Lab Final For Post

Sebastien Parmentier

Fourier Series and RCL Resonances

Introduction Periodic functions Because electricity behaves in harmonic motion in a AC circuit, its behavior is periodic and thus can be expressed and predicted by periodic functions. A function f t is called periodic if there exists a constant T 0 for which, f tT = f t  for any t in the domain of f t , given here as a function of time. Such a constant T is called a period of the function f t . The most familiar periodic functions are sin x, cos x, tan x, etc; x being in the domain of the function f  x  . However, time will be the medium on which the periodicity and the frequency will express themselves. Therefore the functions we will use to describe the behavior of our electric circuits will be in terms of t such as, sin t, cos t, etc; since time is a hugely important factor regarding the behavior of this form of energy. The beauty of a periodic function is that for a given function x = f t  on any closed interval, say, a < t < a + T, we can indeed decipher the behavior of the entire graph of f t just by analyzing only a small part of its domain, because of the periodic repetition of that small portion, and indeed, because that small interval can yet predict the behavior of that function as t → ∞ . Furthermore, if T is a period of the function f t , then the numbers 2T , 3T , 4T , ... are also periods. Better yet, because of the beautiful property of periodicity, because of its great predictability, we can even say that, when inspecting these functions, f t= f tT = f t2T= f t3T=... and

f t= f t−T = f t−2T= f t−3T=...

Therefore, if T is a period, so is nT for any positive n integer. And if a period exists, it is therefore not unique. Lastly, if f t is integrable on any interval of period T, then it is integrable on any interval of the same length, and the value of the integral must be the same. Harmonics The simplest periodic function, and the one of greatest importance for our experiment, is x= Asin t where amplitude A, angular frequency  and the initial phase (or “phase angle”)  are constants. The period of such a harmonic is given by T =2 / , since for any t,

[

A sin  t

 ]

2  

= Asin[t2 ] = Asin t

I will not extend further this presentation of a simple harmonic motion: one can read further my painstaking derivation of Hook's law and functions for systems involving springs in my first lab of this semester (Simple Harmonic Motion). Indeed I'm not here to write a book but a laboratory! Therefore, let's just jump right at this next great tool that will help us in explaining, solving and predicting

harmonic motions: trigonometric polynomials and series. Since I have demonstrated – I hope successfully – that one can evaluate the behavior of an harmonic motion for the whole domain - ∞ < t < ∞ by analyzing x t  at a ≤ t ≤ anT , we can start using this well-known formula from trigonometry, Asin  t = Acos t sin   sin t cos  Then, setting

a = Asin ,

b = Acos 

we can convince ourselves that every harmonic can be expressed by the form, a cos t  b sin t

To prove this (we probably must, alas, since most of the books out there do not take the time to do this) ( I want to thank here my now retired professor of Trigonometry, Calculus, and also quite an active contributor to many college Calculus books, Dr Montano, from Riverside Community College, in California, who gave the curious students great insights in mechanics. Yes, the guy already peppered his trigonometry tests with many circular motion problems and basic harmonic functions, but he taught these topics brilliantly... still, you can bet that many students truly wanted, as he used to say, “to kneel in the corner of the classroom, put their face inside their hands and cry gently” after most of his tests. The following are sampled from notes taken in his class, although I will use  instead of  , and pick different constants more adapted to our problems), A=

 a 2b 2 ,

sin  =

a = A

a ,  a b 2 2

cos  =

b = a

b  a b 2 2

for which  can be easily found. Therefore, for now on, we will use the form a cos t  b sin t , to express harmonics. For example,



2sin 3t



 = 3

 3 cos3t  sin 3t

Finally now, I do realize the scope and usefulness of his lectures! I should probably send him a mail...

Fourier series Let's again consider this harmonic, a n cos n t  bn sin n  t

 for n =1,2,3,4,. ..

Since an integral multiple of a period is again a period, and if one cares to express the behavior of the harmonic function for the whole interval t during which the motion occurs, one can express x t  by, n=∞

x t  = A 

∑  a n cos  n t n=1

 b n sin n T  

where A is a constant. X(t) is in fact a function of period T, since it is a sum of functions of period T. (The addition of a constant does not destroy periodicity; in fact, a constant can be regarded as a

function for which any number is a period). Therefore, and as Jean-Baptiste Fourier was the first to show, in order to give a mechanical interpretation to all this, we can represent a complicated oscillatory motion x t  , as a sum of individual oscillations which are particularly simple. Fourier's technique of solving immensely difficult problems by simply morphing them into simple trigonometric series turned the field of mathematic upside down in the 18th century. For a while, Fourier's techniques were hugely challenged, even mocked by his mathematician colleagues as lacking rigor, before his stunningly insightful techniques got widely accepted. Now, suppose the function x t  of period T has the expansion a x t  = 0  2

n=∞

∑  a n cos n t n=1

b n sin n  t 

where we denote the constant term by a 0 /2 , which is also called the average value of the function. Note: Before I derive further, I must assume a few things; for the sake of mathematical rigor, and because I am trying to keep this presentation short. (which means I won't prove a thing further!) We must assume here that the series expressed above and the ones that will soon follow can be integrated term by term. We will assume that for all these series the integral of the sum equals the sum of the integrals – it is therefore assumed that x(t) is integrable -, and most of all: by theorem, for a Fourier series to deserve its full Fourier glory, “it must be able to be expanded in a trigonometric series which converges uniformly on the whole real axis”. Since Fourier series and its applications are new to me, it is quite hard for me to come up with the exact words as to sum up and explain the purpose and the condition of this transformation. Thus, I would like to quote this paragraph from Wolfram's Mathworld, that may put it in much better terms: “A Fourier series converges to the function (equal to the original function at points of continuity or to the average of the two limits at points of discontinuity)

if the function satisfies so-called Dirichlet conditions. Dini's test gives a condition for the convergence of Fourier series. As a result, near points of discontinuity, a "ringing" known as the Gibbs phenomenon, can occur. For a function periodic on an interval instead of be used to transform the interval of integration from to

, a simple change of variables can .”

Well... this is all way above my head right now, but I'm sure I will learn and understand this jargon as soon as I will sit in my Stony Brooks' E&M class; which I certainly won't escape if I look forward to a

physics degree! Anyway. With those conditions fixed, we can therefore resume our derivation. Let me pick, as we did in class, a period in the interval 0≤t≤T , rather than a≤t≤aT , for the sake of simplicity and in order to help us in getting rid of some terms.... Then, integrating from zero to T, we obtain, (eq 1): T

∫ x t dt

=

0

a0 T ∫ dt  2 0

n=∞

∑

n=1



T

T

a n∫ cos n  t dt  b n∫ sin n  t dt 0

0



For any integer n≠o , , integrating a full period of a sin or cosine function solidly centered at x(t) = 0 always yields zero. It may be hard to see how this happens on the interval 0≤t≤T for those who slept during calculus classes; but let us for a second integrate these functions, say, from −≤t≤ . It is now easy to see that 

∫ cos n t  dt

[

=

−

sin n  t  n

]

−

= 0



and 

∫ sin n  t dt = −

[

−cos n t  n

]

−

= 0



Therefore, all the integrals in (eq 1) vanish, so that T

∫ x t dt 0

=

a0 T 2

Next, let's come back to (eq 1) and multiply both sides of this expression by cos m T  we obtain T

a0 T ∫ x tcos  mt dt = 2 ∫ cos m t  dt 0 0 n=∞



∑

n=1



T

T

a n∫ cos n t cos m t dt = bn∫ sin n t cos m t dt 0

0



By what I have demonstrated earlier, the first integral on the right vanishes. Since the functions of the system are pairwise orthogonal (1, cos ωt, sin ωt, cos 2ωt, sin 2ωt, ..., cos nωt, sin nωt, and so on) , I'll be more than happy to hand to all the other terms a pink slip as well - except one. Thus, the only integral that remains is the coefficient of a n (assuming that n = m), which is therefore given by T

∫ a n cos 2  n t dt 0

= T

which can be re-expressed as, T

a n∫ 0

1cos 2n t dt = T 2

The cosine term contributes nothing between the given limits, but the “one -half” part gives us T /2. Thus we can arrive, finally, to the following identity: T

an

2 = ∫ x t cos  n t dt T 0

bn

2 = ∫ x tsin n  t dt T 0

and T

For a square graph for which exactly half of the period is centered on the x-axis (not on the time-axis), all values of bn will equal zero. Therefore, only an needs to be evaluated for our experiment, since we have chosen to fix the square function in a way that the first half of its period will be centered between -T/4 and T/4. Furthermore, I have also decided to elevate the whole shebang so the whole peak to peak amplitude will stand between x(t) = 0 and x(t) = A0 which correspond to the magnitude of the full peakto-peak amplitude of the first resonance. You will see very soon why this setting does make the whole thing way more digestible... Lastly, we won't be messing around and theorize with a square function of amplitude set between -1 and 1, because there is no way to compare these results with the ones that we will soon obtain using our data... unless we normalize this whole thing. But that's a whole different thing.

Goals and Procedure. The goal of our experiment was to predict and analyze the resonances of a driven AC circuit (consisting of a generator, an inductor, a capacitor and a resistance), connected to an oscilloscope. We won't explain here the phenomenon of resonance in a driven RLC circuit: this work has already been done in past Physics II and III labs. The goal of this experiment is not to analyze resonances but to predict their amplitudes, while being introduced to Fourier series as a brand new tool for solving this problem and set these resonances into different type of functions such as square function, which permit one to visualize all possible resonances – or the “series” of all these resonances which depend on given resonant frequencies - inside one single graph. After having plugged the whole loop of the circuit into the channel one of the oscilloscope, we connected another smaller loop, this one involving only the resistance, into the channel two of that same oscilloscope. We then tweak around with the driven frequency in order to find our first resonance. Mr Breeden theorized that we should have found this resonance at 21 kHz, but we found it at 21.78 kHz instead; surely because of some trace of internal resistance inside the oscilloscope itself. However, 21.78 kHz is what we will have to employ from now on, if we desire to match our theoretical values with the collected data.

We were asked to press the button that switches the oscilloscope into a “square function” mode. We looked for a few more resonances, collected their amplitudes and frequencies. These data will be shown later, in a table, after we have theorized on the values we should have read, using Fourier series. Now, it's time to see if that equation we devised earlier for an will actually work. But we will have to make some changes in the expression on the right side of the equal sign. As I said earlier, I have first raised the whole square function so the whole peak-to-peak amplitude will stand on the t-axis, and since x(t) is only equal to a constant for square function, this makes our calculations much easier, given that we already have some values just begging to do the job: our first peak-to-peak amplitude at 21.78 kHz, and its corresponding amplitude given by A0 = 6.8 V. We have also placed the first half of the period T as to be sitting perfectly symmetrically between -T/4 and T/4. Thus we can avoid integrating from T/4 to 3T/4, since the value of this second interval is zero... but we will have to multiply this whole equation nonetheless by 2 in order to take account of the whole period T. Thus, we have

[

2 an = 2 T

−T /4

∫

A0 cos n t dt

T/4

]

We will integrate by substitution as, u = nt

du = n dt





1 = dt n

hence, we get, 4 1 an = ⋅ T n

−T / 4

∫

T /4

A0 cos n  t n  dt

Solving between this interval, A0

4 T ⋅ T n2

[    sin n  n

T −T − sin n n 4 4

]

= an

Since sine is an odd function we can extract the negative sign from the second argument and simplify

[  ]

2A0 T 2sin n  n n 4

= an

Since  n = 2  f n and T = 1/ f a , we can finally give birth to our final equation: 1

This equation determines for us the amplitude a n associated with any given value of n in the harmonic analysis of x(t).

[  ]

4A 0 n f n sin n 2 fa 1

= an

Now, we shall verify if this equation will suffice to predicting the amplitudes given for their respective resonances.... We know that, for the way we have set up our square function, a0/2 must equal zero. We also know that the equation expressed above will only be greater than zero only for odd integers of n, or 21.78 kHz divided by either 1,3,5,7 etc... Therefore, let's compute and predict the amplitude in Volt that these respective resonances shall yield: Table of theoretical values (from a collected initial value of frequency and amplitude) (see calculations below the graph)

Resonance numbers (odds only)

Frequency in kHz Predicted

Amplitude predicted in Volts (peak-to-peak)

(collected 21.78 kHz divided by n) ( n = 1 was collected, not predicted)

n=1

21.78 ± 1.00*

6.8 ± 0.1

n=3

7.26 ± 1.00

2.88 ± 0.1

n=5

4.36 ± 1.00

1.73 ± 0.1

n=7

3.11 ± 1.00

1.24 ± 0.1

* since an ideal oscilloscope should have given us 21.00 kHz, it is reasonable to set an error margin of 1.00 kHz!

Here's the calculations for the values stated in the table above:

[  [  [ 

]

= 2.88

a3



3 4 6.8 3  7.26x10  sin ⋅ 3 2  21.78X103 

a5



3 4 6.8 5   4.36x10  sin ⋅ 5 2  21.78X10 3

]

= 1.73

a7



3 46.8 7  3.11x10  sin ⋅ 7 2 21.78X103 

]

= 1.24

Of course, if one is interested to know the peak amplitude instead of the peak-to-peak amplitude, one has just to divide these values by 2. We shall thus get, a 3 = 1.44 V , a 5 = 0.87 V , a 7 = 0.62 V . Let's compare these values with the collected data: Collected Data Resonances numbers

Frequencies collected (in kHz)

Amplitudes collected (in Volts)

n=1

21.78 ± 1.00

6.8 ± 0.1

n=3

7.4 ± 1.00

2.8 ± 0.1

n=5

4.48 ± 1.00

1.8 ± 0.1

n=7

3.13 ± 1.00

1.28 ± 0.1

Shall we say, “Bam!”.... Because, these Fourier series seem to do magic here...

Conclusion and Error Analysis Using the Fourier series proved to be an incredibly effective tool in order to predicting any amplitudes at given resonances. In fact, when plotting both our predicted values and our collected values, one can see the precision of that mathematical tool: Am plitude vs. Frequency

8 7

Am plitude (V)

6 5

Column A Linear regression f or Column A Column A Linear regression f or Column A

4 3 2 1 0 0

5

10

15

20

25

Frequency (kHz)

which also shows that the amplitude of a resonance grows linearly according to the magnitude of that corresponding frequency. Since we have compared one predicted value to one collected, we won't have to compute any standard deviations in order to obtain error bars. Instead, a simple percentage error will suffice. Here's our error table: Resonance number

Percent Error for frequency

Percent error for Amplitudes

n=3

1.90%

2.80%

n=5

2.70%

4.00%

n=7

0.60%

3.20%

We took great precaution to take those data, this time around. Therefore I can only attribute those errors to the internal resistance of the oscilloscope, and perhaps the rudimentary circuit we employed for this experiment. Otherwise, the Fourier series have really given us a great satisfaction in predicting solidly the behavior of a system that really required more advanced mathematics to do so. It felt pretty good. I can understand now why this Fourier series and transforms are showing all over the place in many fields of physics but more particularly in those areas where harmonic motions are involved.