Ch. 3: Forced Vibration of 1-dof system
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for the response x(t) by means of Fourier analysis. 3.4 Periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System ( )
) ( )
( ) ( )
( ) 0 0 2 1 1 2 1 2 2 0 0 FBD and assume
, ,
2 Write
in the Fourier series expansion 2 , , , 0
1 1
x n n in t n n T in t n y x F ma mx k y x k x cx k k c mx cx k k x k y m m y t A y t C e y t B t t T T T C y t e dt Be T T ω ω ω ζ ω π ω ∞ =−∞ − − > ⎡ ⎤ = = − − − ⎣ ⎦ + + + + = = = = = = +
≤ ≤ = = ∑ ∑ ∫ ( ) 0 0 0 0 2 2 2 2 2 0 0 0 1 Integration formula: and
1 2 1 , 0 2 2 2 2 T T in t in t ax ax ax ax T T in t in t T T n A dt te dt T T e e e dx c xe dx ax c a a B e A e iA C in t n T T T n in in T T A C B ω ω π π π π π π − − − + = + = − +
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎛ ⎞ ⎢ ⎥ = + − − = ≠ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎛ ⎞ ⎢ ⎥ − − ⎢ ⎥ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ ⎣ ⎦ = + ∫ ∫ ∫ ∫
Ch. 3: Forced Vibration of 1-DOF System 3.4 Periodic Excitation ( ) (
( ) ( )
( ) ( ) ( ) ( ) 0 0 1 0 1 2 1 2 2 0 0 1 2 2 2 0 0 1 2 2 Re cos sin
2 2 1 sin 2 1 Frequency response 1 2 1 1 2 1 1 2 n n n n n n n n n A iA y t B n t i n t n A A y t B n t n H i k k r i r H i n n k k i H n n k k ω ω π ω π ω ζ ω ω ω ζ ω ω ω ω ζ ω ω ∞ = ∞ = ⎡ ⎤ ∴ = + + + ⎢ ⎥ ⎣ ⎦ = + − = ⎡ ⎤ + − +
⎣ ⎦ = ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥ + − + ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ = ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ + − + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∑ ∑ ( )
( ) 0 1 2 2 0 2 2 0 1 1 2 2 , tan 1 1 sin 2
n n ss n n n n H n k k A A x t B H n t H k k n ω ζ ω ω ω ω π − ∞ = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − ⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ = + − + ⎜ ⎟ + ⎝ ⎠ ∑
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. 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 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 is very large in the vicinity of t=a. The impulse of magnitude , which may represent a large force acting over a short period, can be written as 3.5 Non-periodic Excitation ( )
) 0 for
1 t a t a t a dt δ δ ∞ −∞ − = ≠ − = ∫ ( ) t a δ − ˆ F ( )
( ) ˆ F t F t a δ = − 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 by , and integrating w.r.t. time: 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 ( )
f t ( ) t a δ − ( ) ( ) ( ) ( ) ( )
f t t a dt f a t a dt f a δ δ ∞ ∞ −∞ −∞ − = − = ∫ ∫ 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
subject to i.c. 3.5 Non-periodic Excitation ( ) ( )
( ) ( )
mh t ch t kh t t δ + + = ( ) ( ) 0 0, 0 0
h = = Ch. 3: Forced Vibration of 1-DOF System 3.5 Non-periodic Excitation ( ) (
( ) 0 0 0 Get rid of the impulse function by integrating over the duration 0, of the impulse 1 Take limit as 0 and apply the i.c. to evaluate the integral on the left hand side: lim
ε ε ε ε δ ε → + + = = → ∫ ∫ ( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
0 0 0 0 0 0 0 0 0 0 0 lim 0 0, assuming is not continuous lim
lim 0 0, assuming is continuous lim
lim 0 0, assuming is continuous 0 1 h t dt mh t mh h t ch t dt ch t ch h t kh t dt gh t h t mh ε ε ε ε ε ε ε ε ε ε ε + → + → → → → + = = ≠ = = = = = ∴ = ∫ ∫ ∫ 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) Now, we are ready to find the impulse response. The equivalent system is a homogeneous system with i.c. If the system is underdamped, the impulse response is Note that the above i.c. is not the actual i.c. 3.5 Non-periodic Excitation ( ) 0
h m + = ( ) ( )
0 0,
0 1 /
h h m = = ( ) 1 sin , 0 0, 0
n t d d e t t m h t t ζω ω ω − ⎧ ≥ ⎪ = ⎨ ⎪ < ⎩
Ch. 3: Forced Vibration of 1-DOF System 3.5 Non-periodic Excitation Impulse response of underdamped system
Ch. 3: Forced Vibration of 1-DOF System ( )
x t 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 3.5 Non-periodic Excitation LTI system ( )
LTI system ( )
a − ( ) x t a − Hence if the impulse is applied at t=t o , the response is ( ) (
( ) 0 0 0 0 1 sin
, 0,
n t t d d e t t t t m h t t t ζω ω ω − − ⎧ − ≥ ⎪ = ⎨
⎪ < ⎩
Ch. 3: Forced Vibration of 1-DOF System Total response of underdamped MBK with i.c. x(a)=x 0 and v(a)=v 0 subject to the impulse force 3.5 Non-periodic Excitation ( ) ˆ F t a δ − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) 1 2 1 2 1 2 ˆ
sin cos
sin ˆ
sin cos
, ˆ sin cos
n n n n h p t a t a d d d d t a d d d t a n d d d x t x x F e A t a A t a e t a m F e A t a A t a t a m F x t e A t a A t a m ζω ζω ζω ζω ω ω ω ω ω ω ω ζω ω ω ω − − − − − − − − = + = − + − + − ⎧ ⎫ ⎛ ⎞ ⎪ ⎪ = + − + − ≥ ⎜ ⎟ ⎨ ⎬ ⎪ ⎪ ⎝ ⎠ ⎩ ⎭ ⎧ ⎫ ⎛ ⎞ ⎪ ⎪ = − + − + − ⎜ ⎟ ⎨ ⎬ ⎪ ⎪ ⎝ ⎠ ⎩ ⎭ + ( ) ( ) ( ) 1 2 ˆ cos
sin n t a d d d d d F e A t a A t a m ζω ω ω ω ω ω − − ⎧ ⎫ ⎛ ⎞ ⎪ ⎪ + − − − ⎜ ⎟ ⎨ ⎬ ⎪ ⎪ ⎝ ⎠ ⎩ ⎭
Ch. 3: Forced Vibration of 1-DOF System Total response of underdamped MBK with i.c. x(a)=x 0 and v(a)=v 0 subject to the impulse force 3.5 Non-periodic Excitation ( ) ˆ F t a δ − ( ) ( )
( ) ( ) ( ) ( ) ( ) 0 0 1 2 0 2 0 2 1 2 0 1 0 0 0 0 0 Apply i.c. and to solve for and : ˆ
1 and
1 sin
cos ,
n n d d n d t a n d d d x a x x a v A A x A F v A A m F A x A x v m Download 1,63 Mb. Do'stlaringiz bilan baham:
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