0.1 Lecture 2: introduction to systems  (Page 3/8)

For the resistor, if i(t) is bounded then so is v(t), but for the capacitance this is not true. Consider i(t) = u(t) then v(t) = tu(t) which is unbounded.

5/ Linear systems

for all $x_1(t)$ , $x_2(t)$ , a, and b.

6/ Time-invariant systems

for all x(t) and τ.

7/ Linear and time-invariant (LTI) systems

• Many man-made and naturally occurring systems can be modeled as LTI systems.

• Powerful techniques have been developed to analyze and to characterize LTI systems.

• The analysis of LTI systems is an essential precursor to the analysis of more complex systems.

Problem — Multiplication by a time function

A system is defined by the functional description

• Is this system linear?

• Is this system time-invariant?

Solution — Multiplication by a time function

Let

$\{\begin{array}{c}{y}_1(t)=g(t)x_1(t)\\ y_2(t)=g(t)x_2(t)\end{array}$

By definition the response to

$x(t)={\text{ax}}_1(t)+{\text{bx}}_2(t)$

Is

$y(t)=g(t)({\text{ax}}_1(t)+{\text{bx}}_2(t))$

This can be rewritten as

$y(t)=\text{ag}(t)x_1(t)+\text{bg}(t)x_2(t)$

$y(t)={\text{ay}}_1+{\text{by}}_2(t)$

Therefore, the system is linear.

Now suppose that $x_1(t)=\text{x}(t)$ and $x_2(t)=\text{x}(\text{t -}\tau )$ , and the response to these two inputs are $y_1(t)$ and $y_2(t)$ , respectively. Note that

$y_1(t)=y(t)=g(t)x(t)$

And

Therefore, the system is time-varying.

Problem — Addition of a constant

Suppose the relation between the output y(t) and input x(t) is y(t) = x(t)+K, where K is some constant. Is this system linear?

Solution — Addition of a constant

Note, that if the input is $x_1(t)+x_2(t)$ then the output will be

Therefore, this system is not linear.

In general, it can be shown that for a linear system if x(t) = 0 then y(t) = 0. Using the definition of linearity, choose a = b = 1 and $x_2={\text{-x}}_1(t)$ then $x(t)={\text{x}}_1(t)+{\text{x}}_2(t)=\text{0}$ and $y(t)={\text{y}}_1(t)+y_2(t)=\text{0}$ .

Two-minute miniquiz problem

Problem 2-1

The system

$y(t)=x^2(t)$

is (choose one):

1. Linear and time-invariant;
2. Linear but not time-invariant;
3. Not linear but time-invariant;
4. Not linear and not time-invariant.

Solution

Note that if $x_2(t)={\mathrm{2x}}_1(t)$ then $y_2(t)=({\mathrm{2x}}_1(t){)}^2={\mathrm{4y}}_1(t)$

Hence, this system is nonlinear.

Note that if $x_1(t)=x(t)$ and $x_2(t)=x(t-\tau )$ then $y_1(t)=y(t)$

And $y_2(t)=x^2(t-\tau )=y(t-\tau )$ . Hence, this system is time invariant.

V. LINEAR, ORDINARY DIFFERENTIAL EQUATIONS ARISE FOR A VARIETY OF SYSTEM DESCRIPTIONS

1/ Electric network

Kirchhoff’s current law yields

$i(t)=i_C(t)+i_R(t)+i_L(t)$

The constitutive relations for each element yield $$

$i_C(t)=C\frac{\text{dv}(t)}{\text{dt}}$ $i_R(t)=\frac{v(t)}{R}$ $i_L(t)=\frac{1}{L}{\int }_{-\infty }^{t}v(\tau )\mathrm{d\tau }$

Combining KCL and the constitutive relations yields

$\frac{\text{di}(t)}{\text{dt}}=C\frac{d^{2}v(t)}{{\text{dt}}^2}+\frac{1}{R}\frac{\text{dv}(t)}{\text{dt}}+\frac{v(t)}{L}$

2/ Mechanical system

The simplest possible model of a muscle (the linearized Hill model) consists of the mechanical network shown below which relates the rate of change of the length of the muscle v(t) to the external force on the muscle fe(t).

$f_c$ is the internal force generated by the muscle, K is its stiffness and B is its damping.

The muscle velocity can be expressed as

$v(t)=v_c(t)+v_s(t)$

Combining the muscle velocity equation with the constitutive laws for the elements yields

$v(t)=\frac{f_e(t)-f_c(t)}{B}+\frac{1}{L}\frac{{\text{df}}_e(t)}{\text{dt}}$

3/ First-order chemical kinetics

A reversible, first-order chemical reaction can be represented as follows