proving a property of transfer function in the s-plane and the complex domain

Question

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.

Answer 1

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.