Consider a continuous time signal $x(t)$ that has been pre-filtered by a low-pass filter with a cut-off frequency of $10$kHz. The spectrum of $x(t)$ is estimated by use of the N-point DFT. The desired frequency resolution is $0.1$ Hz. Determine the required value of 'N' and the necessary data length $T_L$.
a) $N\;=\;2^6$, $T_L\;=\;14.33\;s$
b) $N\;=\;2^{18}$, $T_L\;=\;13.1072\;s$
c) $N\;=\;2^{10}$, $T_L\;=\;10.1072\;s$
d) $N\;=\;2^4$, $T_L\;=\;12\;s$
Correct Answer: Option B
(I don't know why this question has been given as star)
(I don't know why this question has been given as star)
Solution:
- In a sampling circuit, the sampler has fixed sampling frequency $F_s$. To avoid aliasing we include a pre-alias filter before the sampler to limit the incoming signal frequency within $\frac{F_s}{2}$. Here the signal $x(t)$ is filtered by a pre-alias filter of cut-off frequency $10$ kHz, hence maximum signal frequency is, $$f_{max}\;=\;10\;kHz$$
- The minimum sampling frequency will be $$F_{s}\;=\;20000\; samples/sec$$
- To process these in DSP processor, we need finite time-domain and frequency domain samples. Hence we sample the spectrum to obtain 'N' samples in $2\pi$ or $F_s$ period. So, the frequency resolution $$\triangle f\;=\;\frac{2\pi}{N}$$ or $$\triangle f\;=\;\frac{F_s}{N}$$
- The number of frequency samples will be $$N\;=\;\frac{F_s}{\triangle f}$$ $$=\frac{20000}{0.1}$$ $$N\;=\;2×10^5$$
- To find these frequency samples from time samples, we use N-point DFT. This 'N' value has to be greater than $2×10^5$. Among the options $2^{18}$ is the only 'N' value which is greater than $2×10^5$
- As we choose the N value as $2^{18}$, number of time samples and frequency samples become $2^{18}$, hence the frequency resolution will change to $$\triangle f\;=\;\frac{20k}{2^{18}}$$ $$=0.07629$$
- Data length $T_L$ is nothing but the inverse of frequency resolution $$T_L=\frac{1}{0.07629}=13.1072\;s$$
