We changed our privacy policy. Read more.
13

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_{-\...


11

Your professor is right, and you're almost right too. The filter is clearly an FIR filter, but because its frequency response can be expressed as a geometric series, a recursive implementation is possible. If you write the transfer function as a rational function you get $$H(z)=2\frac{1-z^{-12}}{1+z^{-2}}\tag{1}$$ which is almost the same as you got, apart ...


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

No. The impulse response and frequency response of an LTI system are related by the Fourier transform, which is one-to-one.


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

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

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

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

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

The intuitive answer is that an impulse in time at t=0 contains all frequencies of equal magnitude, so applying an impulse to an LTI system is the same as applying all frequencies at once, thus the result is the response of the system to all frequencies, i.e., the frequency response. For a real world example, you can find the total frequency response of a ...


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 ...


4

You calculating FFT only from two samples. You need to pad your impulse response with zeros to get a valid result. So in MATLAB that would be: N = 1024; % Number of points to evaluate at % Create the vector of angular frequencies at one more point. % Filter itself b=[1,-1]; [h_f, w_f] = freqz(b, 1); figure grid on hold on plot(w_f, abs(h_f), 'or') % MATLAB ...


4

I am afraid that it is rather impossible without a proper hardware. Sweep sine is ok as a general method, but you would either need: Reference transducer with known (preferably) linear frequency response. Then you can find the difference between those two. Signal actuator with known frequency response, and then you can also find the difference between ...


4

For the given system you can write down the input-output relation as $$y[k]=\frac14\left(x[k]+2x[k-1]+x[k-2]\right)\tag{1}$$ because $T$ (or $z^{-1}$) denotes a delay element, which delays its input by one sample interval. The $\mathcal{Z}$-transform of (1) is (assuming zero initial conditions) $$Y(z)=\frac14\left(X(z)+2X(z)z^{-1}+X(z)z^{-2}\right)=\frac{...


4

The magnitude of the frequency response will remain unchanged if you reflect any poles outside the unit circle - these are the ones causing instability - back inside the circle. I.e., a pole $p$ (with $|p|>1$) is replaced by the new pole $\tilde{p}=1/p^*$, where $*$ denotes complex conjugation. This will not change the magnitude of the frequency response (...


4

You just shifted the low pass filter to the right, so you generated a complex-valued filter, as you've observed. Multiplying a real-valued impulse response with a complex exponential naturally results in a complex-valued impulse response. What you actually have to do is shift the spectrum to the right and to the left: $$h_{BP}[n]=h_{LP}[n]e^{jn\omega_0}+h_{...


4

As far as I can tell, the book is not wrong. (By the way, you're studying a 3-point moving average filter here, and it's useful to know about such filters.) Your 1st question: That center dot on the Mag. and Phase curves serves no purpose that I can see. I suggest you not worry about that center dot. Your 2nd question: The Phase curve is generated by ...


Only top voted, non community-wiki answers of a minimum length are eligible