# Group delay and phase delay of a filter

Maybe this question does not belong here but here, but since this webpage is for electric circuits in general and not for LTI systems that is why I post it here.

I have to compute the group delay and the frequency-response phase (which I don't know if it is the same as the phase delay) and I would like to know if for the frequency-response is enough to use P(H(w)) in PSpice, if someone is familiarized with it.

Also, consider the following phase response of a low-pass filter: Where the simulation is up to 10Hz, but now if I perform it for 900MHz: where I have cropped it up to 10Hz again. Is exactly the same circuit but the simulation is computed for a bigger frequency (as I have said) but the graph is totally different. Shouldn't be the same shape for both graphs? And which one of them should be the best option if the following specification mask is chosen? where $\omega_p = 1\;rad/s$, $\omega_{a1} =\;4.08 rad/s$ and $\omega_{a2} =\;4.36 rad/s$. My guess is that the first one is the best option since the managed frequencies are pretty low.

The schematic in PSpice is the following one: which corresponds to a Butterworth filter of order $n=3$

and the parameters of the simulation (frequency domain): for the first one (for the other one End Frequency = 900meg). The stimulus has the AC option equal to 1

• The 2nd graph is plotted as milli degrees in case you hadn't spotted that. – Andy aka May 2 '18 at 12:41
• @Andyaka That is another thing, the values are different even though the circuit is the same, and in reality after multiplying the expression by (3.1416/180) it represents radians – Martín May 2 '18 at 12:44
• @Martin, No, you're spice simulation is wrong. "mhz" unit is MILLLIhertz. if you want "Megahertz" you have to specify "meg" – pgvoorhees May 2 '18 at 13:02
• @pgvoorhees Yes, the simulation goes up to 900MHz, I have specified 900meg in the simulation profile, but I have cropped the image up to 10Hz to compare with the other image – Martín May 2 '18 at 13:16
• Alright, just making sure. The next thing to do is post your full schematic and simulation parameters. – pgvoorhees May 2 '18 at 13:23

## 1 Answer

When you do a linear sweep, the number of points are spaced out evenly over the entire frequency range. So, in the first example, Point #1 is at 0.001hz, the next point is at 0.001999hz. In the second example, Point #1 is at 0.001hz, the second point is at ~90kHz; you will never see anything useful 10Hz. You want to change to one of the logarithmic sweeps which will perform the number of points / decade (or octave).