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For a given transfer function (in the s-plane) we've a general form that looks like:

$$\text{H}\left(\text{s}\right)=\frac{\text{X}\left(\text{s}\right)}{\text{Y}\left(\text{s}\right)}\tag1$$

This kind of transfer function that I studying right now, is in the context of electrical engineering. Now, in the study I focus on passive components, like capacitors, inductors and resistors. Using the complex impedance of those components, that gives me:

  1. Resistor: $$\underline{\text{Z}}_\text{R}=\text{R}\space\left[\Omega\right]\tag2$$
  2. Capacitor: $$\underline{\text{Z}}_\text{C}=\frac{1}{\text{j}\omega\text{C}}\space\left[\Omega\right]\tag3$$ Where $\text{j}^2=-1$
  3. Inductor: $$\underline{\text{Z}}_\text{L}=\text{j}\omega\text{L}\space\left[\Omega\right]\tag4$$ Where $\text{j}^2=-1$

For studying them in the s-plane I can just substitute:

$$\text{s}=\text{j}\omega\space\space\space\to\space\space\space\underline{\text{H}}\left(\text{j}\omega\right)=\frac{\underline{\text{X}}\left(\text{j}\omega\right)}{\underline{\text{Y}}\left(\text{j}\omega\right)}\tag5$$

Because the inductor and capacitor don't involve any real part.

Now, to find an interesting point I want to solve:

$$\Re\left\{\underline{\text{H}}\left(\text{j}\omega\right)\right\}=\Im\left\{\underline{\text{H}}\left(\text{j}\omega\right)\right\}\tag6$$

And I want the solution to $(6)$ to be the same as the solution to the poles and zeros of the s-plane transfer function.

Question: How can I prove for which transfer functions this hold?! That setting the real and imaginary part of the complex transfer function equal give me the same result for $\omega$ as when I solve the poles and zeros of the s-plane transfer function (then using $\left|\text{s}\right|=\omega$).


EXAMPLE WHERE IT DOES NOT WORK:

When we have this transfer function:

$$\underline{\text{H}}\left(\text{j}\omega\right)=\frac{\text{R}_2+\text{j}\omega\text{L}}{\text{R}_1+\text{R}_2+\text{j}\omega\text{L}}\space\space\space\to\space\space\space\text{H}\left(\text{s}\right)=\frac{\text{R}_2+\text{s}\text{L}}{\text{R}_1+\text{R}_2+\text{s}\text{L}}\tag7$$

Now when I solve $\omega$ out of:

$$\Re\left\{\frac{\text{R}_2+\text{j}\omega\text{L}}{\text{R}_1+\text{R}_2+\text{j}\omega\text{L}}\right\}=\Im\left\{\frac{\text{R}_2+\text{j}\omega\text{L}}{\text{R}_1+\text{R}_2+\text{j}\omega\text{L}}\right\}\space\Longleftrightarrow\space\tag8$$ $$\omega=\frac{\text{R}_1\pm\sqrt{\text{R}_1^2-4\cdot\text{R}_2\cdot\left(\text{R}_1-\text{R}_2\right)}}{2\cdot\text{L}}\tag9$$

That gives me different values for $\omega$ when I do it for the poles and zeros in the s-plane:

  1. $$\text{R}_2+\text{s}\text{L}=0\space\Longleftrightarrow\space\left|\text{s}\right|=\omega=\left|-\frac{\text{R}_2}{\text{L}}\right|=\frac{\text{R}_2}{\text{L}}\tag{10}$$
  2. $$\text{R}_1+\text{R}_2+\text{s}\text{L}=0\space\Longleftrightarrow\space\left|\text{s}\right|=\omega=\left|-\frac{\text{R}_1+\text{R}_2}{\text{L}}\right|=\frac{\text{R}_1+\text{R}_2}{\text{L}}\tag{11}$$

And by, for example:

$$\underline{\text{H}}\left(\text{j}\omega\right)=\frac{\text{j}\omega\text{L}}{\text{R}+\text{j}\omega\text{L}}\tag{12}$$

There it will work.

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    $\begingroup$ Find the frequencies where $\angle H(j\omega)=45^{\circ}$ $\endgroup$ – polfosol Dec 28 '16 at 11:38
  • $\begingroup$ What do you mean by "the solution to the poles and zeros of the s-plane"? poles and zeros are just complex values. "solution" does not make much sense. Also, the relation between the question and the explanation about resistor/capacitor/inductor cannot be understood. Off-topic but $\Omega$ is only used for R. Others have different units. $\endgroup$ – msm Dec 28 '16 at 11:41
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    $\begingroup$ @polfosol That's not the way because there is another point that gives me Real=Imaginary at $225^\circ$ $\endgroup$ – Jan Dec 28 '16 at 12:21
  • $\begingroup$ Since j is dimensionless the others do indeed have units of Ohms otherwise the total impedance formula wouldn't work $\endgroup$ – Triatticus Dec 28 '16 at 12:23
  • $\begingroup$ @msm English is not my first language (it is Dutch xD). But I mean that the poles and zeros must give the same omega as the solution to real equals imaginary. And officially you're right because I've to use the absolute value of the impedance to use $\Omega$ but I choose it this way to make it clear that $\text{s}=\text{j}\omega$ because it does not contain a real part $\endgroup$ – Jan Dec 28 '16 at 12:24
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Note that every transfer function can be written in the form of: $$H(j\omega)=\frac{X(j\omega)}{Y(j\omega)}=\frac{A+j\omega B}{C+j\omega D}$$ where $A,B,C,D$ are real functions of $\omega$. If the system has a zero at $\omega_0$ then: $$X(j\omega_0)=0\implies H(j\omega_0)=0\implies\Re\{H\}|_{\omega_0}=\Im\{H\}|_{\omega_0}=0$$ So the desired property always holds for the system's zeros.

By multiplying both numerator and denominator of $H$ into $C-j\omega D$, we can safely say that if $\Re\{H\}=\Im\{H\}$ then: $$AC+\omega^2 BD=\omega(BC-AD)$$ The system's poles are the solution of $C+j\omega D=0$. So we have to investigate these two equations simultaneously: $$\begin{cases} C(\omega)+j\omega D(\omega)=0\\ AC+\omega^2 BD=\omega(BC-AD) \end{cases} \label{*}\tag{*}$$ From the first equation, it is evident that either $C(0)=0$ or $|\omega|=\left|\frac CD\right|$ which implies $\omega^2=\frac{C^2}{D^2}$. Putting this into the second equation results in: $$AC+\frac{BC^2}{D}=\pm\left(AC-\frac{BC^2}{D}\right)$$ and note that only one of $+$ or $−$ can be the correct sign based on the location of the pole.

In conclusion, if the system has a pole in $\omega=0$ then $C(0)=0$ implies $A(0)C(0)=0$ and the $\eqref{*}$ conditions hold.

If the system has a pole at $\omega_1\ne 0$ and $C(\omega_1)\ne 0$ then only one of these two equations must hold (based on the location of $\omega_1$): $$\begin{align} A(\omega_1)C(\omega_1)&=0\\ B(\omega_1)C(\omega_1)&=0 \end{align}$$ which means either $A(\omega_1)=0$ or $B(\omega_1)=0$. In your second example, we have $A\equiv 0$.

Also note that since $\omega_1$ is a pole, if $C(\omega_1)=0$ then we must have $D(\omega_1)=0$ and the conditions of $\eqref{*}$ are satisfied.

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  • $\begingroup$ First of all thanks for your answer. But this is not what I mean in my question. Can you look in my edited question, I've edit a counterexample in it, where it does not work. $\endgroup$ – Jan Dec 28 '16 at 14:56
  • $\begingroup$ So, $\text{A}$ is a function of $\omega$ so we can write $\text{A}\left(\omega\right)$ the same for $\text{B}$,$\text{C}$ and $\text{D}$. Then why does it not hold for my first example? $\endgroup$ – Jan Dec 28 '16 at 17:46
  • $\begingroup$ @JanEerland Because in your first example, $A(\omega)=R_2$ and $C(\omega)=R_1+R_2$ $\endgroup$ – polfosol Dec 28 '16 at 17:48
  • $\begingroup$ Maybe I'm blind, but I don't see how this answers my question. Because when can I say, you can use both ways and when can I say you can't use one of those two? If I understand it correct then one of these equations has to hold, namely: $\text{A}\left(\omega\right)\cdot\text{C}\left(\omega\right)=0$ or $\text{B}\left(\omega\right)\cdot\text{C}\left(\omega\right)=0$? $\endgroup$ – Jan Dec 28 '16 at 17:58
  • $\begingroup$ @Downvoter please complete your feedback by stating what's wrong $\endgroup$ – polfosol Dec 28 '16 at 18:00

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