18
One thing that really helped me understand poles and zeros is to visualize them as amplitude surfaces. Several of these plots can be found in A Filter Primer. Some notes:
It's probably easier to learn the analog S plane first, and after you understand it, then learn how the digital Z plane works.
A zero is a point at which the gain of the transfer ...
12
I think there are actually 3 questions in your question:
Q1: Can I derive the frequency response given the poles of a (linear time-invariant) system?
Yes, you can, up to a constant. If $s_{\infty,i}$, $i=1,\ldots,N,$ are the poles of the transfer function, you can write the transfer function as
$$H(s)=\frac{k}{(s-s_{\infty,1})(s-s_{\infty,2})\ldots (s-s_{\...
12
Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$,
the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$
In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is
$$\begin{align}
y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\
&= \exp(j2\pi ft)\int_{-\...
10
Yes, you can do this with an LMS equalizer which uses the Wiener-Hopf equation to determine the least squared solution to the filter that would compensate for your channel, using the known transmit and receive sequences. The channel is the unknown being solved, and the tx and rx sequences are known.
BOTTOM LINE:
Here is the Matlab function with error ...
8
For a start, any non-linear system will not have an easily-identifiable frequency response. So, it's really a nonsensical question. I intend no offense; nonsensical questions are often the most enlightening!
However one way to try to answer your question is to assume that the LTI filter involved is the mean (rather than the median) of the windowed data.
...
8
No. The impulse response and frequency response of an LTI system are related by the Fourier transform, which is one-to-one.
8
Yes for 2D signals you can take a 2D fft, and if the 2D signal is represented in the time domain, then its fft is represented in the frequency domain. 2D FFT's have many other interesting applications, for example image creation in synthetic aperture radar (SAR), where an inverse 2D FFT of radar reflections results in the creation of an image.
If your ...
7
Shortly, we have two kind of basic responses: time responses and frequency responses. Time responses test how the system works with momentary disturbance while the frequency response test it with continuous disturbance. Time responses contain things such as step response, ramp response and impulse response. Frequency responses contain sinusoidal responses.
...
7
If you know that the system is linear and time invariant, the easiest method (assuming that you have no noise added in the process) is to let the system act on an impulse function. The Fourier transform of the output is the frequency response of the system.
7
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 ...
7
What you do in step 1 is simply truncate the infinite impulse response to approximate it by an FIR filter. If you use sufficiently many filter taps, the approximation becomes arbitrarily accurate. This means that the resulting FIR filter approximates the magnitude and the phase characteristic of the original IIR filter. So with this approach the phase will ...
7
To answer this you need to understand what is a pole and what is a zero of a transfer function.
Let's look at a simple 2 poles 2 zeros filter (also called biquad filter) transfer function :
$$
H(z) = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1 z^-1 +a_2 z^{-2}}
$$
This can be factored as :
$$\begin{align}
H(z) &= \frac{ b_0 \, (1-q_1 z^{-1})(1-q_2 z^{-1})}{(...
7
The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the argument. Using this note that
$$
|e^{-j\Omega \lambda}| = 1
$$
which is why it "disappeared".
6
FDLS requires a causal frequency response. Your prototype frequency response has zero phase everywhere, which is most definitely not causal.
An IIR filter order of 50 is humongous. When FDLS has too many poles and zeroes available, it "tries" to cancel excess poles with excess zeroes. Unfortunately, due to numerical limitations, the cancellation is often ...
6
As Dilip pointed out in the comment above, you can get the impulse response using the inverse Fourier transform. However, a slightly easier method might be to use the Laplace domain instead; it's more amenable to easy inverse transforming via transform tables. First, recall that the frequency response is really just the $s$-plane transfer function evaluated ...
6
You're definitely on the right track. The way you're trying to solve the problem is the best and simplest. You just need to realize that you need to evaluate the magnitude and phase of the frequency response just for one frequency, namely the frequency of the sinusoidal input signal:
$$y[n]=\left|H(e^{j\omega_0})\right|\sin\left(n\omega_0+\phi(\omega_0)\...
6
This has absolutely nothing to do with causality. The frequency response of a real-valued filter (i.e., one with a real-valued impulse response) is (conjugate) symmetric, i.e., the negative frequencies are redundant. That's why it is sufficient to show the frequency response at non-negative frequencies only.
You can easily see that symmetry as follows. The ...
5
In previous sections of the book, the fact that a discrete-time signal's spectrum is periodic may have been mentioned. It can be described formally as follows:
$$X(e^{j\omega})=\frac1{T} \sum_{k=-\infty}^{\infty}X_C\biggr(j\biggr(\frac{\omega}{T}-\frac{2\pi k}{T}\biggr)\biggr)$$
being $X_C(j\omega)$ the Fourier Transform of the continuous signal, and $T=1/...
5
The analytic way is to substitute the variable $z$ by $e^{j\omega}$ to get the frequency response $H(\omega)$ (with $\omega = \frac{2 \pi f}{F_s}$) - that is to say, the frequency response is the $z$ transform evaluated on the unit circle.
Note that matlab has a built-in function for plotting the frequency response straight from filter coefficients (freqz), ...
5
What you're looking for is called a pruned DFT. In principle, it is possible to calculate a subset of outputs from a DFT using fewer mathematical operations. In practice, however, existing highly-optimized FFT implementations like FFTW are designed for full-output transforms. You'll find in many cases, unless you're only concerned with a very small ...
5
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 an output $y(t)=|H(f_0)|e^{j(2\pi f_0t+\angle H(f_0))}$. It is common to divide the frequency response in two, the gain $|H(f)|$ and the phase $\angle H(f)$.
...
5
I found the following in Charles Therrien's "Discrete Random Signals and Statistical Signal Processing" in one of the Appendicies.
Say you have the function $Q(a)$ you wish to minimize such that $C(a)=0$, where $C(a)$ may be complex valued and $a$ may be a complex vector. The constraint really represents two real-valued constraints.
$$C_r(a)=0,\qquad
C_i(a)...
5
Frequency response has two parts: amplitude response and the phase response. Both of these are represented as a complex signal when you get the response from freqz. In order to plot the amplitude response you need to use abs. Otherwise I doubt it only shows you the real part which I think is what you see in the first figure. Note that when dealing with the ...
5
Assume you have the signal
$$x(t)=a+b\cos(\omega_0t)\tag{1}$$
with some non-zero real-valued constants $a$ and $b$. Now remove DC and the negative frequencies to obtain
$$x'(t)=\frac{b}{2}e^{j\omega_0t}\tag{2}$$
Can you calculate the value $a$ from $(2)$?
5
Because you don't have background in wireless communications, I will try to answer as simply as possible.
What is the time unit of the converted CIR?
Just time unit. It can be second, ms, us, etc. They are convertible, aren't they?
How much seconds are they (the 30 impulse responses) apart?
In the case that CSI-pilots are equally seperated, impulses ...
5
$$1 - e^{-4j\omega} = e^{-2j\omega}(e^{2j\omega} - e^{-2j\omega}) \tag{1}$$
Now,
$$ \sin(2\omega) = \frac{e^{2j\omega} - e^{-2j\omega}}{2j} \tag{2}$$
Equation 2 is a consequence of Euler's formula. Multiply and divide by $2j$ in (1) and use the identity (2) in equation 1 we have:
$$1 - e^{-4j\omega} = 2je^{-2j\omega}\sin(2\omega) \tag{3}$$
Now $j = e^{...
5
In general there is no straightforward analytical solution. As you know, you need to solve
$$\left|H(e^{j\omega_c})\right|=\frac{1}{\sqrt{2}}\tag{1}$$
for $\omega_c$, where it is assumed that the maximum filter gain equals $1$.
For Butterworth filters, the specified cut-off frequency always equals the $3\textrm{ dB}$ frequency. This is not the case for other ...
5
You can equalize magnitude and phase simultaneously by defining a desired complex frequency response
$$D(\omega)=M(\omega)e^{j\phi(\omega)}\tag{1}$$
with magnitude $M(\omega)$ and phase $\phi(\omega)$ chosen such that they compensate for the given magnitude and phase distortions.
An FIR filter approximating $(1)$ can be designed by using the following error ...
4
First of all, your transfer function has a pole at $z=-3$, which means your filter is unstable and you will probably see your output going to infinity if your process anything with it.
However, in general, you can process an impulse with your difference equation and do an FFT (or freqz) of the resulting impulse response. That should give the same result as ...
4
The key ingredient is that the base functions of the Fourier transform $\exp(i \omega t)$ are eigenfunctions of LTI systems. That means the LTI system can be represented as a diagonal linear operator in the Fourier basis. Or in other words: To apply an LTI system in frequency domain, you just multiply their frequency responses. And applying an LTI system to ...
Only top voted, non community-wiki answers of a minimum length are eligible
