In Fig. 1, a circuit with a lossless impedance Z, three
resistors, and an ideal current source, is shown. We assume
that the resistor Rh has temperature Th, the resistor Rm
has temperature Tm, and the resistor Ra has no temperature.
The Johnson-Nyquist noise [15], [16] in the resistors Rh and
Rm will “heat up” the impedance Z. The resistance Rm is
the resistance of a passive (voltage) measurement device.
Its noise will add heat to the circuit and give measurement
noise. The resistance Rh models losses of energy to, and
noise from, the environment and the current source (these
are lumped together here). The resistance Ra is introduced to
model energy losses that occur when the current i is applied.
The noisy resistors will provide uncertain energy (“heat
energy”) to the states of Z. The problem addressed in this
paper is the following: Given noisy voltage measurement of
v, how should we choose the current i so as to maximize
the extracted (expected) amount of energy −
t
0 va(s)i(s)ds?
When this problem is solved, we have constructed a heat
engine that transfers heat energy from the resistors into
“useful energy” in the current source. We return to and solve
In Fig. 1, a circuit with a lossless impedance Z, three
resistors, and an ideal current source, is shown. We assume
that the resistor Rh has temperature Th, the resistor Rm
has temperature Tm, and the resistor Ra has no temperature.
The Johnson-Nyquist noise [15], [16] in the resistors Rh and
Rm will “heat up” the impedance Z. The resistance Rm is
the resistance of a passive (voltage) measurement device.
Its noise will add heat to the circuit and give measurement
noise. The resistance Rh models losses of energy to, and
noise from, the environment and the current source (these
are lumped together here). The resistance Ra is introduced to
model energy losses that occur when the current i is applied.
The noisy resistors will provide uncertain energy (“heat
energy”) to the states of Z. The problem addressed in this
paper is the following: Given noisy voltage measurement of
v, how should we choose the current i so as to maximize
the extracted (expected) amount of energy −
t
0 va(s)i(s)ds?
When this problem is solved, we have constructed a heat
engine that transfers heat energy from the resistors into
“useful energy” in the current source. We return to and solve
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