How to Compute Magnitude and Phase Functions of Transfer Functions 1 – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this control engineering and control theory tutorial, we explain how to correctly compute the magnitude and phase functions of transfer function. The YouTube tutorial is given below.

Problem of Computing Phase and Magnitude of Transfer Function

In this tutorial, we solve the following problem. Compute magnitude and phase functions of a transfer function

(1) $\begin{align*}W(s)=\frac{s+5}{s+10}\end{align*}$

Solution

The first step is to derive the sinusoidal transfer function. The sinusoidal transfer function is obtained by substituting $s$ by $j\omega$ in (1). Here, $j$ is the imaginary unit and $\omega$ is the radial frequency. From (1), we have

(2) $\begin{align*}W(j\omega)=\frac{j\omega+5}{j\omega+10}\end{align*}$

The expression (2) is a complex number since it is a ratio of two complex numbers. The next step is to write this complex number in the polar form. We do it by expressing the complex numbers in the numerator and denominator in the polar form:

(3) $\begin{align*}W(j\omega)=\frac{\sqrt{5^{2}+\omega^{2}}\cdot e^{j\arctan (\frac{\omega}{5})} }{\sqrt{10^{2}+\omega^{2}} \cdot e^{j\arctan (\frac{\omega}{10})} }\end{align*}$

From the last expression, we have

(4) $\begin{align*}W(j\omega)=\sqrt{\frac{25+\omega^{2}}{100+\omega^{2}}}\cdot e^{j(\arctan(\frac{\omega}{5})-\arctan(\frac{\omega}{10}) )}\end{align*}$

The magnitude function, denoted by $M(\omega)$ , is the modulus of the sinusoidal transfer function:

(5) $\begin{align*}M(j\omega)=\sqrt{\frac{25+\omega^{2}}{100+\omega^{2}}}\end{align*}$

The phase function, denoted by $\phi(\omega)$ , is the angle of the sinusoidal transfer function:

(6) $\begin{align*}\phi (\omega)= \arctan(\frac{\omega}{5})-\arctan(\frac{\omega}{10}) \end{align*}$

Problem of Computing Phase and Magnitude of Transfer Function

Solution

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