# Tag Info

15

Mathematically we can easily show that the z-transform of a unit cycle delay is $z^{-1}$, Just like the s-transform of a time delay $\tau$ is $e^{-\tau s}$. I would like to add some additional insights that may lead to a more intuitive understanding: Why z?? NOTE: To really understand how the z-transform and unit delays are related (intuitively), it really ...

5

For example a delay of 0.75 sampling periods would have an impulse response like this (red squares sampled from the blue delayed sinc): Time (horizontal axis) unit is sampling periods. That kind of a filter is not causal as there are non-zero samples at negative times. None of the samples is zero, no matter how far left or right you go. That's what they ...

4

Given some digital signal x[n], it's z-transform is defined by, $$X(z) = \sum_{k}(x[k].z^{-k})$$ When a system equation is $$H(z) = z^{-1}$$ It essentially means the output signal is Y(z) = H(z)*X(z) \\ i.e. Y(z) = \sum_{k}(x[k].z^{-k-1}) \\or\\ Y(z) = \sum_{l}(x[l-1].z^{-l}) \end{...

4

You are right that a distributed system could be "something like a transmission line". Note that the system $$y(t)=x(t-T)\tag{1}$$ is a simple model of a transmission line, where just a frequency-independent delay $T$ is taken into account, and the attenuation is neglected. Note that lumped electrical systems, described by resistors, capacitors and ...

3

I faced the same problem in the past. Perhaps there is a way without adding a delay but I haven't found it. You need to realize that your 3 first solutions (delay after vq, delay at the delta_freq and delay after the frequency) will yield the same result as omega_g is a constant and because your PI controller has fixed coefficients. Anyway, place the ...

2

The ideal delay has a frequency response of: $$H(e^{j\omega}) = e^{-j\omega D}$$ this has impulse response $$h(n) = \mbox{sinc}(n - D) = \frac{\sin(\pi(n-D)}{\pi(n-D)}.$$ For $D$ an integer, this just becomes: $$h(n) = \delta(n-D)$$ where $\delta$ is the Kronecker delta. That means there is no resampling, so I am not sure where you get that from (it's ...

2

With $A(q^{-1},t)$ as given in the formula you get $$A(1,t)=\prod_{k=1}^n(1-2\cos(k\hat\omega_0(t))+1)\tag{1}$$ and $$A(\rho,t)=\prod_{k=1}^n(1-2\cos(k\hat\omega_0(t))\rho+\rho^2)\tag{2}$$ So $K(t)$ is simply the quotient of (1) and (2).

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