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Unformatted text preview: for determining the steady state response of a system, its application is F (ω ) X (ω ) extremely simple. It is merely a H (ω ) multiplication in the frequency domain. X (ω ) = H (ω ) × F (ω ) Lecture Notes -37- 06/16/06 12:25 PM Mechanical Vibrations I Complex Exponential vs. Sine/Cosine Relationship
The relationship between the complex exponential form and the sine/cosine form of the can be developed as follows: Starting with the complex exponential form for the forcing function, f (t ) = Fe jωt + F *e − jωt
and recalling the Euler identity, eiθ = cosθ + i sin θ
the relationship is easily developed. (Remember that F is complex.) Expanding the expression for f (t ) ,
f (t ) = ( FR + jFI )(cos ω t + j sin ω t ) + ( FR − jFI )(cos ω t − j sin ω t ) Collecting real and imaginary terms yields,
f (t ) = ( FR cos ω t − FI sin ω t ) + j ( FI cos ω t + FR sin ω t ) + ( FR cos ω t − FI sin ω t ) − j ( FI cos ω t + FR sin ω t ) Observing that the imaginary terms cancel and that the real terms are equivalent yields,
f (t ) = 2 FR cos ω t − 2 FI sin ω t Comparing with a traditional differential equations solution of f (t ) = A cos ω t + B sin ω t shows that A = 2 FR and B = −2 FI . Despite the use of complex exponentials in the development, clearly the resulting time domain waveform is real. P hase [deg] FRF Characteristics
This figure presents a typical single degree-of-freedom frequency response function. The FRF is present two sided (both positive and negative frequency) to clearly show that the FRF for positive and negative frequencies are complex conjugates. m = 5 kg ; c = 20 Ns/m ; k = 1000 N/m 180 90 0 -90 10
-2 M agnitude [m /N] 10 -3 10 -4 10 -5 -20 -15 -10 -5 H (ω ) = H * (−ω ) 0 5 Frequency [Hz ] 10 15 20 Lecture Notes -38- 06/16/06 12:25 PM Mechanical Vibrations I Because of this complex conjugate relationship, usually only the positive frequency part is plotted. Predicted Response*
It is important to remember that although the FRF is defined for all frequencies, the steady-state response (output) for a linear system will always be at the same frequency as the excitation (input). This prediction comes by manipulating the definition of the frequency response function, as shown. X (ω ) = H (ω ) × F (ω )
For a single degree-of-freedom system, the predicted steady-state response has the following form. X (ω ) = F (ω ) − mω + jω c + k
2 Peak Response
Another important aspect of the frequency response function involves the frequency of maximum response. It will be designated by ω p . Starting with the single degree-of-freedom frequency response function, H (ω ) = 1 −mω + jω c + k
2 it is possible to calculate the frequency of maximum (peak) response. Formulating the magnitude of the frequency response yields,
H (ω ) = 1 ( k − mω ) + (ω c )
2 2 2 In order to identify the frequency of maximum response, it is necessary to evaluate the derivative of the frequency response magnitude with res...
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani - Hyderabad.
- Fall '11