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How can I calculate the value of \$R\$ and \$C\$ for an AM envelope detection circuit.

Knowing that the carrier wave is \$20\mathrm{KHz}\$, \$1\mathrm{Vpp}\$ sine wave, and the message signal is a \$32\mathrm{Hz}\$, \$2\mathrm{Vpp}\$ triangular wave with a \$3\mathrm{V}\$ DC offset:

demonstration

I used the low pass fillter equation (\$f_c=\frac{1}{2\pi R C}\$) to calculate \$R_1\$ and \$C_1\$ and it was fine ~100 duplicate. The result was this:

enter image description here

Is this the correct method to use?.

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the cutoff frequency of RC filter is:

$$f_0 = \frac{1}{2\pi R C}$$

you make :

  1. \$f_0 << f_c\$ (carrier frequency)
  2. \$f_0 >> f_d\$ (maximum data frequency)

The optimal value:

$$f_0 = \sqrt{\left(f_c \times f_d\right)} $$

Example:

\$f_c = 20\mathrm{kHz}\$, \$f_d = 32\mathrm{Hz}\$, then:

$$f_0 = \sqrt{\left(20000 \times 32\right)} = 800 \mathrm{Hz}$$

Which can be achieved with for example \$C = 39\mathrm{nF}\$ and \$R=5.1\mathrm{k\Omega}\$.

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  • \$\begingroup\$ I suppose the data signal is sine wave, if it is square wave you must calculate for 5 fd or more ==> f0 = sqrt(fc * 5 * fd) \$\endgroup\$ – ir.imad Oct 22 '15 at 1:07
  • \$\begingroup\$ ok thank you , so we get f0 = 800Mhz , so what equation did you use to calculate r=5.1k and c=39nf the Rc filter equation or the T=r*c (time constant) and T=1/f ?. \$\endgroup\$ – Bilal Oct 22 '15 at 13:21
  • \$\begingroup\$ Time const. T = R * C ; cutoff frequency f0 = 1/ (2 PI R C) where [PI=3.141]. you talk above about 20 kHz and 32 Hz. \$\endgroup\$ – ir.imad Oct 24 '15 at 21:45
  • \$\begingroup\$ Time const. T = R * C ; cutoff frequency f0 = 1/ (2 PI R C) where [PI=3.141]. you talk above about 20 kHz and 32 Hz. for 800 MHz you need more complicated circuits . sorry, i am not a specialist in RF applications \$\endgroup\$ – ir.imad Oct 24 '15 at 21:56
  • \$\begingroup\$ @ir.imad For 800MHz you need a different time constant. Same circuit. \$\endgroup\$ – Marquis of Lorne Jul 12 '16 at 0:15

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