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# Tag Info

• 4,105
Accepted

### Why does reversing the order of these two transfer functions give me different outputs?

If the input is a unit step, then the output of the first block in system 1 is not zero, but it is a Dirac delta impulse $\delta(t)$. Intuitively, the derivative is infinite at $t=0$ because of the ...
• 79.2k
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### Basic Questions on Wiener Filtering

The task is to filter x(t) when given y(t), where y(t) = x(t) + n(t). Great but first we need to build an appropriate filter. At this point: No. The task is to filter $y[n]$ to achieve $x[n]$. Your ...
• 26.6k
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### Poles and zeros of a transfer function

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 ...
• 4,004
Accepted

### Transfer function intuition

Normally, in electrical enginnering, we apply the term "transfer function" and "filter" to an operation that belongs in the class we call Linear Time-Invariant systems (LTI). Sometimes you might read ...
Accepted

### Is this system causal or not?

Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial ...
• 79.2k
Accepted

### Transfer function with blackbox modelling is too slow compared to real expectation

In your original code you defined the sampling interval to be $T_s=0.05$ (idd1 = iddata(ddout, ddin, 0.05);). Yet, according to the data file, the time step between ...
• 79.2k
Accepted

### How accurate is the dominant poles approximation in higher order control systems?

It depends entirely on how close the less dominant poles are to the dominant poles. A simple way to understand what is happening is consider poles on the real negative axis for continuous time systems:...
• 36.1k
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### Estimate the Transfer Function of an Unknown System

Here's the way I think about a discrete Wiener Filter Consider a sequence of observations $\mathbf{y} \in \Re^n$ Form a matrix from the input $\mathbf{x} \in \Re^{n+r-1}$ by shifting columns one ...
• 2,880
Accepted

### What is the relationship between poles and system stability?

The two are both true, but they are for different cases. Case 1 is true for continuous-time systems, and the transform is the Laplace transform and the variable is the derivative operator, $s$. Case ...
• 21.7k

### transfer function of wiener filter

In general, an ideal Wiener filter (i.e. a non-causal filter with an infinitely long impulse response) has the following frequency response $$W(\omega)=\frac{S_{dx}(\omega)}{S_x(\omega)}\tag{1}$$ ...
• 79.2k
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### Deconvolution Using Response to an Heavy Side

If we can assume no noise (Or the SNR is very high) you can get the response by applying the inverse filter in frequency domain. Lets say $y [n]$ are the signal samples. Given $x [n]$ the samples ...
• 39.3k

### How frequency response related to a transfer function

An LTI system's "frequency response" tells you how the system acts on the amplitude and phase of a sinusoidal input. If the frequency response is $H(f)$, then an input $x(t)=e^{j2\pi f_0t}$ produces ...
• 13.6k
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### determining type of filter given its pole zero plot

You'd have to figure out the frequency response of the filter. Here are two methods. I prefer Method 2 because it's quick and dirty, and you don't really care about the exact gain values in the ...
• 4,004

### determining type of filter given its pole zero plot

In this answer I'll try to show you how to qualitatively evaluate a given pole-zero plot by just looking at it. Of course, this method has its limits, but for relatively simple pole-zero plots you can ...
• 79.2k
Accepted

### Why do we assume zero mean noise in sensor data?

The brutally honest answer here is: The noise is considered zero-mean because that's what the author decided to do. Without looking deeper into the signal model employed, it's impossible to answer. ...
• 25.8k

### A question about the meaning of pole in time domain

Let $H(s)$ be a transfer function of the form $$H(s) = \frac{1}{s-p}$$ where $p$, which is a pole of $H(s)$, can be written as a complex number $a+jb$. Taking the inverse Laplace transform of $H(s)$ ...
• 927

### Can someone explain waveshaping to me?

In the audio domain, waveshaping is simply applying a memoryless nonlinear function to an input signal. $$y(t) = g\big( x(t) \big)$$ The waveshaping function, $g(x)$, is most often a continuous ...

### What is the zero in this transfer function?

$$2s+1=2\left(s+\frac12\right)$$ That's all I can say.
• 79.2k

### How to realize Poles and zeros at infinity??especially through transfer function?

It's actually quite straightforward: positive powers of $s$ (or, in discrete-time, $z$), correspond to poles at infinity. Negative powers give you zeros at infinity. Let's look at some examples. In ...
• 79.2k
Accepted

Basically the transfer function is given by (Continuous time case): $$H \left( j \omega \right) = K \frac{ \left( j \omega - {q}_{1} \right) \left( j \omega - {q}_{2} \right) }{ \left( j \omega - {p}... • 39.3k 5 votes ### Filter odd or even harmonics with notch or inverse notch filter What you are looking for are what we, in the audio space, call comb filters. Comb filters may or may not have a feedback path, just like FIR and IIR filters. In fact, there is a generalized theory ... 4 votes Accepted ### Output of a system given it's transfer function and input (beginner) For an LTI system, output y(t) is given by$$y(t) = h(t)\otimes x(t)$$Where x(t) is input and h(t) is impulse response of the system. The operator \otimes represents convolution. Convolution ... • 447 4 votes Accepted ### Block diagram transfer function of a line The transfer function of a line is 1. And hence you can reduce your blocks into a single block with transfer function$$\frac{C(s)}{R(s)} = \frac{G(s)}{1+G(s)}$$where,$$G(s) = \frac{K}{s(s+2)(s+...
• 447
Your filter is the all-poles IIR, this simplifies things a bit. Normally you can write transfer function in following form: $H(z)=\dfrac{\sum_{i=0}^{P}b_{i}z^{-i}}{\sum_{j=0}^{Q}a_{j}z^{-j}}$ Going ...