ME 563
Mechanical Vibrations
Lecture #17
Frequency Response Functions
1
Sinusoidal Response
When we considered a co-sinusoidal input force,
we obtained the following amplitude/phase information:
xp(t)=Xpcos(ωt+φp )
We can also use Laplace transforms to find these amplitude
and phase relationships.
2
Transfer Functions
When we considered a co-sinusoidal input force,
we obtained the following amplitude/phase information:
3
Frequency Response Functions
To obtain the frequency response function, the transfer
function is evaluated at s=jω
Then the amplitude and phase are found by calculating the
modulus and argument of this complex function of ω:
4
Sinusoidal Solution
So in conclusion we can state the following:
In words, a simple-harmonic sinusoidal excitation force
applied to a linear time-invariant system produces a simpleharmonic sinusoidal response in the steady state at a different
amplitude and phase than the excitation force.
Keywords: linear, time-invariant, sinusoidal, steady state,
amplitude, phase
5
Example (Accelerometer)
6
Example (Accelerometer)
where z(t)=xo(t)-xb(t)
H(ω ) =
where H (ω ) =
Z (ω )
−M
=
−ω 2 X b (ω ) K − Mω 2 + jωC
M
[K − Mω
2 2
]
and ∠H (ω ) = −180 o − tan−1
€
+ [ωC ]
2
=
1/ω n2
2
   2 2 

ω
ω
1−    + 2ς 
  ω n    ω n 
ωC
o
−1
=
−180
−
tan
K − Mω 2
2ς
ω
ωn
 ω 2
1−  
ωn 
7
Example (Accelerometer)
Amplitude distortion
Phase distortion
8
Example (Exhaust system)
9
Frequency Response Functions
How do we get the velocity frequency response function or the
acceleration frequency response function given?
Thinking back to the Laplace transform, we simply multiply
by jω: