What I mean by passivity of impulse response is that the energy of impulse response for this system is less than the input impulse signal energy.

For example, in a discrete-time system, considering the impulse response, the input is Kronecker delta function $δ[n]$, and the output is the impulse response $h[n]$. Suppose that $$ h[n]=\sqrt2/2 , n=1,2 $$ and $h[n]$ is 0 for any other components. In discrete-time system, We calculate the energy of the signal with $$ Energy = \sum_{n=-\infty}^{\infty} |h(n)|^2 $$ Therefore, the energy of the impulse response is 1, and the input signal energy($δ[n]$) is also 1, which means the system is passive for this input signal.

Then considering another input signal $δ[n] + δ[n-1]$, the energy of this input is 2. The output can be derived by convolution of impulse response and this input signal. We get: $$ y[n]=\sqrt2/2 , n=1,3 $$ $$ y[n]=\sqrt2 , n=2 $$ and $y[n]$ is 0 for any other components. The energy of this output is 3, larger than the input energy. Therefore, It seems that the passive impulse response doesn't necessarily make a system passive. But if a system is passive, the impulse response of the system should always be passive. Am I right with this conclusion? or can anyone give a rigorous proof for this?

  • $\begingroup$ Note that the input signals in your example are $\delta[n]$ and $\delta[n]+\delta[n-1]$, rather than $\delta[0]$ and $\delta[0]+\delta[1]$, because the latter are just numbers instead of sequences. $\endgroup$
    – Matt L.
    Feb 15, 2016 at 9:04

1 Answer 1


Let me clarify some definitions first. There is no such thing as a "passive impulse response", there are only passive systems. If you like, you can call the impulse response of a passive LTI system "passive", but that's not common usage. From what I understand from your question, you probably mean by a "passive impulse response" an impulse response $h[n]$ that satisfies

$$\sum_{n=-\infty}^{\infty}|h[n]|^2\le 1\tag{1}$$

i.e., an impulse response with energy not greater than $1$. However, as you've shown yourself in an example, this is not sufficient for the corresponding system to be passive. Let's use the following definition of passivity [1]:

A discrete-time system is said to be passive if, for every finite energy input sequence $x[n]$, the output sequence $y[n]$ has, at most, the same energy, i.e., $$\sum_{n=-\infty}^{\infty}|y[n]|^2\le\sum_{n=-\infty}^{\infty}|x[n]|^2\tag{2}$$

It is more straightforward to investigate the passivity of a linear and time-invariant discrete-time system in the frequency domain than in the time domain. Let $H(e^{j\omega})$ be the frequency response corresponding to the impulse response $h[n]$:


Since the Fourier transforms of the input and output sequences are related by


the condition

$$|Y(e^{j\omega})|^2\le |X(e^{j\omega})|^2\quad\text{for all }\omega\tag{5}$$

holds if and only if $H(e^{j\omega})$ satisfies

$$|H(e^{j\omega})|\le 1\quad\text{for all }\omega\tag{6}$$

Integrating $(5)$ gives

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}|Y(e^{j\omega})|^2\le \frac{1}{2\pi}\int_{-\pi}^{\pi}|X(e^{j\omega})|^2d\omega\tag{7}$$

from which it follows via Parseval's theorem that


must hold. So it is not condition $(1)$ which makes a system passive, but condition $(6)$.

A causal and stable transfer function with real coefficients satisfying $(6)$ is called a bounded real transfer function [2]. This type of transfer function is important for the realization of low-sensitivity digital filters.

[1] S.K. Mitra, Digital Signal Processing, A Computer-Based Approach, 2nd ed., p. 70

[2] P.P. Vaidyanathan and Mitra, S.K., Low Passband Sensitivity Digital Filters: A Generalized Viewpoint and Synthesis Procedures, Proc. IEEE, vol. 72, no. 4, April 1984, pp. 404-423.


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