Complete Question
A sensor output was acquired for 32 seconds at a rate of 200 Hz and spectral analysis was performed using FFT. If the data set was split into 5 segments (each 6.4 seconds long), what is the resulting:
a)F minimum
b) F maximum
c) Frequency resolution
Answer:
a) [tex]Fmax=100Hz[/tex]
b) [tex]Fmin=0Hz[/tex]
c) [tex]F_r=0.0313[/tex]
Step-by-step explanation:
From the question we are told that:
Time [tex]t=32sec[/tex]
Frequency[tex]F=200Hz[/tex]
Segments [tex]\mu=5[/tex]
Generally the equation for Frequency Range is mathematically given by
[tex]Fmax=\frac{F}{2}[/tex]
[tex]Fmax=100Hz[/tex]
Therefore
a) [tex]Fmax=100Hz[/tex]
b) [tex]Fmin=0Hz[/tex]
c)
Generally the equation for Frequency Resolution is mathematically given by
[tex]F_r=\frac{F}{N}[/tex]
Where
N=The Total dat points
N=Sampling Frequency *Time
[tex]N=200*32[/tex]
[tex]N=6400[/tex]
Therefore
[tex]F_r=\frac{200}{6400}[/tex]
[tex]F_r=0.0313[/tex]
