Time domin of AM

[Hero999]

3.4kHz is 1.7% of 200kHz which I don't think is unreasonable, even if the transducer does have a sharp resonant peak, although he needs to know what he's looking for.

 

[JimB]

Am I missing something here?

The thread started off asking about AM using a 4066.

So, where did all this talk about piezo transducers and SSB come from?

JimB

 

[Space Varmint]

Am I missing something here?

The thread started off asking about AM using a 4066.

So, where did all this talk about piezo transducers and SSB come from?

JimB

I know. Theoretically you would need no bandwidth. It is the amplitude of the output signal.

 

[audioguru]

He is trying to make a sound system that projects modulated ultrasonic waves at an object or at a person. Then the sound demodulates when it hits and appears to come from the object or inside the person's head.

It is used in some museums and multi-language theaters. The statues and paintings "talk". English hits me but French hits the guy sitting beside me.

 

[Space Varmint]

He is trying to make a sound system that projects modulated ultrasonic waves at an object or at a person. Then the sound demodulates when it hits and appears to come from the object or inside the person's head.

It is used in some museums and multi-language theaters. The statues and paintings "talk". English hits me but French hits the guy sitting beside me.

No kidding? That actually works? I think I have heard of it but thought they might be BSing a bit.

 

[Hayato]

No kidding? That actually works? I think I have heard of it but thought they might be BSing a bit.

Yes, that works. As long as you shoot the modulated wave and the carrier to the person, then you are going to have beating frequency.

 

[Space Varmint]

Yes, that works. As long as you shoot the modulated wave and the carrier to the person, then you are going to have beating frequency.

That's scarey. What frequency if you don't mind my asking? I'm afraid they may have plans like that for the HAARP.

 

[audioguru]

The systems that are sold do not reveal the frequency of their custom-made high power arrays, with wide bandwidth and special pre-distortion to reduce the distortion that is heard.

 

[Hayato]

That's scarey. What frequency if you don't mind my asking? I'm afraid they may have plans like that for the HAARP.

For example, if you have a speaker at 20kHz and another at 23kHz, with both placed on a wall facing to you in a angle of 45°, you are going to have a 3 kHz of beating frequency where the sound waves from each speaker make a intersection.

The concept is similar to the IF from superhet receivers.

 

[Space Varmint]

For example, if you have a speaker at 20kHz and another at 23kHz, with both placed on a wall facing to you in a angle of 45°, you are going to have a 3 kHz of beating frequency where the sound waves from each speaker make a intersection.

The concept is similar to the IF from superhet receivers.

Well yes. You must have a carrier frequency in mind. I think the brain waves operate in the VLF range pretty much. It concerns me with the HAARP system because they use these freqs to communicate with submarines and also to super charge the ionosphere and by heating it up they are able to move the jet stream. There are multiple functions to it all based on Tesla technology.

 

[Hero999]

I know. Theoretically you would need no bandwidth. It is the amplitude of the output signal.

No, for SSB the required bandwidth is equal to the maximum frequency transmitted, for telephone quality, the bandwidth needs to be at least 3.4kHz. If the carrier is 200kHz, the transmitted spectrum would range from 200kHz to 203.4kHz.

EDIT: This is assuming the upper side band is used, if you use the lower side band the spectrum will be 196.6kHz to 200kHz.

 

[JimB]

He is trying to make a sound system that projects modulated ultrasonic waves at an object or at a person.

OK, but that information is not obvious in this thread.

JimB

 

[Hayato]

I know. Theoretically you would need no bandwidth. It is the amplitude of the output signal.

That's a missinterpretation of AM/DSB-SC/SSB...

Although you have a fixed frequency, your amplitude varies on the rate of the modulating signal, so you gonna have a bandpass bandwith = 2*base band bandwidth (base band bandwidth = modulating signal bandwidth), for AM/DSB and bandpass bandwith =~ base band bandwidth for SSB.

 

[audioguru]

OK, but that information is not obvious in this thread.

JimB

Because the OP posted many threads about this same project.

 

[audioguru]

That's a missinterpretation of AM/DSB-SC/SSB...

Space Varmint posted schematics and recordings of his SSB transmitter and receiver. He used narrow bandwidth crystals to select the sideband.

The audio bandwidth was shown to be only a couple of hundred Hz or less and voices sounded like ducks quacking or like a morse code tone, completely unintelligible.

 

[Space Varmint]

No, for SSB the required bandwidth is equal to the maximum frequency transmitted, for telephone quality, the bandwidth needs to be at least 3.4kHz. If the carrier is 200kHz, the transmitted spectrum would range from 200kHz to 203.4kHz.

EDIT: This is assuming the upper side band is used, if you use the lower side band the spectrum will be 196.6kHz to 200kHz.

You mean 2KHz. But that is merely a symptom of shifting capacitances possibly in active devices. I'm saying theoretically there should be no bandwidth. You are not shooting for bandwidth because it is amplitude you are after. Like 100% modulation means the carrier will increase to twice the power level of the unmodulated carrier.

 

[Hayato]

You mean 2KHz. But that is merely a symptom of shifting capacitances possibly in active devices. I'm saying theoretically there should be no bandwidth. You are not shooting for bandwidth because it is amplitude you are after. Like 100% modulation means the carrier will increase to twice the power level of the unmodulated carrier.

No. Theoretically there is Bandwidth.

AM = (k + a*m(f))*cos(2*pi*fc*t).

You are going to have a m(t)*cos(2*pi*fc*t) -> The m(t) signal will be offset to the fc frequency in the spectrum.

Observe:



In this case you have the carrier = 200 kHz, and the modulating = 5 kHz.
As you can see, considering 0.2 (~-14 dB) = zero, the bandwidth is 10 kHz (from 195 kHz to 205 kHz).

 

[Space Varmint]

No. Theoretically there is Bandwidth.

AM = (k + a*m(f))*cos(2*pi*fc*t).

You are going to have a m(t)*cos(2*pi*fc*t) -> The m(t) signal will be offset to the fc frequency in the spectrum.

Observe:

View attachment 38860

In this case you have the carrier = 200 kHz, and the modulating = 5 kHz.
As you can see, considering 0.2 (~-14 dB) = zero, the bandwidth is 10 kHz (from 195 kHz to 205 kHz).

That's what is. Theoritically there should be none. I can transmit an AM signal that is less that 100 Hz wide. Frequency bandwidth is not a function of AM it is a consequence of the electronics being used to transmit the signal. Consider the active device being used such as a modulator. What happens is the junction will expand and contract with varying levels of power. This is in effects a varying capacitance which causes an accidental frequency swing. Any deviation from center frequency is actually FM. So AM as it is today is actually AM and FM. But we detect the amplitude variations because the signal is primarily AM.

 

[Hayato]

That's what is. Theoritically there should be none. I can transmit an AM signal that is less that 100 Hz wide. Frequency bandwidth is not a function of AM it is a consequence of the electronics being used to transmit the signal. Consider the active device being used such as a modulator. What happens is the junction will expand and contract with varying levels of power. This is in effects a varying capacitance which causes an accidental frequency swing. Any deviation from center frequency is actually FM. So AM as it is today is actually AM and FM. But we detect the amplitude variations because the signal is primarily AM.

Forget about the modulator, the junctions, transistors.

Think about math only.

For example, a DSB-SC signal:

s(t) = m(t)*cos(2*pi*f*t), makes sense for you? You have a base-band signal, m(t), that controls a carrier cos(2.pi.f.t);

When m(t) = 0, then s(t) =0.
When carrier = 0, then s(t) is also 0.

Now let's just take the spectrum with fourier transform:

Fourier is a linear operator, so:
S(f) = M(f)->convolution->1/2 * [D(f - fc) + D(f + fc)], where D is the Dirac Impulse. As a cosine is a pure wave it will have only 1 component at fc, that's why you have an impulse deslocated to fc.

A convolution of any function with an impulse results in that function with the argument replaced by the impulse function's argument.

S(f) = 1/2 * [M(f+fc) + M(f-fc)].

Observe that you have a copy of the baseband signal, M(f), deslocated to +fc and -fc. (In any moment I considered junctions, transistors...).

Think about another POV, you have the fixed frequency carrier, but this fixed frequency carrier has its aplitude varying on the rate of m(t), so actually you have 2 signals over there, not just 1.

When you see an DSB on a oscilloscope, you can actually see the carrier and the modulating 2 signals.

So you have the fixed frequency, and an ampliude that changes with time.


It does not happens with FM, you just can't see the modulating signal, just the frequency changing.

Because you have a fixed amplitude, and a frequency that changes with the integral of the modulating signal amplitude and a modulation index k Hz/V.

s(t) = cos[2.pi.(fc.t + k.∫m(t)dt)]

Plotting a FM spectrum is complex, because it is infinite (in theory) you need Bessel functions to plot. Because you just have the fixed amplitude, but the phase and the frequency keeps changing all the time.

 

[Space Varmint]

Fair enough. I was not thinking about the time it took for the amplitude to vary. I guess your right.