Assume that you have 200 (=N) samples of a real signal x(t) sampled at f_s = 20H

Question

Assume that you have 200 (=N) samples of a real signal x(t) sampled at f_s = 20Hz. Let X(k) denote the DFT of x(n). a) Determine the total duration (=T_0) of the observed signal. b) Find the maximum frequency (f_max) that x(t)can have so that there will be no aliasing effect due to sampling? c) Assuming that N = 200 is the DFT length, what is the frequency resolution between two adjacent DFT values in digital (Delta Ohm) and in analog (Delta f)? d) What analog (f) and digital (Ohm) frequencies do the DFT component X(10) correspond to? c) Given, X(10) = -1 - j2. Determine X(190) and X(1010). f) If the desired frequency resolution (Delta f) is 0.02Hz, find the Record Length (T_0). How many more samples or zeroes are needed to achieve this Delta f = 0.02Hz.

Explanation / Answer

1)

sampling frequency is 20Hz

i.e sample taken at every 1/20 s

toatl sample taken is 200

thus toatal duration is 200*1/20=10 s

2) for no aliasing

2fm=fs

2fm=20

fm=10Hz

3)

frequency resolution = fs/N=20/200=0.1Hz in analog

frequency resolution = 2pi/fs=2pi/20=pi/10rad/s in digital

4)

K=10 analog frequency = 10* frequency resolution = 1 Hz

Digital frequency= 10* frequency resolution = 10*pi/10= pi rad/s

5) since x(n) is real

X(k)*=X(N-K)

for k= 10 N=200

X(190)=-1+2j

also X(N)=X(N+K)=X(2N+K) ------

thus X(10)=X(210)=X(410)=---=X(1010)=-1-2j

6)

0.02=20/N N= 20/0.02=1000

Total length= 1000/20=50 s

we need to add extra 800 samples

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