2 Applying Fourier series to solve a differential equation

The following Task which is quite long will provide useful practice in applying Fourier series to a practical problem. Essentially you should follow Steps 1 to 3 above carefully.

Task!

The problem is to find the steady state response y ( t ) of a spring/mass/damper system modelled by

m d 2 y d t 2 + c d y d t + k y = F ( t ) (4)

where F ( t ) is the periodic square wave function shown in the diagram.

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Step 1 : Obtain the Fourier series of F ( t ) noting that it is an odd function:

The calculation is similar to those you have performed earlier in this Workbook.

Since F ( t ) is an odd function and has period 2 t 0 so that ω = 2 π 2 t 0 = π t 0 , it has Fourier coefficients:

b n = 2 t 0 ∫ 0 t o F 0 sin n π t t 0 d t n = 1 , 2 , 3 , … = 2 F 0 t 0 t 0 n π − cos n π t t 0 0 t 0 = 2 F 0 n π ( 1 − cos n π ) = 4 F 0 n π n  odd 0 n  even

so F ( t ) = 4 F 0 π ∑ n = 1 ∞ sin n ω t n (where the sum is over odd n only). Step 2(a) :

Since each term in the Fourier series is a sine term you must now solve (4) to find the steady state response y n to the n  th harmonic input: F n ( t ) = b n sin n ω t n = 1 , 3 , 5 , …

From the basic theory of linear differential equations this response has the form

y n = A n cos n ω t + B n sin n ω t (5)

where A n and B n are coefficients to be determined by substituting (5) into (4) with F ( t ) = F n ( t ) . Do this to obtain simultaneous equations for A n and B n :

We have, differentiating (5),

y n ′ = n ω ( − A n sin n ω t + B n cos n ω t ) y n ″ = ( n ω ) 2 ( − A n cos n ω t − B n sin n ω t )

from which, substituting into (4) and collecting terms in cos n ω t and sin n ω t ,

( − m ( n ω ) 2 A n + c n ω B n + k A n ) cos n ω t + ( − m ( n ω ) 2 B n − c n ω A n + k B n ) sin n ω t = b n sin n ω t

Then, by comparing coefficients of cos n ω t and sin n ω t , we obtain the simultaneous equations:

( k − m ( n ω ) 2 ) A n + c ( n ω ) B n = 0 (6)

− c ( n ω ) A n + ( k − m ( n ω ) 2 ) B n = b n (7)

Step 2(b) :

Now solve (6) and (7) to obtain A n and B n :

A n = − c ω n b n ( k − m ω n 2 ) 2 + ω n 2 c 2 (8)

B n = ( k − m ω n 2 ) b n ( k − m ω n 2 ) 2 + ω n 2 c 2 (9)

where we have written ω n for n ω as the frequency of the n  th harmonic

It follows that the steady state response y n to the n  th harmonic of the Fourier series of the forcing function is given by (5). The amplitudes A n and B n are given by (8) and (9) respectively in terms of the systems parameters k , c , m , the frequency ω n of the harmonic and its amplitude b n . In practice it is more convenient to represent y n in the so-called amplitude/phase form:

y n = C n sin ( ω n t + ϕ n ) (10)

where, from (5) and (10),

A n cos ω n t + B n sin ω n t = C n ( cos ϕ n sin ω n t + sin ϕ n cos ω n t ) .

Hence

C n sin ϕ n = A n C n cos ϕ n = B n

so

tan ϕ n = A n B n = c ω n ( m ω n 2 − k ) 2 (11)

C n = A n 2 + B n 2 = b n ( m ω n 2 − k ) 2 + ω n 2 c 2 (12)

Step 3 :

Finally, use the superposition principle, to state the complete steady state response of the system to the periodic square wave forcing function:

y ( t ) = ∑ n = 1 ∞ y n ( t ) = ∑ n = 1 ( n  odd ) C n ( sin ω n t + ϕ n ) where C n and ϕ n are given by (11) and (12).

In practice, since b n = 4 F 0 n π it follows that the amplitude C n also decreases as 1 n . However, if one of the harmonic frequencies say ω n ′ is close to the natural frequency k m of the undamped oscillator then that particular frequency harmonic will dominate in the steady state response. The particular value ω n ′ will, of course, depend on the values of the system parameters k and m .