BIOE 205

Lecture 15

Reading material: Chapter 12 of CSSB

We finally come to the third and last major thematic portion of the course with this lecture. In this lecture we discuss the basic strategies and theory of analysis of simple mechanical and electrical networks. We start with a review of the basic laws and principles of mechanics.

  1. Mechanical systems
    1. Damped systems or ζ0\zeta \neq 0.
    2. Undamped systems or ζ=0\zeta = 0.
  2. Electrical systems
    1. Analogues with mechanical systems
    2. Impedance, a generalized resistance

Mechanical systems

The most studied mechanical system is probably the mass-spring-damper system. Why do we worry so much about mass-spring systems, you may be wondering. We research these systems because they can be used to represent or simplify a wide range of engineering issues. Several of these consist of:

In the sections that follow, we'll first go through a mass spring system's fundamental parts and how to set up the equations. Masses, springs, and dashpots (or dampers) are the three major parts of mass spring systems. The equation(s) of motion for the system are derived using Newton's second law along with the equations for the spring and friction (dashpot) forces. Let's have a look at a really straightforward mass spring system we already have seen before:

We should already know from PHYS 211 that the force exerted by the spring is in opposition to motion and dependent on the total displacement from its equilibrium position. The spring constant kk tells us how much force is generated per unit displacement and thus takes units N/m. On the other hand, the force exerted by the dashpot is proportional the rate of change of displacement, i.e. velocity. Therefore this term is also often called a viscous damper (in relation to viscous liquids like a can of thick paint). Both these forces along with the applied external force are related to the acceleration experienced by the object (by Newton's second law of motion).

mx¨=Fextkxcx˙ m \ddot{x} = F_{ext} -kx - c \dot x

As a preview of BIOE 420, often this second order system is written as two coupled first order differential equations by the introduction of state variables z1=xz_1=x and z2=z1˙z_2 = \dot{z_1} which are displacement and velocity respectively as:

ddt[z1z2]=[01k/mc/m][z1z2]+[01/m]Fext \dfrac{d}{dt} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -k/m & -c/m \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}F_{ext}

Most mechanical systems we will concern ourselves with are different arrangements of the above basic components.

Answer: Derivation left as an exercise to recall PHYS 211. Setting z1=x1,z3=x2,z5=xaz_1=x_1, z_3=x_2, z_5=x_a we get,
[z˙1m1z˙2z3m2z˙4z˙5maz˙6]=[010000(k1+k2)(c1+c2)k2c200000100k2c2(k2+ka)(c2+ca)kac100000100kacakaca][z1z2z3z4z5z6]+F \begin{bmatrix} \dot z_1 \\ m_1 \dot z_2 \\ z_3 \\ m_2 \dot z_4 \\ \dot z_5 \\ m_a \dot z_6 \end{bmatrix} = \begin{bmatrix} 0 &1 &0 &0 &0 &0 \\ -(k_1+k_2) &-(c_1+c_2) &k_2 &c_2 &0 &0 \\ 0 &0 &0 &1 &0 &0\\ k_2 &c_2 &-(k_2+k_a) & -(c_2 + c_a) &k_a &c_1 \\ 0 &0 &0 &0 &0 &1\\ 0 &0 &k_a &c_a &-k_a &-c_a \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ z_4 \\ z_5 \\ z_6 \end{bmatrix} + \vec{F}
where
F=[00000Fext]T\vec{F} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & F_{ext} \end{bmatrix}^T

To get (1) above written in the standard second order ODE format we are already familiar with, i.e.

x¨+2ζωnx˙+ωn2x=F \ddot{x} + 2 \zeta \omega_n \dot x + \omega_n ^2 x = F

we need to set:

ωn2=km,ζ=c2km,andF=Fextm \omega _n ^2 = \dfrac{k}{m}, \qquad \zeta = \dfrac{c}{2\sqrt{km}}, \qquad \textrm{and} \qquad F = \dfrac{F_{ext}}{m}

This puts our system in the form that we have examined in detail in the previous lecture - Lecture 14. In particular, there we discussed the how different values of ζ\zeta lead to qualitatively different responses from the system and we examined those responses using a demonstration. Recall that these were the cases of: underdamped, overdamped and critically damped systems.

⚠️ Note
Note: If those cases or terms seem alien to you then you should stop here and go catch up with the earlier lecture.

In this lecture note we further discuss two more qualitatively different responses that can come from such systems which have to do with the relative values of ωn\omega_n and the forcing frequency ω\omega of the input function F(ω)F(\omega).

Forced motion means there is some sort of outside force acting on the masses (i.e. motion independent of the spring or damping forces). Let’s modify the above mass-spring system slightly to include a forcing term i.e. something driving the mass that is driven separately from the mass:

Note that FF above can be either a function for force or displacement of the right endpoint.

Solution: Left as an homework exercise where FF is a function for the displacement.

In the following we assume that the reader is familiar with second order differential equations of the inhomogenous type (see MATH 285 syllabus); i.e. with forcing term present. Let the forcing function be sinusoidal with frequency ω\omega. Then we have four cases to discuss:

Damped systems or ζ0\zeta \neq 0.

Quite simply, in the first instance with damping, we know that the homogeneous solution will be a decaying sinusoid, and we know that the specific solution will be a sine/cosine from the method of undetermined coefficients. The homogeneous solution is referred to as the transient solution since it degrades/fades with time. The particular solution is known as the steady-state solution since it consists just of a sine or cosine that never ends.

In general, you should picture the solution's appearance as tt\to \infty when you hear the term "steady state solution". We have already examined what happens with different values of ζ\zeta the damping coefficient and so we don't repeat it here.

Undamped systems or ζ=0\zeta = 0.

The cases of real interest for this lecture note are what happens without damping in the presence of a forcing function.

General case, i.e. ωωn \omega \neq \omega_n

In this case the frequency of the forcing function is very different from the natural frequency of the system. This case is similar to the previous one, except that absent any damping neither the homogeneous and nor the particular solutions decay (they are both sinusoids with distinct frequencies). Thus the full solution will be the superposition or addition of two sinusoids.

Beats, i.e. ωωn\omega \approx \omega_n.

This is where ωωn\omega \approx \omega_n and there’s no damping, so both the homogeneous and particular solutions are sinusoids. Moreover, the frequencies are close enough such that we have periodic cancellations between the homogeneous and particular solutions. An example solution of such a case is there in your homework. It is actually possible to show that when ωωn\omega \approx \omega_n, the solution will be a superposition of two sine waves with very similar frequencies leading to the presence of a high frequency component and a low frequency component in the solution. However,this requires a lot of algebra to show and isn’t going to significantly improve your life, so we don’t go into it here.

Resonance, i.e ω=ωn\omega = \omega_n.

This last case, as the title says is the case of resonance. Recall that for a inhomogeneous system:

d2xdt2+ω2x=F(t) \dfrac{d^2x}{dt^2} + \omega^2 x = F(t)
the homogenous solution (i.e. obtained when F(t)0F(t) \equiv 0) is given by:
xh(t)=C1sin(ωt)+C2cos(ωt) x_h (t) = C_1 \sin (\omega t) + C_2 \cos (\omega t)
and we need to solve for the particular solution using the method of undetermined coefficients. But now F(t)=sin(ωt)F(t) = \sin(\omega t) is a match for a homogenous basis function which results in a modification
xp(t)=C11sin(ωt)+C22tcos(ωt) x_p(t) = C_{11} \sin (\omega t) + C_{22} t \cos (\omega t)
which means the particular solutions grow with time. Thus, in the case of resonance, the solution "blows up".

⚠️ Note
Note: If you have forgotten MATH 285 or did not end up taking it this semester that is fine, BIOE 205 won't test you on differential equations material, we will leave that up to the math professor. Nevertheless, you should see these two solution behaviors show up in your homework via Laplace transforms.

Electrical systems

We assume everyone is familiar with PHYS 212 material regarding basic electrical components like resistors, capacitors and inductors and Kirchoff's current (KCL) and voltage laws (KVL). In this section, we start by discussing how the differential equations arising from electrical circuits have analogues with the mechanical systems we were just discussing.

Analogues with mechanical systems

The basic relations to remember here from PHYS 212 are the equations for the voltage drops across the resistor, capacitor and inductor which are presented below in their most common form:

VR=IR,VC=q/C=1CIdt,VL=LdIdt V_R = IR, \quad \qquad V_C = q/C = \frac{1}{C} \int I dt , \quad \qquad V_L = L \dfrac{dI}{dt}

To start consider the basic RLC circuit shown below:

We can derive the differential equation governing this circuit via an application of KVL using the familiar rules above:

V1=IR+LdIdt+1CIdt V_1 = IR + L \dfrac{dI}{dt} + \dfrac{1}{C} \int I dt

however if we switch the basic unit from current to charge qq such that I=dqdtI=\dfrac{dq}{dt} then we can put the above equation in the form

Lq¨=V1Rq˙qC L \ddot{q} = V_1 - R \dot{q} - \dfrac{q}{C}

which we say is analogous to the mass-spring-damper equation (1) because the variable substitutions qx,V1Fext,Rcq \to x, V_1 \to F_{ext}, R \to c and 1Ck\dfrac{1}{C} \to k allow us to go between the two equations.

In fact we can derive more analogies based on these observations as summarized in the below table. The key points are:

Summary table

TypeBasic unitDerivativeStorageEnergyDissipationDriverInertiaEnergy
MechanicalDisp.[1]VelocitySpringPotentialFrictionForceMassKinetic
Equationsxxx˙\dot xkxk xkx22\dfrac{kx^2}{2}cx˙c\dot xFFMx¨M\ddot{x}mx˙22\dfrac{m\dot{x}^2}{2}
ElectricalChargeCurrentCapacitorElectricResistorVoltageInductorMagnetic
Equationsqqq˙\dot qqC\dfrac{q}{C}q22C\dfrac{q^2}{2C}Rq˙R\dot qVVLq¨L\ddot{q}Lq˙22\dfrac{L\dot{q}^2}{2}
⚠️ Note
Note: The analogies as summarized above is a particular form of similarity that can be derived by equating applied external force in the mechanical system with the potential difference that drives an electrical circuit. Such an analogy is called the Force-Voltage analogy. We could get a different analogy by equating the applied external force with a current source. Such analogies are called Force-Current analogies.
Solution: Discussed in class, see lecture slides.

Impedance, a generalized resistance

We are familiar with the notion of calling the obstruction of the motion of electrons through a material as the characteristic resistance of a material. Naturally, conductors have very low resistance and insulators have extremely high resistance. Generally speaking, this applies to electrical circuit elements regardless of whether we are dealing with a DC (direct current) circuit or an AC circuit (alternating current) circuit. However, when it comes to inductors and capacitors, recall that their behavior is far more interesting in the AC case than in the DC case. Indeed, for example, with DC current, uncharged capacitors act like short circuits (transient phase) and fully charged capacitors (steady state) act like open circuits. Similarly, inductors in steady state can be considered short circuits and vice versa. It is precisely when the electric field changes continuously across the dielectric in a capacitor or within an inductor that our familiar equations

V=LdIdtandI=CdVdt V = L\dfrac{dI}{dt} \qquad \textrm{and} \qquad I = C\dfrac{dV}{dt}

hold. It turns out that each of these circuit elements offer their own obstruction to the flow of AC, albeit in a frequency dependent manner. This obstruction term is called reactance.

In fact, (3) tells us even more. The first of these tells us that the voltage across the inductor is zero whenever the slope of the II vs. tt (time) graph is horizontal and that VV is maximized when the rate of change of current is maximized. Thus if I=sin(t)I=sin(t) then VV is a cosine and vice versa. A similar relationship can also be seen to hold for the second equation above. This leads us to the following conclusions:

Moreover, for an inductor, faster the current switches direction (consider AC) the larger the voltage drop and so is its contribution to the voltage drop in a circuit (i.e reactance) is directly proportional to the frequency ω\omega of the applied AC. Conversely for a capacitor, faster the electric field switches, the more is the current. Thus its opposition to current (or reactance) is inversely proportional to the AC frequency. Thus we get the following definitions of inductive and capacitive reactance (denoted using a XX):

XL=LωandXC=1ωC X_L = L \omega \qquad \textrm{and} \qquad X_C = \dfrac{1}{\omega C}

Coupling this with inherent/internal Ohmic losses (i.e. pure resistance RR) and the fact that the current/voltage relationship are out of phase with each other we defined the impedance for each of these elements as:

ZL=LωjandXC=1jωC Z_L = L\omega j \qquad \textrm{and} \qquad X_C = \dfrac{1}{j\omega C}

which dovetails nicely with our adopted notation for the Laplace domain variable s=jωs = j \omega to give us the usual definition of impedances for the capacitor and inductor:

ZL=LsandZC=1Cs Z_L = L s \qquad \textrm{and} \qquad Z_C = \dfrac{1}{Cs}

Solution: From the form of the generalized Ohm's law: V=IZV=IZ the ZZ term should appropriately add or subtract phase from the current to give the right relationship with velocity. Considering the Argand plane

we see that ZLZ_L should add a phase of π/2\pi/2 whereas ZCZ_C subtracts a phase of π/2\pi/2. This meshes well with the bullet points above since:

This is another example where algebra and math is simplified using "imaginary" numbers.

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[1] Displacement: Just trying to make the table fit. 😁
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