# 0 2 om ega rads ec 4 6 8 10 note that in addition to

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online