9

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 ...


8

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

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

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

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 ...


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 ...


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

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

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: [...] ...


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

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 ...


4

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 ...


4

To be completely honest, I thought the theory behind the Z-transform was kinda opaque in college too. In hindsight, taking a course in complex analysis would have made it clearer. And I too dislike the notational conventions that seem to be used for this stuff. Strictly speaking, the usual convention here is that $x[n]$ denotes a discrete-time sequence $n\...


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 $$H(z) = \frac{(z-Z1)(z-Z2)...(z-Zm)}{(z-P1)(z-P2)...(z-Pn)}$$ ...


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

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 ...


4

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 ...


4

The actual computation for the inverse Z transform is $$\frac1{z-0.5}=\frac1z\cdot\frac1{1-\frac{0.5}{z}}=z^{-1}\sum_{k=0}^{\infty}(0.5)^k z^{-k}=\sum_{n(=k+1)=1}^\infty (0.5)^{n-1}z^{-n}$$ From that you see that the series is one-sided, $h(n)=0$ for $n<1$. This should remove most of the infinities that you stumbled upon. Note that this geometric ...


4

The S-transform allows you to deal with differential equations in an algebraic manner - so they become easier to solve. Since continuous/analog filters consist of integrators and differentiators the S-transform is therefore a natural way to deal with these systems. The z-transform provides an algebraic way of dealing with finite difference systems and ...


4

Actually, the Z transform is not really a proper transform, just a re-interpretation of the sequence of samples as coefficients of a formal Laurent series. In some cases the formal Laurent series converges, if it does, it does so on an annular region in the complex plane. For useful signals (stable, summable, exponentially decaying) this annulus contains ...


4

A transfer function $H(Z)$ can be written as $H(Z)=\frac{Y(Z)}{X(Z)}$. Then, your $H(Z)$ can be written as $\frac{Y(Z)}{X(Z)}=1-\cos\theta~Z^{-1}+Z^{-2}$ or $Y(Z)=X(Z)(1-\cos\theta~ Z^{-1}+Z^{-2})$ Now taking the inverse $Z$transform, we get the difference equation as $y(n)=x(n)-\cos\theta~x(n-1)+x(n-2)$. I hope this help you.


4

You are right that the (bilateral) Laplace transform can be interpreted as the Fourier transform of $e^{-\sigma t}f(t)$. However, I think that the significance of the Laplace transform only becomes clear when $s=\sigma+j\omega$ is viewed as a complex variable because then we can study the analytic properties of the system function. E.g., electrical networks ...


4

The first result is correct, the second is wrong. You're right that $x_i[n]=x[n-1]$ corresponds to $z^{-1}X(z)$. And if you replace $z$ by $1/z$ you need to replace $n$ by $-n$. But $x_i[-n]$ is simply $x[-n-1]$, and not $x[-n+1]$. Note that you don't replace the whole argument of $x[n-1]$ by its negative value, but you just replace $n$ by $-n$.


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