FSK has a large bandwidth requirement and its minimum bandwidth is around \$4 B_T\$, where \$ B_T \$ is the bit rate.

In a coherent FSK system the signals \$s_1(t)\$ and \$s_2(t)\$ representing bit 1 and bit 0 respectively are given by:

\$ s_1(t) = A_C . \cos(2\pi (f_c + \frac{\Delta f}{2})t) \$

\$ s_1(t) = A_C . \cos(2\pi (f_c - \frac{\Delta f}{2})t) \$

Assuming \$f_c > \Delta f \$

Show that the correlation coefficient between \$s_1(t)\$ and \$s_2(t)\$ is given by:

\$ \rho = \frac{\int_0^{T_B}{s_1(t)s_2(t)dt}}{\int_0^{T_B}{s_1^2(t)dt}} \$

is approximately given by:

\$ \rho \approx sinc(2\Delta fT_B) \$

How do I approach this question? I can do the integral but I don't know how the integral is derived, the squared looks like a matched filter or power calculation \$(V^2/R)\$?

  • \$\begingroup\$ Do these 4Xdatarate apply to MFSK? \$\endgroup\$ – analogsystemsrf Apr 21 '18 at 2:11
  • \$\begingroup\$ @analogsystemsrf That was just my comment. It happens when the two spectra can overlap at 2/T. So to your question no, that equation applies for only a binary signal. \$\endgroup\$ – Lewis Apr 21 '18 at 2:16
  • \$\begingroup\$ Did U learn Correlation functions yet? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 21 '18 at 6:24
  • \$\begingroup\$ @TonyStewartEEsince1975 autocorrelation, yes! Should I use that? \$\endgroup\$ – Lewis Apr 21 '18 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.