Chapter 8: Frequency Response Methods

Core Concepts & Definitions
Deriving the Steady-State Output
Frequency Response — Summary Formula
Worked Example: First-Order System
Frequency Response Plot Types
Bode Diagrams — Structure
Basic Factors of a Transfer Function
Factor 1: Gain K
Factor 2 & 3: Integral and Derivative Factors
Factor 4: First-Order Factors
Factor 5: Quadratic Factors
General Procedure for Bode Diagrams
Worked Example: T(s) = (s+1)/[(s+5)(s+2)]
Using Bode Diagrams to Find the Output
Worked Example: G(s) with Integral + Quadratic Factors
Resonant Frequency and Resonant Peak
Worked Example: Mechanical System from Bode Plot
Low-Pass Filter
Cutoff Frequency and Bandwidth
Bandwidth vs. Damping Ratio Trade-off
Desired Outcomes — Study Question Worked Solution
Key Facts to Remember

1. Core Concepts & Definitions

Frequency Response — the steady-state response of a stable, linear, time-invariant (LTI) system to a sinusoidal input.

Property	Statement
Input	$r(t) = A\sin(\omega t)$
Output (steady-state)	$y_{ss}(t) = Y\sin(\omega t + \phi)$
Output amplitude	$Y = A
Phase shift	$\phi = \angle T(j\omega)$
Frequency response function	Evaluate transfer function at $s = j\omega$

Key insight: The system changes only the amplitude and phase of the input sinusoid. Frequency $\omega$ is unchanged.

2. Deriving the Steady-State Output

$T(s) = \frac{q(s)}{(s+s_1)(s+s_2)\cdots(s+s_n)}, \quad R(s) = \frac{A\omega}{s^2+\omega^2}$

Partial fractions: transient terms decay, leaving:

$y_{ss}(t) = ae^{-j\omega t} + \bar{a}e^{j\omega t} = A|T(j\omega)|\sin(\omega t+\phi)$

where $a = -\frac{AT(-j\omega)}{2j}$ , $\bar{a} = \frac{AT(j\omega)}{2j}$ .

3. Frequency Response — Summary Formula

$y_{ss}(t) = A|T(j\omega)|\sin(\omega t + \phi)$

$|T(s=j\omega)| = \frac{|Y(j\omega)|}{|R(j\omega)|} \rightarrow \text{Amplitude ratio}$

$\phi = \angle T(s=j\omega) = \tan^{-1}\left(\frac{\text{Im}\{T(j\omega)\}}{\text{Re}\{T(j\omega)\}}\right) \rightarrow \text{Phase shift}$

4. Worked Example: First-Order System

System: $G(s) = \frac{K}{\tau s+1}$ , input $r(t)=A\sin(\omega t)$

Magnitude: $|T(j\omega)| = \frac{K}{\sqrt{1+\tau^2\omega^2}}$
Phase: $\phi = -\tan^{-1}(\tau\omega)$ (note the minus sign!)

$y_{ss}(t) = A\frac{K}{\sqrt{1+\tau^2\omega^2}}\sin(\omega t - \tan^{-1}(\tau\omega))$

5. Frequency Response Plot Types

Plot Type	x-axis	y-axis
Bode diagram	$\log(\omega)$	Magnitude (dB) and Phase (deg)
Nyquist (polar) plot	$\text{Re}\{T(j\omega)\}$	$\text{Im}\{T(j\omega)\}$
Nichols plot	Phase (deg)	Magnitude (dB)

6. Bode Diagrams — Structure

Magnitude plot: $20\log|T(j\omega)|$ (dB) vs. $\log(\omega)$
Phase plot: $\phi$ (degrees) vs. $\log(\omega)$

One decade = one unit on $\log(\omega)$ axis. Key advantage: multiplication of factors becomes addition in dB.

7. Basic Factors of a Transfer Function

Factor	Standard Form	Corner Freq
Gain	$K$	—
Integral	$1/s^n$	—
Derivative	$s^n$	—
1st-order (denom)	$1/(1+j\omega\tau)^n$	$1/\tau$
1st-order (num)	$(1+j\omega\tau)^n$	$1/\tau$
Quadratic (denom)	$\left(1+2\zeta\frac{j\omega}{\omega_n}+(\frac{j\omega}{\omega_n})^2\right)^{-n}$	$\omega_n$

Standard form rule: use $\frac{1}{\tau s+1}$ , not $\frac{1}{s+1/\tau}$ .

Example: T(s) = s(s+1) / [(s+7)(s^2+5s+4)]

$T(s) = \frac{1}{28}\cdot\frac{s(1+s)}{\left(1+\frac{s}{7}\right)\left(1+2\cdot\frac{5}{4}\cdot\frac{s}{2}+(\frac{s}{2})^2\right)}$

Factor	Type	Parameters
$K=1/28$	Gain	Mag = $20\log(1/28)$
$s$	Derivative	slope +20 dB/dec
$(1+s)$	1st-order num	$\tau_1=1$
$(1+s/7)^{-1}$	1st-order denom	$\tau_2=1/7$
Quadratic $^{-1}$	2nd-order denom	$\zeta=5/4$ , $\omega_n=2$

8. Factor 1: Gain K

Property	Value
Magnitude	$20\log K$ dB (horizontal line)
Phase	$0°$

9. Factor 2 & 3: Integral and Derivative Factors

Integral $1/s^n$

Property	Value
Magnitude	$-20n\log\omega$ (slope $-20n$ dB/dec)
Phase	$-n \times 90°$ (constant)

Table for $n=1$ :

$\omega$	Mag (dB)
0.1	+20
1	0
10	−20
100	−40

Derivative $s^n$

Property	Value
Magnitude	$+20n\log\omega$ (slope $+20n$ dB/dec)
Phase	$+n \times 90°$ (constant)

Physical example: $V(s)=sX(s)$ . If $x(t)=\sin(3t)$ then $v(t)=3\cos(3t)$ : amplitude $\times 3$ , phase $+90°$ .

10. Factor 4: First-Order Factors

Denominator $(1+j\omega\tau)^{-1}$

Magnitude: $\text{Mag} = -20\log\sqrt{1+\omega^2\tau^2}$

Region	Asymptote
$\omega \ll 1/\tau$	0 dB (flat)
$\omega \gg 1/\tau$	$-20\log(\omega\tau)$ , slope $-20$ dB/dec
$\omega = 1/\tau$	Exact = $-3.01$ dB (max error)

Phase: $\phi = -\tan^{-1}(\omega\tau)$

$\omega$	$\phi$
$\omega \ll 1/\tau$	$\approx 0°$
$\omega = 1/\tau$	$-45°$ (inflection point)
$\omega \gg 1/\tau$	$\approx -90°$

Numerator $(1+j\omega\tau)^{+1}$ : mirror image, phase $0°$ to $+90°$ , slope $+20$ dB/dec.

Exponent $n$ : multiply slope and phase by $n$ .

11. Factor 5: Quadratic Factors

$T(j\omega) = \frac{1}{1+2\zeta\frac{j\omega}{\omega_n}+(\frac{j\omega}{\omega_n})^2}$

$\text{Mag} = -20\log\sqrt{\left(1-\frac{\omega^2}{\omega_n^2}\right)^2+\left(2\zeta\frac{\omega}{\omega_n}\right)^2}$

Frequency	Asymptote
$\omega \ll \omega_n$	0 dB (flat)
$\omega = \omega_n$	exact = $-20\log(2\zeta)$
$\omega \gg \omega_n$	slope $-40$ dB/dec

Phase: $\phi = -\tan^{-1}\left(\frac{2\zeta(\omega/\omega_n)}{1-(\omega/\omega_n)^2}\right)$

$\omega$	$\phi$
$\ll \omega_n$	$0°$
$= \omega_n$	$-90°$ (independent of $\zeta$ )
$\to \infty$	$-180°$

12. General Procedure for Bode Diagrams

Rewrite $T(s)$ in standard form (leading “1” in each factor).
Identify all factors and their corner frequencies.
Draw each factor’s Bode plot individually.
Sum magnitudes (dB) and phases (degrees).
Track overall phase with a table.

Phase Tracking Table Template

Factor	Phase Begin	Phase End
Gain $K$	$0°$	$0°$
$(j\omega)^{-1}$	$-90°$	$-90°$
$(1+j\omega\tau)^{+1}$	$0°$	$+90°$
$(1+j\omega\tau)^{-1}$	$0°$	$-90°$
Quadratic $^{-1}$	$0°$	$-180°$
Overall	Sum	Sum

13. Worked Example: T(s) = (s+1)/[(s+5)(s+2)]

$T(s) = \frac{1}{10}\cdot\frac{(1+j\omega)}{(1+j\omega/5)(1+j\omega/2)}$

#	Factor	Corner Freq
1	Gain $K=1/10$ , Mag $= -20$ dB	—
2	$(1+j\omega)$ numerator	$1/\tau=1$
3	$(1+j\omega/5)^{-1}$	$1/\tau=5$
4	$(1+j\omega/2)^{-1}$	$1/\tau=2$

Magnitude behavior:

$\omega<1$ : flat at $-20$ dB
$1<\omega<2$ : $+20$ dB/dec
$2<\omega<5$ : flat
$\omega>5$ : $-20$ dB/dec

num = [1 1];
den = [1 7 10];
sys = tf(num, den);
bode(sys)

Phase table: Overall $0° \to -90°$ (one $+90°$ and two $-90°$ terms cancel to net $-90°$ )

14. Using Bode Diagrams to Find the Output

Given: $x(t) = 10sin(7t)$ , $T(s) = \frac{s+1}{(s+5)(s+2)}$

Interpolate asymptotic magnitude at $omega=7$ :

$p_1 \approx -14 \text{ dB at }\omega=2, \quad p_2 \approx -17 \text{ dB at }\omega=7$

$|T(7j)| \approx 0.14, \quad Y = 1.4, \quad \phi \approx -0.8 \text{ rad}$

$\boxed{y_{ss}(t) \approx 1.4\sin(7t - 0.8)}$

Exact (mathematical):

$(7j+5)(7j+2) = -39+49j$

$|T(7j)| = \frac{\sqrt{50}}{\sqrt{74}\cdot\sqrt{53}} \approx 0.113, \quad \phi = -46.64° = -0.814 \text{ rad}$

$\boxed{y_{ss}(t) = 1.13\sin(7t - 0.814)}$

15. Worked Example: G(s) with Integral + Quadratic Factors

$G(s) = \frac{10(s+3)}{s(s+2)(s^2+s+2)} = \frac{7.5(1+j\omega/3)}{(j\omega)(1+j\omega/2)(1+2\cdot\frac{\sqrt{2}}{4}\cdot\frac{j\omega}{\sqrt{2}}+(\frac{j\omega}{\sqrt{2}})^2)}$

#	Factor	Type	Parameters
1	$K=7.5$	Gain	17.5 dB
2	$(j\omega)^{-1}$	Integral	$\phi=-90°$
3	$(1+j\omega/3)$	1st-order num	$1/\tau=3$
4	$(1+j\omega/2)^{-1}$	1st-order denom	$1/\tau=2$
5	Quadratic $^{-1}$	2nd-order denom	$\omega_n=\sqrt{2}$ , $\zeta=\sqrt{2}/4$

Overall phase: $-90° \to -270°$

16. Resonant Frequency and Resonant Peak

Resonance: maximum amplitude when input frequency $\approx$ natural frequency.

$\omega_r = \omega_n\sqrt{1-2\zeta^2}$

$M_r = |T(j\omega_r)| = \frac{1}{2\zeta\sqrt{1-\zeta^2}}$

Condition	Result
$\zeta \to 0$	$\omega_r \to \omega_n$ , $M_r \to \infty$
$\zeta = 0.707$	$\omega_r = 0$
$\zeta > 0.707$	No resonant peak
$\zeta < 0.707$	Resonance occurs

17. Worked Example: Mechanical System from Bode Plot

Given: spring-mass-damper, $m=15$ kg. Bode: peak at $\omega_r=9$ , $\phi=-90°$ at $\omega=10$ .

a) $\omega_n = 10$ rad/s (from $\phi=-90°$ ). Using $\omega_r = \omega_n\sqrt{1-2\zeta^2}$ : $\zeta = 0.3$

b) $k = m\omega_n^2 = 15\times100 = 1500$ N/m; $b = 2\zeta\omega_n m = 90$ N·s/m

c) Output responses ( $x(t) = \sin(\omega t)$ ):

$\omega$ (rad/s)	Mag	$\phi$	Output $y(t)$
1	0 dB	$\approx 0°$	$\sin(t)$ — perfect tracking
10	$\approx+4$ dB	$-90°$	$1.58\sin(10t-\pi/2)$
20	$\approx-10$ dB	$\approx-155°$	$0.32\sin(20t-2.7)$

18. Low-Pass Filter

$T(j\omega) = \frac{1}{1+j\omega\tau}$ — a first-order system acts as a low-pass filter.

| Frequency | $|T(j\omega)|$ | Behavior | | --- | --- | --- | | $\omega \ll 1/\tau$ | $\approx 1$ | Perfect duplication | | $\omega \gg 1/\tau$ | $\to 0$ | Signal eliminated |

Application: removes high-frequency noise from sensor signals.

19. Cutoff Frequency and Bandwidth

Cutoff frequency $\omega_b$ : where magnitude drops 3 dB from DC.

$|T(j\omega_b)| = \frac{1}{\sqrt{2}} \approx 0.707 \times |T(j0)|$

For first-order systems: $\omega_b = 1/\tau$

Larger bandwidth → higher-frequency signals tracked → faster response.

20. Bandwidth vs. Damping Ratio Trade-off

Fixed $\zeta=0.5$ , comparing $\omega_n=10$ vs $\omega_n=30$ :

Metric	$\omega_n=10$	$\omega_n=30$
Max overshoot	Same	Same
Rise time $T_r$	Larger	Smaller
Bandwidth	Smaller	Larger

Three rules for first-order: Smaller $\tau$ → faster response → larger bandwidth.

Damping trade-off:

$\zeta$	Bandwidth	Overshoot
Low ( $\zeta < 0.7$ )	Large	High
High ( $\zeta > 0.7$ )	Small	Low

21. Desired Outcomes — Study Question Worked Solution

Problem A: Nano-Probe AFM System

System: $k_1=15$ , $k_2=10$ , $b=6$ , $m=1$ . Input $u(t)$ , Output $x(t)$ .

EoM (superposition): $m\ddot{x} = k_2 u - (k_1+k_2)x - b\dot{x}$

$\boxed{T(s) = \frac{X(s)}{U(s)} = \frac{10}{s^2+6s+25}}$

Standard form: $T(s) = \frac{0.4}{(s/5)^2+2(0.6)(s/5)+1}$ , so $\omega_n=5$ , $\zeta=0.6$ , Gain $= -7.96$ dB.

num = 10;
den = [1, 6, 25];
sys = tf(num, den);
bode(sys)

Approximate output for $u(t)=3\sin(7t)$ :

From asymptote at $\omega=7$ (slope $-40$ dB/dec from $-7.96$ dB at $\omega_n=5$ ):

$x = -7.96 + (-40)\cdot\log(7/5) \approx -13.8 \text{ dB} \implies |x| = 3\times10^{-13.8/20} \approx 0.613$

$\phi \approx -120° \approx -2.09 \text{ rad}$

$\boxed{x_{ss}(t) \approx 0.613\sin(7t-2.09)}$

Exact output for $u(t)=3\sin(7t)$ :

$T(7j) = \frac{10}{-24+42j}, \quad |T(7j)| = \frac{10}{\sqrt{576+1764}} \approx 0.2067$

$\phi = -(180°-60.3°) = -119.7° \approx -2.09 \text{ rad}$

$\boxed{x_{ss}(t) = 0.62\sin(7t-2.09)}$

Problem B: Extracting K and τ from a Bode Plot

Given: $T(s) = \frac{K}{\tau s+1}$ .

Finding $\tau$ : Phase $= -45°$ at $\omega = 1/\tau$ . From plot: $1/\tau = 0.05$ rad/s → $\tau = 20$ s.

Finding $K$ : At low frequencies $|T| \approx K$ . From plot: $20\log K = 40$ dB → $K = 100$ .

$\boxed{K = 100, \quad \tau = 20 \text{ s}}$

22. Key Facts to Remember

Quick Reference Table

Factor	Mag slope	Phase range
Gain $K$	0 dB/dec	$0°$
$(j\omega)^{-1}$	$-20$ dB/dec	$-90°$ (constant)
$(j\omega)^{+1}$	$+20$ dB/dec	$+90°$ (constant)
$(1+j\omega\tau)^{-1}$	0 → $-20$ dB/dec	$0° \to -90°$
$(1+j\omega\tau)^{+1}$	0 → $+20$ dB/dec	$0° \to +90°$
Quadratic $^{-1}$	0 → $-40$ dB/dec	$0° \to -180°$

Corner Frequency Facts

At $\omega=1/\tau$ : exact magnitude $= -3.01$ dB; phase $= -45°$ .
At $\omega=\omega_n$ : phase $= -90°$ always (independent of $\zeta$ ).

Resonance

$\omega_r = \omega_n\sqrt{1-2\zeta^2}$ (slightly below $\omega_n$ ).
$M_{p\omega} = \frac{1}{2\zeta\sqrt{1-\zeta^2}}$ (resonant peak). Only exists for $\zeta < 0.707$ .

Reading Parameters from Bode Plot

$\tau$ : read $\omega$ where $\phi = -45°$ → $\tau = 1/\omega$ .
$\omega_n$ : read $\omega$ where $\phi = -90°$ (2nd-order).
$\zeta$ : use $\omega_r = \omega_n\sqrt{1-2\zeta^2}$ .
$K$ : low-frequency magnitude → $K = 10^{\text{Mag(dB)}/20}$ .

Bandwidth & Speed

Larger $\omega_b$ → faster response (smaller $T_r$ , $T_p$ , $T_s$ ).
Larger $\omega_n$ → larger bandwidth (same $\zeta$ ).
Smaller $\tau$ → larger bandwidth (first-order).
Underdamped ( $\zeta<0.707$ ): large bandwidth but high overshoot.

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