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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
Problem — Multiplication by a time function
A system is defined by the functional description
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
${y}_{2}(t)=g(t)x(t-\tau )\ne y(t-\tau )$
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 $y(t)={\text{x}}_{1}(t)+{x}_{2}(t)+\text{K}\ne {\text{y}}_{1}(t)+{y}_{2}(t)=({x}_{1}(t)+K)+({x}_{2}(t)+K)\text{.}$
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):
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
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