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4

The problem is that you statement: "The output signal will be like this:" is rather wrong. What you have shown there is the output of a rectifier circuit. I don't know where that came from but it is NOT the output of the given circuit. You get that output waveform if you replace the R with a diode, but in that case you no longer have a Linear Circuit.


4

That's called ringing, and it's an effect of inductance and capacitance in the circuit (often just parasitic). Picture it this way...the wire from the driver to the receiver has intrinsic inductance, and most digital inputs have [gate] capacitance. Once the output starts driving the line, current builds up in the (inductive) trace until the capacitance ...


2

You are passing what appears to be 60 Hz through a 100 nF capacitor. That capacitor will have an impedance of: - $$|X_C| = \dfrac{1}{2\pi f C}$$ Plug in the numbers and you get 26.5 kohm so, if you need very little attenuation of the AC signal the resistor value needs to be many, many times the value of 26.5 kohm. In your last scenario the equivalent ...


2

Trigger circuit works OK @ 10 MHz but doesn't quite work at lower frequencies. This kind of fault is very often due to degraded electrolytic capacitors in such old equipment. Many fixes have been accomplished with a scattergun approach: replace every electrolytic capacitor in the trigger signal chain. Or you can start with C501, C502, C503, C508 in Unit 3 ...


2

It is certainly not specific to an oscilloscope measurement. Fast edges relative to transmission line length and any unterminated or poorly terminated line results in (real) ringing at a frequency related to the length of the line. Rule of thumb is that the unterminated length in FR4 is about 1.5 * period of oscillation (in ns), so your relatively long ~...


1

You have a massive power MOSFET with a lot of capacitance. If you reduce the size of the MOSFET you can go faster, or change the type to a different technology than silicon. A cheap one that should show a lot (like 5 or 10:1) improvement is the FDD1600N10ALZ. They're cheap enough, but you can run some simulations to see what the predicted performance is. ...


1

You're creating a high pass filter between your capacitor and whatever resistance you're connecting to Vref with. In the first circuit, I can't see the impedance to Vref, but it's apparently not zero. In the second circuit, it's the output impedance of U_2; again, it's unknown to me. I might be able to estimate them if I knew the frequency of your test ...


1

The transfer function of the shown circuit is equal to: $$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{R}}=\frac{1}{1+\text{sCR}}\tag1$$ Now, when: $$\text{v}_\text{in}\left(t\right)=\hat{\text{u}}\sin\left(\omega t+\varphi\...


1

the 1553 bus is mainly used in the space industry, a low signal is equivalent to a negative voltage not zero, if you have a look onto your specification you will see what's is acceptable as the range could be quite wide. Same for a high signal it should be a positive voltage within a specific range. the peak to peak voltage must be >18V


1

This bus is Manchester encoded. This means that one symbol (bit) is encoded as a low transitioning to a high, or a high transitioning to a low. These correspond to 0 and 1 respectively. Therefore, if you want to send: 1 0 0 1 0 1 1 1 the sequence is: HLLHLHHLLHHLHLHL This encoding moves the center frequency up to the bit rate, eliminating the low ...


1

Maybe this picture will help understanding Fig. 2 and Fig. 4... ... and this - the need of a bipolar supply:


1

In 1) the output is Voltage across resistor "R", so it is referenced to Vdd. Your schematic shows it referenced to ground, which is incorrect. If Vdd is constant, both will be the same AC output with a different DC offset. However if Vdd is variable or noisy, referencing the output to Vdd will reject most of the noise, whereas referencing the output to GND ...


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