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) $Fmax=100Hz$

b) $Fmin=0Hz$

c) $F_r=0.0313$

Step-by-step explanation:

From the question we are told that:

Time $t=32sec$

Frequency $F=200Hz$

Segments $\mu=5$

Generally the equation for Frequency Range is mathematically given by

$Fmax=\frac{F}{2}$

$Fmax=100Hz$

Therefore

a) $Fmax=100Hz$

b) $Fmin=0Hz$

Generally the equation for Frequency Resolution is mathematically given by

$F_r=\frac{F}{N}$

Where

N=The Total dat points

N=Sampling Frequency *Time

$N=200*32$

$N=6400$

Therefore

$F_r=\frac{200}{6400}$

$F_r=0.0313$

6 0

3 years ago

What its the exact and estimate of -3.7 +(-6.8) and i need work

hodyreva [135]

Exact:

-3.7 - 6.8 = -10.5

Estimate:

-4 - 7 = -11

6 0

3 years ago

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