lecture5 (amplifier noise etc)(1)

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Methods of Experimental Physics Semester 1 2005

Lecture 5: Fourier Transforms and Differential Equations

5.1 Mass-Spring System

y

x

D k

m

F(t)

Figure 1: A general mass spring system

Lets examine a mass-spring system and consider its response to an external force applied to the mass. Ageneral diagram of such a system is given in Figure 1 above. The first job is to write down the governingequation and then to solve them using our knowledge. We will first do this the way that you are probablyrelatively familiar with, and then I am going to solve it using a Fourier transform approach. I hope thatyou will see that the Fourier transform approach makes life much easier.

We will assume the the force from the spring can be expressed as FSPRING = k(x y) where x and yare measured from the equilibrium points given in the diagram above. The damping can be expressedas FDAMPING = D(x y) i.e. it is proportional to the velocities at the end of the springs, and opposed

to the velocity that we attempt to impose upon it. To make life simple lets assume that y = 0. So usingNewtons laws we write:

mx = Ftotal = F kx Dx (1)

Conventionally one can also rewrite the equations using the substitutions:

2f0 =

k

m(2)

Q =2f0m

D(3)

D/m =2f0

Q(4)

where f0 is called the natural frequency or resonant frequency of the system, Q is called the Q-factor ofthe system and tells us how long the energy persists in the system (i.e. it is actually 2 times the numberof cycles before the energy has fallen to 1/e of some initial value. We will use these substitutions below.

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Lecture 5: Fourier Transforms and Differential Equations 5-2

5.1.1 Solving the mass-spring system in the time domain

This is a bit of a nightmare so what I am going to do is assume that there is no external force i.e.F(t) = 0 and then we will solve this. It is horrible enough and you will get a feel for it. Later we willsolve it using the Fourier transform method and we can do it in the full glory.

Lets assume that the solution can be written as x(t) = Cet - this allows for both an oscillatory solutionas well as a damped exponential solution. If we substitute this and its derivatives into Eq. 1 then weobtain: 2 + + (2f0)

2 = 0 after cancelling a factor of Cet. This polynomial has a two roots:

=

2

2

4 (2f0)2 (5)

We should distinguish three different cases: low damping ( < 4f0), heavy damping ( > 4f0), andcritical damping ( = 4f0)

5.1.1.1 Low Damping

In this case we see that the argument of the square root function is negative and thus is complex(i.e. since x(t) = Cet this means x(t) will be oscillatory). Let us write =

2 2fd where

fd =

f02

2

162 . There are thus two solutions and all linear combinations of these two linear solutions

are also solutions i.e.:

x(t) = exp

t

2

[C1 exp (2fd) + C2 exp(2fd)] (6)

If one requires that the solution is real (as we probably should for physical reasons) then C2 = C1 . If

one thinks about it a little you can see that this linear combination of oppositely rotating exponentialsolutions can be rewritten as:

Aeit + Aeit = 2(A) cos(t) + 2(A) sin(t) = 2|A| cos(t + ) where = arctan [(A)/(A)] (7)

so the net result of this is an exponentially damped oscillatory solution with an arbitrary phase.

2 4 6 8

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

Figure 2: A lightly damped sinwave

5.1.1.2 Heavy Damping

This is the case when both of the solutions for are real and negative. Doing the same maths as before

we get to a solution that looks like:

x(t) = C1et(12) + C2e

t(1+2) (8)


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0.2 0.4 0.6 0.8

0.2

0.4

0.6

0.8

1

Figure 3: Heavily damped mass-spring system: time in units of the oscillator period

where 1 = /2 and 2 =

2

4 4f022 By imposing boundary conditions we obtain the specificvalues for C1 and C2. If we impose x(0) = 1 and = 8f0 we obtain Figure 3. We note that if thedamping is really high then the system is essentially exponentially damped.

5.1.1.3 Critical Damping

By choosing = 4f0 we obtain only a single solution to the problem, yielding x(t) = Cexp(2

t). Thisonly gives us one arbitrary constant where we need two (as this is a second order d.e.) in order to satisfythe full range of boundary conditions. The correct solution is:

x(t) = (C1 + 2C2f0t)e2f0t = (C1 + C2t/2)e

t/2 (9)

Figure 4 shows a much quicker damping than the heavily damped oscillator shown above in Figure 3!

0.2 0.4 0.6 0.

0.2

0.4

0.6

0.8

1

Figure 4: Critically damped: time in units of the period of the oscillator

5.1.2 Forced Vibration

Now lets turn to the possibility of forced vibration of the mass. After we start to drive the device therewill be some transient behaviour while the device comes into a steady-state with the driving force. Wewill ignore this and only be interested in the steady-state solution.


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Lets use a complex notation for the both the applied force, F exp(t) where F = F0 exp() and themovement of the object is given by x exp(t). Of course we want a real solution in both cases but wewill apply this at the end. If the guiding equation (Eq. 1) is: mx = F kxD x then rearranging weget:

x =F

4m2 (f20 f2 + f/(2))

(10)

This equation is termed a response function or a transfer function as it relates the behaviour of theposition of the mass to the driving acceleration (F/m) i.e.

R =1

42 (f20 f2 + f/(2))

(11)

If we want to examine the form of this equation we could either plot the real and imaginary componentsor alternatively the magnitude and phase. The magnitude of the function is:

|R|2 =1

164(f20 f2)2 + 2(2f)2

(12)

To find the phase lets assume that the response function can be written as R = exp() where is

obviously equal to |R| given in Eq. 12 while is yet to be found:

x = RF

m= F0

ee

m=

F0m

e(+) (13)

Lets not forget that the displacement should be real so we will take the real part of this to yield:

( x(t)eit) = x(t) =F0m

cos(+ + t) (14)

Now since exp() = R/ = cos() + sin() by definition, we are able to derive the phase from:

tan() =(R/)

(R/)

= f

2(f2

0 f2

)

(15)

An example for three different systems of the magnitude and phase response is given in Figure 5. It

10 20 30

-45

-40

-35

-30

-25

(a) magnitude response on log scale

10 20 30

-175

-150

-125

-100

-75

-50

-25

(b) phase response ()

Figure 5: Three systems of varying losses

can be seen that there is a strong amplification of motion at the so-called resonant frequency, in fact theamplification is equal to Q factor. The phase is seen to always be negative - that is, the motion of themass is either in synchrony with the applied force if it is applied below resonance, or it lags the force byup to 180 for frequencies well above resonance.


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5.1.3 Solving the mass-spring system in the frequency domain

Lets now use the Fourier method to solve the differential equation. One will see that it is possible toobtain the transfer function of response function very easily. Regaining the time response from theresponse function can be quite difficult but often one is not interested in that. All of the informationis contained within the response function. I note that the notation is going to get awful if we use the

logical variable for the time variation of force (i.e. F(t) for the time variation and F(f) for the frequencydomain). Anyway, to get around this problem, lets use n(t) andN(f) (for Newton) for the Force variable.Returning to Eq. 1 we can write Newtons law as:

mx + Dx + kx = n(t) (16)

Taking the Fourier Transform on both side we get:

m(2f)2X(f) + 2fDX(f) + kX(f) = N(f) (17)

where we have defined X(f) x(t) and N(f) n(t) as the Fourier conjugates of the time domainfunctions.

(1m

k(2f)2 + 2f

D

k)kX(f) = N(f) (18)

Reusing our definitions from above in Eqs 2-4 (2f0 =

km and Q =

2f0mD ) we get:

(1f2

f20+

f

f0Q)kX(f) = N(f) (19)

and thus:

X(f) =1

k

1

1 f2

f20

+ ff0QN(f) (20)

which is the equivalent of Eq. 10 obtained earlier. If we rewrite this equation as X(f) = G(f)N(f) thenthe relating function, G(f), is termed the transfer function, response function or sometimes the impulse

function.This final terminology comes about from considering the situation of applying an impulse of force to asystem i.e. n(t) = (t). The Fourier conjugate of the impulse function is N(f) = 1 and thus triviallyX(f) = G(f) i.e. the systems response to an impulse is indeed its transfer function. The final bit ofconfusing nomenclature is that the time domain version of the transfer function is called the impulseresponse of the system. This is really just the obvious conclusion to the earlier remarks - if we hit thesystem with a short response then the Fourier transform of the resulting motion is just the transferfunction, and so its motion considered in the time domain is the impulse response of the system.

I note also that using this model we see three regimes of behaviour

1. when f f0, |G(f)| 1/k, 0 (spring regime)

2. when f = f0, |G(f)| = Q/k , /2 (resonance)

3. when f f0, |G(f)| =1m

1

2f

2, (free mass)

lecture5 (amplifier noise etc)(1)

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