Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/
One of the most important reasons why we care about sinusoids is that signals can be
represented as sums of sinusoids. The technique that can transform any wave form into a
sum of sinusoids representation is called Fourier analysis. Fourier series analysis can show that
it can be represented as a sum of sinusoids. Circuits have to be linear for this to happen.
The response of circuits to sinusoids as a function of frequency is called the frequency
response of the circuit. As the input frequency of the amplifier increases, the amplitude of the output will decrease. Not only that, the phase will also change.
A very simple circuit example with a sinusoidal input is used - an RC network- a series connection of a resistor and capacitor. This mimics the input of the RC circuit that is part of the amplifier. So the amplifier looks like this- it has a gate capacitance, CGS. GS, and a the resistor, R. which is some parasitic resistance of the wires.
The input voltage is equal to some amplitude VI cosine of omega t, and for t greater than or equal to 0. There will be three ways of approaching the problem - the first being the most difficult - using differential equations. The third being the easiest - almost totally done by inspection.
So the first step, as is our usual practice, is to set up the differential equation by the node method.
So the current leaving the node in this direction is v minus v_I divided by R. And the current heading down this direction is C dv/dt. And the currents must sum to 0 by the node method.
So multiply the whole thing, both sides by R, and shuffle things around. So we get RC dv dt. plus v minus v_I equals 0. So what I want to do is move v_I to the right-hand side and write it like this.
And then I've been given that v_I is V_i cosine of omega t.
So now I am ready for the second step, which is to find the particular solution to the sinusoid, v_P.
This is where things become messy and is not a path to be taken. Instead we will try an exponential input as it gives rise to a nice solution. However we are not sure how this would relate to the answer at the moment.
So for that sneaky input V_i e raised to st let's go back and do our usual thing. We'll try a solution of v_PS ( where s stands for sneaky) given by V_p e raised to st. We will eventually get a solution where V_p is equal to V_i divided by 1 plus sRC. Replacing s with j omega, v_PS will be V_i divide by 1 plus j omega RC times e raised to j omega t. V_i divide by 1 plus j omega, RC is the complex amplitude.
Based on Euler Relations, the real part of the sneaky input is equal to the the input, V_i cosine of omega t. Then based on the inverse superposition argument the real output can be found by
taking the real part of the sneaky output. So V_p would be V_i divide by 1 plus j omega RC. But first we will try to work out the magnitude and phase of the complex number expression in the bracket. With sound knowledge of complex numbers, the particular solution can be found.
Next would be to find the homogeneous solution which would be v_H equals A e raised to minus t
divided by RC.
The fourth step would be to find the total solution which would be v_P + v_H. With the initial conditions A can be found.
The particular solution is the sinusoidal steady state which matters more than the transient state. Steps 3 and 4 were not relevant.
A block diagram of the approach is shown and then after that a summary but there is a simpler approach by inspection.
A magnitude plot and a phase plot can be drawn for the output. This is the frequency response.
Finally a summary of lecture 19 and a preview of what is coming up next.
A very simple circuit example with a sinusoidal input is used - an RC network- a series connection of a resistor and capacitor. This mimics the input of the RC circuit that is part of the amplifier. So the amplifier looks like this- it has a gate capacitance, CGS. GS, and a the resistor, R. which is some parasitic resistance of the wires.
The input voltage is equal to some amplitude VI cosine of omega t, and for t greater than or equal to 0. There will be three ways of approaching the problem - the first being the most difficult - using differential equations. The third being the easiest - almost totally done by inspection.
So the first step, as is our usual practice, is to set up the differential equation by the node method.
So the current leaving the node in this direction is v minus v_I divided by R. And the current heading down this direction is C dv/dt. And the currents must sum to 0 by the node method.
So multiply the whole thing, both sides by R, and shuffle things around. So we get RC dv dt. plus v minus v_I equals 0. So what I want to do is move v_I to the right-hand side and write it like this.
And then I've been given that v_I is V_i cosine of omega t.
So now I am ready for the second step, which is to find the particular solution to the sinusoid, v_P.
This is where things become messy and is not a path to be taken. Instead we will try an exponential input as it gives rise to a nice solution. However we are not sure how this would relate to the answer at the moment.
So for that sneaky input V_i e raised to st let's go back and do our usual thing. We'll try a solution of v_PS ( where s stands for sneaky) given by V_p e raised to st. We will eventually get a solution where V_p is equal to V_i divided by 1 plus sRC. Replacing s with j omega, v_PS will be V_i divide by 1 plus j omega RC times e raised to j omega t. V_i divide by 1 plus j omega, RC is the complex amplitude.
Based on Euler Relations, the real part of the sneaky input is equal to the the input, V_i cosine of omega t. Then based on the inverse superposition argument the real output can be found by
taking the real part of the sneaky output. So V_p would be V_i divide by 1 plus j omega RC. But first we will try to work out the magnitude and phase of the complex number expression in the bracket. With sound knowledge of complex numbers, the particular solution can be found.
Next would be to find the homogeneous solution which would be v_H equals A e raised to minus t
divided by RC.
The fourth step would be to find the total solution which would be v_P + v_H. With the initial conditions A can be found.
The particular solution is the sinusoidal steady state which matters more than the transient state. Steps 3 and 4 were not relevant.
A block diagram of the approach is shown and then after that a summary but there is a simpler approach by inspection.
A magnitude plot and a phase plot can be drawn for the output. This is the frequency response.
Finally a summary of lecture 19 and a preview of what is coming up next.
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