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This is a problem which was asked in paper recently. The final answer they gave was option(A).

Since, in frequency domain we can multiply signals. I assumed option (A) and started drawing signals ahead. Is my beta signal wrong ? I cannot proceed further from that part as it will require alot of calculation and I am afraid this is not the right way. I am not looking for solution here, just guide me the way to solve this.

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Lets check whether option A is the correct choice or not.

  • When signal x(t) goes through system A, the magnitudes of all frequency components get multiplied by +j or -j correspondingly. Hence, the output response alpha will look like:

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  • The magnitude of alpha response get multiplied by j in the next step.

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  • This response beta is then added with the original signal x(t) in the next step. The response beta is just a -ve mirror image of x(t) around -fc band. Hence all negative frequency components are diminished to zero after addition. But the band of frequencies around +fc are added up and doubled in magnitude.

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  • Now if you look at the output response y(t), it is nothing but the frequency shifted version of gamma. Hence the function of system B would be to frequency shift the gamma by -fc, with unity gain. Time to take a look at the following properties of fourier transform.

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From the second property, we can infer that gamma(t) has to be multipled by e^jwt to obtain y(t). Hence the function of system B would be just that. Now we know: $$e^{jwt} = cos(wt) + j sin(wt)$$ So the corresponding frequency domain representation can be obtained as:

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Therefore, in the frequency domain, e^jwt should have to look like:

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Ha finally but not yet !! This aint the transfer function of B. This is just the fourier transform of e^jwt. We wont get y(jw) by multiplying e^jwt's frequency response with gamma . It has to be instead convolved in frequency domain. In fact, we cannot characterise B with a transfer function at all. Because it is NOT an LTI System. It is a time varying system, as system B multiplies gamma(t) by a function of t. So whoever asked this question might have wrongly taken e^jwt's fourier transform as the "transfer function" of B.

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  • \$\begingroup\$ After drawing gamma can we multiply point to point in gamma and B and finally say output will be 0? I multiplied point to point in alpha to each point in A and got beta which was correct. What is wrong in this logic bcz it worked for block A? \$\endgroup\$ – Nikhil Kashyap Dec 21 '17 at 13:33
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    \$\begingroup\$ Actually we are not multiplying there, we are convolving two responses there. Because Multiplication in time domain corresponds to convolution in frequency domain. I agree its misleading, because transfer function of B should be the response with which input gamma(jw) is multiplied in frequency domain to get the output Y(jw). Whoever put this question, made the last part misleading by simply putting the fourier transform of e^jwt, which is actually multiplied in the time domain. \$\endgroup\$ – Mitu Raj Dec 21 '17 at 15:27
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    \$\begingroup\$ B is a frequency shifting system and such a system is a time varying and hence non LTI system. So it cannot be characterized by a transfer function in frequency domain. Really misleading question indeed. \$\endgroup\$ – Meenie Leis Dec 21 '17 at 19:23
  • \$\begingroup\$ Yes. Will edit to add this info. \$\endgroup\$ – Mitu Raj Dec 21 '17 at 19:31
  • \$\begingroup\$ please take a look electronics.stackexchange.com/questions/347346/… \$\endgroup\$ – Nikhil Kashyap Dec 30 '17 at 12:12
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Note that alpha is imaginary, when multiplied by j you get a real valued pair of sidebands with the one around -fc being negative and the one around fc being positive (this is beta).

When you add the original signal the -fc sideband cancels and the +fc sideband doubles in amplitude, this is gamma.

Block B then must be an image reject mixer to convert down to baseband.

This is a classic SSB phasing method setup.

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