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    Group delay of LTI system

    Found the solution by asking some friend. If someone wants to know it, let me know.
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    Partial Fraction Expansion (Inverse Z-Transform)

    Like here: http://dspcan.homestead.com/files/Ztran/zinvpart.htm" [Broken]
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    Partial Fraction Expansion (Inverse Z-Transform)

    Homework Statement H(z) = \frac{6-z^{-1}}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}} = k + \frac{A}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}} where A = (6-z^{-1}) is evaluated at z^{-1}=-2. Homework Equations Partial fraction expansion. The Attempt at a Solution Why is z^{-1} set equal to -2? I...
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    Group delay of LTI system

    Homework Statement Show that the group delay of an LTI discrete-time system characterized by a frequency response H(e^{j\omega}) can be expressed as \tau(\omega)= Re\left\{\frac{j\frac{dH(e^{j\omega}}{d\omega}}{H(e^{j\omega}}\right\} . Homework Equations The...
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    Sum of a finite exponential series

    Ah, I see the problem now. Thanks!
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    Sum of a finite exponential series

    Okay, thank you. For me, it is not about the sign in the exponent. I do not see why we have to multiply by the term in front of the fraction. But I think I rewrote the equation in the wrong way. Can you give me your steps?
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    LaTeX Software for viewing/writing LaTeX

    Try Texmaker. Very intuitive software!
  8. E

    Sum of a finite exponential series

    Yes, I've noticed that it starts there. That's why I thought it can be rewritten as \frac{1−e^{-j\omega(2N+1)}}{1−e^{−jω}}, but the solution states that this fraction is multiplied by e^{−jωN}.
  9. E

    Sum of a finite exponential series

    Homework Statement Given is \sum_{n=-N}^{N}e^{-j \omega n} = e^{-j\omega N} \frac{1-e^{-j \omega (2N+1)}}{1 - e^{-j\omega}}. I do not see how you can rewrite it like that. Homework Equations Sum of a finite geometric series: \sum_{n=0}^{N}r^n=\frac{1-r^{N+1}}{1-r} The Attempt at a...
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