For the circuit shown (which uses an ideal op-amp)

a) Find the transfer function H(s)=E_{out}(s)/E_{in}(s)
in terms of R_{f}, R_{i}, and C.

b) Find an expression for the unit step response in the time domain.

c) Sketch the step response (there are no component values, so you can
only show the shape of the response). Give expressions for the initial
value (t=0^{+}), final value (t=∞) and time constant in terms of circuit
parameters.

a)

b)

c) Initial
value=-R_{f}/R_{i}, final value=0, τ=CR_{i}.

This problem explores how the circuit behaves if an op-amp has finite gain. Therefore do not assume that the two input terminals of the op-amp are at the same voltage though you may still assume no current into the inputs. The last part of the problem explore difficulties that ensue if the circuit does not have negative feedback.

A real op-amp not only has negative feedback, it is also not infinitely fast. We represent this finite speed by giving it a transfer function of the form

where A_{0 }is a large positive quantity (typically 10^{5} or more) and
τ_{0} is a
positive number (typically on the order of 0.1 seconds). The output is
defined to be

where E_{+} and E_{-} are the voltages at the inverting
and non-inverting terminals.

a) For the circuit shown, find an expression for the transfer function
H(s)=E_{out}(s)/E_{in}(s). It should be expressed in terms of a ratio of
polynomials in positive powers of s.

b) What is the time constant of the circuit (i.e., H(s)) if Z_{i}=Z_{f}, A_{0}=10^{6}
and τ_{0}=0.1? Remember, you
can get the time constant of a system by looking at the denominator of the
transfer function. Note that, with feedback, the circuit is *much*
faster (smaller time constant) than the op-amp alone.

c) Find H(s) if the circuit is built as shown below (with
the op-amp terminals switched). What fundamental problem does this
create? This is a circuit with positive feedback. Hint: you can re-use your previous
work if you just make A_{0} negative.

a)

b)
τ
≈ 2e-7 (i.e., 200 nSec - much faster than the op amp alone). The speed goes up dramatically, but the gain of the amplifer drops from 10^{6} to approximately 2. When using negative feedback in this circuit we trade speed (the circuit with feedback is much faster) for gain (the circuit with feedback has much lower gain).

c) We can use the same equation as before but change the
sign in front of A_{0}. The value of τ becomes negative, so the
homogenous solution grows exponentially with time (this is a behavior that
is not seen in circuits with only capacitors, inductors and resistors).
In this case we have positive feedback instead of negative feedback.

In class (gears on a bicycle) we saw how the effective load changes when gears are being used. By varying the turns
ratio (N_{1}/N_{2}) we can match the apparent load in a circuit (R'_{load}) to
the source load (R_{s}) to maximize power transfer. See
image below - the circuit on the right is equivalent to the one on the left
(i.e., currents and voltages are equal). Show that

.

Recall that

where e_{1} and i_{1
}are the voltage across and current through the left coils (and R'_{load}),
and likewise for e_{2} and i_{2} and the left coils (and R_{load}).
You can start by writing equations for R_{load} in terms of e_{2}
and i_{2}, and for R'_{load} in terms of e_{1} and i_{1}.

For the transformer (as for gears and levers) power in is equal to power out. A lever trades force for velocity (the input can be low force with high velocity, with an output at higher force but lower velocity); a transformer trades current for voltage.

If the 8 ohm load is connected
directly to the source (shown below) - you may assume V_{source} is 1 volt.

- Find the power dissipated in R
_{s}(call this P_{Rs}), - Find tthe power dissipated in R
_{load}, (call this P_{Rload}), - Find tthe total power, P
_{tot,}and - Find tthe efficiency, η, which is the power dissipated in the load divided by the total power. This is a measure of how well power is transferred from the source to the load.
- Determine the turns ratio (N
_{1}/N_{2}) needed for a transformer to make the 8 Ohm load appear as 600 Ohms. - Repeat a-d with the 600 Ohm apparent load. Note that although total power supplied by the source decreases by half the efficiency increases enough so that the power dissipated in the load increases by a factor of about 20.

Write equations for the currents summed at the nodes labeled 1 and 2.
Write the currents in terms of i_{a}, e_{1} and e_{2}.

a) For the circuit shown, draw over the mechanical system to find the force-current analog.

b) Draw free body diagrams and write equations of motion.

c) Rewrite the equations of motion substituting in the analogous force-current quantities.

d) Verify that the equations from part c are the same as the equations obtained by summing currents at the nodes labeled 1 and 2.

a) Draw over electrical system (background), replacing mechanical elements with their force-current analogs.

Redraw.

b)

c)

d) They are the same.

a) Find the dual of the circuit shown by drawing over the circuit and substituting the appropriate dual elements.

b) Write the node equations for the original circuit (shown above)circuit and substitute dual quantities.

c) Verify that the equations from part b are appropriate loop equations for part a.

a) Get the dual by drawing over the original system and substituting dual elements.

b)

c)