# Not sure how to apply a filter

**URL:** <https://discourse.julialang.org/t/not-sure-how-to-apply-a-filter/50265>\
**Category:** Signal and Image Processing\
**Tags:** dsp\
**Created:** [November 16, 2020, 11:56pm UTC](https://discourse.julialang.org/t/not-sure-how-to-apply-a-filter/50265 "2020-11-16T23:56:29Z")\
**Posts on this page:** 1\
**Showing post:** 7

<div class="post-metadata">

**Author:** ![contradict](https://avatars.discourse-cdn.com/v4/letter/c/ac91a4/32.png) [@contradict](https://discourse.julialang.org/u/contradict)\
**Post date:** [November 18, 2020, 5:36am UTC](https://discourse.julialang.org/t/not-sure-how-to-apply-a-filter/50265/7 "2020-11-18T05:36:29Z")

</div>

Ah, I think I see now. The fact that `freqs` gives the correct plot (as opposed to `freqz`) means that the Biquad coefficients you have are in the Lapace domain (continuous time, usually represented as a polynomial in `s`). `filt` assumes it’s input is in the Z domain (discrete time, usually represented as a polynomial in `z`). The [Bilinear Transform](https://en.wikipedia.org/wiki/Bilinear_transform#Transformation_of_a_second-order_biquad) can convert coefficients between the two representations.

The (unreleased) latest version of DSP.jl has tools to make this difference explicit, try this out

```julia
### A Pluto.jl notebook ###
# v0.12.10

using Markdown
using InteractiveUtils

# ╔═╡ 84c78010-284a-11eb-1385-cd8bd55f042d
begin
        using Pkg
        Pkg.activate(mktempdir())
        Pkg.add(PackageSpec(name="DSP", rev="master"))
        Pkg.add("Plots")
end

# ╔═╡ 958dba22-2949-11eb-15b6-7f836480cc87
begin
        using DSP
        using Plots
end

# ╔═╡ 620b8630-2871-11eb-06f5-c7d982acd7bd
Tsample=50e-6

# ╔═╡ 83664af0-2861-11eb-084e-6de55511c1dc
filter_s = let
        f1 = 6.31*Tsample*2
        f2 = 1258.9*Tsample*2
        Q1 = 0.71
        f3 = 15.915*Tsample*2
        f4 = 15.915*Tsample*2
        Q2 = 0.64
        SecondOrderSections([
                        Biquad{:s}(4π^2*f2^2,0.0,0.0,2π*f1/Q1, 4π^2*f1^2),
                        Biquad{:s}(0.0,0.0,1.0,2π*f2/Q1, 4π^2*f2^2),
                        Biquad{:s}(0.0, 2π*f4^2/f3, 4π^2*f4^2, 2π*f4/Q2, 4π^2*f4^2)], 1.0)
end

# ╔═╡ b504b86a-2958-11eb-28b5-0df48a71b1b2
 # since frequencies are normalized already, fs is just 2
filter1 = DSP.Filters.bilinear(filter_s, 2)

# ╔═╡ 0938a0a0-28e5-11eb-0e2f-9d28fb0ab8b9
let
        f=10.0
        tr = 0.0:Tsample:10.0
        # x=sin.(2π*(f/4)*tr) + sin.(2π*f*tr) + sin.(2π*4*f*tr)
        x = rand(length(tr))
        perin = periodogram(x; fs=1. / Tsample)
        imax=10000
        imin=10
        p1=plot(freq(perin)[imin:imax], power(perin)[imin:imax], yaxis=:log, xaxis=:log, label="input", leg=false)
        fx = filt(filter1, x)
        perout = periodogram(fx; fs=1. / Tsample)
        p2=plot(freq(perout)[imin:imax], power(perout)[imin:imax], yaxis=:log, xaxis=:log, label="output", leg=false)
        plot(p1, p2, layout=(2,1), link=:x)
end

# ╔═╡ d6b631f0-2863-11eb-1f59-bb236634efcc
let
        fr=1.0:1.0:1200.0
        plot(fr, abs.(freqz(filter1, fr, 1. / Tsample)),
                yaxis=:log,xaxis=:log, title = "filter response", leg=false)
end

# ╔═╡ Cell order:
# ╠═0938a0a0-28e5-11eb-0e2f-9d28fb0ab8b9
# ╠═d6b631f0-2863-11eb-1f59-bb236634efcc
# ╟─620b8630-2871-11eb-06f5-c7d982acd7bd
# ╠═83664af0-2861-11eb-084e-6de55511c1dc
# ╠═b504b86a-2958-11eb-28b5-0df48a71b1b2
# ╠═84c78010-284a-11eb-1385-cd8bd55f042d
# ╠═958dba22-2949-11eb-15b6-7f836480cc87

```

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