Electronic – How to filter RF signals to get the average value

I can't answer this authoritatively but my gut tells me your spec is going to be "very difficult".

In particular, your transition band of 50 MHz is only 0.02 decades at 1 GHz, so you're looking for a drop of 714 dB/decade between your pass band edge and your rejection band. Which implies something like a 71-pole filter, requiring 71 active elements.

For reference, here's what can be done with a reasonable number of elements:

(Graph from TI's App Guide "Op-Amps for Everyone") The graph is in terms of "normalized frequency", meaning you can scale the filter elements in such a way as to make a frequency of "1" on the graph correspond to any frequency you choose, for example 1 GHz in your case.

At lower frequencies, we normally construct multi-pole filters by cascading 1 and 2-pole active sections to obtain some desired response.

At 1 GHz, you may, just be able to do that using rf amplifiers to buffer between stages. But more likely, you'll be stuck falling back on older techniques of constructing an LC ladder to get an approximation of the response you want. The problem with this technique is it tends to make the filter response more sensitive to small variations in the component values, caused by manufacturing differences or temperature sensitivities.

Using microstrip elements, you might have less trouble with L and C variability, but you're likely to find that the range of L and C values required are outside of what can be sensibly constructed in microstrip. In addition, my (very limitted) experience suggests that microstrip filters are only likely to be effective over about an octave frequency range. So if you want a 1 GHz LPF, you might find you get an unwanted blocking band below 500 MHz, or an unwanted pass-band above 2 GHz. In any case you don't want to jump in to designing microstrip filters without access to some kind of reasonable CAD tool. Agilent's ADS or Genesys jump to mind. Genesys would be particularly helpful for you, if you can get access to it, because it provides special tools for generating filter designs given a spec like you've given in your question.

Of course, a combination of lumped and microstrip elements is also possible.

Edit: One reasonable design approach would be to use a tool like Matlab or Octave to see what kind of filter (Butterworth, Chebychev, etc, and how many poles) can come close to meeting your requirements. If you have access to a good library, look for a book with a title like "filter design handbook". This will give you lookup tables for the pole and zero locations of various types of filter of different orders. This will make it "easy" to calculate the response even if you don't have a high-priced tool like Matlab with the right toolbox to get the filter parameters from software.

Then, once you know where you want your poles and zeros, use a tool like ADS, or Genesys, or even SPICE, to design a filter using real L and C elements to create the mathematical response you optimized in Matlab. Then, be sure to do a sensitivity analysis to be sure the response stays in spec under normal variation of the part characteristics. Finally, depending on the L and C values you come up with, decide whether you want to implement some or all of those elements in microstrip instead of with discrete components. If you do decide to use microststrip, then use an rf design tool like ADS or Genesys (those are just two tools I've used myself, but there are others that could do this) to simulate and optimize the microstrip layout to achieve the behavior you want.

Another late note: You can see in the graph that for a Chebychev filter, the slope immediately after cut-off is steeper than the eventual slope of the skirt, so my statement of needing a 71-pole filter is probably too strong. But nonetheless, its clear you need at least 10 poles to meet your spec, and doing that with only passives is very challenging because of the stage-to-stage interactions and the required tight tolerances on the component values.

What does transfer or Vout/Vin mean, and why is it on the y-axis? Because I don't understand this, I can't understand the graph either...

You could make the input signal \$V_{in}\$ anything that you want and solve for \$V_{out}\$. That's fine - it's just a bit of math. However, if you wanted to change \$V_{in}\$ to something else, then you'd have to re-calculate everything. You haven't learned anything; you've just solved the circuit for one signal.

The transfer function of a circuit tells you what \$V_{out}\$ is. It depends on the frequency of your input. If the transfer function is \$H(f)\$, then we can write $$ V_{out} = V_{in} \cdot H(f) $$ and now, if we calculate \$H(f)\$, we know what \$V_{out}\$ looks like for every input frequency.

Why the y-axis also has to do with decibel. I've done some research on decibel, and apparently it's a ratio between an output and input voltage. It's still very confusing to me... I'm going to do some more research on this as soon as I posted this question.

Circuits often deal with a lot of orders of magnitude. Sometimes, you're interested in a signal that's 1 V in amplitude; sometimes, you're looking at 1 uV. That's like multiplying by 0.000001. Yuck - do you like counting zeros?

Decibels are a way of looking at orders of magnitude. Instead of multiplying by 10, you add +20 dB. Now, the difference between 1 V and 1 uV is -120 dB - much easier to read and understand.

How can an analog signal be graphed like this? An analog signal has only one frequency unless it changes periods. So if the frequency of Vin would be 100 Hz, wouldn't there just be a single discrete value when frequency = 100? But instead this graph looks like a continous curve. So what is it showing the frequency of?

This is showing what the output is for any frequency. Let me pretend for a minute that \$f_k\$ = 1000 Hz. Then, look at your graph.

• 100 Hz is to the left of \$f_k\$. That means when you put a 100 Hz signal in, you get out the exact same signal.
• 10 kHz is to the right of \$f_k\$. Now, the output has dropped quite a bit: down to -40 dB. That means the output is 100 times smaller than the input.

Continue this for any other frequency and you get the continuous transfer function.

Replying to some comments,

Why does the influence of the capacitor get smaller as the frequency gets higher?

The charge on a capacitor is $$ Q = CV $$ so the current through the capacitor is $$ I = C\frac{dV}{dt} $$ Think about what happens when you change the input frequency.

• At low frequencies, a sine wave doesn't change very fast, so \$\frac{dV}{dt}\$ is small, and the capacitor doesn't let much current through it.
• At high frequencies, \$\frac{dV}{dt}\$ is big, so \$I\$ can be big, too. Now, the capacitor lets a lot of current through, and the output voltage gets lower (the \$V = IR\$ drop across the resistor gets big).