---
date: '2024-01-09'
description: frequency domain analysis using laplace transforms, transfer functions, stability analysis via poles, partial fraction expansion, and impedance calculations.
id: Frequency Domain
modified: 2026-06-05 15:08:43 GMT-04:00
tags:
  - sfwr3dx4
title: Frequency Domain and a la carte.
created: '2024-01-09'
published: '2024-01-09'
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---
See also [[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/intro.pdf|introduction slides]] and [[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/frequency_domain.pdf|frequency domain slides]]

Open-loop versus closed-loop

Transient and steady-state response

Stability

- Total response = Natural response + Forced response
  - Natural response (homogeneous solution): evolution of system due to initial conditions
  - Forced response (particular solution): evolution of system due to input

Control objects:

- Stabilize the system
- Produce the desired transient response
- Decrease/eliminate steady-state error
- Make system “robust” to withstand disturbances and variations in parameters
- Achieve optimal performance

## [[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/Block Diagrams|Block diagrams]] representation of a system

```mermaid
stateDiagram-v2
  direction LR

  [*] --> System: r(t)
  System --> End: c(t)
```

_System as linear differential equation_

![[thoughts/Laplace transform]]

### Properties

$$
\begin{aligned}
& f(0-)\text{: initial condition just before 0} \\[12pt]
& \textbf{Linearity:} \quad \mathcal{L}\{k_1 f_1(t) \pm k_2 f_2(t)\} = k_1 F_1(s) \pm k_2 F_2(s) \\[12pt]
& \textbf{Differentiation:} \\
& \quad \mathcal{L}\left\{\frac{df(t)}{dt}\right\} = sF(s) - f(0^-) \\
& \quad \mathcal{L}\left\{\frac{d^2f(t)}{dt^2}\right\} = s^2 F(s) - sf(0^-) - f'(0^-) \\[12pt]
& \textbf{Frequency Shifting:} \quad \mathcal{L}\{e^{-at}f(t)\} = F(s + a) \\
\end{aligned}
$$

### Transfer function

$n^{th}$ order _linear, time-invariant_ (LTI) differential equation:

$$
a_n \frac{d^n c(t)}{dt^n} + a_{n-1} \frac{d^{n-1} c(t)}{dt^{n-1}} + \cdots + a_0 c(t) = b_m \frac{d^m r(t)}{dt^m} + b_{m-1} \frac{d^{m-1} r(t)}{dt^{m-1}} + \cdots + b_0 r(t)
$$

_takes Laplace transform from both side_

$$
\begin{aligned}
& a_n s^n C(s) + a_{n-1} s^{n-1} C(s) + \cdots + a_0 C(s) \text{ and init terms for } c(t) \\
& = b_m s^m R(s) + b_{m-1} s^{m-1} R(s) + \cdots + b_0 R(s) \text{ and init terms for } r(t) \\
\end{aligned}
$$

_assume initial conditions are zero_

$$
\begin{aligned}
(a_n s^n + a_{n-1} s^{n-1} + \cdots + a_0)C(s) &= (b_m s^m + b_{m-1} s^{m-1} + \cdots + b_0)R(s) \\[8pt]
\frac{C(s)}{R(s)} &= G(s) = \frac{b_m s^m + b_{m-1} s^{m-1} + \cdots + b_0}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_0}
\end{aligned}
$$

> \[!tip\] Transfer function
>
> $$
> G(s)=\frac{C(s)}{R(s)}
> $$

Q: $G(s) = \frac{1}{S+2}$. Input: $u(t)$. What is $y(t)$ ?

$$
\begin{aligned}
Y(s) &= G(s)\cdot u(s) \rightarrow Y(s)=\frac{1}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2} = \frac{1}{2\cdot{s}} - \frac{1}{2\cdot{(s+2)}} \\
y(t) &= -\frac{1}{2}(1-e^{-2t})u(t)
\end{aligned}
$$

## Partial fraction expansion

$$
\begin{aligned}
F(s) &= \frac{N(s)}{D(s)} \\[8pt]
N(s) &: m^{th} \text{ order polynomial in } s \\
D(s) &: n^{th} \text{ order polynomial in } s \\
\end{aligned}
$$

### Decomposition of $\frac{N(s)}{D(s)}$

1. **Divide if improper**: $\frac{N(s)}{D(s)}$ such that $\text{degree of }N(s) \leq \text{degree of } D(s)$ such that $\frac{N(s)}{D(s)} = \text{a polynomial } + \frac{N_1(s)}{D(s)}$
2. **Factor Denominator**: into factor form
   $$
   (ps+q)^m \text{ and } (as^2+bs+c)^n
   $$
3. **Linear Factors**: $(ps+q)^m$ such that:
   $$
   \sum_{j=1}^{m}\frac{A_j}{(ps+q)^j}
   $$
4. **Quadratic Factors**: $(as^2+bs+c)^n$ such that
   $$
   \sum_{j=1}^{n}{\frac{B_j s+C_j}{(as^2+bs+c)^j}}
   $$
5. **Determine Unknown**

## Stability analysis using Root of $D(s)$

> \[!tip\] roots of `D(s)`
>
> roots of $D(s)$ as **poles**

$$
G(s) = \frac{N(s)}{D(s)} = \frac{N(s)}{\prod_{j=1}^{n}(s+p_j)} = \sum_{j=1}^{n}{\frac{A_j}{s+p_j}}
$$

> $p_i$ can be imaginary

Solving for $g(t)$ gives

$$
g(t) = \sum_{j=1}^{n}{\mathcal{L}^{-1}\{\frac{A_j}{(s+p_j)}\}} = \sum_{j=1}^{n}{A_je^{-p_jt}}
$$

### stability analysis

> \[!tip\] Important
>
> If $\sigma_i > 0$ then pole is in the left side of imaginary plane, and system is ==**stable** ==

### Complex root

For poles at $s=\sigma_i \pm j\omega$ we get

$$
\frac{\alpha + j\beta}{s + \sigma_i + j\omega_i} + \frac{\alpha - j\beta}{s + \sigma_i - j\omega_i}
$$

Wants to be on LHP for time-function associated with $s$ plane to be _stable_

## Impedance of Inductor

$$
Z(s) = \frac{V(s)}{I(s)} = Ls
$$

since the voltage-current relation for an inductor is $v(t) = L\frac{di(t)}{dt}$

## Impedance of Capacitor

$$
Z(s) = \frac{V(s)}{I(s)} = \frac{1}{Cs}
$$

since the voltage-current relation for a capacitor is $v(t) = \frac{1}{C} \int_0^{t}{i(\tau) d\tau}$