9

In PSpice m and M are thousandths. You need to use Meg.


7

There are quite a few advantages. Taking a look at a typical superhet (up to the IF): The input signal at the RF input is small (as low as -122dBm in some narrow band voice systems I have worked on - that is about 6.3fW) To amplify a signal at a high RF (say a few GHz) is expensive compared to doing that amplification at a lower frequency. A few dB of RF ...


6

Note that this answer is skewed toward analog radio reception. The rules are different for software-defined radios, and for digital services. The biggest drawback to direct conversion is sideband suppression. If you use a single mixer, a signal at \$f_c + f_s\$ is indistinguishable from a signal at \$f_c - f_s\$, where \$f_c\$ is the carrier and \$f_s\$ ...


5

That filters has 4 sections, which if we number them 0 to 3 ... 0 is coupled to 1 by mixed coupling which is mostly inductive 1 is coupled to 2 by capacitive coupling through Cc 2 is coupled to 3 as 0 is to 1 Using all one type of coupling results in a filter with the shape tilted noticably one way or the other. Mixing the coupling type results in a more ...


5

This specific application Presumably you want to power the op amps with the (single) 5 V DC supply, so your main requirements are: an op amp that can operate with a 5 V supply or less an op amp which has a common mode range that extends to the negative rail (these are marketed as "single supply" op amps) so that your signal can swing as low as the ground ...


5

You can use modified nodal analysis to solve for all unknown node voltages and unknown currents. Once you get the node voltage, you can find the transfer function. For the analysis, I denote node and current as in the picture below. Now you can write KCL for every node and a constraint by OpAmp. You can get 7 equations: Then you can solve 7 equations to ...


5

It would be better to have 560 ohms to ground, and use only one transistor, switching a value of 3k (should be 2.9k) to ground. That way most of the current flows in a real resistor, which minimises any distortion coming from the transistor. 2N7002 looks fine, any MOSFET should do as long as the gate is driven properly. Unexpectedly, bipolar transistors ...


5

These two circuits can provide similar bandpass frequency response. However, the lower-frequency cutoff must be considerably lower than the high-frequency cutoff. This means that the frequency span of the pass-band must be wide, if these two circuits are to provide similar results. For bandpass response with narrow pass-band, the LC circuit is king. Note ...


4

A square wave of 200 Hz contains odd harmonics all the way to infinity. The 25th harmonic of 200 Hz is 5 kHz hence you see the band pass filter extracting this harmonic and greatly amplifying it. Here's a picture of the spectrum of a band pass filter on the 17th harmonic (nearest I could find): - Picture taken from Acoustics and Psychoacoustics: ...


4

The LM358 and LM741 do not have enough bandwidth to do what you want. At 200kHz the LM358 can only manage about 10dB of gain. It also has poor slew rate and a high phase shift at this frequency, resulting in bad filter performance and distorted output. If you don't have access to better opamps then you will have to use 'passive' filtering. Conventional IF ...


4

Firstly, "laser" tag doesn't (usually) use lasers - and for good reason: Lasers are dangerous and can blind people. Use with caution. Laser tag normally operates using a focused infra-red beam. Basically a powerful and focused infra-red remote control (like you have for the TV). This is usually backed up with normal focused white light (from a white LED) ...


4

Your transfer function can be obtained in different ways and one of them could be by cascading two stages with a buffer in-between. Your original expression follows the form \$H(s)=H_0\frac{1+\frac{s}{\omega_{z1}}}{1+\frac{s}{\omega_{p1}}}\frac{1+\frac{s}{\omega_{z2}}}{1+\frac{s}{\omega_{p2}}}\$ If you design a first stage that introduces a pole-zero, you ...


4

Not sure what kind of simulator that is, but a common pitfall is: 1M = 1 milli (Ohm, Volt, Ampere) and 1Meg = 1 Mega (Ohm, Volt, Ampere). Because they don't care about capitalization they can't make out a difference between 1M and 1m so both end up being 1 milli.


4

Yes, direct-conversion receivers exist, but they require special care, especially with certain kinds of modulation. For example, with SSB modulation, in order to reject the unwanted sideband, your baseband demodulator must be able to distinguish between "positive frequency" and "negative frequency". This is not trivial, and is only really practical using ...


4

Direct conversion is conceptually simple, but requires considerable engineering to do it right. Besides Dave's & Tim's answers, there is a subtle pernicious potential problem with direct conversion... Most mixers (even doubly-balanced ones) leak local oscillator power to both RF port and IF port. Power leaking backwards through to the RF port to the ...


4

When you get unrealisable component values in your theoretical design, it's a sign that you need to change your filter topology. That second stage series resonant tuned circuit has an impedance that's too high, which means a series capacitance that's too small. Amongst your options are (depending on the particular frequencies) a) Reduce the impedance of ...


3

Your LCR resonator has losses, so it is a decaying sinusoid wave at 5 MHz. Before it decays to zero, it is re-excited by the negative-going edge of your 1 MHz square wave. Then re-excited by the next positive-going edge of your 1 MHz square wave....and so on. If you wish to see more of the exponentially-decaying envelope, try changing the frequency of your ...


3

Why is the center frequency of a band-pass filter is given by the geometric average of the two cutoff frequencies instead of arithmetic average? Because its the ratios that are relevant, not the increments. For example, if you have a bandpass filter from 2 kHz to 20 kHz, it covers a 10:1 range. The center is then half way between these in ratio terms, ...


3

After some experimentation I finally solved the riddle: It was noise affecting the feedback loop. After shrinking the feedback resistors by a factor of ten and increasing the capacitors by the same factor, I get a nice and even ringing response at different tunings! Also the noise floor is significantly lower.


3

Perhaps the easiest way to determine this without using any math is to simulate it in a program such as LTSpice. If you aren't familiar with this tool, I strongly encourage you to look into how to use it, but I think a general overview is outside the scope of this answer. To simulate it, you must find a Spice model for the opamp in question (TI conveniently ...


3

These answers are obviously correct but they represent the brute-force approach. Nothing wrong here but considering all these lines, it is likely that mistakes or typos get in the way of the correct answer. Furthermore, the final expression does not give insight in case a design goal exists: what is the mid-band gain? An easier approach consists of using ...


3

If total BW is defined as the 0.707 amplitude threshold ( -3dB) and on the lower half BW averages to -35kHz at 965 KHz, we might approximate it as a BW of 70kHz with a Q= 1000/70= 14 or inversely 1/Q=7.1% Given a 10% 90% rise time,T = 0.35/f(-3dB) we expect a burst envelope rise time of 0.35/70kHz = 5 us or approximately 5 cycles at center f. The center f ...


3

But it IS working, at least within the limitations of your opamp. The peaks in your response curves fall about where they should. In fact, if you draw the asymptotes to your response curves, they hit the peak response level exactly where they should, as shown here for the low band filter, V(n002): You can also see that the lower -3 dB point for V(n013) ...


3

How do these formulas come to be? Start with a generalized picture of your circuit: simulate this circuit – Schematic created using CircuitLab From the usual equations for an op-amp in negative feedback you can find, $$ v_o = -Z_1\frac{v_i}{Z_2} $$ Now substitute in the impedances for the elements in your design and you will eventually be able to ...


2

Well a simple answer is that you can't make a 100 Hz centre frequency with a 3 dB bandwidth of 200 Hz because you crash into DC. You have to treat it logarithmically. Having said that a lot of bandpass filters are very "tight" and numerically there is little difference between cetre frequency being bang in the middle or \$\sqrt{f_1.f_2}\$


2

In the following, I try to describe the way for deriving the wanted formula (geometric mean value). Start with the classical 2nd-order bandpass function (involving the parameter pole quality factor Qp and pole frequency wp). Replace the variable w by wc (3dB cut-off)and - at the same time - set the magnitude of the transfer function to A=Amax/SQRT(2). As a ...


2

I would probably use a switched capacitor filter. There are numerous offerings available and they are very easy to drive. There is also a universal filter - I am sure more types are available.


2

You can build a pretty good narrow bandpass filter from a typical quad op-amp package and some resistors and capacitors. If you want to use fewer components than that, you could look at one of the specialised filter ICs on the market, such as TI's UAF42. Opamps will probably work out cheaper though. The requirement for binary output might be harder. You'...


2

Your design is a very basic and might work but since it has no feedback it is not so predictable in how it will behave in practice. The "proper" engineer's way of designing this is by use of feedback. You basically make a crude amplifier with a high gain and enough bandwidth and use feedback to get the gain you actually want. Unfortunately this design ...


2

Note that the mounting terminals marked "C" are insulated from the case, while the terminals at the bottom that have two coils attached to them are grounded to the case. Electrically speaking, the grounded end of the coil is the "bottom", and that area has the lowest impedance (relatively low voltage and high current), which is a good match for the input ...


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