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A (discrete) LTI system has the frequency response

$$ H(e^{j\omega}) = \frac{1-1.25e^{-jw}}{1-0.8e^{-j\omega}} $$

Show that $|H(e^{jw})|=G^2$, where G is a constant. Determine the constant $G$.

Expanding the exponentials to the form $e^{j\theta} = cos\theta + jsin\theta$, gives the following expression:

$$ \frac{(1 - 1.25\cos{\omega}) + j1.25\sin{\omega}}{(1 - .8\cos{\omega})+ j0.8\sin{\omega}} $$

the square of the magnitude is

$$ \frac{1-2.5\cos{\omega}+2.25}{1 - 1.6\cos{\omega}+0.64} $$

Clearly, putting in different values of $\omega$ will give different values of squared magnitudes in the expression above.

I feel that my approach is not correct; where did I go wrong?

Shukant Pal
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  • The numerator for the squared magnitude is not correct. Correctly computing it, results in a numerator that is a multiple of the denominator. Maybe it would be easier to do computations using $8/10$ and $10/8$ in place of $0.8$ and $1.25$, respectively. – Stelios Jul 05 '20 at 04:51
  • @Stelios I see I squared 1.25 wrong (thinking it was 1.5 instead). Thanks for the hint. – Shukant Pal Jul 05 '20 at 16:06

1 Answers1

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Your calculation of the squared magnitude is wrong because $(1.25)^2\neq 2.25$. The magnitude of the given transfer function is indeed constant. This can be shown in a general way as follows:

$$\begin{align}H(z)&=\frac{1-az^{-1}}{1-\frac{1}{a}z^{-1}}\\&=a\frac{1-az^{-1}}{a-z^{-1}}\\&=-az\frac{P(z^{-1})}{P(z)}\tag{1}\end{align}$$

On the unit circle $z=e^{j\omega}$ we have $|z|=1$ and $|P(z)|=|P(z^{-1})|$, and, consequently, $|H(z)|=|a|$. Hence, for the given example we obtain $G=a=1.25$.

Matt L.
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