# Tag Info

13

You have created an ideal (infinite bandwidth OPAMP, infinite output voltage etc...) model which is conditionally stable & executed it. look at your y-axis, it has reach 1.5x$10^{59}$ HUGE!. This is what is causing an exception... its run out of resolution to represent this unstable "oscilator" IF a closer to reality OPAMP is used (+-15V, gain ...

10

It does oscillate, you can see that it does on the right side of the plot. If you zoom in on the left side you should also see the oscillation but at a much smaller amplitude. Your misconception comes from the fact that you first have to understand the theory of operation of oscillators. I suggest reading about the Barkhausen stability criterion. This ...

8

My experience of Matlab to HDL (using AccelDSP which Xilinx then bought) has not been that great. One of my problems has always been that Matlab didn't offer any pleasing way to instantiate a module (entity in VHDL terms, object in OO terms) more than once if it had state inside. You had to pass the old state in as a parameter and the new state would come ...

7

I'm afraid you're up against a fundamental limitation of real, causal physical systems. You're asking for a filter that can distinguish between signals that have periods of 12 to 16 seconds, but only allowing it to "look at" a 2-second segment (1/6 to 1/8 of the period) of the waveform in question. It simply isn't possible to get no phase shift and low delay ...

7

Your model is not built in simpowersystem environment so you can not use the voltage measurement block. The correct way to measure the voltage, is by using "voltage sensor" from Simscape>foundation library> Electrical> Electrical sensors. And then use "PS-Simulink Converter" to convert the physical signal to a simulink signal. And most importantly, use "...

6

Matlab is largely equal to Mathematical is largely equal to Maple. Just because you are more familiar with one does not make it "better". The reason Matlab/Mathematica/Maple are popular are because they are much, much easier to get started in then C++/most programming languages. They come ready-to-go with everything you need to draw graphs, do differential ...

6

Write the TF as $\small G(s)=\large\frac{ab}{(s+a)(s+b)}$, where $a$ and $b$ are real and distinct. This factorises to: $\small G(s)=\large\frac{ab}{(b-a)} \large(\frac{1}{s+a}-\frac{1}{s+b}\large)$ From which the inverse LT, i.e. the unit impulse response, is: $g(t)= \large\frac{ab}{(b-a)} \normalsize (e^{-at}-e^{-bt})$ Differentiating and ...

6

There is no problem here at all. The measured step response is in perfect agreement with the simulation. It's just tricks of scaling and windowing that is making it appear that anything is amiss. Here, I reproduced your circuit in LTspice: I then matched the timing of the pulse to what is shown on your oscilloscope, and also set the voltage and time ...

6

The mistake you are making is not seeing that there is a time delay lag in the digital signal - this causes you to believe that the noise is mainly related to the input signal - try aligning the digital version of the signal with the analogue signal and you will find a point where there is very little relationship between Q-noise and original input (time ...

6

LTspice solves nonlinear differential equations numerically. It's not an AI system that can figure out intelligently whether a diode is conducting or not, and thus solve your homework questions or whatever. In general I think that models that are not "smooth" will cause convergence problems. So a perfect resistor is fine, but a perfect diode would have ...

5

There is a little uncertainty in my mind what the OP is asking for so I'll try and generalize a little. Look at the drawing below (top half) - it shows three examples of bode plots for 2nd order systems where damping has a few values: - Below the three bode plots are examples of the pole zero diagram and how it relates to the bode plot. Mathematically they ...

5

I found the mistake in my code: when the Matlab's Filter Design & Analisys Tool generates the cofficients, there are even the Gains for the every biquad section. Before the value goes out from one section, it must be multiplied with this gain. So the right code is the following: input = AmplitudeF8.'; delayline_lp = [0,0,0,0,0,0,0,0,0,0]; ...

5

It has started to oscillate. The amplitude grows exponentially. Simulation is stopped when some current, voltage or internal variable reaches the limit of the available number range. Take a couple of zener diodes connected in series having the anodes against each other. Insert that limiting circuit in parallel with one of the capacitors. The amplitude does ...

5

Just to add yet another post on this. Here's a circuit that should actually work, using a quad LT1631 and a single power supply rail: simulate this circuit – Schematic created using CircuitLab The output frequency should be close to about $\frac{1}{2\pi R C}$ or near $15.9\:\textrm{kHz}$ for the given values of $R=1\:\textrm{k}\Omega$ and $C=... 5 There's a double integrator in the TF that inverse transforms to a ramp, which is what you're seeing added to the sinusoidal response to the sinusoidal input. 5 The problem is you are shorting the supply. What you have is the NEUTRAL of the supply (associated with the SimPowerSystems reference) and the NEGATIVE of the rectifier commomed together. Essentially all phases are shorted via one diode. In practice a rectifier will produce a mean voltage who's differential voltage is$\ \frac{3\cdot \sqrt{6}}{\pi}$... 5 It is not a matter of simplicity. Rather it is attention to details by stating the critical performance specs THEN meeting them simply. Your first task in ANY DESIGN is to define ALL requirements. Then break it down into more detailed specs in a written checklist! cost vs qty, R&D time, size or space, performance, reliability, stability, tolerances ... 5 You have 4 times as many cycles in the time -5:ts:5. If your y and g signals had the same number of cycles and different frequencies, then the expected result would have appeared I believe. In short, g is not time compressed version of y as we can verify by simply counting the number of cycles in each signal. Time domain comparing time compressed signal and ... 4 It is not correct to have 0dB represent "off". dB is a relative scale and 0 dB, effectively, means "the same as some reference value", and your reference value can't be silence. A more common implementation of such a control would be to have 0dB represent maximum volume and negative values represent lower output. This is what you see on all professional ... 4 You need to add a dot in front of the divide symbol (this means array divide as opposed to matrix divide): n=linspace(0,29,30); h =1./(1+n); figure stem (n,h); Resulting Graph: To find out more, type help . - you will get something like this (this is only part of the result, as it's rather long...): >> help . Operators and special characters. ... 4 The following diagram block diagram taken from the xDSL Modem/DSLAM page at Texas Instruments illustrates what you'll find in a typical ADSL modem: So while your general assumption that they contain an ADC and can interface to a PC via USB or Ethernet is correct a few likely challenges are: There is some coupling and filtering on the analog side of things.... 4 You use one of the on-chip PLLs (there are four of them in your device) to multiply the 50 MHz input clock to something higher, such as 200 MHz. That becomes the sample rate you use for the VGA output. 4 The slow drift may be due to leakage currents in your sensor. Since the input is open-circuit (theoretically infinite impedance), even a tiny leakage current can cause a significant offset voltage. The simplest solution is to put some resistance across the sensor. 1 MΩ will likely swamp the leakage signal. 10 MΩ may work too. It depends on ... 4 Does this example prove to you that the spectrum spreads ? The spectral plot uses dB (I find that more convenient) on the Y-axis, so consider it a logarithmic scale. You can see that I need to add A LOT of noise to the phase. I think this has to do with the fact that the noise I add is white noise. In practical solutions this noise usually has a 1/f ... 4 I have done something very similar in the past. My aim was to transfer an image (*.bmp) from PC to FPGA (internal BRAM), and send it back to PC after the watermarking process. As previously mentioned, UART is your best bet. Implement a UART in FPGA or use an existing design. For Xilinx, look at this design provided with the Picoblaze. It is well documented ... 4 It varies with the situation. In the real world, there will be some transmission line effects, mainly filtering out higher frequencies but also some reflections. I found a circuit lab schematic/simulation someone named signality created with a lumped-element model of a transmission line, and I've slightly modified it. simulate this circuit – ... 4 For a continuous time system the impulse function can be used to plot the impulse response. The system can be specified using zero-pole (zpk), state-space (ss) or transfer function (tf) notation. For example: sys = tf[1, [1 1]); impulse(sys); 4 By way of a simple illustration, say you have a system that is, essentially, an integrator and require to close the loop around it to establish some form of feedback control. The OLTF is: \small G(s)= \frac{1}{s} There's a pole at$\small s=0$, and if a step were applied, the response would be a ramp$ (\frac{1}{s^2})\$. The CLTF is:$\$\small \frac{...

4

The output of the fft function will go from 0 Hz to fs/2, then -fs/2 to almost 0. If you ignore the 2nd half (for purely real signals), the following works well: For n samples at a sampling rate of fs, you can create a frequency axis, f with the following formula: f = (0:n-1)/n*fs This gives you freqs from fs/2 to almost fs, but if you overlook that... ...

4

I assume that your quantity beta is the maximum momentary frequency offset per the maximum baseband signal frequency fm. That beta is called "modulation index" and it exists in the approximated formula of the needed transmission bandwidth which is the uppermost of your bandwidth formula options. In narrow band FM the modulation index is much ...

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