The WPD has been fabricated and tested and shows good agreement between simulated and measured results and the proposed design has desirable characteristics for LTE and GSM applications. The proposed WPD has a small size of 33.8 mm × 27 mm (0.42 λg × 0.33 λg), where λg is the guided wavelength at the operating frequency of 1.8 GHz. The proposed modified WPD has a wide 20 dB stopband (from 2.54 GHz to 13.48 GHz) and filters the second to seventh harmonics with attenuation levels of greater than 20 dB. According to the obtained results, the input return loss (| S 11|), output return loss (| S 22|), output insertion loss (| S 21|) and isolation (| S 32|) are better than 34.2 dB, 26.2 dB, 3.52 dB and 31.2 dB, respectively. A lowpass filter (LPF) structure is utilized in both branches of the power divider to provide harmonic suppression. In order to improve on the current state of the art, a modified microstrip WPD is proposed that exhibits a substantially improved stopband and high isolation.

However, these WPDs can also exhibit poor out-of-band performance while requiring a large footprint. Oh yes, I also think I'll post something deeper at some point, but I still have a few articles about simple one pole filters in mind.Conventional Wilkinson power dividers (WPDs) can provide acceptable performance close to the nominal center frequency. With a bit of oversampling we get those straight lines to match our continuous signal pretty well! Again, here's a graph that shows how the s and y values of our filter corresond to trapezoidal integration:Īlso note that the depicted sample sequence is a little exaggerated. The integrated current over the capcitor is our filter output. Like above, don't confuse our filter output with the function that is integrated. But then again, once you get the hang of it, solving implicitely isn't all that difficult: we had to forgo the unit delay that's so very helpful with the Euler method. So, here we go with the equation for a trapezoidal one pole lowpass filter:Īgain, I used array notation to show that we implicitely need to solve y by a function of y itself, i.e. By solving our filter output implicitly we manage to correctly 'predict' the output sample with having only 'half the state' at hand. Marvelously, the grey area matches the area below the perfectly continuous signal much better than the previous method.īut how do we get there? - As I wrote in my previous blog entry, at first glance we need to look into the future. This is where the trapezoidal rule comes in.Ĭheck this out! - Instead of using a staircase analogy, we draw a straight line from one sample point to the other. We can hardly expect to get really close results for future samples when we base our computation on those staircase equivalent blocks. Now, what we also see: That integrated area is really pretty far off from that below the perfectly continuous signal. Here's an adapted graph that shows how our filter output (grey area) corresponds to the current updates (function values) which are taken for the calculation of the next step: the grey area) is the output voltage of a filter for each sample step. The function we're integrating in a RC filter is the current over the capacitor. The difference to the above graphic of a function being integrated is, in a filter the result output of the next step depends on the integrated area of the current step. The grey area under the sample is then equivalent to 1 * y, and - if it wasn't too obvious - any muliplication by factor 1 is surplus.

When we divide our frequency by the sample rate, we essentially 'normalize' the time constant so that it becomes 1 for each instance. Note that we don't need to multiply y with any time constant 't' whatsoever because the time constant is already 'baked'into our coefficient g when we set it to something like 1 - exp(2 * PI * cutoff/samplerate). This is what we all know from lowpass filters which go like this:Īt each sample instance we calculate a new value based on the input sample and a state that we assume to be equivalent to the previous sample value - which we stretch out to a rectangular block over the whole period of time that passed since the previous sample.

Those blocks are supposed to approximate the integral of that function. The idea is, to integrate a known function y = f(x) we simply sample that function and build square blocks between the sample intervals. So I finally fired up OmniGraffle to sketch some examples for Forward Euler vs.

Qucs low pass filter tutorial