Revision as of 20:45, 9 February 2011

Fourier Transforms

Forest_Electronic_Instrumentation_and_Measurement

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+=Fourier Transforms=
-=Differentiator circuit=
-Consider the effect of a low-pass filter on a rectangular shaped input pulse with a width <math>\tau</math> and period <math>T</math>.
-[[File:TF_EIM_PulsedRCHighpass.png | 400 px]]
-The first thing to consider is what happens to the different parts of the square pulse as it travels to the circuit.
-You know that the circuit, as a high pass filter, will tend to attenuate low frequency (slow changing) voltages and not high frequency (quick changing) voltages.
-This means no DC voltage values pass beyond the capacitor.
-This means that points <math>a</math> and <math>b</math> in the pulse shown below should pass through the circuit.
-[[File:TF_EIM_SquarePulseDifferentiated.png| 200 px]]
-The voltage changes in between point <math>a</math> and DC are passed according to what you select for the break frequency (<math>\omega_B = \frac{1}{RC}</math>)
-In the above pulse picture it was assumed that <math>\tau >> \frac{1}{\omega_B} = RC</math>  or in other words the RC network time constant is a lot smaller than the pulse width.
-If the RC time constant is larger than the pulse width then you will see the voltage stay high.  The rectangular pulse will be output as a square like pulse which is "sagging".
-{| border="3"  cellpadding="20" cellspacing="0"
-| [[File:TF_EIM_PulsedRCHighpassSmallRC.png | 400 px]]|| [[File:TF_EIM_PulsedRCHighpassBigRC.png | 400 px]]
-|-
-|<math> RC << \tau</math> || <math>RC >> \tau</math>
-|}
-== Loop Theorem==
-:<math>V_{in} = V_C + V_R = I X_C + I R</math>
-;How does <math>V_C</math> compare to <math>V_R</math> ?
-:<math>\left | \frac{V_R}{V_C} \right | = \frac{\left | IR \right |}{\left | I X_C\right |}= \frac{R}{\left |  \frac{1}{i \omega C}\right |} = \omega RC</math>
-The maximum value of <math>\omega</math> is determined by the pulse width <math> \tau</math> while <math>T</math> is the period that corresponds the lowest frequency (other than DC).
-:<math>\left | \frac{V_R}{V_C} \right |_{MAX} = \frac{RC}{\tau}</math>
-=== <math>RC << \tau</math> ===
-When <math>RC << \tau</math>
-:<math>\left | \frac{V_R}{V_C} \right |_{MAX} = \frac{RC}{\tau} << 1 \Rightarrow V_R </math> may be ignored.
-:<math>\Rightarrow V_{in} = V_C + V_R \approx V_C = \frac{Q}{C}</math>
-taking the derivative with respect to time yields
-: <math>\frac{d V_{in}}{dt} = \frac{I}{C}</math>
-or
-: <math>I = C \frac{d V_{in}}{dt}</math>
-The loop theorem for V_{out} :
-:<math>V_{out} = IR = \left ( C \frac{d V_{in}}{dt} \right ) R = RC \frac{d V_{in}}{dt}</math>
-The output of the above circuit when <math>RC << \tau</math> will be proportional to the derivative of the input AC source with respect to time.
-=Integrator=
-[[File:TF_EIM_PulsedRCLowpass.png | 400 px]]
-The above circuit is a low pass filter.  When a square pulse hits it all of the low frequency components will pass through and the high frequency components will be attenuated.  This essentially makes the pulse smooth as shown below.
-[[File:TF_EIM_SquarePulse.png| 200 px]]
-The above is referred to as an integration circuit.
-==Loop theorem==
-:<math>V_{in} = IR + \frac{Q}{C}</math>
-:: <math>= R \frac{dQ}{dt} + \frac{Q}{C}</math>
-This differential equation may be written as
-: <math>\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_{in}}{R}</math>
-:<math>\Rightarrow Q = CV_{in} \left (1 - e^{-t/RC} \right)</math>
-:<math>V_{out} = \frac{Q}{C} = V_{in}\left (1 - e^{-t/RC} \right)</math>
-== <math>RC >> \tau</math> ==
-If <math>RC >> \tau</math>
-then
-::<math>t/RC << 1</math>
-:<math>e^{-t/RC} \approx 1- t/RC</math>
-:<math>\Rightarrow V_{out} = V_{in}\left (1 - e^{-t/RC} \right) = V_{in} \frac{t}{RC} = \frac{1}{RC} \int_0^t V_{in} dt</math>  The circuit appears to be integrating the input voltage.
-If <math>RC << \tau</math> then very little integration is done because the pulse is change at a low frequency compared to the RC constant.
-=Pulse Sharpening=
-The circuit below is a way to decrease the rise time of an input pulse at the expense of attenuating the pulse.  The output pulse is a "sharpened" version of the input pulse and attenuated.
-[[File:TF_EIM_PulseSharpnr.png| 200 px]]
-==Junction Rule==
-If we apply the junction rule for the currents in the above circuit
-Then
-:<math>I_1 +I_2 = I</math>
-:<math>I_1 = \frac{V_{in}-V_{out}}{R_1}</math>
-:<math>I_2 = \frac{d}{dt} C\left( V_{in}-V_{out}\right) : Q =C \Delta V = C\left( V_{in}-V_{out}\right)</math>
-:<math>I =  \frac{V_{out}}{R}</math>
-:<math>\Rightarrow</math>
-::<math>\frac{V_{in}-V_{out}}{R_1} + \frac{d}{dt} C\left( V_{in}-V_{out}\right) =  \frac{V_{out}}{R}</math>
-Collecting terms
-:<math>\frac{V_{in}}{R_1} + \frac{d}{dt} C\left( V_{in}\right) =  \frac{d}{dt} C\left( V_{out}\right)  + \frac{V_{out}}{R} + \frac{V_{out}}{R_1} </math>
-:<math>\frac{d V_{in} }{dt}+ \frac{V_{in}}{CR_1}=  \frac{dV_{out}}{dt}   + \frac{(R+R_1)}{CRR_1} V_{out} </math>
-assuming
-:<math>V_{in} = V_0 \left ( 1-e^{-t/\tau_{in}}\right )</math>
-:<math>\Rightarrow</math>
-::<math>\frac{dV_{out}}{dt}   + \frac{(R+R_1)}{CRR_1} V_{out} = V_0 \left ( \frac{1}{\tau_{in}} - \frac{1}{CR_1}\right )e^{-t/\tau_{in}} + \frac{V_0}{CR_1} </math>
-If <math>CR_1 = \tau_{in}</math>
-Then
-:<math>\frac{dV_{out}}{dt}   + \frac{(R+R_1)}{CRR_1} V_{out} =  \frac{V_0}{CR_1} = Constant</math>
-The above is a first order non-homogeneous differential equation
-To solve you first find a solution to the homogeneous equation (<math>V_{out}^H)</math> and then form a particular solution <math>(V_{out}^P)</math> based on the homogeneous solution
-===Homogeneous Solution===
-:<math>\frac{dV_{out}^H}{dt}   + \frac{(R+R_1)}{CRR_1} V_{out}^H =  0</math>
-:<math>\Rightarrow \frac{dV_{out}^H}{dt}   = - \frac{(R+R_1)}{CRR_1} V_{out}^H \equiv \frac{1}{CR_{\parallel}} V_{out}^H</math>
-:<math> V_{out}^H = D e^{-t/CR_{\parallel}}</math>
-=== Particular Solution===
-:<math>\frac{dV_{out}}{dt}   + \frac{(R+R_1)}{CRR_1} V_{out} =  \frac{V_0}{CR_1} = Constant</math>
-:<math>V_{out} = V_{out}^P + V_{out}^H</math>
-::<math> V_{out}^H = D e^{-t/CR_{\parallel}}</math>
-:<math>\Rightarrow</math>
-::<math>V_{out}^P = \frac{CR_{\parallel}}{CR_1}V_0</math>
-=== Apply Boundary Conditions===
-:<math>V_{out} =\frac{R_{\parallel}}{R_1}V_0 + D e^{-t/CR_{\parallel}}</math>
-:<math>V_{out}(t=0) = 0 \Rightarrow  D = - \frac{R_{\parallel}}{R_1}V_0</math>
-:<math>V_{out} =\frac{R_{\parallel}}{R_1}V_0 (1- e^{-t/CR_{\parallel}})</math>
 [[Forest_Electronic_Instrumentation_and_Measurement]]

Difference between revisions of "TF EIM Chapt3"

Revision as of 20:45, 9 February 2011

Fourier Transforms

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