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I'm having trouble understanding outputs of a BFSK modulator.

$$S_1(t) = cos(πt/T)$$ $$S_2(t) = cos(2πt/T)$$

If the input 0 or 1, how can I calculate the output?

Edit:

There is a formula to calculate the output:

$$Zi(t) = ∫ r(t)S_i(t)dt$$

If the input is \$S_2\$ what will be the outputs and decision? I couldn't find any examples for these formula. Any help is greatly appreciated.

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BFSK uses a pair of discrete frequencies to transmit binary (0s and 1s) information. \$s_1(t)\$ and \$s_2(t)\$ are the discrete frequency signals here. \$s_1(t)\$ has a frequency of \$1/2T\$ and \$s_2(t)\$ has a frequency of \$1/T\$. But your question does not say which signal is sent for input \$0\$.

So if \$s_1(t)\$ is sent for input \$0\$ and \$s_2(t)\$ is sent for input \$1\$, then the output is given by

$$y(t) = \overline{x(t)}\ s_1(t)+x(t)\ s_2(t)$$

where, \$x(t) \in \{0,1\}\$ is the input and \$\overline{x(t)}=1-x(t)\$.

EDIT:
In a FSK demodulator, the received signal, \$r(t)\$ is correlated with signal corresponding to each symbol, \$S_i(t)\$ to get \$Z_i(t)\$.

$$Z_i(t) = \int r(t)\ s_i(t)\ dt\tag1$$ Then the demodulator makes a decision based on the value of \$Z_i(t)\$. The decision is made in favor of i\$^{th}\$ symbol producing maximum correlation product (\$Z_i(t)\$)

So in BFSK, the received signal \$r(t)\$ is correlated with both \$s_1(t)\$ and \$s_2(t)\$. So if 1 be the value of modulating signal, then \$s_2(t)\$ is the received signal and its correlation product with s1(t) and s2(t) are calculated using equation (1). Correlating \$s_2(t)\$ with itself will produce the maximum value and hence the demodulator makes a decision in favor of \$s_2(t)\$.

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  • \$\begingroup\$ Thanks nidhin. I edited the question, could you please take a look. \$\endgroup\$ – Jenny94 Jul 6 '14 at 13:53
  • \$\begingroup\$ I calculated z1(t)=0 and, z2(t)=T/2 when s2(t) is the received signal. So what does this mean? \$\endgroup\$ – Jenny94 Jul 6 '14 at 16:38
  • \$\begingroup\$ @Jenny94 It means that the received signal is more 'related' to s2(t) than s1(t). \$\endgroup\$ – nidhin Jul 6 '14 at 16:58

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