# 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 ...

9

This derivation is a tricky one. The approach suggested before has a flaw. Let me demonstrate this first; then I will give the correct solution. We wish to relate the $\mathcal{Z}$-transform of the downsampled signal, $Y_D(z) = \mathcal{Z}\{x[Mn]\}$, to the $\mathcal{Z}$-transform of the original signal $X(z) = \mathcal{Z}\{x[n]\}$. The wrong way One ...

8

Why is the fourier transform a special case of the laplace transform? The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter $H(s)=\frac{1}{s+1}$ has a ...

8

First it's important to realize that many authors use the terms zero-input response and natural response as synonyms. This convention is used in the corresponding wikipedia article, and for instance also in this book. Even Proakis and Manolakis are not entirely clear about it. In the book you quoted you can find the following sentence on page 97: [...] ...

8

The system $$y[n]=y[n-1]+x[n]\tag{1}$$ is an ideal accumulator, i.e., it computes the cumulative sum of the input samples: $$y[n]=\sum_{k=-\infty}^nx[k]\tag{2}$$ It is in a way analogous to a continuous-time integrator, but this doesn't mean that you will necessarily obtain an ideal integrator by transforming the discrete-time system to a continuous-time ...

8

short answer: all the poles of a causal (right-sided) and stable LTI system must be inside the unit circle whereas all the poles of an acausal (left-sided) and stable LTI system must be outside the unit circle. The explanation: Consider a causal LTI system with right-sided impulse response $h[n]$ whose transfer function can be expressed in terms of ...

8

It's not sufficient to only consider causality, you also need to check whether the inverse system is stable, otherwise it can't be implemented. If $G(z)$ has zeros on the unit circle, it cannot be inverted. If $G(z)$ has no zeros on the unit circle, but if there are zeros outside the unit circle, then there is no causal and stable inverse, because the zeros ...

7

The problem is not sufficiently specified, because the range of admissible values of $n$ is missing. Here I make the assumption that we consider $n>0$. With this assumption we have $$X(z)=\sum_{n=1}^{\infty}x[n]z^{-n}=\sum_{n=1}^{\infty}\frac{z^{-n}}{n^2}\tag{1}$$ And that's the point where we might get stuck, if we didn't have a list of mathematical ...

6

Note that a stable and causal continuous-time transfer function does not need to be strictly proper but only proper, i.e. the degree of the numerator does not exceed the degree of the denominator, but numerator and denominator degree can be equal. E.g. $$H(s)=\frac{as^2+bs+c}{s^2+ds+e}$$ can represent a causal and stable system, as long as its poles are in ...

6

First of all, I think you're reading the wrong books. Almost any basic text on DSP has a chapter on the $\mathcal{Z}$-transform and its significance to describe linear time-invariant (LTI) discrete-time systems. If you're looking for good (and free) books, take a look at this answer. I will not repeat all the details you can find in those books (and in many ...

6

Consider a liner discrete-time system. Assume we can define it in terms of an input-output relation as follows (you can assume a more general model but it is enough for our purpose): $$a_0y[n]+a_{1}y[n-1]+\cdots+a_{N}y[n-N]=b_0x[n]+b_{1}x[n-1]+\cdots+b_{M}x[n-M]\tag{1}$$ When the coefficients $\{a_i\}$ and $\{b_i\}$ are constant, we call it a finite-order ...

6

The "poles-inside-unit-circle" stability criterion only applies to causal systems. Your system is not causal because it uses one sample from the future owing to the $z$ term. The general technique to check for stability involves looking at the regions of convergence (ROC) of $H(z)$. If the ROC includes the unit circle, then the system is stable. See also ...

6

There are a couple reasons. One is that $(1-z^{-1})$ represents $x[n]-x[n-1]$ which is a finite difference over a very small period of time. and that is an approximation to a differentiator. The reciprocal is $$\frac{1}{1-z^{-1}} = \frac{z}{z-1}$$ which is the inverse operator. We normally call the inverse operation of differentiation, we call that "...

5

If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z ...

5

Tricky problem. Not sure I can answer this but here are a few pointers: Direct Form II is the worst biquad for audio processing. The transfer function between the input and the state is given just by the poles. The gain can be really large, I have seen gains of in excess of a 100 dB for reasonable audio filters. This makes real time switching of Direct Form ...

5

I've not seen this notation before. However, it does seem to make sense. The $M$-downsampler is defined by the equation: $$y_D[n] = x[Mn]$$ Its $z$ transform is defined by the equation: \begin{align} Y_D(z) &= \sum_{n=-\infty}^\infty y_D[n]z^{-n} \\ &= \sum_{n=-\infty}^\infty x[Mn] z^{-n} \end{align} Apply a change of variable, letting $n'... 5 The time domain signal (or impulse response) $$h(n)=a^{n}\cos n\theta_0,\quad \theta_0=2\pi\frac{f_0}{f_s},\; n\ge 0$$ is very common: it is a damped sinusoidal function (assuming$|a|<1$) which occurs frequently, because it is one possible response of a second order linear time-invariant system. So concerning your doubt, the cosine part is definitely ... 5 If you have no prior knowledge about the approximate locations of the frequencies, the Chirp Z-transform is of no immediate use to you. The Chirp Z-transform functions like a magnifying glass, so you need to know where you want to look and the Chirp Z-transform will show you the details. I would suggest you use an FFT to get an idea where the frequencies are,... 5 This is related to Chirp Z-transform (CZT) (refer to the Bluestein's algorithm). Using this identity, the CZT can be expressed in terms of a convolution. Hence, it can be efficiently implemented using FFT. 5 Let me show you a simple way to see this property. Assume$x[k]$is a causal sequence and let $$x[\infty]=\lim_{k\rightarrow\infty}x[k]$$ be finite. Then the sequence$x[k]$can be written as $$x[k]=x[\infty]u[k]+y[k]\tag{1}$$ where$u[k]$is the unit step sequence, and$y[k]$is a causal sequence that decays to zero as$k\rightarrow\infty$. Taking the$\...

5

The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform. This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds ...

5

if you define the constant like this: $$x[n] = C ~~~,~~~\text{ for all } n$$ Then its $\mathcal{Z}$ transform $$X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} = \sum_{n=-\infty}^{\infty} C z^{-n}$$ does not converge for any value of $z$. Hence its $\mathcal{Z}$ transform does not exist. Note that $\mathcal{Z}$ transform must be an analytic function with ...

5

Instead of considering a constant signal, let us consider a rectangular window of length $2 \ell + 1$ $$w (n) = \begin{cases} 1 & \text{if } |n| \leq \ell\\ 0 & \text{otherwise} \end{cases}$$ whose $\mathcal Z$-transform is $$W (z) = z^{-\ell} \left( 1 + z + z^2 + \cdots + z^{2 \ell} \right) = \dfrac{1}{z^\ell} \left(\dfrac{z^{2\ell+1} - 1}{z-1}\... 5 Just to avoid a misunderstanding: the \mathcal{Z}-transform is a transform defined for sequences, comparable to the Laplace transform for continuous functions. What you are talking about is not the \mathcal{Z}-transform, but methods for converting analog to digital (actually, discrete-time) systems. [And it doesn't help that one of those conversion ... 5 First of all, it's important to understand that there is no single best way to transform a continuous-time system to a discrete-time system. The method you're using is called backward Euler method, and it is defined by the mapping$$s\leftarrow\frac{1-z^{-1}}{T}\tag{1}$$Note that in (1) you scale by 1/T, where T is the sampling interval (i.e., 1/T ... 5 It's natural consequence of applying a transform to a convolution relation. The output y(t) of an (continuous-time) LTI system is described by a convolution integral :$$y(t) = h(t)\star x(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$And when you apply a Fourier transform on this relation, it turns out to be a multiplication in the transform ... 4 I'll go through it in the z-domain. First, we find the transfer function H_1(z) = \frac{V(z)}{X(z)}. As you noted, in the time domain, x[n] and v[n] are related as follows:$$ v[n] = x[n] + g * v[n - M] $$Take the z-transform of the above and you get:$$ V(z) = X(z) + z^{-M}G(z) V(z) $$taking advantage of the convolution property, which ... 4 No, it is not possible to generate the z-domain transfer function uniquely and solely from the pole-zero plot. The reason is because you can only generate something like:$$ H(z) = K \frac{(z-Z_1)(z-Z_2)...(z-Z_m)}{(z-P_1)(z-P_2)...(z-P_n)} $$from the pole-zero diagram, and there is nothing in that diagram to tell you what the gain term, K, is. 4 Without knowing specifics (ignoring a proportionality constant we'll call K as inidicated in the more compete answer above). Note that K does not affect the dynamic behavior of the system, so it may not be of particular interest. it goes like this... H(z) is a rational function of the form$$H(z) = \frac{(z-Z1)(z-Z2)...(z-Zm)}{(z-P1)(z-P2)...(z-Pn)} ...

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