Assume we have: $$x_c(t)=\sum_{k=-9}^{9} a_k e^{j(2000\pi kt)}$$

and Fourier coefficients out of $$|k|>9$$ is zero. So, it is band-limited.

The sampling frequency is:$$F_s=\frac {6} {10^{-6}} HZ$$ and $$T_s=\frac {10^{-3}} {6}$$

My question: what is the maximum frequency? how we can calculate it?

I know the answer that is $$F_n=\frac {9} {10^{-6}} HZ$$

But how?

  • $\begingroup$ Your question seems to have a couple of mistakes: $n$ is not used in the first summation (maybe you meant to write $n$ instead of $k$?) and the value of $T_s$ is wrong (it's missing a 6 and the exponent should be positive). $\endgroup$ – MBaz Mar 4 '16 at 19:06
  • $\begingroup$ @MBaz, Yes, you are right. sorry. the post has been updated. $\endgroup$ – David Mar 4 '16 at 19:10

What you need to know is that the signal $$s(t)=e^{j2\pi f t}$$ has frequency $f$. At any time $t_0$, $s(t_0$) is a point on the unit circle. As time advances, this point moves on the circle, and it completes $f$ turns in one second.

Knowing this, you can easily see that the maximum frequency in your signal $x_c(t)$ is 9000 Hz. That is because, for $k=9$, you can write the exponent as $j2\pi 9\cdot 1000 t$.

The sampling frequency you mention has no bearing on the signal's maximum frequency. However, if you intend to sample this signal, you need to make sure that the sampling frequency is at least twice the maximum frequency, or $f_s=18000$.

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  • $\begingroup$ thanks but what is $10^{-6}$ in denominator of maximum frequency? $\endgroup$ – David Mar 4 '16 at 20:00
  • $\begingroup$ I would interpret it as saying that the frequency is expressed in MHz. However, I don't know why you wrote it like that. Also, the right answer is 9000, not nine million. $\endgroup$ – MBaz Mar 4 '16 at 23:03
  • $\begingroup$ it is the solution for question 8.1 of Oppenheim's 3rd edition book. So, I do not think it is incorrect. But, you are right. It does not make sense. it is impossible to have 9 million as frequency of the signal. $\endgroup$ – David Mar 5 '16 at 3:10
  • $\begingroup$ @David I don't have ready access to that book, so I can't check. However, you're correct, the answer to the problem as stated is 9000. $\endgroup$ – MBaz Mar 5 '16 at 3:27

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