In the following code example, a 6th-order Butterworth filter is defined over 4-50 Hz and the frequency response plotted. How to use DSP.jl to compute the corresponding zero-phase filter time response?

Using `filt(fil,ht1)`

in code example below, it seems to provide the minimum-phase response, while the workaround using `ifft`

and `fftshift`

outputs what I expect but it is rather convoluted.

```
using DSP, FFTW, Plots
fs = 1000; Δt = 1.0/fs; # sampling frequency [Hz] amd period [s]
fc1, fc2 = 4, 50; # low-cut / high-cut frequencies [Hz]
order = 6; # Butterworth filter order
fil = digitalfilter(Bandpass(fc1,fc2,fs=fs), Butterworth(order));
Hf = freqz(fil,0:1:200,fs); # 0-200 Hz response
plot(20*log10.(abs.(Hf)),title="Frequency Response", label="Butterworth 6th order (4-50 Hz)",
xlabel="Frequency [Hz]", ylabel="Amplitude [dB]")
n = 500;
t = Δt * vec(-n÷2:n÷2-1);
# minimum phase impulse response?
ht1 = [1.0; zeros(n-1,1)]; # impulse at time 0
ht1 = filt(fil,ht1)
ht1 = [zeros(n÷2,1); ht1[1:n÷2]]
plot(t, ht1, title="Impulse Response", label="Min-phase (4-50 Hz)")
# 0-phase filter time response (using ifft)
ht0 = fftshift(ifft(fftshift(abs.(freqz(fil,-n÷2:n÷2-1,fs))))); # over -250 to +250 Hz
plot(t, real.(ht0), title="0-phase filter Time Response", label="Zero-phase (4-50 Hz)",
xlabel="Time [s]", ylabel="Amplitude")
```

Thanks in advance.