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Supposed I have a tansfer function: $$\frac{s+1}{s+2}$$

So the zero of the tansfer function is -1, while the pole is -2.

When we want to know about the frequency characteristics of the transfer function, we just use "jω" to substitute the "s", then we can plot the Bode to know about the frequency characteristics.

As mentioned before, the zero is "s = -1", then I use the "jω" to substitute the "s": "jω = -1". Now I am confused, what does the "jω = -1" mean? How could the frequency ω be negative? Could anyone give me some tips? I am very appreciated for your help!

T L
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3 Answers3

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This is actually a deep dive into math. But basically, the sine or cosine frequency you know is not considered to be mathematically fundamental in the sense that they are not the simplest buildings blocks

What is considered the simplest building block is \$e^{jθ}\$.

Compare the three following cases:

$$e^{jθ} = cosθ+jsinθ$$

$$cos\theta = \frac{e^{jθ}+e^{-jθ}}{2}$$

$$sin\theta = \frac{e^{jθ}-e^{-jθ}}{2j}$$

\$e^{jθ}\$ actually describes something mathematically whereas \$cos\theta\$ and \$sin\theta\$ are just mysterious function names. And when you break cosine or sine up into a mathematical expression it is more complex than \$e^{jθ}\$

You can see from the three comparisons that it takes multiple \$e^{jθ}\$ terms to produce \$cos\theta\$ and \$sin\theta\$. It takes multiple terms because you need to get rid of the cosine or sine that you don't want from Euler's identity, and it requires negative values of \$\theta\$ to do this and that's where the negative frequencies come in. Because if your periodic functions are time dependent, \$\theta\$ is replaced with \$\omega t + \phi\$.

When graphed, \$e^{jθ}\$ looks like a coil in 3D space with the sine or cosine you know being a projection of that.

enter image description here

You should really just read the link that I learned this from which is here:

https://www.dsprelated.com/freebooks/mdft/Complex_Sinusoids.html

It's almost like a 5th dimensional object passing in and out of our 3-dimensional space plus time. Except it's not really passing in and out so much as it is being constrained to our 3D space, being projection.

DKNguyen
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    [This one is prettier](https://www.deviantart.com/woodmath/art/Phasor-Candidate-315348571). – jonk Sep 25 '22 at 02:31
  • Thanks! So it seems that negative frequency is introduced just to be able to represent both the amplitude and phase of a signal on one graphthe. The introduction of negative frequencies appears to have mathematical rather than physical significance? – T L Sep 25 '22 at 03:02
  • @TL It's more like Euler's formula is a really simple helix (when you introduce time.) It's formula is two parts that are in quadrature and they determine the chirality of the helix. If you want to 'flatten' the helix, you need to subtract away one of the dimensions from the helix (not time, of course.) If you do, it's no longer 3D but now 2D with one axis remaining as time. Asking about physical significance is a whole other thing. Depends on how you choose to see the world. No? – jonk Sep 25 '22 at 03:21
  • @TL You could call it a fancy bookkeeping trick if you want. It's not really any different from when you look at a phasor rotating on a unit circle but only pay attention to the real component as what can be directly observed in physical reality. The imaginary component is also there keeping track of what you don't see. – DKNguyen Sep 25 '22 at 03:32
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Really some confusion here. Transfer functions present the behaviour of filters if s is a free complex valued variable. That's a math fact which is proven by people who have researched thoroughly the math of Laplace transforms.

Poles and zeros of your transfer function happen to be s=-2 and s=-1, which are complex numbers with imaginary part = 0. Poles and zeros actually tell what a transfer function is except a constant amount of gain.

The frequency response of a filter can be got by limiting the s in the transfer function to values where the real part is 0. That's commonly written as "s=jω" But you had forgotten it soon and started again to search the zero i.e. where jω = -1. Such zero doesn't exist because jω was just assumed to be an imaginary number, it cannot be -1.

In your confused state you started to think that "jω = -1" means negative frequency. That's another error. That "jω = -1" means "ω = j" which was excluded when one decided that to get the frequency response let s=0+jω, where ω is a real number variable.

But that ω can really have all real values, including negative ones. Negative frequencies must be allowed. And you cannot see what they mean. You have used to live with the fact that frequency is positive. That's true when when calculate the number of oscillations per time unit or actually the angular rotation velocity, when we talk of ω. Except we must allow ω<0 to be able to present rotations to both directions.

You may wonder ω<0 can well be possible with rotations, but voltages go up and down. That's a valid looking point, but unfortunately in math working with frequency responses we must allow negative frequencies to be able to work the phase conditions right. One can find filterings which are different if they are the same with positive frequencies, but differ in negative frequencies. The frequency response stops be unique if we omit negative frequencies.

  • Glad you wrote this. The very first thing that struck me is that jw ne -1. But I didn't want to write at the time and, anyway, you did better than I would have. Thanks for taking it on! And I think you nailed down what's going on with the OP about it, too! – jonk Sep 25 '22 at 03:27
  • Thanks!! This my understanding about your answer. The 's '(s = σ+jω) could be any value. But when we talk about the frequency response, we need to make s = jω, and the ω must be positive in our real world. The negative is meaningless for us because we cannot see what they mean. Do my understangding right? – T L Sep 25 '22 at 03:47
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    You have understood it right. All possible frequencies of up and down swinging voltages can be told with ω equal or greater to zero. That's our intuitive way to think of frequency. But to be able to present different filterings with different transfer functions we must extend the range of ω to contain also negative values. –  Sep 25 '22 at 08:58
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    @TL (Continued) To stay in truth also the Fourier transforms of signals have jω as their free variable. and that transform can be got from the Laplace transform by setting s=jω. The ω must be allowed to get all real values to separate the Fourier transforms of real valued signals and complex valued signals. Complex valued signals in really built systems have 2 wires - one for the real part of the signal and one for the imaginary part. Negative frequency is not the only obscurity the engineering math needs. You'll meet also infinity as "the Dirac delta function". –  Sep 25 '22 at 09:33
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Now I am confused, what does the "jω = -1" mean?

An imaginary quantity cannot equal a real quantity so, \$j\omega\neq-1\$! Ever.

The Laplace variable is \$s=\sigma+j\omega\$.

The crosses on the pole-zero map, shown below, represent poles in the Laplace domain. Circles represent zeros. The pole-zero map can represent either hardware, signals or both. The example shown is based on the transfer function that you provided.

The two crosses on the vertical (imaginary)axis represent a sinusoidal signal. They are both necessary for a single sinusoid. The pole-zero map does not represent phase.

All sinusoids appear on the vertical axis only. To the left are damped sinusoids and to the right are sinusoids increasing in amplitude. So in order to represent the sinusoidal impedances and transfer functions \$\sigma\$ must be \$0\$

For the transfer function: $$G(s)=\frac{s+1}{s+2}$$,

The pole on the \$\sigma\$ (horizontal) axis is \$p=-2+j0\$. The zero is at \$z=-1+j0\$.

All the poles and zeros on the diagram together represent the output signal.

Pole-zero map

I think the confusion, that arises, occurs when the Laplace variable is used to mean different things. When \$s\$ represents a pole it is customary to assign the letter \$p\$ to indicate it is a special value of \$s\$. The letter \$z\$ is the value of \$s\$ that represents a zero.

To transform a transfer function from the Laplace domain into the sinusoidal domain, only values along the \$j\omega\$ axis can be used. So \$\sigma\$ is set to \$0\$.

\$ s=j\omega\$ means \$s=0+j\omega\$. The values of the poles and zeros remain the same but the transfer function is restricted to the \$j\omega\$ axis.

While the transfer function \$G(s)\$ represents hardware, it also represents the ratio of the out put to the input as in: $$\frac{V_{o}(s)}{V_{i}(s)}=G(s)=\frac{s+1}{s+2}$$

It is then transformed to the sinusoidal domain as: $$\frac{V_{o}(j\omega)}{V_{i}(j\omega)}=G(j\omega)=\frac{j\omega+1}{j\omega+2}$$

RussellH
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