Table of Contents

1 Frequency Response

1.1 Introduction

Review of time domain analysis

  • System modeling
  • Time-domain responses systems of control
  • Routh-Hurwitz Criterion and Root locus for stability analysis
  • Time-domain specifications:
    • rise time
    • settling time
    • maximum overshoot
    • steady-state error

1.2 What is Frequency Response: RC-Network

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Consider the above RC-network, whose transfer function is

\begin{align*} \frac{V_0(s)}{V_i(s)}= \frac{1/sC}{R+1/sC} = \frac{1}{\tau s+1},\,\tau=RC. \end{align*}

Assuming \(R = 1k\Omega\), \(C = 10^{-4}F\), then \(\tau = 0.1s\).

1.3 RC-Network: Step Response

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1.4 RC-Network: Response to Sinusoidal Input with \(\omega = 5\)

Assume the input is a sinusoidal wave:

\begin{align*} v_i(t) = 2\sin(5 t+ 30^\circ), \end{align*}

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1.5 RC-Network: Response to Sinusoidal Input with \(\omega = 50\)

Assume the input is a sinusoidal wave:

\begin{align*} v_i(t) = 2\sin(50t+ 30^\circ), \end{align*}

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1.6 Frequency Response

The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input.

1.7 Why Do We Need Frequency Response

  • The frequency response of a complex system can be derived from the response of simple systems (Bode Plot).
  • The stability of the closed-loop system is determined by the frequency response of the open-loop system (Nyquist Stability Criterion)
  • Controller design can be carried out in an intuitive fashion by shaping the frequency response of the open-loop systems (Lag, Lead, PID controller Design)

1.8 How to Get Frequency Response

  • The frequency response can be derived from transfer functions
  • The frequency response of complex systems can be obtained experimentally
  • In fact, for an unknown system, usually we get its transfer function from the frequency response.

1.9 How to get Frequency Response from Transfer Function

Consider a stable system:

\begin{align*} G(s) = \frac{a(s)}{b(s)}=\frac{a(s)}{(s+p_1)\dots(s+p_n)}, \end{align*}

where \(p_i\) are assumed to be distinct poles.

Consider the input

\begin{align*} r(t) = A \exp\left[j\omega t\right] = A\cos(\omega t + \phi)+jA\sin(\omega t+\phi). \end{align*}

whose Laplace transform is

\begin{align*} R(s) = A\exp(j\phi)\frac{1}{s-j\omega}. \end{align*}

The output is given by

\begin{align*} Y(s) &= G(s)R(s) \\ &= \frac{a(s)}{(s+p_1)\dots(s+p_n)}\times \frac{A\exp(j\phi)}{s-j\omega} \\ &=\frac{k_1}{s+p_1}+\frac{k_2}{s+p_2}+\dots+\frac{k_n}{s+p_n} + \frac{\alpha}{s-j\omega}. \end{align*}

If we multiply the LHS and RHS of the equation by \(s-j\omega\) and take the limit \(s\rightarrow j\omega\), we get

\begin{align*} \lim_{s\rightarrow -j\omega}G(s)R(s) = A\exp(j\phi)G(j\omega) = A|G(j\omega)|\times e^{j(\phi + \angle G(j\omega))} = \alpha. \end{align*}

Therefore, the response of the system can be written as

\begin{align*} y(t) &= k_1e^{-p_1t}+\dots+k_ne^{-p_nt} \\ &+ A|G(j\omega)|\cos(\omega t + \phi + \angle G(j\omega))\\ &+j A|G(j\omega)|\sin(\omega t + \phi + \angle G(j\omega)). \end{align*}

If we input \(r(t) = A\sin(\omega t + \phi)\), then the steady-state output will be

\begin{align*} y_{ss}(t) = A|G(j\omega)|\sin(\omega t + \phi + \angle G(j\omega)). \end{align*}
  • The steady-state response of a sinusoidal signal is another sinusoidal signal, where
    1. The frequency is the same.
    2. The amplitude is \(A\times |G(j\omega)|\).
    3. The phase is \(\phi + \angle G(j\omega)\).
  • \(|G (j\omega)|\) and \(\angle G(j\omega)\) are frequency dependent and are referred to as the magnitude and phase responses of the system, respectively.

1.10 Example: RC-Network

Consider the RC-network with transfer function

\begin{align*} G(s) = \frac{1}{0.1s+1}. \end{align*}

Find the steady state output due to the input \(r(t) = 2\sin(\omega t+ 30^\circ)\).

1.11 RC-Network: Solution

  • Notice that\[0.1j\omega + 1 = \sqrt{1+0.01\omega^2}\angle \tan^{-1}(0.1\omega).\]
  • Therefore,\[G(j\omega) = \frac{1}{\sqrt{1+0.01\omega^2}}\angle -\tan^{-1}(0.1\omega).\]
  • Hence, the response to \(r(t)\) is\[y_{ss}(t) = \frac{2}{\sqrt{1+0.01\omega^2}}\sin\left[\omega t+30^\circ -\tan^{-1}(0.1\omega)\right.\]
  • For \(\omega = 5\), \(|G(j5)| = 0.8944\), \(\angle G(j5) = -26.6^\circ\).\[y_{ss}(t) = 1.7888\sin(5t+3.4^\circ).\]
  • For \(\omega = 50\), \(|G(j5)| = 0.1961\), \(\angle G(j5) = -78.7^\circ\).\[y_{ss}(t) = 0.3922\sin(50t-48.7^\circ).\]

1.12 Bode and Nyquist Plot

  • From the analysis above, \(G(j\omega)\) plays a key role in frequency response.
  • To visualize \(G(j\omega)\), the following plots are typically used
    1. Bode Plots: Bode Magnitude Plot + Bode Phase Plot
    2. Nyquist Plots

1.13 Bode Plot

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1.14 Nyquist Plot

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1.15 Real World Example: Voice Coil Motor

Figure-3-HDD-with-a-VCM-actuator.png

Figure-4-Frequency-responses-of-the-VCM-actuator-LDV-range-2-m-m-V.png

1.16 Hendrik Wade Bode(1905 - 1982)

Hendrik_Wade_Bode.png

  • A pioneer of modern control theory and electronic telecommunications.
  • He made important contributions to the design, guidance and control of anti-aircraft systems during World War II
  • During the Cold War, he also made significant contributions to the design and control of missiles and anti-ballistic missiles.
  • Contributions to control system theory and mathematical tools for the analysis of stability of linear systems, inventing Bode plots, gain margin and phase margin.
  • Worked in Bell Lab from 1926 to 1967. Became a professor at Harvard after retiring from Bell Lab.

1.17 Harry Nyquist (1889 - 1976)

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  • Worked in Bell Lab from 1917 to 1954.
  • Received the IEEE Medal of Honor in 1960 for "fundamental contributions to a quantitative understanding of thermal noise, data transmission and negative feedback."
  • Received the National Academy of Engineering's fourth Founder's Medal "in recognition of his many fundamental contributions to engineering."

1.18 Summary

The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input.

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It is also a sinusoidal signal, where

  1. The frequency is the same.
  2. The amplitude is \(A\times |G(j\omega)|\).
  3. The phase is \(\phi + \angle G(j\omega)\).