Frequency Modulation

February 12, 2011

GoLuckyRyan Basic, circuits, Lab Log, others, Working on , , , Leave a comment

Frequency Modulation is encode a message signal on a carrier signal by changing the frequency of it.

the carrier signal has a basis frequency and the form is like:

S_C(t) = A_C cos( \omega_C t )

the message signal can take any form. after the Modulation, the output signal is:

S_O (t) = A_C cos( \omega_C t + \Delta f \int{S_M dt } )

the integrated message signal should be normalized to 1. and the Δf is the range of frequency change.

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The frequency produced by the Gunn Oscillator can be modulated, in order to matching the resonance frequency of the microwave chamber.

there is a adjusting knob on the Gunn Oscillator, which determine the basic frequency of the microwave. and a modulation signal is from the power supply. the frequency of the Gunn Oscillator is ranging from 8.6 GHz to 9.6 GHz.

the modulation signal is a simple linear function with frequency 23ms.

since our modulation frequency is a linear function. thus, the change of  the amplitude will change the output frequency of the microwave.

but the actually frequency modulation is by a Varicap or Varactor, which the capacitance can be changed by applied voltage. by changing the capacitor of the Gunn Oscillator, the output frequency changed.

However, the principle of frequency modulation unchanged.

In microwave engineering,  the Gunn Oscillator with the modulator will be called Voltage-Controlled Oscillator or VCO.

I still not fully understand the mathematic of the Gunn Oscialltor, and how it reacts with modulation signal. since i don’t have the internal structure of the oscillator.

Transmission Line

February 12, 2011

GoLuckyRyan Basic, circuits, others , , , , , 1 Comment

a Transmission Line is any thing used to transmit a electric signal.

in a AC circuit, the voltage does not only variate on time but also on space. For low frequency, the wavelength is long and this can be neglected. But at hight frequency, the variation is significant.

since the voltage is changing from different location. we cut the transmission line in a small sector, and each sector is analog to some circuit elements, no matter the shape of the line. this gives us a easy understanding of what is going on for the voltage and current. But this analogy neglected the effect of temperature, material non-linearity and magnetic hysteresis effect.

the equation of voltage across a section is

V(x) = ( R + i \omega L )\Delta x I(x)+ V(x+\Delta x )

since the resistance and inductance have unit per length. rearrange and take limit of x.

- \frac {d V(x)}{dx} = ( R + i \omega L ) I(x)

the equation of current is

I(x) = ( Q + i \omega C )\Delta x V(x+ \Delta x) + I(x+ \Delta x )

The Q is conductance, which is NOT an invert of resistance in case of AC. both Q and C will draw some current away in AC circuit. take limit gives

- \frac { d I(x) }{dx} = (Q + i \omega C ) V(x)

now we have 2 coupled equations. If we de-couple them, we will have

\frac {d^2 V(x)}{dx^2} = k^2 V(x)

k = \sqrt{ (R+i \omega L ) ( Q + i \omega C ) }

and the current share the same equation. notice that k is a complex number.

the solution are:

V(x) = V_f Exp( - k x ) + V_b Exp( k x)

I(x) = I_f Exp( - k x) + I_b Exp( k x )

where the subscript means forward and backward. from the coupled equation of Current, we can related the voltage and current and find out the impedance.

V(x) = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C } } (I_f Exp(-kx) - I_b Exp(k x)

Thus, we define the Characteristic Impedance for forward wave.

Z_0 = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C }}

the Characteristic Impedance for backward wave is a minus sign. there fore, we can rewrite the current in term of voltage.

I(x) = \frac {1}{Z_0} ( V_f Exp(-kx) - V_b Exp(kx)

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impedance matching

for a load at the end of the transmission line, the wave will get reflected. to see this, we have to consider the load, which imposed another equation. the voltage across the load is:

V(L) = V_f Exp( - k L ) + V_b Exp( k L)

and the current input to the load is:

I(L) = \frac {1}{Z_0} ( V_f Exp(-kL) - V_b Exp(kL)

the current and the voltage is related by:

V(L) = Z_L I(L)

solve it, and find the ratio of :

\frac { V_b }{V_f} = Exp( - 2 k L ) \frac { Z_L - Z_0 }{ Z_L + Z_0 }

since we can do nothing on the exponential, thus, we define a reflection coefficient:

\Gamma_0 = \frac { Z_L - Z_0} {Z_L - Z_0}

for no reflected wave, the impedance of the load and the transmission line should be equal and it is called impedance matching.

the power of the load is:

P = V(L)I*(L) = \frac { 1}{Z_0} ( V_f^2 Exp( - 2 k L ) - V_b^2 Exp( 2k L) + V_f V_b* -V_f* V_b )

when the impedance is matched, the power is maximum.