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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 degreeoffreedom 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 steadystate 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 degreeoffreedom system, the predicted steadystate 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 degreeoffreedom 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.
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