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Ch. 3: Forced Vibration of 1-DOF System 3.0 Outline Harmonic Excitation Frequency Response Function Applications Periodic Excitation Non-periodic Excitation 3.0 Outline Ch. 3: Forced Vibration of 1-DOF System 3.1 Harmonic Excitation Force input function of the harmonic excitation is the harmonic function, i.e. functions of sines and cosines. This type of excitation is common to many system involving rotating and reciprocating motion. Moreover, many other forces can be represented as an infinite series of harmonic functions. By the principle of superposition, the response is the sum of the individual harmonic response. It is more convenient to use the frequency domain technique in solving the harmonic excitation problems. This is because the response to different excitation frequencies can be seen in one graph. 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System Let us focus on the particular solution of mx + cx + kx = F0 cos ωt normalize the equation of motion x + 2ζωn x + ωn2 x = f 0 cos ωt , f 0 = F0 / m f ( t ) = f 0 Re ⎡⎣eiωt ⎤⎦ ∴ solve for z ( t ) from z + 2ζωn z + ωn2 z = f 0 eiωt and the solution is the real part of z ( t ) ; x ( t ) = Re ⎡⎣ z ( t ) ⎤⎦ Assume the solution to have the same form as the forcing function z ( t ) = Z ( iω ) eiωt ( same frequency as the input w/ different mag. and phase ) ( −ω 2 + i 2ζωωn + ωn2 ) Z ( iω ) eiωt = f 0 eiωt f 0 / ωn2 f0 = Z ( iω ) = 2 2 ωn − ω + i 2ζωωn 1 − (ω / ωn )2 + i 2ζω / ωn = F0 2 k ⎡1 − (ω / ωn ) + i 2ζω / ωn ⎤ ⎣ ⎦ 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System z (t ) = F0 iωt iωt = ω e H i F e ( ) 0 k ⎡⎣1 − r 2 + i 2ζ r ⎤⎦ ⎡ ⎤ F0 iωt ∴ x ( t ) = Re ⎢ e ⎥ , r = ω / ωn 2 ⎢⎣ k ⎡⎣1 − r + i 2ζ r ⎤⎦ ⎥⎦ 1 iθ = If H ( iω ) = is the frequency response ω H i e ( ) 2 k ⎡⎣1 − r + i 2ζ r ⎤⎦ ∴ x ( t ) = F0 H ( iω ) cos (ωt + θ ) 1 where H ( iω ) = k (1 − r ) + ( 2ζ r ) 2 2 = magnitude 2 −2ζ r = phase 2 1− r The system modulates the harmonic input by θ = tan −1 the magnitude H ( iω ) and phase H ( iω ) 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System total response = homogeneous soln. + particular soln. Recall the homogeneous solution of the underdamped system xh = Ce −ζωnt cos (ωd t − φ ) or xh = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) ∴ x ( t ) = Ce −ζωnt cos (ωd t − φ ) + F0 H ( iω ) cos (ωt + θ ) or x ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + F0 H ( iω ) cos (ωt + θ ) The initial conditions will be used to determine C , φ or A1 , A2 They will be different from those of free response because the transient term now is partly due to the excitation force and partly due to the initial conditions 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. 1 Compute and plot the response of a spring-mass system to a force of magnitude 23 N, driving frequency of twice the natural frequency and i.c. given by x0 = 0 m and v0 = 0.2 m/s. The mass of the system is 10 kg and the spring stiffness is 1000 N/m. 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System ωn = k / m = 1000 /10 = 10 rad/s ζ = c / ( 2mωn ) = 0 ω = 2 ×10 = 20 rad/s H ( iω ) = 1 1 −3 0.333 10 = = − × k ⎣⎡1 − r 2 + i 2ζ r ⎤⎦ 1000 × (1 − 22 ) x ( t ) = A1 sin ωnt + A2 cos ωnt − 23 × 0.333 ×10−3 cos ωt x ( t ) = ωn A1 cos ωnt − ωn A2 sin ωnt + ω × 23 × 0.333 × 10−3 sin ωt i.c. x ( 0 ) = 0 = A2 − 23 × 0.333 ×10−3 , A2 = 7.667 ×10−3 x ( 0 ) = 0.2 = 10 × A1 , A1 = 0.02 ∴ x ( t ) = 0.02sin10t + 7.667 ×10−3 ( cos10t − cos 20t ) m 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. 2 Find the total response of a SDOF system with m = 10 kg, c = 20 Ns/m, k = 4000 N/m, x0 = 0.01 m, v0 = 0 m/s under an external force F(t) = 100cos10t. 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System ωn = k / m = 20 rad/s ζ = c / ( 2mωn ) = 0.05 r = ω / ωn = 0.5 H ( iω ) = 1 1 = = 332.6 E − 6 − 0.0666 2 2 k ⎣⎡1 − r + i 2ζ r ⎦⎤ 4000 × (1 − 0.5 + i 2 × 0.05 × 0.5 ) x p ( t ) = F0 H ( iω ) cos (ωt + θ ) = 33.26 E − 3cos (10t − 0.0666 ) xh ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) , ωd = ωn 1 − ζ 2 = 19.975 rad/s x ( t ) = e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + F0 H ( iω ) cos (ωt + θ ) x ( t ) = −ζωn e −ζωnt ( A1 sin ωd t + A2 cos ωd t ) + e −ζωnt (ωd A1 cos ωd t − ωd A2 sin ωd t ) − ω F0 H ( iω ) sin (ωt + θ ) 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System ωresponse finally becomes ω, and in phase 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System ωresponse finally becomes ω, and out of phase 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System F0ωnt/(2k) In case of ζ = 0 and ω = ωn , the guess solution of the form x ( t ) = X ( iω ) eiωt = A cos ωt + B sin ωt is invalid. This is because it has the same form as the homogeneous solution. Fω t The correct particular solution is x p ( t ) = 0 n sin ωnt. 2k 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System Beat when the driving frequency is close to natural freq. The total solution can be arranged in the form x (t ) = v0 ωn sin ωnt + x0 cos ωnt + f0 ( cos ωt − cos ωnt ) ωn2 − ω 2 x02ωn2 + v02 ⎛ xω ⎞ 2f ⎛ ω − ω ⎞ ⎛ ωn + ω ⎞ t ⎟ sin ⎜ t⎟ sin ⎜ ωn t + tan −1 0 n ⎟ + 2 0 2 sin ⎜ n v0 ⎠ ωn − ω ωn ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ If the system is at rest in the beginning, = x (t ) = 2 f0 ⎛ ω n − ω ⎞ ⎛ ωn + ω ⎞ t ⎟ sin ⎜ t⎟ sin ⎜ 2 2 ωn − ω ⎝ 2 ⎠ ⎝ 2 ⎠ The response oscillates with frequency ωn + ω inside 2 2f ⎛ ω −ω ⎞ t⎟ the slowly oscillated envelope 2 0 2 sin ⎜ n ωn − ω ⎝ 2 ⎠ ∴ The beat frequency is ωn − ω 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System Beat when the driving frequency is close to natural freq. 3.1 Harmonic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.2 Frequency Response Function The core of the particular solution to the harmonic function is H ( iω ) = 1 ; frequency response function 2 k (1 − r + i 2ζ r ) It specifies how the system responds to harmonic excitation. As a standard, we normalize the frequency response function 1 and then study how it varies as the 2 1 − r + i 2ζ r excitation frequency ω and system parameters ζ ,ωn vary. G ( iω ) = It is indeed more convenient since we already normalized the frequency; r = ω / ωn . So we can now study its variation to ζ and r. For the fixed damping ratio, we plot G ( iω ) with r varies. G ( iω ) has both magnitude and phase ⇒ magnitude and phase plot. Then we repeatedly evaluate G ( iω ) by varying ζ . 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System H ( iω ) = 1 (1 − r ) 2 2 + ( 2ζ r ) 2 Frequency response plot (Bode diagram) ⎛ −2ζ r ⎞ 2 ⎟ 1 r − ⎝ ⎠ θ = tan −1 ⎜ 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System Resonance is defined to be the vibration response at ω=ωn, regardless whether the damping ratio is zero. At this point, the phase shift of the response is –π/2. The resonant frequency will give the peak amplitude for the response only when ζ=0. For0 < ζ < 1/ 2 ,the peak amplitude will be at ω = ωn 1 − 2ζ 2 , slightly before ωn. For ζ ≥ 1/ 2 , there is no peak but the max. value of the output is equal to the input for the dc signal (of course, for this normalized transfer function). 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System Ex. 3 Consider the pivoted mechanism with k=4x103 N/m, l1=0.05 m, l2=0.07 m, l=0.10 m, and m=40 kg. The mass of the beam is 40kg which is pivoted at point O and assumed to be rigid. Calculate c so that the damping ratio of the system is 0.2. Also determine the amplitude of vibration of the steady-state response if a 10 N force is applied to the mass at a frequency of 10 rad/s. 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System l − l1 ⎞ ⎡⎣ ∑ M O = I Oθ ⎤⎦ Fl − mglθ − Mg ⎛⎜ ⎟ θ − c l2θ l2 − k ( l1θ ) l1 2 ⎝ ⎠ ( ) 2 2 ⎡ 2 l + l1 ) ( ⎛ l − l1 ⎞ ⎤ = ⎢ ml + M +M⎜ ⎟ ⎥θ 12 ⎝ 2 ⎠ ⎥⎦ ⎢⎣ 0.1×10 cos10t = 0.5θ + 0.0049cθ + 59.05θ ωn = 15.37, ζ = 0.2 = 0.0049c , c = 627.3 Ns/m 2 × 0.5 × ωn ω = 10, r = 0.6506 H ( iω ) = 1 = 0.02677 − 24.268° 59.05 ( 0.5767 + i 0.26 ) θ ss = 0.02677 cos (10t − 0.424 ) 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System Ex. 4 A foot pedal for a musical instrument is modeled as in the figure. With k=2000 kg/s2, c=25 kg/s, m=25 kg, and F(t)=50cos2πt N, compute the steady-state response assuming the system starts from rest. Use the small angle approximation. 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System ( ) ⎡⎣ ∑ M O = I Oθ ⎤⎦ F × 0.15 − k ( 0.05θ ) × 0.05 − c 0.05θ × 0.1 = m × 0.152 θ 5 100 θ = 50 cos 2π t , positive CW 3.75θ + θ + 6 3 Find the parameters ωn = 2.98, ζ = 0.0373, ω = 2π , r = 2.108 H ( iω ) = 1 = 0.0087 − 177.4° k (1 − r 2 + i 2ζ r ) since ζ ≠ 0, the transient response will die out θ ss = F0 H ( iω ) cos (ωt + θ ) = 0.435cos ( 2π t − 3.096 ) 3.2 Frequency Response Function Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 2 measured acc. ωn z 10 = = = y actual acc. 9.81 (1 − r ) + ( 2ζ r ) 2 2 2 1 (1 − r 2 ) + ( 2ζ r ) 2 2 = 0.962 From the problem statement, ω = 628 rad/s, ωd = ωn 1 − ζ 2 = 628 rad/s r ω = =1 2 ωd 1− ζ ωn = ωd 1− ζ 2 = 758 rad/s = k c , ζ = 0.56 = m 2mωn k = 5745.6 N/m, c = 8.49 Ns/m 3.3 Applications Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation A periodic function is any function that repeats itself in time, called period T. f (t ) = f (t + T ) It is more general than the harmonic function. Here, we will find the response to the input that is a periodic function. The idea is to decompose that periodic input into the sum of many harmonics. The response, by the superposition principle of linear system, is then the sum of the responses of individual harmonic. The response of a harmonic function was studied in section 3.1 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Fourier found the way to decompose the periodic function into sum of harmonic functions (sine & cosine) whose frequencies are multiples of the fundamental frequency. The fundamental frequency is the frequency of the periodic function. 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Fourier series Fourier series in real form: a0 ∞ 2π f ( t ) = + ∑ ( an cos nω0t + bn sin nω0t ), ω0 = 2 n =1 T Fourier coefficients: T 2 an = ∫ f ( t ) cos nω0t dt , n = 0,1, 2,… T 0 T 2 bn = ∫ f ( t ) sin nω0t dt , n = 1, 2,3,… T 0 Fourier series in complex form: 2π T n =−∞ Fourier cofficients (complex): f (t ) = ∞ ∑ Cneinω0t , ω0 = T 1 Cn = ∫ f ( t ) e − inω0t dt , n = … , −2, −1, 0,1, 2,… T 0 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Some properties of Fourier series 1) If f ( t ) is an even function, bn = 0. 2 ) If f ( t ) is an odd function, an = 0. a0 is the average value of f ( t ) over one period. 2 ⎛ ∞ ⎞ 4 ) If f ( t ) is real, Ck = C− k ⇒ f ( t ) = C0 + 2 Re ⎜ ∑ Cn einω0t ⎟ ⎝ n =1 ⎠ 3) 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Frequency spectrum tells how much each harmonic contributes to the periodic function f ( t ) . In real form, the harmonic at nω0 has the amplitude an2 + bn2 In complex form, the harmonic at nω0 has the amplitude 2 Re ( Cn ) Plot of the amplitude of each harmonic vs. its frequency is the (discrete) frequency spectrum. 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Superposition principle of linear system 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Response to harmonic excitation 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System mx + cx + kx = F ( t ) = Cn einω0t From section 3.1, xss = Cn H ( inω0 ) einω0t where H ( inω0 ) = 1 ⎛ ⎛ nω ⎞2 k ⎜1 − ⎜ 0 ⎟ + i 2ζ ⎜ ⎝ ωn ⎠ ⎝ ⎛ nω0 ⎞ ⎞ ⎜ ⎟ ⎟⎟ ⎝ ωn ⎠ ⎠ ⎛ ∞ ⎞ mx + cx + kx = F ( t ) = C0 + 2 Re ⎜ ∑ Cn einω0t ⎟ ⎝ n =1 ⎠ C0 ⎞ ⎛ ∞ by superposition, xss = + 2 Re ⎜ ∑ Cn H ( inω0 ) einω0t ⎟ k ⎠ ⎝ n =1 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System response frequency spectrum excitation frequency spectrum system frequency response 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. Calculate the response of a damped system to the periodic excitation f(t) depicted in the figure by means of the exponential form of the Fourier series. The system damping ratio is 0.1 and the driving frequency is ¼ of the system natural freq. 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Expand f ( t ) as sum of harmonic series f (t ) = ∞ ∑Ce ω in n =−∞ 0t n T T /2 T ⎤ 1 1⎡ 2π − inω0t − inω0t Cn = ∫ f ( t ) e dt = ⎢ ∫ Ae dt + ∫ − Ae − inω0t dt ⎥ , ω0 = T 0 T⎣0 T T /2 ⎦ ⎧ 0, n = even iA ⎡ n ⎪ Cn = 1 − ( −1) ⎤ = ⎨ i 2 A ⎦ nπ ⎣ ⎪⎩− nπ , n = odd ⎛ π⎞ i 2 A inω0t 2 A i⎜⎝ nω0t − 2 ⎟⎠ 4 A ∞ 1 ∴ f ( t ) = ∑ n =odd − ×e = ∑ n =odd ×e = ∑ sin nω0t nπ nπ π n =1,3,… n Gn ( iω ) = Gn ( iω ) = Gn = ω nω0 n 1 r ζ , 0.1, = = = = ωn ω n 4 1 − r 2 + i 2ζ r 1 1 − ( n / 4 ) + i 0.05n 2 1 2 , Gn = tan −1 ⎡1 − ( 0.25n )2 ⎤ + ( 0.05n )2 ⎣ ⎦ 4A ∞ 1 ∴ xss ( t ) = ∑ G sin ( nω0t + Gn ) π n =1,3,… n n −0.05n 1 − ( 0.25n ) 2 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. The cam and follower impart a displacement y(t) in the form of a periodic sawtooth function to the lower end of the system. Derive an expression for the response x(t) by means of Fourier analysis. 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System FBD and assume y > x ⎡⎣ ∑ Fx = max ⎤⎦ mx = k2 ( y − x ) − k1 x − cx mx + cx + ( k1 + k2 ) x = k2 y, ωn = k1 + k2 c , ζ = m 2mωn Write y ( t ) in the Fourier series expansion y (t ) = ∞ ∑Ce ω in n =−∞ n 0t , ω0 = 2π A , y (t ) = B + t, 0 ≤ t ≤ T T T T T T 1 1 1 A Cn = ∫ y ( t ) e − inω0t dt = ∫ Be − inω0t dt + ∫ te − inω0t dt T 0 T 0 T 0T e ax Integration formula: ∫ e dx = + c and a ax e ax ∫ xe dx = a 2 ( ax − 1) + c ax T ⎡ ⎤ 2π ⎡ − in 2π t ⎤ in t − ⎢ ⎥ 2π B⎢e T ⎥ A⎢ e T iA ⎛ ⎞⎥ 1 , n≠0 Cn = ⎢ in t + − − = ⎥ ⎟⎥ 2 ⎜ 2 π T ⎢ −in 2π ⎥ T 2 ⎢ ⎛ T n ⎠ 2π ⎞ ⎝ − in ⎢ ⎥ ⎜ ⎟ ⎣ T ⎦0 T ⎠ ⎣⎝ ⎦0 A C0 = B + 2 T 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System A ⎡ ∞ iA ⎤ ∴ y ( t ) = B + + 2 Re ⎢ ∑ ( cos nω0t + i sin nω0t )⎥ 2 ⎣ n =1 2π n ⎦ A A ∞ 1 y ( t ) = B + − ∑ sin nω0t 2 π n =1 n Frequency response H n ( iω ) = H n ( iω ) = Hn = 1 ( k1 + k2 ) ⎡⎣1 − r 2 + i 2ζ r ⎤⎦ 1 ⎡ ⎛ nω ⎞ 2 ⎛ nω0 ⎞ ⎤ 0 ( k1 + k2 ) ⎢1 − ⎜ ⎟ + i 2ζ ⎜ ⎟ ⎥ ⎢⎣ ⎝ ωn ⎠ ⎝ ωn ⎠ ⎥⎦ 1 2 , ⎛ nω ⎞ −2ζ ⎜ 0 ⎟ ⎝ ωn ⎠ H n = tan −1 2 ⎛ nω0 ⎞ 1− ⎜ ⎟ ⎝ ωn ⎠ ⎛ ⎛ nω ⎞ ⎞ ⎛ ⎛ nω ⎞ ⎞ ( k1 + k2 ) ⎜⎜1 − ⎜ 0 ⎟ ⎟⎟ + ⎜ 2ζ ⎜ 0 ⎟ ⎟ ω ω ⎝ ⎝ n ⎠ ⎠ ⎝ ⎝ n ⎠⎠ k2 ⎛ A ⎞ k2 A ∞ 1 xss ( t ) = ∑ H n sin ( nω0t + H n ) ⎜B+ ⎟− k1 + k2 ⎝ 2 ⎠ π n =1 n 2 2 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.5 Non-periodic Excitation Harmonic and steady-state excitation and response are conveniently described in the frequency domain. For deterministic non-periodic excitation and response, time domain technique is more suitable. We cannot find the repeated pattern that lasts forever (both in the past & future) for the non-periodic excitation. System response to the unit impulse, called the impulse response, will be first studied. Then, this fundamental response will be used to synthesize the response of the LTI system to arbitrary excitation. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Impulse The unit impulse, or Dirac delta function, is defined as δ ( t − a ) = 0 for t ≠ a ∞ ∫ δ ( t − a ) dt = 1 −∞ This means that the unit impulse is zero everywhere except in the neighborhood of t=a. Since the area under the graph δ-t is 1, the value of δ ( t − a ) is very large in the vicinity of t=a. The impulse of magnitudeF̂ , which may represent a large force acting over a short period, can be written as F ( t ) = Fˆ δ ( t − a ) 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System The unit impulse has a useful property called the “sampling property”. Multiplying a continuous function f ( t ) by δ ( t − a ) , and integrating w.r.t. time: ∞ ∞ −∞ −∞ ∫ f ( t ) δ ( t − a ) dt = f ( a ) ∫ δ ( t − a ) dt = f ( a ) which is just the value of f(t) at t=a. This is a way in evaluating integrals involving with impulse. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Impulse response The impulse response, h(t), is the response to the unit impulse, δ(t), applied at t=0 with zero initial conditions. The impulse response is very important since it contains all the system characteristics and can be used to find the response to arbitrary excitation of LTI system via the convolution integral theorem. The impulse response of a 1 DOF MBK system must satisfy mh ( t ) + ch ( t ) + kh ( t ) = δ ( t ) subject to i.c. h ( 0 ) = 0, h ( 0 ) = 0 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Get rid of the impulse function by integrating over the duration ( 0, ε ) of the impulse ε ε 0 0 ∫ ( mh + ch + kh ) dt = ∫ δ ( t ) dt = 1 Take limit as ε → 0 and apply the i.c. to evaluate the integral on the left hand side: ε lim ∫ mh ( t ) dt = lim mh ( t ) = mh ( 0+ ) ≠ 0, assuming h ( t ) is not continuous ε →0 0 ε ε →0 ε 0 lim ∫ ch ( t ) dt = lim ch ( t ) 0 = ch ( 0+ ) = 0, assuming h ( t ) is continuous ε →0 0 ε ε →0 ε ε lim ∫ kh ( t ) dt = lim gh ( 0 ) t 0 = 0, assuming h ( t ) is continuous ε →0 0 ε →0 ∴ mh ( 0+ ) = 1 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Therefore, the effect of a unit impulse at t=0 is to produce equivalent initial velocity (impulse-momentum) h ( 0+ ) = 1/ m Now, we are ready to find the impulse response. The equivalent system is a homogeneous system with i.c. h ( 0 ) = 0, h ( 0 ) = 1/ m If the system is underdamped, the impulse response is ⎧ 1 −ζωnt e sin ωd t , t ≥ 0 ⎪ h ( t ) = ⎨ mωd ⎪ 0, t < 0 ⎩ Note that the above i.c. is not the actual i.c. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Impulse response of underdamped system 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Linear Time Invariant (LTI) system has the characteristic that the shape of the response will not be influenced by the time the input is applied to the system. That is f (t ) f (t − a ) LTI system LTI system x (t ) x (t − a ) Hence if the impulse is applied at t=to, the response is ⎧ 1 −ζωn ( t −t0 ) e sin ωd ( t − t0 ) , t ≥ t0 ⎪ h ( t ) = ⎨ mωd ⎪ 0, t < t0 ⎩ 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Total response of underdamped MBK with i.c. x(a)=x0 and v(a)=v0 subject to the impulse force F̂δ ( t − a ) x ( t ) = xh + x p =e −ζωn ( t − a ) =e −ζωn ( t − a ) x ( t ) = −ζωn e +e Fˆ −ζωn (t − a ) sin ωd ( t − a ) ( A1 sin ωd ( t − a ) + A2 cos ωd ( t − a ) ) + mω e d ⎧⎪⎛ Fˆ ⎨⎜ A1 + mωd ⎩⎪⎝ −ζωn ( t − a ) −ζωn ( t − a ) ⎫⎪ ⎞ ⎟ sin ωd ( t − a ) + A2 cos ωd ( t − a ) ⎬ , t ≥ a ⎠ ⎭⎪ ⎧⎪⎛ Fˆ ⎨⎜ A1 + mωd ⎪⎩⎝ ⎫⎪ ⎞ ⎟ sin ωd ( t − a ) + A2 cos ωd ( t − a ) ⎬ ⎪⎭ ⎠ ⎧⎪ ⎛ ⎫⎪ Fˆ ⎞ ⎟ cos ωd ( t − a ) − ωd A2 sin ωd ( t − a ) ⎬ ⎨ωd ⎜ A1 + ω m ⎪⎩ ⎝ ⎪⎭ d ⎠ 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Total response of underdamped MBK with i.c. x(a)=x0 and v(a)=v0 subject to the impulse force F̂δ ( t − a ) Apply i.c. x ( a ) = x0 and x ( a ) = v0 to solve for A1 and A2 : x0 = A2 ⎛ Fˆ ⎞ v0 = −ζωn A2 + ωd ⎜ A1 + ⎟ ω m d ⎠ ⎝ 1 ⎛ Fˆ ⎞ ∴ A2 = x0 and A1 = ⎜ ζω x + v − ⎟ ωd ⎝ n 0 0 m ⎠ ⎫ −ζωn ( t − a ) ⎧ 1 ∴ x (t ) = e ⎨ (ζωn x0 + v0 ) sin ωd ( t − a ) + x0 cos ωd ( t − a ) ⎬ , t ≥ a ⎩ ωd ⎭ 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Arbitrary Excitation Ideally, arbitrary excitation can be expressed as linear combinations of simpler excitations. The simpler excitations are simple enough that the response is readily available. This concept is exactly used by Fourier. Now, the idea is to regard the arbitrary excitation as a superposition of impulses of varying magnitude and applied at different times. It is used when the excitation can be easily described in time domain. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Consider the excitation F(t). We can imagine that it is constructed from infinite impulses at different times. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Convolution integral theorm Focus on the time interval τ < t < τ + Δτ , at which the impulse of magnitude F (τ ) Δτ is acting. This shifted impulse can be written as F (τ ) Δτδ ( t − τ ) . The response of the LTI system to this particular impulse is Δx ( t ,τ ) = F (τ ) Δτ h ( t − τ ) Since by sampling property F ( t ) = ∑ F (τ ) Δτδ ( t − τ ), τ and the system is linear, the response to F ( t ) is x ( t ) = ∑ F (τ ) Δτ h ( t − τ ) τ t In the limit as Δτ → 0, x ( t ) = ∫ F (τ ) h ( t − τ ) dτ . 0 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Convolution integral theorm The response of the arbitrary excitation is the superposition of shifted impulse responses. Interpretation for the whole range of time; t 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System To obtain h ( t − τ ) from h (τ ) , we need to carry out two operation; shifting and folding. This is another interpretation for the specific time t. The figures show the steps in evaluating the convolution. If we define a new variable λ = t − τ , then τ = t − λ and dτ = −d λ. With the change of the integration limits, 0 t t 0 x ( t ) = ∫ F ( t − λ ) h ( λ )( −d λ ) = ∫ F ( t − τ ) h (τ ) dτ That is the convolution is symmetric in F ( t ) and h ( t ) . To decide which formula to use depends on the nature of F ( t ) and h ( t ) . It is obvious that if the excitation F ( t ) or the impulse response h ( t ) is too complicated, we may be unable to evaluate the closed form solution of the convolution integral. The excitation may not at all be written as functions of time. In these cases, the integration must be carried out numerically. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Interpretation for the specific time; t 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. Determine the response of the underdamped MBK to the unit step input. 1 0 u(t) t 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System t x ( t ) = ∫ F (τ ) h ( t − τ ) dτ 0 F (τ ) = u (τ ) and h ( t − τ ) is the system impulse response shifted by t and mirrored about the vertical axis. If t < 0, F (τ ) h ( t − τ ) = 0 because of no overlap If t > 0, F (τ ) h ( t − τ ) = h ( t − τ ) ∴ x ( t ) = 0, t < 0 t ∴ x ( t ) = ∫ h ( t − τ ) dτ , t > 0 0 Let t − τ = λ. Hence dτ = − d λ t t 1 −ζωn λ sin ωd λd λ e mωd 0 ∴ x (t ) = ∫ h ( λ ) d λ = ∫ 0 eiωd λ − e − iωd λ e ax ax Substitute sin ωd λ = and use ∫ e dx = +c 2i a ⎛ ⎞⎤ ζω 1⎡ ∴ x ( t ) = ⎢1 − e −ζωnt ⎜ cos ωd t + n sin ωd t ⎟ ⎥ , t > 0 k⎣ ωd ⎝ ⎠⎦ 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System Ex. Find the undamped response for the sinusoidal pulse force shown using zero i.c. 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System t x ( t ) = ∫ F (τ ) h ( t − τ ) dτ 0 ⎛ 2π ⎞ ⎛π ⎞ F (τ ) = F0 sin ⎜ τ ⎟ = F0 sin ⎜ τ ⎟ T 2 ⎝ 0 ⎠ ⎝ T0 ⎠ 1 h (τ ) = sin ωnτ mωn h ( t − τ ) is the system impulse response shifted by t and mirrored about the vertical axis. If t < 0, F (τ ) h ( t − τ ) = 0 because of no overlap ∴ x ( t ) = 0, t < 0 3.5 Non-periodic Excitation Ch. 3: Forced Vibration of 1-DOF System ⎛π ⎞ 1 If 0 < t < T0 , F (τ ) h ( t − τ ) = F0 sin ⎜ τ ⎟ × sin ωn ( t − τ ) ⎝ T0 ⎠ mωn t t ⎛π ⎞ F0 sin x ( t ) = ∫ F (τ ) h ( t − τ ) dτ = ⎜ τ ⎟ sin ωn ( t − τ ) dτ ∫ mωn 0 ⎝ T0 ⎠ 0 using the relation sin α sin β = ∴ x (t ) = 1 ⎡cos (α − β ) − cos (α + β ) ⎤⎦ and some arrangements 2⎣ F0 π ω 2 ω ω t r t t T ω ω − < < = , r = , k = m sin sin , 0 where ( ) n n 0 ωn T0 k (1 − r 2 ) t T0 0 0 If t > T0 , x ( t ) = ∫ F (τ ) h ( t − τ ) dτ = ∫ F (τ ) h ( t − τ ) dτ = ∴ x (t ) = { t ∫ F ( t − τ ) h (τ ) dτ t −T0 } F0 [sin ωt − r sin ωnt ] − ⎡⎣sin ω ( t − T0 ) − r sin ωn ( t − T0 )⎤⎦ , t > T0 2 k (1 − r ) superposition of the out-of-phase shifted sine trains 3.5 Non-periodic Excitation