, and initial phase $\theta$, then the output of that LTI system, $y(t)$, is

$$ y(t) = |H(j\omega)| A \cos(\omega t + \theta + \phi) $$

which is a sinusoid of the very same frequency $\omega$ but with amplitude scaled by $|H(j\omega)|$ and phase shifted by $\phi$ radians where

$$ \phi = \arg\big\{ H(j\omega) \big\} $$

so the amplitude of the sinusoid is boosted by the magnitude of the complex $H(s)$ and the phase of the sinusoid is shifted by the angle of the complex $H(s)$ where $s=j\omega$.


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