Tag Info

38

Bode plot Take a typical 2nd order low pass filter amplitude response bode plot: - 3-D picture introducing pole-zero diagram Here's the bigger picture of that response when combined with a pole zero diagram: - Traditional pole zero diagram Looking down from above onto the 3-D picture shows the traditional pole zero diagram: - Pole zero geometry and |H(...

20

When you put something to your ear reproducing standard stereo recordings, you don't want a flat frequency response because the head-related transfer function that normally comes into play for a sound source much further away looks very different when the source is against your ear. Let me quote you a couple of paragraphs from a book: Of all the ...

15

A circuit's transfer function is a fully mathematical model that can be used to derive the frequency response and phase response (both together are called the bode plot). However the same isn't true in reverse - you can't always derive the TF from the bode plot. Sometimes you can but not always. So, the frequency response is a subset of the bode-plot and ...

13

Any repeating waveform can be constructed from a series of sines of integer multiples of the fundamental frequency of the waveform. These are the harmonics. Using Fourier expansion with cycle frequency f over time t, an ideal square wave with an amplitude of 1 can be represented as an infinite sum of sinusoidal waves of odd harmonics of the fundamental. ...

12

Although this question has been answered I just wanted to add something for those seeking an ideal logarithmic potentiometer law for simulation. A mapping from linear law to logarithmic law can be found in the general form: $$y = a\ b^{x} + c$$ Let this equation function define a mapping from $0\leq x\leq1$ to $0 \leq y \leq 1$, where $a$, $b$ and ...

9

I would expect so, assuming the speaker is capable of vibrating with the same speed. As far as I know each speaker has a range in which it can vibrate (usefully). Mostly bigger speakers are better suited for vibrating at lower speeds (and having more 'pressure', more air that is moved), small speakers can resonate faster (thus higher frequencies). That is ...

9

An ideal speaker will vibrate at whatever frequency of signal you drive it with. However, in reality, there are a few constrainsts. The amplifier output you are driving it with has a frequency range it can amplify. Extremely low frequencies, including DC, and high frequencies will be severely attenuated in the amplifier. The physical construction of a ...

8

Inspecting a circuit to determine if there are zeroes is part of the fast analytical circuits techniques or FACTs as described in my book on the subject. The first thing is to place each energy-storing element in its high-frequency state and check if the stimulus can propagate through the circuit to form a response on the output. See the circuit below: If ...

7

One way to look at this is that the inductor is a frequency-variable resistor (not quite since there's also a phase shift in there, but that's not important for this point). The higher the frequency, the higher the resistance. At the same voltage, the inductor draws less current at higher frequencies. A totally different way to think of this is to ...

7

I agree with Lundin's comment that 1 uF is a bit of a high value, especially if you only use a 1uF capacitor. But to answer your actual question: yes, it will probably just work if your circuit's design isn't sensitive to noise and disturbances on the supply voltage. For example: a LED flasher or a motor driver will probably just work. If you're using this ...

7

What is the reasoning behind this other than measurement errors? tolerances of components wrong theoretical calculation incorrect measurement procedures (wrong impedance scope-probe) test equipment errors (out of tolerance or calibration) human errors (mis-reading numbers / figures) implementation error (such as impedance of signal generator) implementation ...

6

Usually audio taper pots are not logarithmic but a piecewise approximation with only 2 segments. Each segment of the track will be coated with different resistivity material or have a different width than the other segments. I have seen wire-wound taper pots where the former has a gradually changing width to achieve the varying slope. A linear pot may be ...

6

1) Why does the magnitude Bode Plot of the response of a filter NOT approach Infinity at a pole? Try looking at this picture and recognize that poles may exist as infinities in the bode plot but more usually they are "behind" it: - 2) In the attached image, why is Wp (Omega subscript:P) called the pole frequency when the denominator clearly does not ...

6

Does this make any sense? Yes it does, from what I've seen they manufactures can use different silicon sometimes for a dual vs a single or quad part. But without actually seeing the die layout there is no way to really know (you could get some nitric acid and microscope -- be careful-- and find out for sure). I would think the same op amp would have the ...

6

A quick look at the fast analytical techniques or FACTs gives you the transfer function of this guy in the blink of an eye. First, turn the stimulus off or reduce $V_{in}$ to 0 V: replace the source by a short circuit. Then, "look" through the capacitor terminals to determine resistance $R$ between the terminals. It's immediate: $R=R+R=2R$. ...

6

It is due to the opamp's open-loop output impedance. As you can see in Figure 21 of the datasheet, the open-loop output impedance is inductive from about 100Hz to 10kHz. I modeled it using this circuit and got the same results. * sham opamp * GBW is 1MHz * open-loop gain at DC is 240dB * open-loop output resistance is 900Ω // 2mH G1 0 1 vinp vinn 1 C1 1 0 ...

6

If you think more geometrically, then it's not terribly complex (sorry for the pun.) Multiplication of complex numbers is counter-clockwise rotation and scaling. Division is clockwise rotation and inverse scaling. In polar, you have $A=r_a e^{I\theta_a}=r_a\angle\theta_a$ and $\beta=r_b e^{I\theta_b}=r_b\angle\theta_b$. Then multiplication just produces \...

5

As you probably know, $\beta$ or $H_{\text{fe}}$ is in first order determined by the ratio of the doping of base and emitter. This doping ratio is intrinsic (determined during manufacturing), it has no frequency dependency. What does limit the frequency behaviour though is the carrier mobility. This explains why there are Silicon-Germanium (SiGe) based ...

5

The frequency response is a special case of the Laplace transfer function where the transients are assumed to be completely dissipated, leaving the steady state sinusoidal response. Take, as an example, a sinusoid, $\small \sin(\omega t)\rightarrow \dfrac{\omega}{s^2+\omega^2}$, applied to a simple first order lag, $\small G(s)=\dfrac{1}{1+s}$. The ...

5

It looks like there is an actual difference between the OPA890 and the OPA2890. For example, the parametric table on the OPA890's description page shows that the OPA890 has a slightly higher GBW (260 MHz vs. 250 MHz), higher slew rate (500 V/µs vs. 400 V/µs), lower offset drift over temperature (15 µV/C vs. 35 µV/C), etc. The electrical tables in the two ...

5

First I don't understand why you are refering to page 450 of the book; I found your circuit on page 456. Figure 11.23 "High-pass equal-component" (VCVS). In the book they describe two types of Sallen-Key Topology filters, one is the "unit-gain"-version which as the name suggests has Av=1. and the other is the "equal-component"-version (the one you have), ...

5

capacitor does not like the fact that voltage is changed across it since it acts as a short circuit initially I don't know what the capacitor likes or doesn't like, but I think your reasoning is backwards. First, we usually say "the voltage across an ideal capacitor cannot change instantly" rather than say what the capacitor likes or doesn't like (we might ...

5

Why going through a complicated analysis with KVL and KCL then ending stuck with a system of equations to solve? The fast analytical circuits techniques or FACTs are an interesting alternative to follow. They are described in the book I published in 2016. The principle is to chop this 2nd-order circuit into a succession of smaller sketches you can solve ...

5

If you are familiar with the function could you please explain in an answer instead of simply saying that solutions might "exist somewhere on the internet"? Thank you! Not "might exist" but "do exist". Try this site's simulator: - The end goal: to get both V_out and V_in as functions of ω (with the resistor and capacitor values treated as constants)...

5

Here's the short answer: a first-order lowpass and a rectangular-in-frequency filter don't look the same in the frequency domain. So why should they be the same when you take the inverse Fourier transform?

5

Phase margin only applies to the open-loop transfer function of a closed-loop system. A compensator has a phase response, but phase margin doesn't make sense unless you add the transfer functions of the rest of the loop. You would expect the phase of your compensator to start near -90 degrees due to the pole at origin. Once you add your compensator to the ...

5

The maximum output amplitude of an op-amp, above a certain frequency, decreases with frequency. From the LT1128 datasheet we have this graph What this graph tells us is that even though a particular op-amp may have a unity-gain bandwidth of, say, 10 MHz, that only applies to very weak signals.

5

Most recognized suppliers of capacitors (including AVX and Panasonic) do provide extensive data on their websites regarding the important information that isn't included in their data sheets. So, go to your favourite recognized supplier (such as this one from Kemet) and look at the data under the product part number: - And scroll down to you find the graph ...

5

I think you have assumed in your calculations that A is a real number? that way: Real(AB) = AReal(B) and Im(AB) = AIm(B), and thinking intuitively, that if A is much larger than 1, A^2 is that much larger than A, thus {A/A^2 -> 0}. Having known that, the expression: 2Re(BA) + Re(BA)^2 + Im(BA)^2 becomes BA^2 inserting that back in again, you will again ...

5

In SPICE, the AC analysis is a linear analysis. It is assumed that the excitation of the circuit is small enough not to produce nonlinear behavior. If you want to simulate nonlinear behavior you must do a transient analysis. In other SPICE-like simulators (ADS, for example) there may be other analysis options such as harmonic balance that can predict ...

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