# Is the frequency phase-response of a linear filter always decreasing?

Consider a linear electronic filter with a sinusoidal input. Is it always true that the graph of phase vs input frequency is decreasing?

Note that I'm talking about physical passive electronic filters not mathematical constructions. Why is it that all the filters I've seen which are composed of resistors, capacitors, and inductors, have decreasing phase?

• No not necessarily. But for single pole filters, yes. Multiple pole filters with zeros (lead) you can engineer phase to increase/decrease over any desired band of frequency. – docscience Feb 6 '17 at 16:24
• Can you give me a concrete example? – Joshua Benabou Feb 6 '17 at 16:25
• here is a filter example with increasing/decreasing phase electronics.stackexchange.com/q/164336/65409 – docscience Feb 6 '17 at 17:39
• This is BTW one of the most amazing thing in signal processing: a filter can be causal and still have a negative group delay in some frequency range. @docscience : an even simpler example (note the single pole filter!): run freqz([1],[1, -.5]) in Matlab. – user130529 Feb 7 '17 at 8:15
• @claude chuber: I'm talking about physical passive electronic filters not mathematical constructions. Why is it that all the filters I've seen which are composed of resistors, capacitors, and inductors, have decreasing phase. – Joshua Benabou Feb 7 '17 at 9:31

Here is one counter example - hope it works !

Consider a circuit made of two resistors $R$, two coils $L$ and one capacitor $C$ such that $LC=\omega_0^2=1$ and $\frac{1}{R}\sqrt{\frac{L}{C}}=Q=1$, all connected in series. The circuit is powered by $V_{in}=V_0 e^{i\omega t}$ and the out voltage is taken over one coil, one capacitor and the resistor (see scheme).

The transfer function is given by $$\frac{V_{out}}{V_0} = \frac{Z_{R+L+C}}{Z_{R+L+C}+Z_{L+R}}=\frac{\frac{1}{jC\omega}+R+jL\omega}{\frac{1}{jC\omega}+2R+2jL\omega}=\frac{1+ix-x^2}{1+2ix-2x^2}$$

and the phase is first decreasing, then increasing as shown below

• did you find this by working backwards from the transfer function? – Joshua Benabou Feb 9 '17 at 16:02
• Yes. 2nd order filter means second order polynoms in the transfer function. I thought that to get a weebbly wobbly phase behavior, I needed something with the same phase at 0 and infinite frequency, so same max and min order for both numerator and denominator. I then played a bit to avoid the standard circuits - which don't illustrate the behavior you were interested in. Once I found a transfer function, finding the circuit was just using the voltage divider method the other way around. – Pen Feb 10 '17 at 1:26
• Very nice. Could you just add a few lines in your answer describing how you have used the voltage divider method to obtain the circuit from the transfer function. – user130529 Feb 10 '17 at 6:52
• Thanks. That's what I tried to do on the first step of $V_{out}/V_0$. Do you think it should be more explicit ? – Pen Feb 11 '17 at 11:22
• OK, it's fine. One thing I wonder, do you think you could demonstrate with an example (output versus input) the negative group phase delay that occurs in your circuit in the frequency range where the phase slope is positive? That would make your answer even nicer. – user130529 Feb 13 '17 at 17:24

No. In general the phase of minimum-phase filters and mixed-phase transfer functions is not a monotonic function of frequency.

Factorize a general rational transfer function into a ratio of the product of single pole and single zero transfer functions. Witness that the phase of each factor is additive (being the imaginary part of the logarithm).

Now sketch the phase (argument) of a vector joining a point on the imaginary axis ($s=i\,\omega$) as $\omega$ moves from $-\infty$ to $+\infty$.