# Tag Info

1

You can use a 555 timer IC to create a tone generator Here is a page that shows such a circuit but you can find many more from google. You'll need to connect a pot (or a resistor divider) at the output to lower the output level to line in/mic levels. The switch can be used in the power supply of the circuit.

0

Having the Audible Source, connected to the switch , and then to the mic input should do it. Depending on number of wires in the Audio source(Say 2),you might need 2 switches to do this.

2

An antenna picks up aerial waves. A tuner will extract waves in one narrow range of frequencies from the antenna and do something with it, like outputting it to a component in a VCR or older TV set. I have been successful at using antennae for short-range (802.11 range) broadcasting as well, though, when I output a RF signal from a RF amplifier. An antenna ...

0

According to bit format (page 6), there are two frequencies. In case of RF transmission, which is/are the frequency/ies of the encoded signal? The encoded signal (pin SO) comprises two frequencies in three parts. Part A is a marker to indicate that data is about to start it is 500 Hz on a 75% mark space ratio. Part B is the data and it appears to be ...

5

An antenna is an aerial. A wire structure which, to pick up TV signals, is typically mounted at the highest point in a house and connected by a coaxial cable. A tuner is a circuit that selects a particular signal from all those received by the antenna. Typically this is by tuning to a specific frequency. Some small portable TVs do have a small antenna as ...

0

I think you'll find the following article very useful, it describes the effect of higher sampling rate (higher than the manufacturer recommendation) to the ADC resolution in AVR http://www.openmusiclabs.com/learning/digital/atmega-adc/

2

The only worthwhile component of the EM wave created by AC power is the magnetic component and this is really easy to pick-up. Use a coil of wire - maybe ten turns or more with a diameter of say 10cm and then an op-amp amplifier configured as a sallen key low pass filter set at 50Hz. For added high-frequency noise immunity, put a 10uF non-polarized capacitor ...

2

Theoretically, or more precisely, ideally, the answer is no. If $$Z_{in}(\omega) \rightarrow \infty$$ for an ideal op-amp, then there is zero input current for any input voltage of any frequency. But, this isn't surprising or, for that matter, very interesting. Ideal op-amps are an abstraction that, in the appropriate context, are a good approximation ...

3

If something has an infinite impedance then any frequency cannot affect it however, it's never going to be infinite and as frequency rises the small amount of capacitance associated with the input will start to become significant even if the input resistance stays in the Giga ohm range. The frequency won't affect the capacitance but the amplitude of the ...

0

You just said the input impedance is infinite. If it is always infinite, then obviously it it doesn't vary with frequency. This question makes no sense.

2

Here is one method: Use a high quality condenser microphone, and first start by recording the hum that you are trying to identify. Use a spectrum analysis program such as "Spectrogram" to visualize the exact frequencies that you are trying to identify. Once you have identified the frequencies in question, you can then work to create a directional reflector ...

1

It is largely because of fairly arbitrary decisions made by various electrical companies and their spheres of influence in the early years of electrical power production. Japan is an interesting case. See Utility Frequency - history There are no really significant consequences in terms of flickering of public lighting or signals when comparing 50 Hz and ...

0

Because in USA they count in dozens, but in Europe in decades. 60Hz are 5 dozens. 50Hz are 5 decades. 50Hz (and 60Hz) is a trade-off - it is higher enough to allow non-flickering light and acceptably small transformers and low enough to allow the electro-mechanical generators to turn with acceptably low speed.

1

However, we are in the imaginary frequency domain, so what would the bandwidth of this filter be? I'm not sure why you're having difficulty with the fact that the frequency domain signal is, in this case, imaginary. The symmetries of the Fourier transform are usually taught early on in signal processing courses: If the time domain signal is real ...

4

From rad/sec to Hz Usually we work in the frequency domain with Hz. Let me first make translations: Signal $\sin(6t)$ is a signal of pulsation of $6 rad.s^{-1}$, which means it has a frequency $f = \frac{6}{2\pi} \approx0.955 Hz$ Same translation for $\sin(5t)$, it's a $\approx~0.795Hz$ Time view If you subtract them, you would see something ...

0

There seems to be a slight confusion in your understanding of amplitude, phase and frequency: Without this being in the imaginary frequency domain, for $\omega\ge0$ there would be no bandwidth (everything is zero or has a negative amplitude for that frequency). Firstly, your frequencies are not imaginary here. Your frequency-domain function $F1∗F2$ ...

-1

As the other answers mention heat. But also the speed of light is a limiting factor. http://www.google.co.uk/search?q=3+ghz%2F+speed+of+light Assuming the electrons travel at the speed of light at 3ghz the signal can make it 10cms before the next clock tick

3

If you keep raising the clock, eventually you will have power consumption/temperature problems (you probably remember Pentium 4 CPUs which ran fast and hot). You can work around those, of course, but then your cooling system becomes more expensive. Thus, multi-core (and other approaches, like improving the CPU architecture/organization - so much that ...

1

There are a number of related factors at work. For one thing, main memories have really lagged behind CPU speeds; if hypothetically one were running a single core at 50GHZ and it had to wait 10ns for a value to be fetched from RAM [quite fast, actually], would forfeit an opportunity to do 500 clock cycles' worth of useful work. By contrast, if one had 10 ...

0

This a current divider so the greater current is through the lesser impedance. The resistor impedance is constant with frequency. However, the (magnitude of the) capacitor impedance varies with (angular) frequency as $$|Z_C| = \dfrac{1}{\omega C}$$ Can you take it from here?

0

If the frequency is high, meaning current doesn't flow for very long in either direction, the capacitor will suck it up, in the sense that in a short amount of time it can't get charged much before the accumulated charge is removed again. So not much voltage can develop and therefore little current flows through the resistor (the current through resistor ...

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