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(Fourier Transform) What does F(jω) represent in frequency domain?

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spark360z

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I'm trying to understand the concept of Fourier transform and get stuck on something very simple.

My question is, what does F(jω) represent in frequency domain?

If I have a voltage source v(t), v(t) represents voltage as a function of time, therefore I know the voltage at any given time.

But in frequency domain, thing seems to be different.

Let say the transformed signal is V(jω). Let say at ω=2pi(1000), substitute ω in V(jω), suppose I get V(jω)=V(j2pi(1000))=10

What does "10" represent?

Is it an amplitude of a sinusoidal at 1000 Hz? Therefore the signal has sinusoidal 1k Hz frequency amplitude=10 as a component?

I'm not sure about this because the Fourier transform of cosine is Impulse which has amplitude=infinity.

Please help!
 

I'm trying to understand the concept of Fourier transform and get stuck on something very simple.

My question is, what does F(jω) represent in frequency domain?

If I have a voltage source v(t), v(t) represents voltage as a function of time, therefore I know the voltage at any given time.

But in frequency domain, thing seems to be different.

Let say the transformed signal is V(jω). Let say at ω=2pi(1000), substitute ω in V(jω), suppose I get V(jω)=V(j2pi(1000))=10

What does "10" represent?

Is it an amplitude of a sinusoidal at 1000 Hz? Therefore the signal has sinusoidal 1k Hz frequency amplitude=10 as a component?

I'm not sure about this because the Fourier transform of cosine is Impulse which has amplitude=infinity.

Please help!
The Fourier Transform just Transform a signal from the "Time Domain" to "the Frequency Domain".

v(t) is "The Signal in the Time Domain" ≡ F(jω) is "The SAME Signal in the Frequency Domain" !
V(t) is like looking at a signal with an Oscope and F(jω) is looking at the same signal with a Spectrum Analyzer.

Note t = 1/f , and f = ω/2pi.
The "j" ( or s=jω) is because signals have a complex component.

Go look for the proof of e^(-Jω) = Cos(ω) + jSin(ω)
 

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