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Homework Statement:

Someone decides to replace the simple detector with a design based on a phasesensitive detector, a block diagram of which is shown in Figure 2. The signal after the intermediate frequency section can be written as: [tex] v(t) = A(1 + m cos(2 \pi f_m t) ) cos(2 \pi f_i t) + x(t) cos(2 \pi f_i t) + y(t) sin(2 \pi f_i t) [/tex] where [itex] f_m [/itex] is the signal frequency and [itex] f_i [/itex] is the intermediate frequency and [itex] x(t) [/itex] and [itex] y(t) [/itex] are the narrow band noise.
The frequency [itex] f_i >> f_m [/itex] and the phasesensitive detection includes a filter which cuts off all frequencies greater than [itex] 2 f_m [/itex]. Derive an expression for the steady state output from the signal detector.
Relevant Equations:

Low Pass Filter Transfer Function
Trigonometric Identities
Hi,
Here is the figure 2 that the question referred to:
How do I go about a question like this? I have made an attempt, but am not very confident with the method. Note, I have learned about/am aware of Fourier transforms.
My attempt:
So first I thought that the square wave will be at the intermediate frequency as we are trying to demodulate the signal and superheterodyne receivers have components which work at the intermediate frequency. Thus, I found the Fourier series of the square wave to be: [itex] r(t) = \frac{4}{\pi} \left( cos(\omega_i t)  \frac{1}{3} cos(3 \omega_i t) + \frac{1}{5} cos(5 \omega_i t)  .... \right) [/itex].
The design below will basically just multiply [itex] r(t) [/itex] and [itex] v(t) [/itex] together. Due to all the sin and cos terms, when we multiply them we will get lots of addition and subtraction of frequencies. I considered each of the terms in [itex] v(t) [/itex] one by one for the multiplication process.
The first term [itex] v_1 (t) [/itex] (note I have changed from f to [itex] \omega [/itex]) is given by [tex] v_1(t) r(t) = A(1 + m cos(\omega_m t) ) cos(\omega_i t) \times \frac{4}{\pi} cos( \omega_i t) = \frac{2 A}{\pi}(1 + m cos(\omega_m t) )(1 + 2 cos(2 \omega_i t) ) [/tex] I have only considered the first term of the fourier series of the square wave as all the other terms will just produce sums & differences which will be filtered off. After filtering off terms (we can ignore the [itex] cos(2 \omega_i t) [/itex] term), we are left with [itex] \frac{2 A}{\pi}(1 + m cos(\omega_m t) ) \times effect of transfer function [/itex]
The second term becomes [itex] x(t) cos(\omega_i t) cos(\omega_i t) [/itex] and after filtering [itex] \frac{x(t)}{2} \times effect of transfer function [/itex].
The third term becomes [itex] y(t) sin(\omega_i t) cos(\omega_i t) = \frac{y(t)}{2} ( 2 sin(\omega_i t)) [/itex] and this will all get filtered off.
The transfer function is given by [itex] H(j \omega) = \frac{1}{1 + j \omega RC} [/itex]. Taking into account the transfer function, term two (with the x(t) ) remains the same as it is DC. For term 1, we let [itex] \omega = \frac{1}{2RC} [/itex] and thus, we get [itex] H = \frac{2}{\sqrt 5} [/itex] and a phase distortion of [itex]  arctan(\frac{0.5}{1}) [/itex]
Combining these two terms, we get that the output signal is:
[tex] v_{out} (t) = \frac{x(t)}{2} + \frac{2}{\pi} + \frac{2}{\sqrt 5} \frac{2}{\pi} cos(\omega_f t) = \frac{2}{\pi} + \frac{x(t)}{2} + \frac{4}{\pi \sqrt 5} cos(\omega_f t  arctan(0.5)) [/tex]
I feel that I may have done something wrong by treating [itex] x(t) [/itex] like a dc component when perhaps the IF filter in the superheterodyne filter would have made it at [itex] \omega_i [/itex].
Thank you very much in advance for the help.
Here is the figure 2 that the question referred to:
How do I go about a question like this? I have made an attempt, but am not very confident with the method. Note, I have learned about/am aware of Fourier transforms.
My attempt:
So first I thought that the square wave will be at the intermediate frequency as we are trying to demodulate the signal and superheterodyne receivers have components which work at the intermediate frequency. Thus, I found the Fourier series of the square wave to be: [itex] r(t) = \frac{4}{\pi} \left( cos(\omega_i t)  \frac{1}{3} cos(3 \omega_i t) + \frac{1}{5} cos(5 \omega_i t)  .... \right) [/itex].
The design below will basically just multiply [itex] r(t) [/itex] and [itex] v(t) [/itex] together. Due to all the sin and cos terms, when we multiply them we will get lots of addition and subtraction of frequencies. I considered each of the terms in [itex] v(t) [/itex] one by one for the multiplication process.
The first term [itex] v_1 (t) [/itex] (note I have changed from f to [itex] \omega [/itex]) is given by [tex] v_1(t) r(t) = A(1 + m cos(\omega_m t) ) cos(\omega_i t) \times \frac{4}{\pi} cos( \omega_i t) = \frac{2 A}{\pi}(1 + m cos(\omega_m t) )(1 + 2 cos(2 \omega_i t) ) [/tex] I have only considered the first term of the fourier series of the square wave as all the other terms will just produce sums & differences which will be filtered off. After filtering off terms (we can ignore the [itex] cos(2 \omega_i t) [/itex] term), we are left with [itex] \frac{2 A}{\pi}(1 + m cos(\omega_m t) ) \times effect of transfer function [/itex]
The second term becomes [itex] x(t) cos(\omega_i t) cos(\omega_i t) [/itex] and after filtering [itex] \frac{x(t)}{2} \times effect of transfer function [/itex].
The third term becomes [itex] y(t) sin(\omega_i t) cos(\omega_i t) = \frac{y(t)}{2} ( 2 sin(\omega_i t)) [/itex] and this will all get filtered off.
The transfer function is given by [itex] H(j \omega) = \frac{1}{1 + j \omega RC} [/itex]. Taking into account the transfer function, term two (with the x(t) ) remains the same as it is DC. For term 1, we let [itex] \omega = \frac{1}{2RC} [/itex] and thus, we get [itex] H = \frac{2}{\sqrt 5} [/itex] and a phase distortion of [itex]  arctan(\frac{0.5}{1}) [/itex]
Combining these two terms, we get that the output signal is:
[tex] v_{out} (t) = \frac{x(t)}{2} + \frac{2}{\pi} + \frac{2}{\sqrt 5} \frac{2}{\pi} cos(\omega_f t) = \frac{2}{\pi} + \frac{x(t)}{2} + \frac{4}{\pi \sqrt 5} cos(\omega_f t  arctan(0.5)) [/tex]
I feel that I may have done something wrong by treating [itex] x(t) [/itex] like a dc component when perhaps the IF filter in the superheterodyne filter would have made it at [itex] \omega_i [/itex].
Thank you very much in advance for the help.