Outline. Introduction

Transcription

1 Outline Periodic and Non-periodic Dipartimento di Ingegneria Civile e Ambientale, Politecnico di Milano March 5, 4 A periodic loading is characterized by the identity p(t) = p(t + T ) where T is the period of the loading, and ω = π T principal frequency. p is its p(t) p(t + T ) T t Note that a function with period T /n is also periodic with period T.

2 Periodic loadings can be expressed as an infinite series of harmonic functions using the Fourier theorem, e.g., for an antisymmetric loading you can write p(t) = p( t) = j= p j sin jω t = j= p j sin ω j t. The steady-state response of a SDOF system for a harmonic loading p j (t) = p j sin ω j t is known; with β j = ω j /ω n it is: x j,s-s = p j k D(β j, ζ) sin(ω j t θ(β j, ζ)). In general, it is possible to sum all steady-state responses, the infinite series giving the SDOF response to p(t). Due to the asymptotic behaviour of D(β; ζ) (D goes to zero for large, increasing β) it is apparent that a good approximation to the steady-state response can be obtained using a limited number of low-frequency terms. Using Fourier theorem any practical periodic loading can be expressed as a series of harmonic loading terms. Consider a loading of period, its Fourier series is given by p(t) = a + a j cos ω j t + b j sin ω j t, j= j= where the harmonic amplitude coefficients have expressions: a = Tp p(t) dt, as, by orthogonality, Tp o p(t)cosω j dt = o ω j = j ω = j π, a j = Tp p(t) cos ω j t dt, b j = Tp p(t) sin ω j t dt, a j cos ω j t dt = Tp a j, etc etc. Fourier Coefficients Exponential Form If p(t) has not an analytical representation and must be measured experimentally or computed numerically, we may assume that it is possible (a) to divide the period in N equal parts t = /N, (b) measure or compute p(t) at a discrete set of instants t, t,..., t N, with t m = m t, obtaining a discrete set of values p m, m =,..., N (note that p = p N by periodicity). Using the trapezoidal rule of integration, with p = p N we can write, for example, the cosine-wave amplitude coefficients, a j t N p m cos ω j t m m= = N N p m cos(jω m t) = N m= N p m cos m= It s worth to note that the discrete function cos with period N. jm π N jm π N. is periodic The Fourier series can be written in terms of the exponentials of imaginary argument, p(t) = j= P j exp iω j t where the complex amplitude coefficients are given by P j = Tp p(t) exp iω j t dt, j =,..., +. For a sampled p m we can write, using the trapezoidal integration rule and substituting t m = m t = m /N, ω j = j π/ : P j N N m= p m exp( i π j m N ),

3 Undamped Damped We have seen that the steady-state response to the jth sine-wave harmonic can be written as [ x j = b j k βj sin ω j t, β j = ω j /ω n, analogously, for the jth cosine-wave harmonic, [ x j = a j k βj cos ω j t. In the case of a damped, we must substitute the steady state response for both the jth sine- and cosine-wave harmonic, x(t) = a k + k j= + k +( β j ) a j ζβ j b j ( β j ) + (ζβ j ) cos ω j t+ j= +ζβ j a j + ( β j ) b j ( β j ) + (ζβ j ) sin ω j t. Finally, we write [ x(t) = k a + βj j= (a j cos ω j t + b j sin ω j t). As usual, the exponential notation is neater, x(t) = j= P j k exp iω j t ( β j ) + i (ζβ j). Example Example cont. As an example, consider the loading p(t) = max{p sin πt, } a = Tp / a j = Tp / p o sin πt p o sin πt dt = p π, { for j odd = [ p π j for j even, b j = Tp / p o sin πt cos πjt dt sin πjt dt = { p for j = for n >. Assuming ( β = 3/4, from p = p π + π sin ω t 3 cos ω t 5 cos 4ω t... ) with the dynamic amplifiction factors D = etc, we have x(t) = p kπ ( 3 4 ) = 6 7, D = ( 3 4 ) = 4 5, D 4 = (4 3 4 ) = 8, D 6 =... ( + 8π 7 sin ω t cos ω t + 6 cos 4ω t +... ) Take note, these solutions are particular solutions! If your solution has to respect given initial conditions, you must consider also the homogeneous solution.

4 Example cont. Outline of Fourier transform x(t) k π / p o Periodic x Loading xfourier Series x x 4 x 4 -x Extension of to non periodic functions in the Frequency Domain Extension of to non periodic functions in the Frequency Domain t/ Non periodic loadings Non periodic loadings () It is possible to extend the Fourier analysis to non periodic loading. Let s start from the Fourier series representation of the load p(t), + p(t) = P r exp(iω r t), ω r = r ω, ω = π, introducing P(iω r ) = P r and substituting, p(t) = + P(iω r ) exp(iω r t) = ω π + P(iω r ) exp(iω r t). Due to periodicity, we can modify the extremes of integration in the expression for the complex amplitudes, Extension of to non periodic functions in the Frequency Domain If the loading period is extended to infinity to represent the non-periodicity of the loading ( ) then (a) the frequency increment becomes infinitesimal ( ω = π dω) and (b) the discrete frequency ω r becomes a continuous variable, ω. In the limit, for we can then write p(t) = π P(iω) = P(iω) exp(iωt) dω p(t) exp( iωt) dt, which are known as the inverse and the direct s, respectively, and are collectively known as the Fourier transform pair. Extension of to non periodic functions in the Frequency Domain P(iω r ) = +Tp / / p(t) exp( iω r t) dt.

5 SDOF Outline of the Discrete In analogy to what we have seen for periodic loads, the response of a damped SDOF system can be written in terms of H(iω), the complex frequency response function, x(t) = H(iω) P(iω) exp iωt dt, where π H(iω) = [ = [ ( β ) i(ζβ), β = ω. k ( β ) + i(ζβ) k ( β ) + (ζβ) ω n To obtain the response through frequency domain, you should evaluate the above integral, but analytical integration is not always possible, and when it is possible, it is usually very difficult, implying contour integration in the complex plane (for an example, see Example E6-3 in Clough Penzien). Extension of to non periodic functions in the Frequency Domain The Fast Fourier Discrete Discrete () To overcome the analytical difficulties associated with the inverse Fourier transform, one can use appropriate numerical methods, leading to good approximations. Consider a loading of finite period, divided into N equal intervals t = /N, and the set of values p s = p(t s ) = p(s t). We can approximate the complex amplitude coefficients with a sum, P r = Tp p(t) exp( iω r t) dt, that, by trapezoidal rule, is ( ) N t p s exp( iω r t s ) = N p s exp( i πrs N t N N ). s= s= Fourier In the last two passages we have used the relations p N = p, exp(iω r t N ) = exp(ir ω ) = exp(irπ) = exp(i) ω r t s = r ω s t = rs π N = π rs N. Take note that the discrete function exp( i πrs N ), defined for integer r, s is periodic with period N, implying that the complex amplitude coefficients are themselves periodic with period N. P r+n = P r Starting in the time domain with N distinct complex numbers, p s, we have found that in the frequency domain our load is described by N distinct complex numbers, P r, so that we can say that our function is described by the same amount of information in both domains. Fourier

6 () Only N/ distinct frequencies ( N = +N/ N/ ) contribute to the load represen- sin( * (π)/ * s /N), N=, s=,.., sin( * (π)/ * s /N), N=, s=,.., tation, what if the frequency.5 content of the loading has contributions from frequencies higher than ω N/? What -.5 happens is aliasing, i.e., the upper frequencies contributions are mapped to contri- - /4 T butions of lesser frequency. p See the plot above: the contributions from the high frequency sines, when sampled, are indistinguishable from the contributions from lower frequency components, i.e., are aliased to lower frequencies! Fourier The maximum frequency that can be described in the DFT is called the Nyquist frequency, ω Ny = π t. It is usual in signal analysis to remove the signal s higher frequency components preprocessing the signal with a filter or a digital filter. It is worth noting that the resolution of the DFT in the frequency domain for a given sampling rate is proportional to the number of samples, i.e., to the duration of the sample. Fourier The Fast The Fast The operation count in a DFT is in the order of N A Fast Fourier is an algorithm that reduces the operation count. The first and simpler FFT algorithm is the Decimation in Time algorithm by Tukey and Cooley (965). Assume N is even, and divide the DFT summation to consider even and odd indices s N X r = x s e πi N sr, r =,..., N s= N/ = q= N/ x q e πi N (q)r + q= x q+ e πi N (q+)r collecting e πi N r in the second term and letting q N = q N/ Fourier Say that X r = E r + e πi N r O r where E r and O r are the even and odd half-dft s, of which we computed only coefficients from to N/. To get the full sequence we have to note that. the E and O DFT s are periodic with period N/, and. exp( πi(r +N/)/N) = e πi exp( πir/n) = exp( πir/n), so that we can write { E r + exp( πir/n)o r X r = E r N/ exp( πir/n)o r N/ if r < N/, if r N/. Fourier N/ = q= x q e πi N/ qr + e πi N r N/ q= x q+ e πi N/ qr We have two DFT s of length N/, the operations count is hence (N/) = N /, but we have to combine these two halves in the full DFT. The algorithm that was outlined can be applied to the computation of each of the half-dft s when N/ were even, so that the operation count goes to N /4. If N/4 were even...

7 Pseudocode for CT algorithm from cmath i m p o r t exp, p i def fft(x, N): if N = then Y = X else Y = fft(x, N/) Y = fft(x, N/) for k = to N/- Y_k = Y_k + exp( pi i k/n) Y_k Y_(k+N/) = Y_k - exp( pi i k/n) Y_k endfor endif return Y Fourier d e f d _ f f t ( x, n ) : """ D i r e c t f f t o f x, a l i s t o f n = m complex v a l u e s """ r e t u r n _ f f t ( x, n, [ exp ( p i j k /n ) f o r k i n r a n g e ( n / ) ) d e f i _ f f t ( x, n ) : """ I n v e r s e f f t o f x, a l i s t o f n = m complex v a l u e s """ t r a n s f o r m = _ f f t ( x, n, [ exp (+ p i j k /n ) f o r k i n r a n g e ( n / ) ) r e t u r n [ x /n f o r x i n t r a n s f o r m d e f _ f f t ( x, n, t w i d d l e ) : """ D e c i m a t i o n i n Time FFT, t o be c a l l e d by d _ f f t and i _ f f t. x i s t h e s i g n a l t o t r a n s f o r m, a l i s t o f c omplex v a l u e s n i s i t s l e n g t h, r e s u l t s a r e u n d e f i n e d i f n i s not a power o f tw i s a l i s t o f t w i d d l e f a c t o r s, precomputed by t h e c a l l e r r e t u r n s a l i s t o f complex v a l u e s, t o be n o r m a l i z e d i n c a s e o f an i n v e r s e t r a n s f o r m """ i f n == : r e t u r n x # bottom r e a c h e d, DFT o f a l e n g t h v e c x i s x # c a l l f f t w i t h t h e e v e n and t h e odd c o e f f i c i e n t s i n x # t h e r e s u l t s a r e t h e so c a l l e d e v e n and odd DFT s y_ = _ f f t ( x [ : :, n /, tw [ : : ) y_ = _ f f t ( x [ : :, n /, tw [ : : ) # a s s e m b l e t h e p a r t i a l r e s u l t s " i n _ p l a c e " : # s t h a l f o f f u l l DFT i s p u t i n e v e n DFT, nd h a l f i n odd DFT f o r k i n r a n g e ( n / ) : y_ [ k, y_ [ k = y_ [ k + tw [ k y_ [ k, y_ [ k tw [ k y_ [ k Fourier # c o n c a t e n a t e t h e two h a l v e s o f t h e DFT and r e t u r n t o c a l l e r r e t u r n y_+y_ Dynamic () d e f main ( ) : """ Run some t e s t c a s e s """ from cmath i m p o r t cos, sin, p i d e f t e s t i t ( t i t l e, s e q ) : """ u t i l i t y t o f o r m a t and p r i n t a v e c t o r and t h e i f f t o f i t s f f t """ l _ s e q = l e n ( s e q ) Fourier p r i n t " " 5, t i t l e, " " 5 p r i n t "\n". j o i n ( [ " %.6 f : : %.6 f, %.6 f j " % ( a. r e a l, t. r e a l, t. imag ) f o r ) ( a, t ) i n z i p ( seq, i _ f f t ( d _ f f t ( seq, l _ s e q ), l _ s e q ) ) l e n g t h = 3 t e s t i t ( " S q u a r e wave ", [ j ( l e n g t h /) + [.+. j ( l e n g t h / ) ) t e s t i t ( " S i n e wave ", [ s i n ( ( p i k )/ l e n g t h ) f o r k i n r a n g e ( l e n g t h ) ) t e s t i t ( " C o s i n e wave ", [ c o s ( ( p i k )/ l e n g t h ) f o r k i n r a n g e ( l e n g t h ) ) i f name == " main " : main ( ) To evaluate the dynamic response of a linear SDOF system in the frequency domain, use the inverse DFT, N π rs x s = V r exp(i ), s =,,..., N N r= where V r = H r P r. P r are the discrete complex amplitude coefficients computed using the direct DFT, and H r is the discretization of the complex frequency response function, that for viscous damping is H r = k [ ( β r ) + i(ζβ r ) = k while for hysteretic damping is H r = [ k ( βr = ) + i(ζ) k [ ( β r ) i(ζβ r ) ( β r ) + (ζβ r ) [ ( β r ) i(ζ) ( βr. ) + (ζ), β r = ω r ω n. Fourier

8 Some words of caution Loading If you re going to approach the application of the complex frequency response function without proper concern, you re likely to be hurt. Let s say ω =., N = 3, ω n = 3.5 and r = 3, what do you think it is the value of β 3? If you are thinking β 3 = 3 ω/ω n = 3/3.5 { 8.57 you re wrong! r ω r N/ Due to aliasing, ω r = (r N) ω r > N/, note that in the upper part of the DFT the coefficients correspond to negative frequencies and, staying within our example, it is β 3 = (3 3) / If N is even, P N/ is the coefficient corresponding to the Nyquist frequency, if N is odd P N corresponds to the largest positive frequency, while P N+ corresponds to the largest negative frequency. Fourier infinitesimal infinitesimal a short duration load a short duration load An approximate procedure to evaluate the maximum displacement for a short loading is based on the -momentum relationship, m ẋ = [p(t) kx(t) dt. When one notes that, for small t, the displacement is of the order of t while the velocity is in the order of t, it is apparent that the kx term may be dropped from the above expression, i.e., infinitesimal Using the previous approximation, the velocity at time t is ẋ(t ) = m p(t) dt, and considering again a negligibly small displacement at the end of the loading, x(t ), one has x(t t ) p(t) dt sin ω n (t t ). infinitesimal m ẋ p(t) dt. Please note that the above equation is exact for an infinitesimal loading.

9 Undamped SDOF Damped SDOF For an infinitesimal, the -momentum is exactly p(τ) dτ and the response is dx(t τ) = p(τ) dτ sin ω n (t τ), t > τ, and to evaluate the response at time t one has simply to sum all the infinitesimal contributions for τ < t, x(t) = p(τ) sin ω n (t τ) dτ, t >. This relation is known as the, and tacitly depends on initial rest conditions for the system. infinitesimal The derivation of the equation of motion for a generic load is analogous to what we have seen for undamped SDOF, the infinitesimal contribution to the response at time t of the load at time τ is dx(t) = p(τ) mω D dτ sin ω D (t τ) exp( ζω n (t τ)) and integrating all infinitesimal contributions one has t τ x(t) = p(τ) sin ω D (t τ) exp( ζω n (t τ)) dτ, t. mω D infinitesimal Evaluation of, undamped Numerical evaluation of, undamped Using the trig identity sin(ω n t ω n τ) = sin ω n t cos ω n τ cos ω n t sin ω n τ the is rewritten as x(t) = p(τ) cos ω nτ dτ sin ω n t p(τ) sin ω nτ dτ = A(t) sin ω n t B(t) cos ω n t cos ω n t infinitesimal Usual numerical procedures can be applied to the evaluation of A and B, e.g., using the trapezoidal rule, one can have, with A N = A(N τ) and y N = p(n τ) cos(n τ) A N+ = A N + τ (y N + y N+ ). infinitesimal where { A(t) = p(τ) cos ω nτ dτ B(t) = p(τ) sin ω nτ dτ

10 Evaluation of, damped Numerical evaluation of, damped For a damped system, it can be shown that with x(t) = A(t) sin ω D t B(t) cos ω D t A(t) = mω D B(t) = mω D p(τ) exp ζω nτ exp ζω n t cos ω Dτ dτ, p(τ) exp ζω nτ exp ζω n t sin ω Dτ dτ. infinitesimal Numerically, using e.g. Simpson integration rule and y N = p(n τ) cos ω D τ, A N+ = A N exp( ζω n τ)+ τ 3mω D [y N exp( ζω n τ) + 4y N+ exp( ζω n τ) + y N+ N =,, 4, infinitesimal Transfer Functions Transfer Functions The response of a linear SDOF system to arbitrary loading can be evaluated by a convolution integral in the time domain, x(t) = p(τ) h(t τ) dτ, with the unit response function h(t) = mω D exp( ζω n t) sin(ω D t), or through the frequency domain using the Fourier integral x(t) = H(ω)P(ω) exp(i ωt) dω, infinitesimal These response functions, or transfer functions, are connected by the direct and inverse Fourier transforms: H(ω) = h(t) = π h(t) exp( i ωt) dt, H(ω) exp(iωt) dω. infinitesimal where H(ω) is the complex frequency response function.

11 Relationship of transfer functions Relationship of transfer functions We write the response and its Fourier transform: x(t) = X (ω) = p(τ)h(t τ) dτ = [ p(τ)h(t τ) dτ p(τ)h(t τ) dτ exp( i ωt) dt the lower limit of integration in the first equation was changed from to because p(τ) = for τ <, and since h(t τ) = for τ > t, the upper limit of the second integral in the second equation can be changed from t to +, +s X (ω) = lim s s +s s p(τ)h(t τ) exp( iωt) dt dτ infinitesimal Introducing a new variable θ = t τ we have +s +s τ X (ω) = lim p(τ) exp( iωτ) dτ h(θ) exp( iωθ) dθ s s s τ with lim s s τ =, we finally have X (ω) = = P(ω) p(τ) exp( iωτ) dτ h(θ) exp( iωθ) dθ h(θ) exp( iωθ) dθ where we have recognized that the first integral is the Fourier transform of p(t). infinitesimal Relationship of transfer functions Our last relation was X (ω) = P(ω) h(θ) exp( iωθ) dθ but X (ω) = H(ω)P(ω), so that, noting that in the above equation the last integral is just the Fourier transform of h(θ), we may conclude that, effectively, H(ω) and h(t) form a Fourier transform pair. infinitesimal