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So by the DTFT pairs, if the magnitude of the frequency response of a signal is 1, then the autocorrelation is the Kroneker Delta Function.

What if I find that the magnitude of the frequency response of the system is greater than one?

Is the autocorrelation just the constant multiplied by the Kroneker Delta Function?

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Make sure not to confuse the power spectral density (PSD) with the frequency response.

  • PSD is the Fourier transform of the autocorrelation function.
  • Frequency response is the Fourier transform of impulse response.

That aside, your question is about the Fourier transform of a constant $c$. Since $$\mathcal{F}\{1\}=\delta(\omega),$$ considering linearity of the Fourier transform we have $$\mathcal{F}\{(c)1)\}=(c)\delta(\omega)$$

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  • $\begingroup$ I don't think that the OP is confusing frequency response and PSD. He's talking about the deterministic autocorrelation function, which is related to the (squared) magnitude of the frequency response. $\endgroup$
    – Matt L.
    Sep 30, 2016 at 6:44
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The deterministic autocorrelation function of a finite-energy discrete-time sequence $h[n]$ is defined by

$$r_h[n]=\sum_{k=-\infty}^{\infty}h[k]h[k+n]\tag{1}$$

Its (discrete-time) Fourier transform is the squared magnitude of the Fourier transform of $h[n]$:

$$\text{DTFT}\{r_h[n]\}=|H(e^{j\omega})|^2\quad\text{with}\quad\text{DTFT}\{h[n]\}=H(e^{j\omega})\tag{2}$$

From $(2)$, if $|H(e^{j\omega})|=1$, then $r_h[n]=\delta[n]$, as mentioned in your question.

Now in general, if $|H(e^{j\omega})|=c$, $c>0$, then

$$r_h[n]=\text{IDTFT}\{|H(e^{j\omega})|^2\}=\text{IDTFT}\{c^2\}=c^2\delta[n]\tag{3}$$

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  • $\begingroup$ Did you just concluded from $|H(e^{j\omega})|=1$ that $h[n]=\delta[n]$? I don't think this is correct. $\endgroup$
    – msm
    Sep 30, 2016 at 6:57
  • $\begingroup$ @msm: No, I didn't, because that would indeed be incorrect. Did I write this? What I have written is that it is $r_h[n]$ that equals $\delta[n]$ if $|H(e^{j\omega})|=1$. $\endgroup$
    – Matt L.
    Sep 30, 2016 at 7:03
  • $\begingroup$ So won't you refer to the FT of the autocorrelation function ($r_h[n]$) as PSD? $\endgroup$
    – msm
    Sep 30, 2016 at 7:11
  • $\begingroup$ @msm: In the deterministic case, for finite-energy signals, the squared magnitude of the Fourier transform is usually referred to as Energy Spectral Density. Power spectral density is usually used for power signals (with infinite energy), and for random processes. In these latter cases, the definition of the autocorrelation function is also different. $\endgroup$
    – Matt L.
    Sep 30, 2016 at 7:18
  • $\begingroup$ I am fine as long as it is a spectral density (although I don't agree that this term (PSD) is not used in DSP content where most often we deal with energy signals) $\endgroup$
    – msm
    Sep 30, 2016 at 7:31

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