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suppose I have the following discrete time signal $x\left(n\right)=0.7\sin\left(2\pi \times 623\:nT\right)$ where $N=4000$ and $T=\frac{1}{f_s}$, where $f_s=40$ kHz

I pass this signal through a simple IIR filter $ H\left(z\right)\:=\:\frac{0.2}{1-0.3z^{-1}}$

I calculate the power of the ouput signal $P_y\:=\:P_x\:\left|H\left(z\right)\right|^2_2$ basically am using the filter 2-norm in this case. for that my calculations would be as follows:

$P_y\:=\left(0.245\right)\left(\frac{0.2^2}{1-0.3^2}\right)\:=\:0.0108$

However when I develop this in Matlab, I get the output power as 0.0199 Here's my code in Matlab

fs = 40000;
f1 = 623;
N = 4000;
n = 1:N;
b = 0.2;
a = [1 -0.3];
x = 0.7*sin(2*pi*f1*n/fs);
P_x = var(x);
y = filter(b,a,x);
P_y = var(y);

I don't understand what I was doing wrong, would someone please clarify for me? I would appreciate it. Thank you in advance.

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  • $\begingroup$ $H(z)|^2$ is a function of frequency not a single number. How did you come up with your expression for $H(z)|^2$ . The squared magnitude of the denominator is $1+0.3^2- 0.3(z^{-1}+z)$ $\endgroup$
    – Hilmar
    Commented Apr 25, 2021 at 19:33
  • $\begingroup$ Because of the two-norm. That's how i learned it. $\endgroup$
    – JordenSH
    Commented Apr 25, 2021 at 21:10
  • $\begingroup$ The 2-norm of a digital filter is defined as $||H(e^{j\omega})||_2 = (1/2\pi \int_{-\pi}^{\pi}|H(e^{j\omega})|^2 d\omega)^{1/2} = (\sum_n |h(n)|^2)^{1/2}$. The second equation is derived from Parseval's theorem. I think what you mean is just the magnitude $|H(z)|$. $\endgroup$
    – ZR Han
    Commented Apr 26, 2021 at 3:12
  • $\begingroup$ @ZRHan you got it correct, I DID mean the 2-norm. That's what I mean and is getting me confused because I don't know why my matlab code is not giving the same result as the theory of the filter 2-norm. Will you please provide explanation or an answer? $\endgroup$
    – JordenSH
    Commented Apr 26, 2021 at 6:43
  • $\begingroup$ I think what you need is the power of output signal through an IIR filter. The output has a frequency response $Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega})$ and its overall power is $\int_{-\pi}^\pi |Y(e^{j\omega})|^2d\omega = \int_{-\pi}^\pi |X(e^{j\omega})|^2|H(e^{j\omega})|^2d\omega$, which is obviously not equal to the power of input times the 2-norm of the filter. $\endgroup$
    – ZR Han
    Commented Apr 26, 2021 at 7:15

1 Answer 1

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The analog frequency $\Omega=2\pi\times 623$ and the digital frequency is $\omega = \Omega T = \Omega/f_s = 0.03115\pi$

The frequency response at this frequency is $$H(z)|_{z=e^{j\omega}} = H(e^{j0.03115\pi}) = \frac{0.2}{1-0.3 e^{-j0.03115\pi}}$$

and its square of magnitude is

$$ |H(e^{j0.03115\pi})|^2 = \Big|\frac{0.2}{1-0.3 e^{-j0.03115\pi}}\Big|^2 = 0.0812 $$

Now you'll see $P_y = P_x |H(e^{j0.03115\pi})|^2 = 0.0199$

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  • $\begingroup$ This is so useful, thank you very much, but would you tell me how come the 2-norm did not work here? $\endgroup$
    – JordenSH
    Commented Apr 26, 2021 at 16:31
  • $\begingroup$ also, why did the $0.03115\pi $ change to $0.1557\pi $? thank you $\endgroup$
    – JordenSH
    Commented Apr 26, 2021 at 16:33
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    $\begingroup$ @JordenSH Sorry, it's $0.03115\pi$. The 2-norm describes the energy of entire digital filter including all frequencies. But you have a single frequency sine wave input and output. Essentially, it's math. Write down the definition of signal power in the frequency domain step by step and you'll find it out. $\endgroup$
    – ZR Han
    Commented Apr 27, 2021 at 1:00

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