By negative frequency, I refer to Fourier transform. Often, the frequency response of a digital filter is only displayed for positive frequencies. For a linear IIR digital filter, what happens for negative frequencies? Are frequency response for negative frequencies mirror images of what happens for positive frequencies?


1 Answer 1


It may be easier to start with analog signals. If the filter's impulse response $h(t)$ is real, then its frequency response $H(f)$ is conjugate symmetric: $H(-f)=H^*(f)$, where $(\cdot)^*$ indicates complex conjugate. Since $H(f)=|H(f)|e^{j\angle H(f)}$, then conjugate symmetry implies that $|H(f)|=|H(-f)|$ and $\angle H(f) = - \angle H(-f)$. This is true for any kind of filter, FIR or IIR.

So, if the filter input $x(t)$ is $$x(t)=A_0\cos(2\pi f_0t+\phi_0)=\frac{A_0}{2}\left(e^{j2\pi f_0t}e^{j\phi_0}+e^{-j2\pi f_0t}e^{-j\phi_0}\right),$$ then the filter output is

$\begin{align*} y(t) &= \frac{|H(f_0)|A_0}{2}\left(e^{j2\pi f_0t}e^{j(\phi_0+\angle H(f_0))}+e^{-j2\pi f_0t}e^{-j(\phi_0+\angle H(f_0)}\right) \\ &= |H(f_0)|A_0\cos\left(2\pi f_ot+\phi_0+\angle H(f_0)\right). \end{align*}$

As you can see, the amplitude of negative frequencies are affected exactly the same as positive frequencies. The phases are affected in a similar fashion, but with phase of opposite sign for negative frequencies. Note that this is necessary to preserve the symmetries that real signals have in the frequency domain.

In the case of discrete signals and filters, the conclusion regarding negative frequencies is the same. Finding the actual gain and phase at a particular frequency is slightly more difficult, though. The reason is that the DFT of the filter's impulse response defines the filter's gain and phase only at a finite number of frequencies. In general, the filter's gain and phase at an arbitrary frequency $f_0$ can be found by evaluating the filter's transfer function at $z=e^{j2\pi f_0}\,$.

  • $\begingroup$ please edit your answer so that it will be about a discrete time digital filter's impulse response h[n] rather than a continous time LTI system's impulse response h(t)... Also please replace the corresponding fourier transforms as well... $\endgroup$
    – Fat32
    Jun 26, 2015 at 23:32
  • $\begingroup$ @Fat32, I added a few comments on digital signals. I left the explanation using analog signals because I think it is easier to understand for a beginner, and the conclusion is the same as for discrete-time systems. $\endgroup$
    – MBaz
    Jun 28, 2015 at 2:56
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    $\begingroup$ I just want to clarify for future readers that if the filter's impulse response is complex (real and imaginary coefficients), the negative and positive frequencies will not be the same. $\endgroup$ Feb 28, 2017 at 12:56

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