amcommunicationfilterRFwave
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:
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:
Is this the correct method to use?.
A lot of questions but if you are looking for a WWV receiver design here is one and below is one of the circuits contained on the link: -
YES but it's probably not too critical if you are receiving a decent signal. A quarter wave dipole at 10MHz is optimum but this will be 7.5m long so try a couple of metres.
NO, you need to tailor the inductance capacitance ratio to give a decent Q factor but not be susceptible to parasitic components. Try for a resonant circuit that uses no less than 50pF - 50pF and 5060nH resonate at 10MHz - if you half C, double L for same resonance at an improvement in Q of 2:1. BTW it's "M" not "m" in MHz
Tricky, suck it and see
See design linked to for best guess at what you should be aiming for.
WWV uses up to 100% modulated AM (or normal DSB) transmission. Use a diode detector (as per link) to produce the demodulated waveforms.
Note that in your case:
$$\int_{0}^{1}v(t-2)^{2}\:\mathrm{d}t=\int_{2}^{3}v(t)^{2}\:\mathrm{d}t$$
Using this, and the generic formula for a parabola with it's vertex at the origin, we have that: \$v(t-2)=4t^{2}\$, we therefore have that:
$$\int_{2}^{3}(v(t))^{2}\:\mathrm{d}t=\int_{0}^{1}(4t^{2})^{2}\:\mathrm{d}t=\int_{0}^{1}16t^{4}\:\mathrm{d}t=\left[\frac{16t^{5}}{5}\right]_{0}^{1}=\frac{16}{5}$$
I hope this helps you!
EDIT: In order to clarify how I arrived at the equation \$v(t-2)=4t^{2}\$, we can examine the general equation of a parabola, given by: $$y(x)=a(x-b)^{2}+c$$
Where \$a\$ is a parameter which modifies the gradient of the parabola and \$b\$ and \$c\$ determine the vertex (bottom-/top-most point) of the parabola (at \$(x,y)=(b,-c)\$), in this case we have that the vertex of \$v(t-2)\$ is at the point \$(0,0)\$ (imagine shifting the graph 2 units to the left) and therefore \$b=-c=0\$ and therefore we have the equation: $$v(t-2)=a(t-2)^{2}$$
Using the fact that when \$t-2=1\$, \$v(t-2)=4\$, we get \$a=4\$.
Best Answer
the cutoff frequency of RC filter is:
$$f_0 = \frac{1}{2\pi R C}$$
you make :
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}\$.