Superheterodyne Receiver with Phase Sensitive Detection Question

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Homework Statement:

Someone decides to replace the simple detector with a design based on a phase-sensitive 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 phase-sensitive 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:
Screen Shot 2020-04-05 at 11.23.12 PM.png


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.

 

Answers and Replies

  • #2
Joshy
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I'm curious myself and was having hard time following. I'm not strong in this area and have been aiming to learn more- I don't know the answer or if you're approach is wrong or right.

I was looking at ##sin(\omega_i t)cos(\omega_i t)## I think it should be ##{ {1} \over {2} } sin(2\omega_i t)##? I suppose either way it'll still get filtered out anyways.

That DC component sounds right. You already noticed ##cos(\omega t)cos(\omega t)## is going to be ##{ {1} \over {2} }(1+cos(2\omega t))## so that first component is DC without any sinusoid. I would imagine that's the point of the exercise show that you can't get rid of all the noise? I'll post anyways without watching silently... I'm ready to learn more :)

Choosing ##\omega = { {1} \over {2RC} }## did not make sense to me. Why did you do that? Wouldn't it be whatever the frequency ##f_i## is? Forgive me if I'm being silly.
 
  • #3
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Thanks for responding.

I'm curious myself and was having hard time following. I'm not strong in this area and have been aiming to learn more- I don't know the answer or if you're approach is wrong or right.
I apologise about this, I was trying to strike the balance between explanation of steps and length of post - it seems that I did not get this quite right. I will try to add some explanations here (on that note, do you know whether I am able to edit the original post - it doesn't seem to let me?).

The overall method was just to multiply [itex] v(t) [/itex] with [itex] r(t) [/itex], then use identities to get sum and difference terms, and then remove all the ones that would get filtered out while accounting for the effect of the LPF on the remaining terms.

I was looking at ##sin(\omega_i t)cos(\omega_i t)## I think it should be ##{ {1} \over {2} } sin(2\omega_i t)##? I suppose either way it'll still get filtered out anyways.
Yes, you are correct. I did try to give it a read through for errors but must have missed that.

Choosing ##\omega = { {1} \over {2RC} }## did not make sense to me. Why did you do that? Wouldn't it be whatever the frequency ##f_i## is? Forgive me if I'm being silly.
I thought that we choose [itex] \omega_m [/itex] because it is the frequency of the cos term and that is how the filter will affect it. Thus, as the break-point is at [itex] \frac{1}{RC} [/itex], then we are at half of the break-point frequency so [itex] \omega = \frac{1}{2RC} [/itex]. I might very well be wrong myself (I feel that this is part of the problem with me working in the time domain rather than putting it all into the frequency domain - but I feel that this problem was written to be solved in the time domain (the lecture series doesn't ever really use Fourier transforms or make much mention of them, despite us learning about them in other parts of our course).
 
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  • #4
Joshy
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Fair enough. That sounds reasonable. I don't know if it's the right answer, but it's not throwing up immediate red flags for me anymore. It took me a moment to say that ##\omega_{cutoff}=2\pi (2f_m)## when you make ##\omega=2\pi f_m## then the constant is halved. You then switched to phasor form to multiply the amplitudes and add the phase.

I don't know if this would be easier to solve in the frequency domain? If you could do both then why not? Whenever I see multiplication I'm happy where I'm at because I know the other domain is going to be a convolution problem- I'm not a big fan although sanity checks are always pleasant.

I think people who donate to the community can edit their post or they have more time to edit it (gold members). I donated. I forgot how I stumbled upon this site, but I knew right away that I really liked it and didn't mind supporting it. The price seems like a lot at a first glance, but it's perpetual not a subscription or monthly fee I could take the hit once and look away :) those little features really make a big difference.

If I were in your shoes I would ask a moderator if they could change the title of your thread (maybe send them a PM) because when I saw the words Superheterodyne Receiver I was thinking it was an architecture question I would have guessed it's a question about images, interferers, linearity, harmonics and spurious components. Quite honestly the math is the same but it sounds terrifying I'm wondering if this scaring away some other posters who could be more helpful than myself. Maybe something simplified like Fourier Transform on LPF? Just a suggestion.
 
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